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iVinArrow [24]
3 years ago
15

A farmer is raising rabbits. The average body weight in his population is 3kg. The farmer selects the 10 largest rabbits in the

population, whose average body weight is 4kg, and interbreeds them. If the heritability of body weight in the rabbit population is 0.7, what is the expected body weight among the offspring of the selected rabbits?
Biology
1 answer:
lakkis [162]3 years ago
5 0

Answer:

The expected weight increase that is expected of offspring of selected parents is 0.7 kg

Explanation:

Given from the question:

The average body weight in his population of rabbits = 3 kg

heritability of body weight in the rabbit population (h^2)=0.7

Average body weight of the 10 selected largest rabbit =4 kg

(R) = response to selection

selection differential (S)= Average body weight of the 10 selected largest rabbit - The average body weight in his population of rabbits

selection differential (S)= 4-3=1 kg

But heritability of body weight in the rabbit population (h^2)=0.7

Then, Heritability of body weight in the rabbit population (h^2) can be calculated using the formula below;

Heritability of body weight in the rabbit population (h^2)=

response to selection (R) / selection differential (S)

response to selection (R) = Heritability of body weight in the rabbit population (h^2) × selection differential (S)

R= h^2 x S= 0.7 x 1kg= 0.7 kg

The expected weight increase that is expected of offspring of selected parents is 0.7 kg

Average weight of offspring = The average body weight in his population of rabbits + heritability of body weight in the rabbit population (h^2)

Average weight of offspring= 3 kg + 0.7 kg = 3.7kg

Therefore, the Average weight of offspring is 3.7kg and The expected weight increase that is expected of offspring of selected parents is 0.7 kg

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Beginning of term: a) 0.7; b) 0.3;

End of term: a) 0.67; b) 0.33; c) no

Explanation:

The population is in Hardy-Weinberg equilibrium, so the genotypic frequencies are:

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aa = q²

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<h3><u>Beginning of the term</u></h3>

Total population (N): 1000

aa individuals: 90

b)

q² = 90/1000

q= √0.09

q=0.3

The frequency of the a allele is 0.3

a) p + q = 1

p = 1 - 0.3

p= 0.7

The frequency of the A allele is 0.7

<u>The initial number of students with each genotype was:</u>

  • aa = 90
  • AA = p² × N = 0.7 × 1000= 490
  • Aa = 2 × p × q × N = 2 × 0.7 × 0.3 × 1000 = 420

<h3><u>End of the term</u></h3>

Number of students with each genotype:

  • AA = 490 - 280 = 210
  • Aa = 420
  • aa = 0

N = 630

a) The frequecny of the A allele can be calculated as:

p= (2 × AA + Aa)/ 2 × N

Because there are 2 × N number of alleles in the population and the A allele appears twice in the homozygous individuals and once in the heterozyogus.

p= (2 × 210 + 420)/ 2 × 630

p= 0.67

The frequency of the A allele is 0.67

q= 1-p =0.33

The frequency of the a allele is 0.33

c) The population is not at H-W equilibrium at the end of the term, because the observed genotypic frequencies are not those expected by the equilibrium.

Expected AA = p² × N = 0.67² × 630 = 283

Observed AA = 210

Expected Aa = 2pq × N = 2 × 0.67 × 0.33 × 630 = 279

Observed Aa= 420

Expected aa = q² × N = 0.33² × 630 = 69

Observed aa = 0

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