Study Use Probability For Trait Frequency in Biology with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.
All flashcards Flashcard 1: Identify the probability of an offspring being a a aa aa from parents A a × A a Aa \times Aa A a × A a . Answer: 1 4 \frac{1}{4} 4 1 . Punnett square shows a a aa aa outcome in 1 of 4 boxes.
Flashcard 2: Which assumption of Hardy–Weinberg is violated when individuals choose mates by phenotype? Answer: Random mating. Mate choice by traits creates non-random breeding patterns.
Flashcard 3: What is the formula for allele frequency q q q of allele a a a using genotype frequencies? Answer: q = f ( a a ) + 1 2 f ( A a ) q = f(aa) + \frac{1}{2}f(Aa) q = f ( aa ) + 2 1 f ( A a ) . Homozygotes contribute 1 copy, heterozygotes contribute 0.5.
Flashcard 4: Identify the allele frequency p p p if f ( A A ) = 0.36 f(AA)=0.36 f ( AA ) = 0.36 , f ( A a ) = 0.48 f(Aa)=0.48 f ( A a ) = 0.48 , and f ( a a ) = 0.16 f(aa)=0.16 f ( aa ) = 0.16 . Answer: 0.60 0.60 0.60 . p = f ( A A ) + 1 2 f ( A a ) = 0.36 + 0.24 = 0.60 p = f(AA) + \frac{1}{2}f(Aa) = 0.36 + 0.24 = 0.60 p = f ( AA ) + 2 1 f ( A a ) = 0.36 + 0.24 = 0.60
Flashcard 5: What is the probability of showing a recessive phenotype from parents A a × A a Aa \times Aa A a × A a ? Answer: 1 4 \frac{1}{4} 4 1 . Recessive phenotype requires a a aa aa genotype (q 2 q^2 q 2 ).
Flashcard 6: What condition describes Hardy–Weinberg equilibrium in terms of allele frequencies? Answer: Allele frequencies remain constant across generations. No evolutionary forces acting means stable frequencies.
Flashcard 7: Which assumption of Hardy–Weinberg is violated when population size is very small? Answer: Infinitely large population (no genetic drift). Small populations experience random sampling effects.
Flashcard 8: What is the formula for allele frequency q q q of allele a a a using genotype frequencies? Answer: q = f ( a a ) + 1 2 f ( A a ) q = f(aa) + \frac{1}{2}f(Aa) q = f ( aa ) + 2 1 f ( A a ) . Homozygotes contribute 1 copy, heterozygotes contribute 0.5.
Flashcard 9: Which option best explains why drift is stronger in small populations than large ones? Answer: Sampling error is larger when the number of breeders is small. Fewer breeding individuals means larger random variation.
Flashcard 10: Which assumption of Hardy–Weinberg is violated when population size is very small? Answer: Infinitely large population (no genetic drift). Small populations experience random sampling effects.
Flashcard 11: Identify q q q if the recessive genotype frequency is q 2 = 0.09 q^2 = 0.09 q 2 = 0.09 in Hardy–Weinberg. Answer: q = 0.30 q = 0.30 q = 0.30 . Take square root of q 2 q^2 q 2 : 0.09 = 0.30 \sqrt{0.09} = 0.30 0.09 = 0.30
Flashcard 12: Identify the probability of an offspring being A A AA AA from parents A a × A a Aa \times Aa A a × A a . Answer: 1 4 \frac{1}{4} 4 1 . Punnett square shows A A AA AA outcome in 1 of 4 boxes.
Flashcard 13: Which assumption of Hardy–Weinberg is violated when alleles change due to copying errors? Answer: No mutation. New alleles arise spontaneously, changing frequencies.
