BREEDER’S EQUATION FOR PREDICTING
RESPONSE TO SELECTION
1 MATHEMATICAL PROBLEMS ON BREEDER’S EQUATION
1.1 PROBLEM 1: BASIC RESPONSE TO SELECTION
The Breeder’s equation is given by R = h²S, where R is the response to selection, h² is the
narrow-sense heritability, and S is the selection differential. In a population of wheat, the
narrow-sense heritability for grain yield is 0.4, and the selection differential is 500 kg/ha.
Calculate the expected response to selection for grain yield.
Solution:
Given:
• h² = 0.4
• S = 500 kg/ha
Using the Breeder’s equation: R = h²S
R = 0.4 × 500 kg/ha = 200 kg/ha
Therefore, the expected response to selection for grain yield is 200 kg/ha.
1.2 PROBLEM 2: SELECTION INTENSITY
The Breeder’s equation can also be written as R = ih²σ_p, where i is the selection intensity, and
σ_p is the phenotypic standard deviation. In a dairy cattle breeding program, the narrow-sense
heritability for milk yield is 0.3, the phenotypic standard deviation is 1000 kg, and the selection
intensity is 1.4. Calculate the expected response to selection for milk yield.
Solution:
Given:
• h² = 0.3
• σ_p = 1000 kg
• i = 1.4
Using the Breeder’s equation: R = ih²σ_p
R = 1.4 × 0.3 × 1000 kg = 420 kg
Therefore, the expected response to selection for milk yield is 420 kg.
1.3 PROBLEM 3: CALCULATING HERITABILITY
In a selection experiment for plant height in peas, the selection differential was 10 cm, and the
response to selection was 4 cm. Using the Breeder’s equation, calculate the narrow-sense
heritability for plant height in this population.
Solution:
Given:
• S = 10 cm
• R = 4 cm
Using the Breeder’s equation: R = h²S
Rearranging to solve for h²: h² = R / S = 4 cm / 10 cm = 0.4
Therefore, the narrow-sense heritability for plant height in this population is 0.4 or 40
1.4 PROBLEM 4: PREDICTING MULTIPLE GENERATIONS OF SELECTION
Assume you are selecting for increased body weight in a population of chickens. The narrow-
sense heritability for body weight is 0.5, and you can achieve a selection differential of 200 g
per generation. Predict the cumulative response to selection after 3 generations, assuming the
heritability remains constant.
Solution:
Given:
• h² = 0.5
• S = 200 g per generation
• Number of generations = 3
Step 1: Calculate the response to selection per generation R = h²S = 0.5 × 200 g = 100 g
Step 2: Calculate the cumulative response over 3 generations Cumulative R = 3 × 100 g = 300
g
Therefore, the predicted cumulative response to selection after 3 generations is 300 g.
1.5 PROBLEM 5: CORRELATED RESPONSE TO SELECTION
The correlated response (CR_y) in trait y when selecting for trait x is given by the equation:
CR_y = ih_x h_y r_g σ_py
Where i is the selection intensity for trait x, h_x and h_y are the square roots of the
heritabilities for traits x and y respectively, r_g is the genetic correlation between traits x and y,
and σ_py is the phenotypic standard deviation of trait y.
In a tomato breeding program, you are selecting for increased fruit weight (trait x) but are also
interested in the correlated response in sugar content (trait y). Given:
• Selection intensity for fruit weight (i) = 1.2
• Heritability of fruit weight (h²_x) = 0.6
• Heritability of sugar content (h²_y) = 0.4
• Genetic correlation between fruit weight and sugar content (r_g) = -0.3
• Phenotypic standard deviation of sugar content (σ_py) = 0.5 °Brix
Calculate the expected correlated response in sugar content when selecting for increased fruit
weight.
Solution:
Step 1: Calculate the square roots of heritabilities h_x = √(0.6) = 0.775 h_y = √(0.4) = 0.632
Step 2: Apply the correlated response equation CR_y = ih_x h_y r_g σ_py CR_y = 1.2 × 0.775
× 0.632 × (-0.3) × 0.5 °Brix CR_y = -0.088 °Brix
Therefore, when selecting for increased fruit weight, the expected correlated response in sugar
content is a decrease of 0.088 °Brix.
1.6 PROBLEM 1: BASIC RESPONSE TO SELECTION
The Breeder’s equation is given by R = h²S, where R is the response to selection, h² is the
narrow-sense heritability, and S is the selection differential. In a population of wheat, the
narrow-sense heritability for grain yield is 0.4, and the selection differential is 500 kg/ha.
Calculate the expected response to selection for grain yield.
Solution:
Given:
• h² = 0.4
• S = 500 kg/ha
Using the Breeder’s equation: R = h²S
R = 0.4 × 500 kg/ha = 200 kg/ha
Therefore, the expected response to selection for grain yield is 200 kg/ha.
1.7 PROBLEM 2: SELECTION INTENSITY
The Breeder’s equation can also be written as R = ih²σ_p, where i is the selection intensity, and
σ_p is the phenotypic standard deviation. In a dairy cattle breeding program, the narrow-sense
heritability for milk yield is 0.3, the phenotypic standard deviation is 1000 kg, and the selection
intensity is 1.4. Calculate the expected response to selection for milk yield.
Solution:
Given:
• h² = 0.3
• σ_p = 1000 kg
• i = 1.4
Using the Breeder’s equation: R = ih²σ_p
R = 1.4 × 0.3 × 1000 kg = 420 kg
Therefore, the expected response to selection for milk yield is 420 kg.
1.8 PROBLEM 3: CALCULATING HERITABILITY
In a selection experiment for plant height in peas, the selection differential was 10 cm, and the
response to selection was 4 cm. Using the Breeder’s equation, calculate the narrow-sense
heritability for plant height in this population.
Solution:
Given:
• S = 10 cm
• R = 4 cm
Using the Breeder’s equation: R = h²S
Rearranging to solve for h²: h² = R / S = 4 cm / 10 cm = 0.4
Therefore, the narrow-sense heritability for plant height in this population is 0.4 or 40
1.9 PROBLEM 4: PREDICTING MULTIPLE GENERATIONS OF SELECTION
Assume you are selecting for increased body weight in a population of chickens. The narrow-
sense heritability for body weight is 0.5, and you can achieve a selection differential of 200 g
per generation. Predict the cumulative response to selection after 3 generations, assuming the
heritability remains constant.
Solution:
Given:
• h² = 0.5
• S = 200 g per generation
• Number of generations = 3
Step 1: Calculate the response to selection per generation R = h²S = 0.5 × 200 g = 100 g
Step 2: Calculate the cumulative response over 3 generations Cumulative R = 3 × 100 g = 300
g
Therefore, the predicted cumulative response to selection after 3 generations is 300 g.
1.10 PROBLEM 5: CORRELATED RESPONSE TO SELECTION
The correlated response (CR_y) in trait y when selecting for trait x is given by the equation:
CR_y = ih_x h_y r_g σ_py
Where i is the selection intensity for trait x, h_x and h_y are the square roots of the
heritabilities for traits x and y respectively, r_g is the genetic correlation between traits x and y,
and σ_py is the phenotypic standard deviation of trait y.
