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  1. 1-biological-molecules

    1-1-biological-molecules-carbohydrates
    11 主题
  2. 1-2-biological-molecules-lipids
    3 主题
  3. 1-3-biological-molecules-proteins
    5 主题
  4. 1-4-proteins-enzymes
    12 主题
  5. 1-5-nucleic-acids-structure-and-dna-replication
    8 主题
  6. 1-6-atp-water-and-inorganic-ions
    4 主题
  7. 2-cell-structure
    2-1-cell-structure
    7 主题
  8. 2-2-the-microscope-in-cell-studies
    4 主题
  9. 2-3-cell-division-in-eukaryotic-and-prokaryotic-cells
    8 主题
  10. 2-4-cell-membranes-and-transport
    7 主题
  11. 2-5-cell-recognition-and-the-immune-system
    7 主题
  12. 2-6-vaccines-disease-and-monoclonal-antibodies
    6 主题
  13. 3-exchange-and-transport
    3-1-adaptations-for-gas-exchange
    6 主题
  14. 3-2-human-gas-exchange
    10 主题
  15. 3-3-digestion-and-absorption
    5 主题
  16. 3-4-mass-transport-in-animals
    6 主题
  17. 3-5-the-circulatory-system-in-animals
    8 主题
  18. 3-6-mass-transport-in-plants
    6 主题
  19. 4-genetics-variation-and-interdependence
    4-1-dna-genes-and-chromosomes
    7 主题
  20. 4-2-dna-and-protein-synthesis
    6 主题
  21. 4-3-genetic-diversity-mutations-and-meiosis
    7 主题
  22. 4-4-genetic-diversity-and-adaptation
    6 主题
  23. 4-5-species-and-taxonomy
    4 主题
  24. 4-6-biodiversity
    9 主题
  25. 5-energy-transfers-in-and-between-organisms-a-level-only
    5-1-photosynthesis-a-level-only
    5 主题
  26. 5-2-respiration-a-level-only
    7 主题
  27. 5-3-energy-and-ecosystems-a-level-only
    9 主题
  28. 5-4-nutrient-cycles-a-level-only
    4 主题
  29. 6-organisms-respond-to-changes-in-their-environments-a-level-only
    6-1-response-to-stimuli-a-level-only
    12 主题
  30. 6-2-nervous-coordination-a-level-only
    10 主题
  31. 6-3-skeletal-muscles-a-level-only
    6 主题
  32. 6-4-homeostasis-a-level-only
    11 主题
  33. 7-genetics-populations-evolution-and-ecosystems-a-level-only
    7-1-inheritance-a-level-only
    6 主题
  34. 7-2-populations-a-level-only
    3 主题
  35. 7-3-evolution-a-level-only
    5 主题
  36. 7-4-populations-in-ecosystems-a-level-only
    7 主题
  37. 8-the-control-of-gene-expression-a-level-only
    8-1-genetic-mutations-a-level-only
    2 主题
  38. 8-2-regulation-of-gene-expression-a-level-only
    11 主题
  39. 8-3-using-genome-projects-a-level-only
    4 主题
  40. 8-4-gene-technologies-a-level-only
    13 主题
  41. exam-guidance-and-skills
    essay-guidance
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Natural selection: Hardy-Weinberg principle

  • The Hardy-Weinberg principle states that allele frequencies in a population will remain constant from one generation to the next if specific conditions are met

  • The equation can be used to:

    • Calculate allele and genotype frequencies

    • Predict how these frequencies will change across generations

Conditions for the Hardy-Weinberg principle

  • For the Hardy-Weinberg principle to be correctly applied to a population a series of conditions, or assumptions, need to be met

    • Organisms are diploid

    • Organisms reproduce by sexual reproduction only

    • There is no overlap between generations (parents do not mate with offspring)

    • Mating is random

    • The population is large

    • There is no migration, mutation, or selection

      • This would mean no individuals entering the population (immigration) or leaving (emigration)

      • Selection refers to both natural and artificial selection

    • Allele frequencies are equal in both sexes

  • The Hardy-Weinberg principle can be useful when building models and making predictions, but the assumptions listed are very rarely, if ever, all present in nature

Hardy-Weinberg equations

  • If the phenotype of a trait in a population is determined by a single gene with only two alleles (we will use B / b as examples throughout this section) then the population will consist of individuals with three possible genotypes:

    • Homozygous dominant (BB)

    • Heterozygous (Bb)

    • Homozygous recessive (bb)

  •  All frequencies are expressed as proportions (values between 0 and 1)

  • For example:

    • if every individual in the population has the homozygous dominant genotype (BB), its frequency will be 1

    • if half of the population show this genotype, then the frequency will be 0.5

Allele frequencies

  • For a gene with two alleles (B and b):

    • The letter p represents the frequency of the dominant allele (B)

    • The letter q represents the frequency of the recessive allele (b)

    • As there are only two alleles for this gene:

p + q = 1

  • For example:

    • In a population of 100 individuals, there are 200 alleles

    • If there are 120 dominant (B) alleles, the frequency of the dominant allele = 120/200

    • Therefore:

      • p = 120 ÷ 200 = 0.6

      • q = 1 – 0.6 = 0.4

Genotype frequencies

  • Frequency of genotypes can also be represented; this is the proportion of all of the individuals with a particular genotype

  • For a gene with two alleles (B and b):

    • The homozygous dominant (BB) genotype is represented by p2

    • The heterozygous genotype (Bb) is represented by 2pq

    • The homozygous recessive genotype (bb) is represented by q2

  •  As these are all the possible genotypes of individuals in the population, the following equation can be constructed:

p2 + q2 + 2pq = 1

Worked Example

In a population of birds 10% of the individuals exhibit the recessive phenotype of white feathers. Calculate the frequencies of all genotypes.

Solution:

  • We will use F / f to represent dominant and recessive alleles for feather colour

  • Those with the recessive phenotype must have the homozygous recessive genotype, ff

  • Therefore q2 = 0.10 (as 10% of the individuals have the recessive phenotype and q2 represents this)

To calculate the frequencies of the homozygous dominant ( p2 ) and heterozygous ( 2pq ):

Step 1: Find q

Natural Selection Hardy-Weinberg Principle Worked Example equation

Step 2: Find p (the frequency of the dominant allele F). If q = 0.32, and p + q = 1

p + q = 1

p = 1 – 0.32

p = 0.68

Step 3: Find p2 (the frequency of homozygous dominant genotype)

0.682 = 0.46

p2 = 0.46

Step 4: Find 2pq = 2 x (p) x (q)

2 x (0.68) x (0.32) = 0.44

Step 5: Check calculations by substituting the values for the three frequencies into the equation; they should add up to 1

p2 + 2pq + q2 = 1

0.46 + 0.44 + 0.10 = 1

In summary:

  • Allele frequencies:

    • p = F = 0.68

    • q = f = 0.32

  • Genotype frequencies:

    • p2 = FF = 0.46

    • q2 = ff = 0.10

    • 2pq = Ff = 0.44

Examiner Tips and Tricks

Tips for approaching a Hardy-Weinberg calculation question

  1. Start with q²: Always begin by identifying the frequency of the recessive phenotype

    • This is the only genotype you can determine directly (must be homozygous recessive, bb)

  2. Identify known and unknown values:

    • Check what the question gives you (e.g. q² or p)

    • Work out what it asks for (e.g. 2pq)

    • Use the appropriate equation(s) to connect them

  3. Don’t confuse the equations with the principle:

    • The equations estimate allele and genotype frequencies

    • The principle assumes no change in allele frequencies between generations (i.e. genetic equilibrium)