Flashcard 14: Identify the expected heterozygote frequency if p = 0.70 p=0.70 p = 0.70 and q = 0.30 q=0.30 q = 0.30 . Answer: 2 p q = 0.42 2pq = 0.42 2 pq = 0.42 . Calculate 2 × 0.70 × 0.30 = 0.42 2 \times 0.70 \times 0.30 = 0.42 2 × 0.70 × 0.30 = 0.42
Flashcard 15: Identify whether a population is evolving if p p p changes from 0.40 0.40 0.40 to 0.45 0.45 0.45 in one generation. Answer: Yes; a change in allele frequency indicates evolution. Any change in allele frequency indicates evolutionary forces.
Flashcard 16: What is the probability interpretation of 2 p q 2pq 2 pq in Hardy–Weinberg equilibrium? Answer: Probability a randomly chosen individual is heterozygous A a Aa A a . Probability of getting one A A A and one a a a allele.
Flashcard 17: Identify the most likely outcome if selection strongly favors allele A A A over many generations. Answer: Allele A A A increases in frequency; a a a decreases. Consistent selection pressure drives directional change.
Flashcard 18: Identify p p p if q = 0.30 q = 0.30 q = 0.30 for a two-allele Hardy–Weinberg population. Answer: p = 0.70 p = 0.70 p = 0.70 . Since p + q = 1 p + q = 1 p + q = 1 : p = 1 − 0.30 = 0.70 p = 1 - 0.30 = 0.70 p = 1 − 0.30 = 0.70
Flashcard 19: Which Hardy–Weinberg term corresponds to the expected frequency of homozygous recessive a a aa aa ? Answer: q 2 q^2 q 2 . Probability of inheriting a a a from both parents is q × q q \times q q × q .
Flashcard 20: What is mutation as a source of variation? Answer: Random DNA change that creates new alleles. Introduces novel alleles into the gene pool.
Flashcard 21: What is the founder effect? Answer: Genetic drift when a new population is started by few individuals. Small founding group may not represent original diversity.
Flashcard 22: Identify the expected A A AA AA genotype frequency if p = 0.70 p=0.70 p = 0.70 in Hardy–Weinberg. Answer: p 2 = 0.49 p^2 = 0.49 p 2 = 0.49 . Calculate p 2 = ( 0.70 ) 2 = 0.49 p^2 = (0.70)^2 = 0.49 p 2 = ( 0.70 ) 2 = 0.49
Flashcard 23: What is the formula for allele frequency p p p of allele A A A using genotype frequencies? Answer: p = f ( A A ) + 1 2 f ( A a ) p = f(AA) + \frac{1}{2}f(Aa) p = f ( AA ) + 2 1 f ( A a ) . Homozygotes contribute 1 copy, heterozygotes contribute 0.5.
Flashcard 24: What is the definition of genotype frequency in a population? Answer: Proportion of individuals with a specific genotype. Measured by counting individuals with each genotype.
Flashcard 25: What is the probability interpretation of p p p in Hardy–Weinberg equilibrium? Answer: Probability a randomly chosen allele from the gene pool is A A A . Represents sampling probability from the allele pool.
Flashcard 26: Identify the allele frequency p p p if f ( A A ) = 0.36 f(AA)=0.36 f ( AA ) = 0.36 , f ( A a ) = 0.48 f(Aa)=0.48 f ( A a ) = 0.48 , and f ( a a ) = 0.16 f(aa)=0.16 f ( aa ) = 0.16 . Answer: 0.60 0.60 0.60 . p = f ( A A ) + 1 2 f ( A a ) = 0.36 + 0.24 = 0.60 p = f(AA) + \frac{1}{2}f(Aa) = 0.36 + 0.24 = 0.60 p = f ( AA ) + 2 1 f ( A a ) = 0.36 + 0.24 = 0.60
Flashcard 27: What is the Hardy–Weinberg genotype frequency equation? Answer: p 2 + 2 p q + q 2 = 1 p^2 + 2pq + q^2 = 1 p 2 + 2 pq + q 2 = 1 . Expansion of ( p + q ) 2 (p + q)^2 ( p + q ) 2 for random mating probabilities.
