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State the Hardy-Weinberg equation and define each term. In a population of 1000 individuals, 360 are homozygous dominant (AA), 480 are heterozygous (Aa), and 160 are homozygous recessive (aa). Calcula

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Here is the solution to question 6.

Part 6

Part a

State the Hardy-Weinberg equation and define each term. (04 marks)

The Hardy-Weinberg equations are:

p+q=1p + q = 1 p2+2pq+q2=1p^2 + 2pq + q^2 = 1

Where:

  • pp: Represents the frequency of the dominant allele (e.g., A).
  • qq: Represents the frequency of the recessive allele (e.g., a).
  • p2p^2: Represents the frequency of the homozygous dominant genotype (e.g., AA).
  • q2q^2: Represents the frequency of the homozygous recessive genotype (e.g., aa).
  • 2pq2pq: Represents the frequency of the heterozygous genotype (e.g., Aa).

Part b(i)· 4 steps

Allele frequencies of A and a. (04 marks)

In a population of 1000 individuals, 360 are homozygous dominant (AA), 480 are heterozygous (Aa), and 160 are homozygous recessive (aa). Calculate the

Total individuals = 1000
Number of AA = 360
Number of Aa = 480
Number of aa = 160

  1. Calculate the total number of alleles in the population.

    Total alleles = 2×Total individuals=2×1000=20002 \times \text{Total individuals} = 2 \times 1000 = 2000

  2. Calculate the number of A alleles.

    Number of A alleles = (2×Number of AA)+(1×Number of Aa)(2 \times \text{Number of AA}) + (1 \times \text{Number of Aa})
    Number of A alleles = (2×360)+(1×480)=720+480=1200(2 \times 360) + (1 \times 480) = 720 + 480 = 1200

  3. Calculate the frequency of allele A (pp).

    p=12002000=0.6p = \frac{1200}{2000} = 0.6

  4. Calculate the frequency of allele a (qq).

    q=1−p=1−0.6=0.4q = 1 - p = 1 - 0.6 = 0.4

    The allele frequencies are:

    • Frequency of A (pp): 0.6\boxed{\text{0.6}}
    • Frequency of a (qq): 0.4\boxed{\text{0.4}}

Part b(ii)· 1 step

Genotype frequencies of AA, Aa, and aa. (04 marks)

  1. Calculate the frequency for each genotype.

    Frequency of AA=3601000=0.36\text{Frequency of AA} = \frac{360}{1000} = 0.36 Frequency of Aa=4801000=0.48\text{Frequency of Aa} = \frac{480}{1000} = 0.48 Frequency of aa=1601000=0.16\text{Frequency of aa} = \frac{160}{1000} = 0.16

    The genotypic frequencies are:

    • Frequency of AA: 0.36\boxed{\text{0.36}}
    • Frequency of Aa: 0.48\boxed{\text{0.48}}
    • Frequency of aa: 0.16\boxed{\text{0.16}}

Part b(iii)· 2 steps

Phenotype frequencies of dominant and recessive traits. (04 marks)

  1. Calculate the frequency of the dominant phenotype (AA + Aa).

    Frequency of dominant phenotype=Frequency of AA+Frequency of Aa=0.36+0.48=0.84\text{Frequency of dominant phenotype} = \text{Frequency of AA} + \text{Frequency of Aa} = 0.36 + 0.48 = 0.84

  2. Calculate the frequency of the recessive phenotype (aa).

    Frequency of recessive phenotype=Frequency of aa=0.16\text{Frequency of recessive phenotype} = \text{Frequency of aa} = 0.16

    The phenotype frequencies are:

    • Frequency of dominant trait: 0.84\boxed{\text{0.84}}
    • Frequency of recessive trait: 0.16\boxed{\text{0.16}}

Part b(iv)· 3 steps

Number of homozygous dominant, heterozygous, and homozygous recessive individuals in the next generation, assuming the population is in Hardy-Weinberg equilibrium. (04 marks)

Using the allele frequencies p=0.6p = 0.6 and q=0.4q = 0.4 from part (i) and the total population of 1000 individuals:

  1. Calculate the expected frequency of homozygous dominant (AA) individuals (p2p^2).

    p2=(0.6)2=0.36p^2 = (0.6)^2 = 0.36

    Number of AA individuals = 0.36×1000=3600.36 \times 1000 = 360

  2. Calculate the expected frequency of heterozygous (Aa) individuals ($2pq$).

    2pq=2×0.6×0.4=0.482pq = 2 \times 0.6 \times 0.4 = 0.48

    Number of Aa individuals = 0.48×1000=4800.48 \times 1000 = 480

  3. Calculate the expected frequency of homozygous recessive (aa) individuals (q2q^2).

    q2=(0.4)2=0.16q^2 = (0.4)^2 = 0.16

    Number of aa individuals = 0.16×1000=1600.16 \times 1000 = 160

    The number of individuals in the next generation, assuming Hardy-Weinberg equilibrium, are:

    • Homozygous dominant (AA): 360 individuals\boxed{\text{360 individuals}}
    • Heterozygous (Aa): 480 individuals\boxed{\text{480 individuals}}
    • Homozygous recessive (aa): 160 individuals\boxed{\text{160 individuals}}
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