In a tomato breeding program, you are selecting for increased fruit weight (trait x) but are also
interested in the correlated response in sugar content (trait y). Given:
• Selection intensity for fruit weight (i) = 1.2
• Heritability of fruit weight (h²_x) = 0.6
• Heritability of sugar content (h²_y) = 0.4
• Genetic correlation between fruit weight and sugar content (r_g) = -0.3
• Phenotypic standard deviation of sugar content (σ_py) = 0.5 °Brix
Calculate the expected correlated response in sugar content when selecting for increased fruit
weight.
Solution:
Step 1: Calculate the square roots of heritabilities h_x = √(0.6) = 0.775 h_y = √(0.4) = 0.632
Step 2: Apply the correlated response equation CR_y = ih_x h_y r_g σ_py CR_y = 1.2 × 0.775
× 0.632 × (-0.3) × 0.5 °Brix CR_y = -0.088 °Brix
Therefore, when selecting for increased fruit weight, the expected correlated response in sugar
content is a decrease of 0.088 °Brix.
1.11 PROBLEM 1: BASIC RESPONSE TO SELECTION
The Breeder’s equation is given by R = h²S, where R is the response to selection, h² is the
narrow-sense heritability, and S is the selection differential. In a population of wheat, the
narrow-sense heritability for grain yield is 0.4, and the selection differential is 500 kg/ha.
Calculate the expected response to selection for grain yield.
Solution:
Given:
• h² = 0.4
• S = 500 kg/ha
Using the Breeder’s equation: R = h²S
R = 0.4 × 500 kg/ha = 200 kg/ha
Therefore, the expected response to selection for grain yield is 200 kg/ha.
1.12 PROBLEM 2: SELECTION INTENSITY
The Breeder’s equation can also be written as R = ih²σ_p, where i is the selection intensity, and
σ_p is the phenotypic standard deviation. In a dairy cattle breeding program, the narrow-sense
heritability for milk yield is 0.3, the phenotypic standard deviation is 1000 kg, and the selection
intensity is 1.4. Calculate the expected response to selection for milk yield.
Solution:
Given:
• h² = 0.3
• σ_p = 1000 kg
• i = 1.4
Using the Breeder’s equation: R = ih²σ_p
R = 1.4 × 0.3 × 1000 kg = 420 kg
Therefore, the expected response to selection for milk yield is 420 kg.
1.13 PROBLEM 3: CALCULATING HERITABILITY
In a selection experiment for plant height in peas, the selection differential was 10 cm, and the
response to selection was 4 cm. Using the Breeder’s equation, calculate the narrow-sense
heritability for plant height in this population.
Solution:
Given:
• S = 10 cm
• R = 4 cm
Using the Breeder’s equation: R = h²S
Rearranging to solve for h²: h² = R / S = 4 cm / 10 cm = 0.4
Therefore, the narrow-sense heritability for plant height in this population is 0.4 or 40
1.14 PROBLEM 4: PREDICTING MULTIPLE GENERATIONS OF SELECTION
Assume you are selecting for increased body weight in a population of chickens. The narrow-
sense heritability for body weight is 0.5, and you can achieve a selection differential of 200 g
per generation. Predict the cumulative response to selection after 3 generations, assuming the
heritability remains constant.
Solution:
Given:
• h² = 0.5
• S = 200 g per generation
• Number of generations = 3
Step 1: Calculate the response to selection per generation R = h²S = 0.5 × 200 g = 100 g
Step 2: Calculate the cumulative response over 3 generations Cumulative R = 3 × 100 g = 300
g
Therefore, the predicted cumulative response to selection after 3 generations is 300 g.
1.15 PROBLEM 5: CORRELATED RESPONSE TO SELECTION
The correlated response (CR_y) in trait y when selecting for trait x is given by the equation:
CR_y = ih_x h_y r_g σ_py
Where i is the selection intensity for trait x, h_x and h_y are the square roots of the
heritabilities for traits x and y respectively, r_g is the genetic correlation between traits x and y,
and σ_py is the phenotypic standard deviation of trait y.
In a tomato breeding program, you are selecting for increased fruit weight (trait x) but are also
interested in the correlated response in sugar content (trait y). Given:
• Selection intensity for fruit weight (i) = 1.2
• Heritability of fruit weight (h²_x) = 0.6
• Heritability of sugar content (h²_y) = 0.4
• Genetic correlation between fruit weight and sugar content (r_g) = -0.3
• Phenotypic standard deviation of sugar content (σ_py) = 0.5 °Brix
Calculate the expected correlated response in sugar content when selecting for increased fruit
weight.
Solution:
Step 1: Calculate the square roots of heritabilities h_x = √(0.6) = 0.775 h_y = √(0.4) = 0.632
Step 2: Apply the correlated response equation CR_y = ih_x h_y r_g σ_py CR_y = 1.2 × 0.775
× 0.632 × (-0.3) × 0.5 °Brix CR_y = -0.088 °Brix
Therefore, when selecting for increased fruit weight, the expected correlated response in sugar
content is a decrease of 0.088 °Brix.
1.16 PROBLEM 1: BASIC RESPONSE TO SELECTION
The Breeder’s equation is given by R = h²S, where R is the response to selection, h² is the
narrow-sense heritability, and S is the selection differential. In a population of wheat, the
narrow-sense heritability for grain yield is 0.4, and the selection differential is 500 kg/ha.
Calculate the expected response to selection for grain yield.
Solution:
Given:
• h² = 0.4
• S = 500 kg/ha
Using the Breeder’s equation: R = h²S
R = 0.4 × 500 kg/ha = 200 kg/ha
Therefore, the expected response to selection for grain yield is 200 kg/ha.
1.17 PROBLEM 2: SELECTION INTENSITY
The Breeder’s equation can also be written as R = ih²σ_p, where i is the selection intensity, and
σ_p is the phenotypic standard deviation. In a dairy cattle breeding program, the narrow-sense
heritability for milk yield is 0.3, the phenotypic standard deviation is 1000 kg, and the selection
intensity is 1.4. Calculate the expected response to selection for milk yield.
Solution:
Given:
• h² = 0.3
• σ_p = 1000 kg
• i = 1.4
Using the Breeder’s equation: R = ih²σ_p
R = 1.4 × 0.3 × 1000 kg = 420 kg
Therefore, the expected response to selection for milk yield is 420 kg.
1.18 PROBLEM 3: CALCULATING HERITABILITY
In a selection experiment for plant height in peas, the selection differential was 10 cm, and the
response to selection was 4 cm. Using the Breeder’s equation, calculate the narrow-sense
heritability for plant height in this population.
Solution:
Given:
• S = 10 cm
• R = 4 cm
Using the Breeder’s equation: R = h²S
Rearranging to solve for h²: h² = R / S = 4 cm / 10 cm = 0.4
Therefore, the narrow-sense heritability for plant height in this population is 0.4 or 40
1.19 PROBLEM 4: PREDICTING MULTIPLE GENERATIONS OF SELECTION
Assume you are selecting for increased body weight in a population of chickens. The narrow-
sense heritability for body weight is 0.5, and you can achieve a selection differential of 200 g
per generation. Predict the cumulative response to selection after 3 generations, assuming the
heritability remains constant.
Solution:
Given:
• h² = 0.5
• S = 200 g per generation
• Number of generations = 3
Step 1: Calculate the response to selection per generation R = h²S = 0.5 × 200 g = 100 g
Step 2: Calculate the cumulative response over 3 generations Cumulative R = 3 × 100 g = 300
g
Therefore, the predicted cumulative response to selection after 3 generations is 300 g.