Flashcard 28: Identify the expected frequency of carriers A a Aa A a if q 2 = 0.01 q^2 = 0.01 q 2 = 0.01 in Hardy–Weinberg. Answer: 2 p q = 0.18 2pq = 0.18 2 pq = 0.18 . If q 2 = 0.01 q^2 = 0.01 q 2 = 0.01 , then q = 0.10 q = 0.10 q = 0.10 , p = 0.90 p = 0.90 p = 0.90 , 2 p q = 0.18 2pq = 0.18 2 pq = 0.18
Flashcard 29: What is the probability interpretation of p 2 p^2 p 2 in Hardy–Weinberg equilibrium? Answer: Probability a randomly chosen individual is A A AA AA . Independent probability of getting A A A from both parents.
Flashcard 30: Identify p p p if q 2 = 0.16 q^2 = 0.16 q 2 = 0.16 in Hardy–Weinberg equilibrium. Answer: p = 0.60 p = 0.60 p = 0.60 . If q 2 = 0.16 q^2 = 0.16 q 2 = 0.16 , then q = 0.40 q = 0.40 q = 0.40 , so p = 0.60 p = 0.60 p = 0.60
Flashcard 31: Which assumption of Hardy–Weinberg is violated when many individuals enter or leave a population? Answer: No migration (no gene flow). Movement changes local allele frequencies through mixing.
Flashcard 32: Which Hardy–Weinberg term corresponds to the expected frequency of homozygous dominant A A AA AA ? Answer: p 2 p^2 p 2 . Probability of inheriting A A A from both parents is p × p p \times p p × p .
Flashcard 33: Which Hardy–Weinberg term corresponds to the expected frequency of heterozygotes A a Aa A a ? Answer: 2 p q 2pq 2 pq . Two ways to get A a Aa A a : A A A from mom, a a a from dad or vice versa.
Flashcard 34: What is natural selection in terms of trait frequency? Answer: Nonrandom change in trait frequency due to differential fitness. Favorable traits increase, unfavorable traits decrease.
Flashcard 35: Identify whether allele frequency can change by chance alone in a small population. Answer: Yes; genetic drift can change allele frequency by chance. Random sampling of gametes causes frequency fluctuations.
Flashcard 36: Identify the probability of an offspring being A a Aa A a from parents A a × A a Aa \times Aa A a × A a . Answer: 1 2 \frac{1}{2} 2 1 . Punnett square shows A a Aa A a outcome in 2 of 4 boxes.
Flashcard 37: What is meant by microevolution? Answer: Change in allele frequencies within a population over time. Evolution within populations, not between species.
Flashcard 38: What condition describes Hardy–Weinberg equilibrium in terms of allele frequencies? Answer: Allele frequencies remain constant across generations. No evolutionary forces acting means stable frequencies.
Flashcard 39: Which assumption of Hardy–Weinberg is violated when alleles change due to copying errors? Answer: No mutation. New alleles arise spontaneously, changing frequencies.
Flashcard 40: Which assumption of Hardy–Weinberg is violated when survival differs by genotype? Answer: No natural selection. Differential survival/reproduction changes allele frequencies.
Flashcard 41: Which option best describes how gene flow usually affects differences between populations? Answer: It reduces differences by making allele frequencies more similar. Migration homogenizes allele frequencies across populations.
Flashcard 42: What is the probability interpretation of p 2 p^2 p 2 in Hardy–Weinberg equilibrium? Answer: Probability a randomly chosen individual is A A AA AA . Independent probability of getting A A A from both parents.
Flashcard 43: Identify p p p if q = 0.30 q = 0.30 q = 0.30 for a two-allele Hardy–Weinberg population. Answer: p = 0.70 p = 0.70 p = 0.70 . Since p + q = 1 p + q = 1 p + q = 1 : p = 1 − 0.30 = 0.70 p = 1 - 0.30 = 0.70 p = 1 − 0.30 = 0.70
Flashcard 44: What is directional selection? Answer: Selection favoring one extreme phenotype, shifting the mean. Population mean moves toward the favored extreme.
