Use Probability for Trait Frequency - Biology

  1. Identify the probability of an offspring being aaaa from parents Aa×AaAa \times Aa.
    14\frac{1}{4}. Punnett square shows aaaa outcome in 1 of 4 boxes.
  2. Which assumption of Hardy–Weinberg is violated when individuals choose mates by phenotype?
    Random mating. Mate choice by traits creates non-random breeding patterns.
  3. What is the formula for allele frequency qq of allele aa using genotype frequencies?
    q=f(aa)+12f(Aa)q = f(aa) + \frac{1}{2}f(Aa). Homozygotes contribute 1 copy, heterozygotes contribute 0.5.
  4. Identify the allele frequency pp if f(AA)=0.36f(AA)=0.36, f(Aa)=0.48f(Aa)=0.48, and f(aa)=0.16f(aa)=0.16.
    0.600.60. p=f(AA)+12f(Aa)=0.36+0.24=0.60p = f(AA) + \frac{1}{2}f(Aa) = 0.36 + 0.24 = 0.60
  5. Which assumption of Hardy–Weinberg is violated when population size is very small?
    Infinitely large population (no genetic drift). Small populations experience random sampling effects.
  6. Identify qq if the recessive genotype frequency is q2=0.09q^2 = 0.09 in Hardy–Weinberg.
    q=0.30q = 0.30. Take square root of q2q^2: 0.09=0.30\sqrt{0.09} = 0.30
  7. Which assumption of Hardy–Weinberg is violated when alleles change due to copying errors?
    No mutation. New alleles arise spontaneously, changing frequencies.
  8. Identify the expected heterozygote frequency if p=0.70p=0.70 and q=0.30q=0.30.
    2pq=0.422pq = 0.42. Calculate 2×0.70×0.30=0.422 \times 0.70 \times 0.30 = 0.42
  9. What is the founder effect?
    Genetic drift when a new population is started by few individuals. Small founding group may not represent original diversity.
  10. Identify the expected AAAA genotype frequency if p=0.70p=0.70 in Hardy–Weinberg.
    p2=0.49p^2 = 0.49. Calculate p2=(0.70)2=0.49p^2 = (0.70)^2 = 0.49
  11. What is the formula for allele frequency pp of allele AA using genotype frequencies?
    p=f(AA)+12f(Aa)p = f(AA) + \frac{1}{2}f(Aa). Homozygotes contribute 1 copy, heterozygotes contribute 0.5.
  12. What is the definition of genotype frequency in a population?
    Proportion of individuals with a specific genotype. Measured by counting individuals with each genotype.
  13. What is the probability interpretation of pp in Hardy–Weinberg equilibrium?
    Probability a randomly chosen allele from the gene pool is AA. Represents sampling probability from the allele pool.
  14. What is the Hardy–Weinberg genotype frequency equation?
    p2+2pq+q2=1p^2 + 2pq + q^2 = 1. Expansion of (p+q)2(p + q)^2 for random mating probabilities.
  15. Which Hardy–Weinberg term corresponds to the expected frequency of homozygous dominant AAAA?
    p2p^2. Probability of inheriting AA from both parents is p×pp \times p.
  16. What is natural selection in terms of trait frequency?
    Nonrandom change in trait frequency due to differential fitness. Favorable traits increase, unfavorable traits decrease.
  17. Identify whether allele frequency can change by chance alone in a small population.
    Yes; genetic drift can change allele frequency by chance. Random sampling of gametes causes frequency fluctuations.
  18. What is meant by microevolution?
    Change in allele frequencies within a population over time. Evolution within populations, not between species.
  19. What condition describes Hardy–Weinberg equilibrium in terms of allele frequencies?
    Allele frequencies remain constant across generations. No evolutionary forces acting means stable frequencies.
  20. Which assumption of Hardy–Weinberg is violated when survival differs by genotype?
    No natural selection. Differential survival/reproduction changes allele frequencies.
  21. What is the probability interpretation of p2p^2 in Hardy–Weinberg equilibrium?
    Probability a randomly chosen individual is AAAA. Independent probability of getting AA from both parents.
  22. Identify pp if q=0.30q = 0.30 for a two-allele Hardy–Weinberg population.
    p=0.70p = 0.70. Since p+q=1p + q = 1: p=1−0.30=0.70p = 1 - 0.30 = 0.70
  23. What is directional selection?
    Selection favoring one extreme phenotype, shifting the mean. Population mean moves toward the favored extreme.
