Question

In garden pea plants, tall stem length (T) is dominant over short stem length (t). In an ideal garden pea plant population exhibiting Hardy–Weinberg equilibrium, if allele T has a frequency of 0.88, what percentage of the population is recessive for this trait? Round your answer to the nearest whole number.

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There is a lack of dominance in black (B) and white (W) horse coat colours. An offspring produced from a black horse and a white horse has a blue-roan coat colour (BW). When the blue-roan coat is closely examined, both black and white hair can be found. In an ideal horse population exhibiting Hardy–Weinberg equilibrium, 24 out of every 150 foals have a white coat. Calculate the frequency of the black horse coat allele in the population. Express your answer using two significant digits.

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Grey eyes (b) are recessive to brown eyes (B) in rabbits. In an ideal rabbit population exhibiting Hardy–Weinberg equilibrium, if allele b has a frequency of 0.48, what percentage of the population is heterozygous for this trait? Round your answer to the nearest whole number.

Answer 1

Hardy-Weinberg equilibrium:

· Hardy-Weinberg equation describes: The most possible distribution of the population genotype of the population, if the frequencies of the alleles are known.

· If allele frequencies are p and q, then the genotype frequencies are p2, q2, 2pq.

· A population is considered to be in Hardy-Weinberg equilibrium when: In a randomly mating population, allele frequency and genotype frequency remain constant.

· Thus, under Hardy-Weinberg equilibrium:

p+ q=1 and

p2+2pq + q2 = 1.

1.

Frequency of dominant allele T= p = 0.88

Frequency of recessive allele t = q = 1- 0.88 = 0.12

• Genotype frequency with recessive trait (tt) = q2 = 0.0144
• Thus, percentage of population with recessive trait = 0.0144 x 100 = 1.44% = 1 % (approx)

2.

• Total = 150
• Numbers with white coats = 24
• Genotype frequency for white coat = 24/ 150 = 0.16 = q2
• Frequency of W (white allele) = q = 0.4
• Frequency of black = 1-0.4 = 0.6

3.

• Frequency of recessive b allele = q = 0.48
• Frequency of dominant B alllele = p =1- q = 1-0.48 =0.52
• Frequency of heterozygous = 2pq = 2. 0.48 x 0.52 = 0.4992
• Percentage of heterozygous = 0.4992 x 100 = 49.92% = 50% (approx)