Flashcard 45: What is the probability interpretation of q 2 q^2 q 2 in Hardy–Weinberg equilibrium? Answer: Probability a randomly chosen individual is a a aa aa . Independent probability of getting a a a from both parents.
Flashcard 46: Identify the expected genotype frequencies if p = 0.50 p=0.50 p = 0.50 and q = 0.50 q=0.50 q = 0.50 in Hardy–Weinberg. Answer: p 2 = 0.25 p^2=0.25 p 2 = 0.25 , 2 p q = 0.50 2pq=0.50 2 pq = 0.50 , q 2 = 0.25 q^2=0.25 q 2 = 0.25 . Equal allele frequencies produce maximum heterozygosity.
Flashcard 47: Which assumption of Hardy–Weinberg is violated when many individuals enter or leave a population? Answer: No migration (no gene flow). Movement changes local allele frequencies through mixing.
Flashcard 48: What is gene flow? Answer: Movement of alleles between populations via migration and breeding. Immigration and emigration transfer alleles between populations.
Flashcard 49: What is selection pressure? Answer: Environmental factor that affects survival or reproduction. External conditions that determine survival and reproduction.
Flashcard 50: What is the Hardy–Weinberg genotype frequency equation? Answer: p 2 + 2 p q + q 2 = 1 p^2 + 2pq + q^2 = 1 p 2 + 2 pq + q 2 = 1 . Expansion of ( p + q ) 2 (p + q)^2 ( p + q ) 2 for random mating probabilities.
Flashcard 51: What is stabilizing selection? Answer: Selection favoring intermediate phenotypes, reducing variation. Extreme variants are eliminated, average is favored.
Flashcard 52: Identify the allele frequency q q q if f ( A A ) = 0.36 f(AA)=0.36 f ( AA ) = 0.36 , f ( A a ) = 0.48 f(Aa)=0.48 f ( A a ) = 0.48 , and f ( a a ) = 0.16 f(aa)=0.16 f ( aa ) = 0.16 . Answer: 0.40 0.40 0.40 . q = 1 − p = 1 − 0.60 = 0.40 q = 1 - p = 1 - 0.60 = 0.40 q = 1 − p = 1 − 0.60 = 0.40
Flashcard 53: Identify the expected genotype frequencies if p = 0.50 p=0.50 p = 0.50 and q = 0.50 q=0.50 q = 0.50 in Hardy–Weinberg. Answer: p 2 = 0.25 p^2=0.25 p 2 = 0.25 , 2 p q = 0.50 2pq=0.50 2 pq = 0.50 , q 2 = 0.25 q^2=0.25 q 2 = 0.25 . Equal allele frequencies produce maximum heterozygosity.
Flashcard 54: Identify p p p if q 2 = 0.16 q^2 = 0.16 q 2 = 0.16 in Hardy–Weinberg equilibrium. Answer: p = 0.60 p = 0.60 p = 0.60 . If q 2 = 0.16 q^2 = 0.16 q 2 = 0.16 , then q = 0.40 q = 0.40 q = 0.40 , so p = 0.60 p = 0.60 p = 0.60
Flashcard 55: Identify q q q if the recessive genotype frequency is q 2 = 0.09 q^2 = 0.09 q 2 = 0.09 in Hardy–Weinberg. Answer: q = 0.30 q = 0.30 q = 0.30 . Take square root of q 2 q^2 q 2 : 0.09 = 0.30 \sqrt{0.09} = 0.30 0.09 = 0.30
Flashcard 56: Identify the allele frequency q q q if f ( A A ) = 0.36 f(AA)=0.36 f ( AA ) = 0.36 , f ( A a ) = 0.48 f(Aa)=0.48 f ( A a ) = 0.48 , and f ( a a ) = 0.16 f(aa)=0.16 f ( aa ) = 0.16 . Answer: 0.40 0.40 0.40 . q = 1 − p = 1 − 0.60 = 0.40 q = 1 - p = 1 - 0.60 = 0.40 q = 1 − p = 1 − 0.60 = 0.40
Flashcard 57: Identify the probability of an offspring being a a aa aa from parents A a × A a Aa \times Aa A a × A a . Answer: 1 4 \frac{1}{4} 4 1 . Punnett square shows a a aa aa outcome in 1 of 4 boxes.