1.20 PROBLEM 5: CORRELATED RESPONSE TO SELECTION
The correlated response (CR_y) in trait y when selecting for trait x is given by the equation:
CR_y = ih_x h_y r_g σ_py
Where i is the selection intensity for trait x, h_x and h_y are the square roots of the
heritabilities for traits x and y respectively, r_g is the genetic correlation between traits x and y,
and σ_py is the phenotypic standard deviation of trait y.
In a tomato breeding program, you are selecting for increased fruit weight (trait x) but are also
interested in the correlated response in sugar content (trait y). Given:
• Selection intensity for fruit weight (i) = 1.2
• Heritability of fruit weight (h²_x) = 0.6
• Heritability of sugar content (h²_y) = 0.4
• Genetic correlation between fruit weight and sugar content (r_g) = -0.3
• Phenotypic standard deviation of sugar content (σ_py) = 0.5 °Brix
Calculate the expected correlated response in sugar content when selecting for increased fruit
weight.
Solution:
Step 1: Calculate the square roots of heritabilities h_x = √(0.6) = 0.775 h_y = √(0.4) = 0.632
Step 2: Apply the correlated response equation CR_y = ih_x h_y r_g σ_py CR_y = 1.2 × 0.775
× 0.632 × (-0.3) × 0.5 °Brix CR_y = -0.088 °Brix
Therefore, when selecting for increased fruit weight, the expected correlated response in sugar
content is a decrease of 0.088 °Brix.
1.21 PROBLEM 1: BASIC RESPONSE TO SELECTION
The Breeder’s equation is given by R = h²S, where R is the response to selection, h² is the
narrow-sense heritability, and S is the selection differential. In a population of wheat, the
narrow-sense heritability for grain yield is 0.4, and the selection differential is 500 kg/ha.
Calculate the expected response to selection for grain yield.
Solution:
Given:
• h² = 0.4
• S = 500 kg/ha
Using the Breeder’s equation: R = h²S
R = 0.4 × 500 kg/ha = 200 kg/ha
Therefore, the expected response to selection for grain yield is 200 kg/ha.
1.22 PROBLEM 2: SELECTION INTENSITY
The Breeder’s equation can also be written as R = ih²σ_p, where i is the selection intensity, and
σ_p is the phenotypic standard deviation. In a dairy cattle breeding program, the narrow-sense
heritability for milk yield is 0.3, the phenotypic standard deviation is 1000 kg, and the selection
intensity is 1.4. Calculate the expected response to selection for milk yield.
Solution:
Given:
• h² = 0.3
• σ_p = 1000 kg
• i = 1.4
Using the Breeder’s equation: R = ih²σ_p
R = 1.4 × 0.3 × 1000 kg = 420 kg
Therefore, the expected response to selection for milk yield is 420 kg.
1.23 PROBLEM 3: CALCULATING HERITABILITY
In a selection experiment for plant height in peas, the selection differential was 10 cm, and the
response to selection was 4 cm. Using the Breeder’s equation, calculate the narrow-sense
heritability for plant height in this population.
Solution:
Given:
• S = 10 cm
• R = 4 cm
Using the Breeder’s equation: R = h²S
Rearranging to solve for h²: h² = R / S = 4 cm / 10 cm = 0.4
Therefore, the narrow-sense heritability for plant height in this population is 0.4 or 40
1.24 PROBLEM 4: PREDICTING MULTIPLE GENERATIONS OF SELECTION
Assume you are selecting for increased body weight in a population of chickens. The narrow-
sense heritability for body weight is 0.5, and you can achieve a selection differential of 200 g
per generation. Predict the cumulative response to selection after 3 generations, assuming the
heritability remains constant.
Solution:
Given:
• h² = 0.5
• S = 200 g per generation
• Number of generations = 3
Step 1: Calculate the response to selection per generation R = h²S = 0.5 × 200 g = 100 g
Step 2: Calculate the cumulative response over 3 generations Cumulative R = 3 × 100 g = 300
g
Therefore, the predicted cumulative response to selection after 3 generations is 300 g.
1.25 PROBLEM 5: CORRELATED RESPONSE TO SELECTION
The correlated response (CR_y) in trait y when selecting for trait x is given by the equation:
CR_y = ih_x h_y r_g σ_py
Where i is the selection intensity for trait x, h_x and h_y are the square roots of the
heritabilities for traits x and y respectively, r_g is the genetic correlation between traits x and y,
and σ_py is the phenotypic standard deviation of trait y.
In a tomato breeding program, you are selecting for increased fruit weight (trait x) but are also
interested in the correlated response in sugar content (trait y). Given:
• Selection intensity for fruit weight (i) = 1.2
• Heritability of fruit weight (h²_x) = 0.6
• Heritability of sugar content (h²_y) = 0.4
• Genetic correlation between fruit weight and sugar content (r_g) = -0.3
• Phenotypic standard deviation of sugar content (σ_py) = 0.5 °Brix
Calculate the expected correlated response in sugar content when selecting for increased fruit
weight.
Solution:
Step 1: Calculate the square roots of heritabilities h_x = √(0.6) = 0.775 h_y = √(0.4) = 0.632
Step 2: Apply the correlated response equation CR_y = ih_x h_y r_g σ_py CR_y = 1.2 × 0.775
× 0.632 × (-0.3) × 0.5 °Brix CR_y = -0.088 °Brix
Therefore, when selecting for increased fruit weight, the expected correlated response in sugar
content is a decrease of 0.088 °Brix.
1.26 PROBLEM 1: BASIC RESPONSE TO SELECTION
The Breeder’s equation is given by R = h²S, where R is the response to selection, h² is the
narrow-sense heritability, and S is the selection differential. In a population of wheat, the
narrow-sense heritability for grain yield is 0.4, and the selection differential is 500 kg/ha.
Calculate the expected response to selection for grain yield.
Solution:
Given:
• h² = 0.4
• S = 500 kg/ha
Using the Breeder’s equation: R = h²S
R = 0.4 × 500 kg/ha = 200 kg/ha
Therefore, the expected response to selection for grain yield is 200 kg/ha.
1.27 PROBLEM 2: SELECTION INTENSITY
The Breeder’s equation can also be written as R = ih²σ_p, where i is the selection intensity, and
σ_p is the phenotypic standard deviation. In a dairy cattle breeding program, the narrow-sense
heritability for milk yield is 0.3, the phenotypic standard deviation is 1000 kg, and the selection
intensity is 1.4. Calculate the expected response to selection for milk yield.
Solution:
Given:
• h² = 0.3
• σ_p = 1000 kg
• i = 1.4
Using the Breeder’s equation: R = ih²σ_p
R = 1.4 × 0.3 × 1000 kg = 420 kg
Therefore, the expected response to selection for milk yield is 420 kg.
1.28 PROBLEM 3: CALCULATING HERITABILITY
In a selection experiment for plant height in peas, the selection differential was 10 cm, and the
response to selection was 4 cm. Using the Breeder’s equation, calculate the narrow-sense
heritability for plant height in this population.
Solution:
Given:
• S = 10 cm
• R = 4 cm
Using the Breeder’s equation: R = h²S
Rearranging to solve for h²: h² = R / S = 4 cm / 10 cm = 0.4
Therefore, the narrow-sense heritability for plant height in this population is 0.4 or 40
1.29 PROBLEM 4: PREDICTING MULTIPLE GENERATIONS OF SELECTION
Assume you are selecting for increased body weight in a population of chickens. The narrow-
sense heritability for body weight is 0.5, and you can achieve a selection differential of 200 g
per generation. Predict the cumulative response to selection after 3 generations, assuming the
heritability remains constant.