  24. What is the probability interpretation of q2q^2 in Hardy–Weinberg equilibrium?
    Probability a randomly chosen individual is aaaa. Independent probability of getting aa from both parents.
  25. Identify the expected genotype frequencies if p=0.50p=0.50 and q=0.50q=0.50 in Hardy–Weinberg.
    p2=0.25p^2=0.25, 2pq=0.502pq=0.50, q2=0.25q^2=0.25. Equal allele frequencies produce maximum heterozygosity.
  26. Which assumption of Hardy–Weinberg is violated when many individuals enter or leave a population?
    No migration (no gene flow). Movement changes local allele frequencies through mixing.
  27. What is gene flow?
    Movement of alleles between populations via migration and breeding. Immigration and emigration transfer alleles between populations.
  28. What is stabilizing selection?
    Selection favoring intermediate phenotypes, reducing variation. Extreme variants are eliminated, average is favored.
  29. Identify pp if q2=0.16q^2 = 0.16 in Hardy–Weinberg equilibrium.
    p=0.60p = 0.60. If q2=0.16q^2 = 0.16, then q=0.40q = 0.40, so p=0.60p = 0.60
  30. Identify the allele frequency qq if f(AA)=0.36f(AA)=0.36, f(Aa)=0.48f(Aa)=0.48, and f(aa)=0.16f(aa)=0.16.
    0.400.40. q=1−p=1−0.60=0.40q = 1 - p = 1 - 0.60 = 0.40
  31. Which option best explains why drift is stronger in small populations than large ones?
    Sampling error is larger when the number of breeders is small. Fewer breeding individuals means larger random variation.
  32. What is selection pressure?
    Environmental factor that affects survival or reproduction. External conditions that determine survival and reproduction.
  33. What is genetic drift?
    Random change in allele frequencies due to chance events. Sampling error in small populations causes frequency fluctuations.
  34. Which option best describes how gene flow usually affects differences between populations?
    It reduces differences by making allele frequencies more similar. Migration homogenizes allele frequencies across populations.
  35. Identify whether a population is evolving if pp changes from 0.400.40 to 0.450.45 in one generation.
    Yes; a change in allele frequency indicates evolution. Any change in allele frequency indicates evolutionary forces.
  36. What is the bottleneck effect?
    Genetic drift after a sharp population size reduction. Surviving individuals may not represent original diversity.
  37. What is fitness in evolutionary biology?
    Relative reproductive success of a genotype or phenotype. Higher fitness means more offspring in next generation.
  38. Which Hardy–Weinberg term corresponds to the expected frequency of homozygous recessive aaaa?
    q2q^2. Probability of inheriting aa from both parents is q×qq \times q.
  39. What is mutation as a source of variation?
    Random DNA change that creates new alleles. Introduces novel alleles into the gene pool.
  40. Identify the expected frequency of carriers AaAa if q2=0.01q^2 = 0.01 in Hardy–Weinberg.
    2pq=0.182pq = 0.18. If q2=0.01q^2 = 0.01, then q=0.10q = 0.10, p=0.90p = 0.90, 2pq=0.182pq = 0.18
  41. What equation must allele frequencies satisfy for a two-allele gene?
    p+q=1p + q = 1. All alleles must sum to 100% of the gene pool.
  42. Identify the expected aaaa genotype frequency if q=0.30q=0.30 in Hardy–Weinberg.
    q2=0.09q^2 = 0.09. Calculate q2=(0.30)2=0.09q^2 = (0.30)^2 = 0.09
  43. What is the definition of phenotype frequency in a population?
    Proportion of individuals showing a specific trait (phenotype). Observed by counting individuals displaying each trait.
  44. Identify the most likely outcome if selection strongly favors allele AA over many generations.
    Allele AA increases in frequency; aa decreases. Consistent selection pressure drives directional change.
  45. What is the probability of showing a recessive phenotype from parents Aa×AaAa \times Aa?
    14\frac{1}{4}. Recessive phenotype requires aaaa genotype (q2q^2).
  46. Which Hardy–Weinberg term corresponds to the expected frequency of heterozygotes AaAa?
    2pq2pq. Two ways to get AaAa: AA from mom, aa from dad or vice versa.
  47. What is the probability interpretation of 2pq2pq in Hardy–Weinberg equilibrium?
    Probability a randomly chosen individual is heterozygous AaAa. Probability of getting one AA and one aa allele.

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