Flashcard 58: Which option best explains why drift is stronger in small populations than large ones? Answer: Sampling error is larger when the number of breeders is small. Fewer breeding individuals means larger random variation.
Flashcard 59: What is selection pressure? Answer: Environmental factor that affects survival or reproduction. External conditions that determine survival and reproduction.
Flashcard 60: What is genetic drift? Answer: Random change in allele frequencies due to chance events. Sampling error in small populations causes frequency fluctuations.
Flashcard 61: Identify the expected A A AA AA genotype frequency if p = 0.70 p=0.70 p = 0.70 in Hardy–Weinberg. Answer: p 2 = 0.49 p^2 = 0.49 p 2 = 0.49 . Calculate p 2 = ( 0.70 ) 2 = 0.49 p^2 = (0.70)^2 = 0.49 p 2 = ( 0.70 ) 2 = 0.49
Flashcard 62: What is the probability interpretation of q 2 q^2 q 2 in Hardy–Weinberg equilibrium? Answer: Probability a randomly chosen individual is a a aa aa . Independent probability of getting a a a from both parents.
Flashcard 63: Which option best describes how gene flow usually affects differences between populations? Answer: It reduces differences by making allele frequencies more similar. Migration homogenizes allele frequencies across populations.
Flashcard 64: Identify whether a population is evolving if p p p changes from 0.40 0.40 0.40 to 0.45 0.45 0.45 in one generation. Answer: Yes; a change in allele frequency indicates evolution. Any change in allele frequency indicates evolutionary forces.
Flashcard 65: What is meant by microevolution? Answer: Change in allele frequencies within a population over time. Evolution within populations, not between species.
Flashcard 66: What is the bottleneck effect? Answer: Genetic drift after a sharp population size reduction. Surviving individuals may not represent original diversity.
Flashcard 67: What is the definition of phenotype frequency in a population? Answer: Proportion of individuals showing a specific trait (phenotype). Observed by counting individuals displaying each trait.
Flashcard 68: What is fitness in evolutionary biology? Answer: Relative reproductive success of a genotype or phenotype. Higher fitness means more offspring in next generation.
Flashcard 69: Which Hardy–Weinberg term corresponds to the expected frequency of homozygous recessive a a aa aa ? Answer: q 2 q^2 q 2 . Probability of inheriting a a a from both parents is q × q q \times q q × q .
Flashcard 70: What is mutation as a source of variation? Answer: Random DNA change that creates new alleles. Introduces novel alleles into the gene pool.
Flashcard 71: What is the definition of genotype frequency in a population? Answer: Proportion of individuals with a specific genotype. Measured by counting individuals with each genotype.
Flashcard 72: What is natural selection in terms of trait frequency? Answer: Nonrandom change in trait frequency due to differential fitness. Favorable traits increase, unfavorable traits decrease.
Flashcard 73: Which assumption of Hardy–Weinberg is violated when individuals choose mates by phenotype? Answer: Random mating. Mate choice by traits creates non-random breeding patterns.
Flashcard 74: Identify the probability of an offspring being a a aa aa from parents A a × a a Aa \times aa A a × aa . Answer: 1 2 \frac{1}{2} 2 1 . Test cross produces 50% heterozygous, 50% recessive.