Solution:
Given:
• h² = 0.5
• S = 200 g per generation
• Number of generations = 3
Step 1: Calculate the response to selection per generation R = h²S = 0.5 × 200 g = 100 g
Step 2: Calculate the cumulative response over 3 generations Cumulative R = 3 × 100 g = 300
g
Therefore, the predicted cumulative response to selection after 3 generations is 300 g.
1.30 PROBLEM 5: CORRELATED RESPONSE TO SELECTION
The correlated response (CR_y) in trait y when selecting for trait x is given by the equation:
CR_y = ih_x h_y r_g σ_py
Where i is the selection intensity for trait x, h_x and h_y are the square roots of the
heritabilities for traits x and y respectively, r_g is the genetic correlation between traits x and y,
and σ_py is the phenotypic standard deviation of trait y.
In a tomato breeding program, you are selecting for increased fruit weight (trait x) but are also
interested in the correlated response in sugar content (trait y). Given:
• Selection intensity for fruit weight (i) = 1.2
• Heritability of fruit weight (h²_x) = 0.6
• Heritability of sugar content (h²_y) = 0.4
• Genetic correlation between fruit weight and sugar content (r_g) = -0.3
• Phenotypic standard deviation of sugar content (σ_py) = 0.5 °Brix
Calculate the expected correlated response in sugar content when selecting for increased fruit
weight.
Solution:
Step 1: Calculate the square roots of heritabilities h_x = √(0.6) = 0.775 h_y = √(0.4) = 0.632
Step 2: Apply the correlated response equation CR_y = ih_x h_y r_g σ_py CR_y = 1.2 × 0.775
× 0.632 × (-0.3) × 0.5 °Brix CR_y = -0.088 °Brix
Therefore, when selecting for increased fruit weight, the expected correlated response in sugar
content is a decrease of 0.088 °Brix.
1.31 PROBLEM 1: BASIC RESPONSE TO SELECTION
The Breeder’s equation is given by R = h²S, where R is the response to selection, h² is the
narrow-sense heritability, and S is the selection differential. In a population of wheat, the
narrow-sense heritability for grain yield is 0.4, and the selection differential is 500 kg/ha.
Calculate the expected response to selection for grain yield.
Solution:
Given:
• h² = 0.4
• S = 500 kg/ha
Using the Breeder’s equation: R = h²S
R = 0.4 × 500 kg/ha = 200 kg/ha
Therefore, the expected response to selection for grain yield is 200 kg/ha.
1.32 PROBLEM 2: SELECTION INTENSITY
The Breeder’s equation can also be written as R = ih²σ_p, where i is the selection intensity, and
σ_p is the phenotypic standard deviation. In a dairy cattle breeding program, the narrow-sense
heritability for milk yield is 0.3, the phenotypic standard deviation is 1000 kg, and the selection
intensity is 1.4. Calculate the expected response to selection for milk yield.
Solution:
Given:
• h² = 0.3
• σ_p = 1000 kg
• i = 1.4
Using the Breeder’s equation: R = ih²σ_p
R = 1.4 × 0.3 × 1000 kg = 420 kg
Therefore, the expected response to selection for milk yield is 420 kg.
1.33 PROBLEM 3: CALCULATING HERITABILITY
In a selection experiment for plant height in peas, the selection differential was 10 cm, and the
response to selection was 4 cm. Using the Breeder’s equation, calculate the narrow-sense
heritability for plant height in this population.
Solution:
Given:
• S = 10 cm
• R = 4 cm
Using the Breeder’s equation: R = h²S
Rearranging to solve for h²: h² = R / S = 4 cm / 10 cm = 0.4
Therefore, the narrow-sense heritability for plant height in this population is 0.4 or 40
1.34 PROBLEM 4: PREDICTING MULTIPLE GENERATIONS OF SELECTION
Assume you are selecting for increased body weight in a population of chickens. The narrow-
sense heritability for body weight is 0.5, and you can achieve a selection differential of 200 g
per generation. Predict the cumulative response to selection after 3 generations, assuming the
heritability remains constant.
Solution:
Given:
• h² = 0.5
• S = 200 g per generation
• Number of generations = 3
Step 1: Calculate the response to selection per generation R = h²S = 0.5 × 200 g = 100 g
Step 2: Calculate the cumulative response over 3 generations Cumulative R = 3 × 100 g = 300
g
Therefore, the predicted cumulative response to selection after 3 generations is 300 g.
1.35 PROBLEM 5: CORRELATED RESPONSE TO SELECTION
The correlated response (CR_y) in trait y when selecting for trait x is given by the equation:
CR_y = ih_x h_y r_g σ_py
Where i is the selection intensity for trait x, h_x and h_y are the square roots of the
heritabilities for traits x and y respectively, r_g is the genetic correlation between traits x and y,
and σ_py is the phenotypic standard deviation of trait y.
In a tomato breeding program, you are selecting for increased fruit weight (trait x) but are also
interested in the correlated response in sugar content (trait y). Given:
• Selection intensity for fruit weight (i) = 1.2
• Heritability of fruit weight (h²_x) = 0.6
• Heritability of sugar content (h²_y) = 0.4
• Genetic correlation between fruit weight and sugar content (r_g) = -0.3
• Phenotypic standard deviation of sugar content (σ_py) = 0.5 °Brix
Calculate the expected correlated response in sugar content when selecting for increased fruit
weight.
Solution:
Step 1: Calculate the square roots of heritabilities h_x = √(0.6) = 0.775 h_y = √(0.4) = 0.632
Step 2: Apply the correlated response equation CR_y = ih_x h_y r_g σ_py CR_y = 1.2 × 0.775
× 0.632 × (-0.3) × 0.5 °Brix CR_y = -0.088 °Brix
Therefore, when selecting for increased fruit weight, the expected correlated response in sugar
content is a decrease of 0.088 °Brix.
1.36 PROBLEM 1: BASIC RESPONSE TO SELECTION
The Breeder’s equation is given by R = h²S, where R is the response to selection, h² is the
narrow-sense heritability, and S is the selection differential. In a population of wheat, the
narrow-sense heritability for grain yield is 0.4, and the selection differential is 500 kg/ha.
Calculate the expected response to selection for grain yield.
Solution:
Given:
• h² = 0.4
• S = 500 kg/ha
Using the Breeder’s equation: R = h²S
R = 0.4 × 500 kg/ha = 200 kg/ha
Therefore, the expected response to selection for grain yield is 200 kg/ha.
1.37 PROBLEM 2: SELECTION INTENSITY
The Breeder’s equation can also be written as R = ih²σ_p, where i is the selection intensity, and
σ_p is the phenotypic standard deviation. In a dairy cattle breeding program, the narrow-sense
heritability for milk yield is 0.3, the phenotypic standard deviation is 1000 kg, and the selection
intensity is 1.4. Calculate the expected response to selection for milk yield.
Solution:
Given:
• h² = 0.3
• σ_p = 1000 kg
• i = 1.4
Using the Breeder’s equation: R = ih²σ_p
R = 1.4 × 0.3 × 1000 kg = 420 kg
Therefore, the expected response to selection for milk yield is 420 kg.
1.38 PROBLEM 3: CALCULATING HERITABILITY
In a selection experiment for plant height in peas, the selection differential was 10 cm, and the
response to selection was 4 cm. Using the Breeder’s equation, calculate the narrow-sense
heritability for plant height in this population.