Flashcard 75: Identify the expected a a aa aa genotype frequency if q = 0.30 q=0.30 q = 0.30 in Hardy–Weinberg. Answer: q 2 = 0.09 q^2 = 0.09 q 2 = 0.09 . Calculate q 2 = ( 0.30 ) 2 = 0.09 q^2 = (0.30)^2 = 0.09 q 2 = ( 0.30 ) 2 = 0.09
Flashcard 76: Which Hardy–Weinberg term corresponds to the expected frequency of homozygous dominant A A AA AA ? Answer: p 2 p^2 p 2 . Probability of inheriting A A A from both parents is p × p p \times p p × p .
Flashcard 77: Identify the expected frequency of carriers A a Aa A a if q 2 = 0.01 q^2 = 0.01 q 2 = 0.01 in Hardy–Weinberg. Answer: 2 p q = 0.18 2pq = 0.18 2 pq = 0.18 . If q 2 = 0.01 q^2 = 0.01 q 2 = 0.01 , then q = 0.10 q = 0.10 q = 0.10 , p = 0.90 p = 0.90 p = 0.90 , 2 p q = 0.18 2pq = 0.18 2 pq = 0.18
Flashcard 78: What equation must allele frequencies satisfy for a two-allele gene? Answer: p + q = 1 p + q = 1 p + q = 1 . All alleles must sum to 100% of the gene pool.
Flashcard 79: Identify the expected a a aa aa genotype frequency if q = 0.30 q=0.30 q = 0.30 in Hardy–Weinberg. Answer: q 2 = 0.09 q^2 = 0.09 q 2 = 0.09 . Calculate q 2 = ( 0.30 ) 2 = 0.09 q^2 = (0.30)^2 = 0.09 q 2 = ( 0.30 ) 2 = 0.09
Flashcard 80: Identify the expected heterozygote frequency if p = 0.70 p=0.70 p = 0.70 and q = 0.30 q=0.30 q = 0.30 . Answer: 2 p q = 0.42 2pq = 0.42 2 pq = 0.42 . Calculate 2 × 0.70 × 0.30 = 0.42 2 \times 0.70 \times 0.30 = 0.42 2 × 0.70 × 0.30 = 0.42
Flashcard 81: What is the probability interpretation of p p p in Hardy–Weinberg equilibrium? Answer: Probability a randomly chosen allele from the gene pool is A A A . Represents sampling probability from the allele pool.
Flashcard 82: Identify whether allele frequency can change by chance alone in a small population. Answer: Yes; genetic drift can change allele frequency by chance. Random sampling of gametes causes frequency fluctuations.
Flashcard 83: Identify the probability of an offspring being A a Aa A a from parents A A × a a AA \times aa AA × aa . Answer: 1 1 1 . All offspring inherit A A A from one parent, a a a from other.
Flashcard 84: Which assumption of Hardy–Weinberg is violated when survival differs by genotype? Answer: No natural selection. Differential survival/reproduction changes allele frequencies.
Flashcard 85: What is the definition of phenotype frequency in a population? Answer: Proportion of individuals showing a specific trait (phenotype). Observed by counting individuals displaying each trait.
Flashcard 86: Identify the most likely outcome if selection strongly favors allele A A A over many generations. Answer: Allele A A A increases in frequency; a a a decreases. Consistent selection pressure drives directional change.
Flashcard 87: What is the probability of showing a recessive phenotype from parents A a × A a Aa \times Aa A a × A a ? Answer: 1 4 \frac{1}{4} 4 1 . Recessive phenotype requires a a aa aa genotype (q 2 q^2 q 2 ).
Flashcard 88: Which Hardy–Weinberg term corresponds to the expected frequency of heterozygotes A a Aa A a ? Answer: 2 p q 2pq 2 pq . Two ways to get A a Aa A a : A A A from mom, a a a from dad or vice versa.
Flashcard 89: What is the probability interpretation of 2 p q 2pq 2 pq in Hardy–Weinberg equilibrium? Answer: Probability a randomly chosen individual is heterozygous A a Aa A a . Probability of getting one A A A and one a a a allele.