Solution:
Given:
• S = 10 cm
• R = 4 cm
Using the Breeder’s equation: R = h²S
Rearranging to solve for h²: h² = R / S = 4 cm / 10 cm = 0.4
Therefore, the narrow-sense heritability for plant height in this population is 0.4 or 40
1.39 PROBLEM 4: PREDICTING MULTIPLE GENERATIONS OF SELECTION
Assume you are selecting for increased body weight in a population of chickens. The narrow-
sense heritability for body weight is 0.5, and you can achieve a selection differential of 200 g
per generation. Predict the cumulative response to selection after 3 generations, assuming the
heritability remains constant.
Solution:
Given:
• h² = 0.5
• S = 200 g per generation
• Number of generations = 3
Step 1: Calculate the response to selection per generation R = h²S = 0.5 × 200 g = 100 g
Step 2: Calculate the cumulative response over 3 generations Cumulative R = 3 × 100 g = 300
g
Therefore, the predicted cumulative response to selection after 3 generations is 300 g.
1.40 PROBLEM 5: CORRELATED RESPONSE TO SELECTION
The correlated response (CR_y) in trait y when selecting for trait x is given by the equation:
CR_y = ih_x h_y r_g σ_py
Where i is the selection intensity for trait x, h_x and h_y are the square roots of the
heritabilities for traits x and y respectively, r_g is the genetic correlation between traits x and y,
and σ_py is the phenotypic standard deviation of trait y.
In a tomato breeding program, you are selecting for increased fruit weight (trait x) but are also
interested in the correlated response in sugar content (trait y). Given:
• Selection intensity for fruit weight (i) = 1.2
• Heritability of fruit weight (h²_x) = 0.6
• Heritability of sugar content (h²_y) = 0.4
• Genetic correlation between fruit weight and sugar content (r_g) = -0.3
• Phenotypic standard deviation of sugar content (σ_py) = 0.5 °Brix
Calculate the expected correlated response in sugar content when selecting for increased fruit
weight.
Solution:
Step 1: Calculate the square roots of heritabilities h_x = √(0.6) = 0.775 h_y = √(0.4) = 0.632
Step 2: Apply the correlated response equation CR_y = ih_x h_y r_g σ_py CR_y = 1.2 × 0.775
× 0.632 × (-0.3) × 0.5 °Brix CR_y = -0.088 °Brix
Therefore, when selecting for increased fruit weight, the expected correlated response in sugar
content is a decrease of 0.088 °Brix.
1.41 PROBLEM 1: BASIC RESPONSE TO SELECTION
The Breeder’s equation is given by R = h²S, where R is the response to selection, h² is the
narrow-sense heritability, and S is the selection differential. In a population of wheat, the
narrow-sense heritability for grain yield is 0.4, and the selection differential is 500 kg/ha.
Calculate the expected response to selection for grain yield.
Solution:
Given:
• h² = 0.4
• S = 500 kg/ha
Using the Breeder’s equation: R = h²S
R = 0.4 × 500 kg/ha = 200 kg/ha
Therefore, the expected response to selection for grain yield is 200 kg/ha.
1.42 PROBLEM 2: SELECTION INTENSITY
The Breeder’s equation can also be written as R = ih²σ_p, where i is the selection intensity, and
σ_p is the phenotypic standard deviation. In a dairy cattle breeding program, the narrow-sense
heritability for milk yield is 0.3, the phenotypic standard deviation is 1000 kg, and the selection
intensity is 1.4. Calculate the expected response to selection for milk yield.
Solution:
Given:
• h² = 0.3
• σ_p = 1000 kg
• i = 1.4
Using the Breeder’s equation: R = ih²σ_p
R = 1.4 × 0.3 × 1000 kg = 420 kg
Therefore, the expected response to selection for milk yield is 420 kg.
1.43 PROBLEM 3: CALCULATING HERITABILITY
In a selection experiment for plant height in peas, the selection differential was 10 cm, and the
response to selection was 4 cm. Using the Breeder’s equation, calculate the narrow-sense
heritability for plant height in this population.
Solution:
Given:
• S = 10 cm
• R = 4 cm
Using the Breeder’s equation: R = h²S
Rearranging to solve for h²: h² = R / S = 4 cm / 10 cm = 0.4
Therefore, the narrow-sense heritability for plant height in this population is 0.4 or 40
1.44 PROBLEM 4: PREDICTING MULTIPLE GENERATIONS OF SELECTION
Assume you are selecting for increased body weight in a population of chickens. The narrow-
sense heritability for body weight is 0.5, and you can achieve a selection differential of 200 g
per generation. Predict the cumulative response to selection after 3 generations, assuming the
heritability remains constant.
Solution:
Given:
• h² = 0.5
• S = 200 g per generation
• Number of generations = 3
Step 1: Calculate the response to selection per generation R = h²S = 0.5 × 200 g = 100 g
Step 2: Calculate the cumulative response over 3 generations Cumulative R = 3 × 100 g = 300
g
Therefore, the predicted cumulative response to selection after 3 generations is 300 g.
1.45 PROBLEM 5: CORRELATED RESPONSE TO SELECTION
The correlated response (CR_y) in trait y when selecting for trait x is given by the equation:
CR_y = ih_x h_y r_g σ_py
Where i is the selection intensity for trait x, h_x and h_y are the square roots of the
heritabilities for traits x and y respectively, r_g is the genetic correlation between traits x and y,
and σ_py is the phenotypic standard deviation of trait y.
In a tomato breeding program, you are selecting for increased fruit weight (trait x) but are also
interested in the correlated response in sugar content (trait y). Given:
• Selection intensity for fruit weight (i) = 1.2
• Heritability of fruit weight (h²_x) = 0.6
• Heritability of sugar content (h²_y) = 0.4
• Genetic correlation between fruit weight and sugar content (r_g) = -0.3
• Phenotypic standard deviation of sugar content (σ_py) = 0.5 °Brix
Calculate the expected correlated response in sugar content when selecting for increased fruit
weight.
Solution:
Step 1: Calculate the square roots of heritabilities h_x = √(0.6) = 0.775 h_y = √(0.4) = 0.632
Step 2: Apply the correlated response equation CR_y = ih_x h_y r_g σ_py CR_y = 1.2 × 0.775
× 0.632 × (-0.3) × 0.5 °Brix CR_y = -0.088 °Brix
Therefore, when selecting for increased fruit weight, the expected correlated response in sugar
content is a decrease of 0.088 °Brix.
1.46 PROBLEM 1: BASIC RESPONSE TO SELECTION
The Breeder’s equation is given by R = h²S, where R is the response to selection, h² is the
narrow-sense heritability, and S is the selection differential. In a population of wheat, the
narrow-sense heritability for grain yield is 0.4, and the selection differential is 500 kg/ha.
Calculate the expected response to selection for grain yield.
Solution:
Given:
• h² = 0.4
• S = 500 kg/ha
Using the Breeder’s equation: R = h²S
R = 0.4 × 500 kg/ha = 200 kg/ha
Therefore, the expected response to selection for grain yield is 200 kg/ha.
1.47 PROBLEM 2: SELECTION INTENSITY
The Breeder’s equation can also be written as R = ih²σ_p, where i is the selection intensity, and
σ_p is the phenotypic standard deviation. In a dairy cattle breeding program, the narrow-sense
heritability for milk yield is 0.3, the phenotypic standard deviation is 1000 kg, and the selection
intensity is 1.4. Calculate the expected response to selection for milk yield.
Solution:
Given:
• h² = 0.3
• σ_p = 1000 kg
• i = 1.4
Using the Breeder’s equation: R = ih²σ_p
R = 1.4 × 0.3 × 1000 kg = 420 kg
Therefore, the expected response to selection for milk yield is 420 kg.
1.48 PROBLEM 3: CALCULATING HERITABILITY
In a selection experiment for plant height in peas, the selection differential was 10 cm, and the
response to selection was 4 cm. Using the Breeder’s equation, calculate the narrow-sense
heritability for plant height in this population.
Solution:
Given:
• S = 10 cm
• R = 4 cm
Using the Breeder’s equation: R = h²S
Rearranging to solve for h²: h² = R / S = 4 cm / 10 cm = 0.4
Therefore, the narrow-sense heritability for plant height in this population is 0.4 or 40
1.49 PROBLEM 4: PREDICTING MULTIPLE GENERATIONS OF SELECTION
Assume you are selecting for increased body weight in a population of chickens. The narrow-
sense heritability for body weight is 0.5, and you can achieve a selection differential of 200 g
per generation. Predict the cumulative response to selection after 3 generations, assuming the
heritability remains constant.
Solution:
Given:
• h² = 0.5
• S = 200 g per generation
• Number of generations = 3
Step 1: Calculate the response to selection per generation R = h²S = 0.5 × 200 g = 100 g
Step 2: Calculate the cumulative response over 3 generations Cumulative R = 3 × 100 g = 300
g
Therefore, the predicted cumulative response to selection after 3 generations is 300 g.
1.50 PROBLEM 5: CORRELATED RESPONSE TO SELECTION
The correlated response (CR_y) in trait y when selecting for trait x is given by the equation:
CR_y = ih_x h_y r_g σ_py
Where i is the selection intensity for trait x, h_x and h_y are the square roots of the
heritabilities for traits x and y respectively, r_g is the genetic correlation between traits x and y,
and σ_py is the phenotypic standard deviation of trait y.
In a tomato breeding program, you are selecting for increased fruit weight (trait x) but are also
interested in the correlated response in sugar content (trait y). Given:
• Selection intensity for fruit weight (i) = 1.2
• Heritability of fruit weight (h²_x) = 0.6
• Heritability of sugar content (h²_y) = 0.4
• Genetic correlation between fruit weight and sugar content (r_g) = -0.3
• Phenotypic standard deviation of sugar content (σ_py) = 0.5 °Brix
Calculate the expected correlated response in sugar content when selecting for increased fruit
weight.
Solution:
Step 1: Calculate the square roots of heritabilities h_x = √(0.6) = 0.775 h_y = √(0.4) = 0.632
Step 2: Apply the correlated response equation CR_y = ih_x h_y r_g σ_py CR_y = 1.2 × 0.775
× 0.632 × (-0.3) × 0.5 °Brix CR_y = -0.088 °Brix
Therefore, when selecting for increased fruit weight, the expected correlated response in sugar
content is a decrease of 0.088 °Brix.
1.51 PROBLEM 1: BASIC RESPONSE TO SELECTION
The Breeder’s equation is given by R = h²S, where R is the response to selection, h² is the
narrow-sense heritability, and S is the selection differential. In a population of wheat, the
narrow-sense heritability for grain yield is 0.4, and the selection differential is 500 kg/ha.
Calculate the expected response to selection for grain yield.
Solution:
Given:
• h² = 0.4
• S = 500 kg/ha
Using the Breeder’s equation: R = h²S
R = 0.4 × 500 kg/ha = 200 kg/ha
Therefore, the expected response to selection for grain yield is 200 kg/ha.
1.52 PROBLEM 2: SELECTION INTENSITY
The Breeder’s equation can also be written as R = ih²σ_p, where i is the selection intensity, and
σ_p is the phenotypic standard deviation. In a dairy cattle breeding program, the narrow-sense
heritability for milk yield is 0.3, the phenotypic standard deviation is 1000 kg, and the selection
intensity is 1.4. Calculate the expected response to selection for milk yield.
Solution:
Given:
• h² = 0.3
• σ_p = 1000 kg
• i = 1.4
Using the Breeder’s equation: R = ih²σ_p
R = 1.4 × 0.3 × 1000 kg = 420 kg
Therefore, the expected response to selection for milk yield is 420 kg.
1.53 PROBLEM 3: CALCULATING HERITABILITY
In a selection experiment for plant height in peas, the selection differential was 10 cm, and the
response to selection was 4 cm. Using the Breeder’s equation, calculate the narrow-sense
heritability for plant height in this population.
Solution:
Given:
• S = 10 cm
• R = 4 cm
Using the Breeder’s equation: R = h²S
Rearranging to solve for h²: h² = R / S = 4 cm / 10 cm = 0.4
Therefore, the narrow-sense heritability for plant height in this population is 0.4 or 40
1.54 PROBLEM 4: PREDICTING MULTIPLE GENERATIONS OF SELECTION
Assume you are selecting for increased body weight in a population of chickens. The narrow-
sense heritability for body weight is 0.5, and you can achieve a selection differential of 200 g
per generation. Predict the cumulative response to selection after 3 generations, assuming the
heritability remains constant.
Solution:
Given:
• h² = 0.5
• S = 200 g per generation
• Number of generations = 3
Step 1: Calculate the response to selection per generation R = h²S = 0.5 × 200 g = 100 g
Step 2: Calculate the cumulative response over 3 generations Cumulative R = 3 × 100 g = 300
g
Therefore, the predicted cumulative response to selection after 3 generations is 300 g.
1.55 PROBLEM 5: CORRELATED RESPONSE TO SELECTION
The correlated response (CR_y) in trait y when selecting for trait x is given by the equation:
CR_y = ih_x h_y r_g σ_py
Where i is the selection intensity for trait x, h_x and h_y are the square roots of the
heritabilities for traits x and y respectively, r_g is the genetic correlation between traits x and y,
and σ_py is the phenotypic standard deviation of trait y.
In a tomato breeding program, you are selecting for increased fruit weight (trait x) but are also
interested in the correlated response in sugar content (trait y). Given:
• Selection intensity for fruit weight (i) = 1.2
• Heritability of fruit weight (h²_x) = 0.6
• Heritability of sugar content (h²_y) = 0.4
• Genetic correlation between fruit weight and sugar content (r_g) = -0.3
• Phenotypic standard deviation of sugar content (σ_py) = 0.5 °Brix
Calculate the expected correlated response in sugar content when selecting for increased fruit
weight.
Solution:
Step 1: Calculate the square roots of heritabilities h_x = √(0.6) = 0.775 h_y = √(0.4) = 0.632
Step 2: Apply the correlated response equation CR_y = ih_x h_y r_g σ_py CR_y = 1.2 × 0.775
× 0.632 × (-0.3) × 0.5 °Brix CR_y = -0.088 °Brix
Therefore, when selecting for increased fruit weight, the expected correlated response in sugar
content is a decrease of 0.088 °Brix.
1.56 PROBLEM 1: BASIC RESPONSE TO SELECTION
The Breeder’s equation is given by R = h²S, where R is the response to selection, h² is the
narrow-sense heritability, and S is the selection differential. In a population of wheat, the
narrow-sense heritability for grain yield is 0.4, and the selection differential is 500 kg/ha.
Calculate the expected response to selection for grain yield.
Solution:
Given:
• h² = 0.4
• S = 500 kg/ha
Using the Breeder’s equation: R = h²S
R = 0.4 × 500 kg/ha = 200 kg/ha
Therefore, the expected response to selection for grain yield is 200 kg/ha.
1.57 PROBLEM 2: SELECTION INTENSITY
The Breeder’s equation can also be written as R = ih²σ_p, where i is the selection intensity, and
σ_p is the phenotypic standard deviation. In a dairy cattle breeding program, the narrow-sense
heritability for milk yield is 0.3, the phenotypic standard deviation is 1000 kg, and the selection
intensity is 1.4. Calculate the expected response to selection for milk yield.
Solution:
Given:
• h² = 0.3
• σ_p = 1000 kg
• i = 1.4
Using the Breeder’s equation: R = ih²σ_p
R = 1.4 × 0.3 × 1000 kg = 420 kg
Therefore, the expected response to selection for milk yield is 420 kg.
1.58 PROBLEM 3: CALCULATING HERITABILITY
In a selection experiment for plant height in peas, the selection differential was 10 cm, and the
response to selection was 4 cm. Using the Breeder’s equation, calculate the narrow-sense
heritability for plant height in this population.
Solution:
Given:
• S = 10 cm
• R = 4 cm
Using the Breeder’s equation: R = h²S
Rearranging to solve for h²: h² = R / S = 4 cm / 10 cm = 0.4
Therefore, the narrow-sense heritability for plant height in this population is 0.4 or 40
1.59 PROBLEM 4: PREDICTING MULTIPLE GENERATIONS OF SELECTION
Assume you are selecting for increased body weight in a population of chickens. The narrow-
sense heritability for body weight is 0.5, and you can achieve a selection differential of 200 g
per generation. Predict the cumulative response to selection after 3 generations, assuming the
heritability remains constant.
Solution:
Given:
• h² = 0.5
• S = 200 g per generation
• Number of generations = 3
Step 1: Calculate the response to selection per generation R = h²S = 0.5 × 200 g = 100 g
Step 2: Calculate the cumulative response over 3 generations Cumulative R = 3 × 100 g = 300
g
Therefore, the predicted cumulative response to selection after 3 generations is 300 g.
1.60 PROBLEM 5: CORRELATED RESPONSE TO SELECTION
The correlated response (CR_y) in trait y when selecting for trait x is given by the equation:
CR_y = ih_x h_y r_g σ_py
Where i is the selection intensity for trait x, h_x and h_y are the square roots of the
heritabilities for traits x and y respectively, r_g is the genetic correlation between traits x and y,
and σ_py is the phenotypic standard deviation of trait y.
In a tomato breeding program, you are selecting for increased fruit weight (trait x) but are also
interested in the correlated response in sugar content (trait y). Given:
• Selection intensity for fruit weight (i) = 1.2
• Heritability of fruit weight (h²_x) = 0.6
• Heritability of sugar content (h²_y) = 0.4
• Genetic correlation between fruit weight and sugar content (r_g) = -0.3
• Phenotypic standard deviation of sugar content (σ_py) = 0.5 °Brix
Calculate the expected correlated response in sugar content when selecting for increased fruit
weight.
Solution:
Step 1: Calculate the square roots of heritabilities h_x = √(0.6) = 0.775 h_y = √(0.4) = 0.632
Step 2: Apply the correlated response equation CR_y = ih_x h_y r_g σ_py CR_y = 1.2 × 0.775
× 0.632 × (-0.3) × 0.5 °Brix CR_y = -0.088 °Brix
Therefore, when selecting for increased fruit weight, the expected correlated response in sugar
content is a decrease of 0.088 °Brix.
1.61 PROBLEM 1: BASIC RESPONSE TO SELECTION
The Breeder’s equation is given by R = h²S, where R is the response to selection, h² is the
narrow-sense heritability, and S is the selection differential. In a population of wheat, the
narrow-sense heritability for grain yield is 0.4, and the selection differential is 500 kg/ha.
Calculate the expected response to selection for grain yield.
Solution:
Given:
• h² = 0.4
• S = 500 kg/ha
Using the Breeder’s equation: R = h²S
R = 0.4 × 500 kg/ha = 200 kg/ha
Therefore, the expected response to selection for grain yield is 200 kg/ha.
1.62 PROBLEM 2: SELECTION INTENSITY
The Breeder’s equation can also be written as R = ih²σ_p, where i is the selection intensity, and
σ_p is the phenotypic standard deviation. In a dairy cattle breeding program, the narrow-sense
heritability for milk yield is 0.3, the phenotypic standard deviation is 1000 kg, and the selection
intensity is 1.4. Calculate the expected response to selection for milk yield.
Solution:
Given:
• h² = 0.3
• σ_p = 1000 kg
• i = 1.4
Using the Breeder’s equation: R = ih²σ_p
R = 1.4 × 0.3 × 1000 kg = 420 kg
Therefore, the expected response to selection for milk yield is 420 kg.
1.63 PROBLEM 3: CALCULATING HERITABILITY
In a selection experiment for plant height in peas, the selection differential was 10 cm, and the
response to selection was 4 cm. Using the Breeder’s equation, calculate the narrow-sense
heritability for plant height in this population.
Solution:
Given:
• S = 10 cm
• R = 4 cm
Using the Breeder’s equation: R = h²S
Rearranging to solve for h²: h² = R / S = 4 cm / 10 cm = 0.4
Therefore, the narrow-sense heritability for plant height in this population is 0.4 or 40
1.64 PROBLEM 4: PREDICTING MULTIPLE GENERATIONS OF SELECTION
Assume you are selecting for increased body weight in a population of chickens. The narrow-
sense heritability for body weight is 0.5, and you can achieve a selection differential of 200 g
per generation. Predict the cumulative response to selection after 3 generations, assuming the
heritability remains constant.
Solution:
Given:
• h² = 0.5
• S = 200 g per generation
• Number of generations = 3
Step 1: Calculate the response to selection per generation R = h²S = 0.5 × 200 g = 100 g
Step 2: Calculate the cumulative response over 3 generations Cumulative R = 3 × 100 g = 300
g
Therefore, the predicted cumulative response to selection after 3 generations is 300 g.
1.65 PROBLEM 5: CORRELATED RESPONSE TO SELECTION
The correlated response (CR_y) in trait y when selecting for trait x is given by the equation:
CR_y = ih_x h_y r_g σ_py
Where i is the selection intensity for trait x, h_x and h_y are the square roots of the
heritabilities for traits x and y respectively, r_g is the genetic correlation between traits x and y,
and σ_py is the phenotypic standard deviation of trait y.
In a tomato breeding program, you are selecting for increased fruit weight (trait x) but are also
interested in the correlated response in sugar content (trait y). Given:
• Selection intensity for fruit weight (i) = 1.2
• Heritability of fruit weight (h²_x) = 0.6
• Heritability of sugar content (h²_y) = 0.4
• Genetic correlation between fruit weight and sugar content (r_g) = -0.3
• Phenotypic standard deviation of sugar content (σ_py) = 0.5 °Brix
Calculate the expected correlated response in sugar content when selecting for increased fruit
weight.
Solution:
Step 1: Calculate the square roots of heritabilities h_x = √(0.6) = 0.775 h_y = √(0.4) = 0.632
Step 2: Apply the correlated response equation CR_y = ih_x h_y r_g σ_py CR_y = 1.2 × 0.775
× 0.632 × (-0.3) × 0.5 °Brix CR_y = -0.088 °Brix
Therefore, when selecting for increased fruit weight, the expected correlated response in sugar
content is a decrease of 0.088 °Brix.
1.66 PROBLEM 1: BASIC RESPONSE TO SELECTION
The Breeder’s equation is given by R = h²S, where R is the response to selection, h² is the
narrow-sense heritability, and S is the selection differential. In a population of wheat, the
narrow-sense heritability for grain yield is 0.4, and the selection differential is 500 kg/ha.
Calculate the expected response to selection for grain yield.
Solution:
Given:
• h² = 0.4
• S = 500 kg/ha
Using the Breeder’s equation: R = h²S
R = 0.4 × 500 kg/ha = 200 kg/ha
Therefore, the expected response to selection for grain yield is 200 kg/ha.
1.67 PROBLEM 2: SELECTION INTENSITY
The Breeder’s equation can also be written as R = ih²σ_p, where i is the selection intensity, and
σ_p is the phenotypic standard deviation. In a dairy cattle breeding program, the narrow-sense
heritability for milk yield is 0.3, the phenotypic standard deviation is 1000 kg, and the selection
intensity is 1.4. Calculate the expected response to selection for milk yield.
Solution:
Given:
• h² = 0.3
• σ_p = 1000 kg
• i = 1.4
Using the Breeder’s equation: R = ih²σ_p
R = 1.4 × 0.3 × 1000 kg = 420 kg
Therefore, the expected response to selection for milk yield is 420 kg.
1.68 PROBLEM 3: CALCULATING HERITABILITY
In a selection experiment for plant height in peas, the selection differential was 10 cm, and the
response to selection was 4 cm. Using the Breeder’s equation, calculate the narrow-sense
heritability for plant height in this population.
Solution:
Given:
• S = 10 cm
• R = 4 cm
Using the Breeder’s equation: R = h²S
Rearranging to solve for h²: h² = R / S = 4 cm / 10 cm = 0.4
Therefore, the narrow-sense heritability for plant height in this population is 0.4 or 40
1.69 PROBLEM 4: PREDICTING MULTIPLE GENERATIONS OF SELECTION
Assume you are selecting for increased body weight in a population of chickens. The narrow-
sense heritability for body weight is 0.5, and you can achieve a selection differential of 200 g
per generation. Predict the cumulative response to selection after 3 generations, assuming the
heritability remains constant.
Solution:
Given:
• h² = 0.5
• S = 200 g per generation
• Number of generations = 3
Step 1: Calculate the response to selection per generation R = h²S = 0.5 × 200 g = 100 g
Step 2: Calculate the cumulative response over 3 generations Cumulative R = 3 × 100 g = 300
g
Therefore, the predicted cumulative response to selection after 3 generations is 300 g.
1.70 PROBLEM 5: CORRELATED RESPONSE TO SELECTION
The correlated response (CR_y) in trait y when selecting for trait x is given by the equation:
CR_y = ih_x h_y r_g σ_py
Where i is the selection intensity for trait x, h_x and h_y are the square roots of the
heritabilities for traits x and y respectively, r_g is the genetic correlation between traits x and y,
and σ_py is the phenotypic standard deviation of trait y.
In a tomato breeding program, you are selecting for increased fruit weight (trait x) but are also
interested in the correlated response in sugar content (trait y). Given:
• Selection intensity for fruit weight (i) = 1.2
• Heritability of fruit weight (h²_x) = 0.6
• Heritability of sugar content (h²_y) = 0.4
• Genetic correlation between fruit weight and sugar content (r_g) = -0.3
• Phenotypic standard deviation of sugar content (σ_py) = 0.5 °Brix
Calculate the expected correlated response in sugar content when selecting for increased fruit
weight.
Solution:
Step 1: Calculate the square roots of heritabilities h_x = √(0.6) = 0.775 h_y = √(0.4) = 0.632
Step 2: Apply the correlated response equation CR_y = ih_x h_y r_g σ_py CR_y = 1.2 × 0.775
× 0.632 × (-0.3) × 0.5 °Brix CR_y = -0.088 °Brix
Therefore, when selecting for increased fruit weight, the expected correlated response in sugar
content is a decrease of 0.088 °Brix.
1.71 PROBLEM 1: BASIC RESPONSE TO SELECTION
The Breeder’s equation is given by R = h²S, where R is the response to selection, h² is the
narrow-sense heritability, and S is the selection differential. In a population of wheat, the
narrow-sense heritability for grain yield is 0.4, and the selection differential is 500 kg/ha.
Calculate the expected response to selection for grain yield.
Solution:
Given:
• h² = 0.4
• S = 500 kg/ha
Using the Breeder’s equation: R = h²S
R = 0.4 × 500 kg/ha = 200 kg/ha
Therefore, the expected response to selection for grain yield is 200 kg/ha.
1.72 PROBLEM 2: SELECTION INTENSITY
The Breeder’s equation can also be written as R = ih²σ_p, where i is the selection intensity, and
σ_p is the phenotypic standard deviation. In a dairy cattle breeding program, the narrow-sense
heritability for milk yield is 0.3, the phenotypic standard deviation is 1000 kg, and the selection
intensity is 1.4. Calculate the expected response to selection for milk yield.
Solution:
Given:
• h² = 0.3
• σ_p = 1000 kg
• i = 1.4
Using the Breeder’s equation: R = ih²σ_p
R = 1.4 × 0.3 × 1000 kg = 420 kg
Therefore, the expected response to selection for milk yield is 420 kg.
1.73 PROBLEM 3: CALCULATING HERITABILITY
In a selection experiment for plant height in peas, the selection differential was 10 cm, and the
response to selection was 4 cm. Using the Breeder’s equation, calculate the narrow-sense
heritability for plant height in this population.
Solution:
Given:
• S = 10 cm
• R = 4 cm
Using the Breeder’s equation: R = h²S
Rearranging to solve for h²: h² = R / S = 4 cm / 10 cm = 0.4
Therefore, the narrow-sense heritability for plant height in this population is 0.4 or 40
1.74 PROBLEM 4: PREDICTING MULTIPLE GENERATIONS OF SELECTION
Assume you are selecting for increased body weight in a population of chickens. The narrow-
sense heritability for body weight is 0.5, and you can achieve a selection differential of 200 g
per generation. Predict the cumulative response to selection after 3 generations, assuming the
heritability remains constant.
Solution:
Given:
• h² = 0.5
• S = 200 g per generation
• Number of generations = 3
Step 1: Calculate the response to selection per generation R = h²S = 0.5 × 200 g = 100 g
Step 2: Calculate the cumulative response over 3 generations Cumulative R = 3 × 100 g = 300
g
Therefore, the predicted cumulative response to selection after 3 generations is 300 g.
1.75 PROBLEM 5: CORRELATED RESPONSE TO SELECTION
The correlated response (CR_y) in trait y when selecting for trait x is given by the equation:
CR_y = ih_x h_y r_g σ_py
Where i is the selection intensity for trait x, h_x and h_y are the square roots of the
heritabilities for traits x and y respectively, r_g is the genetic correlation between traits x and y,
and σ_py is the phenotypic standard deviation of trait y.
In a tomato breeding program, you are selecting for increased fruit weight (trait x) but are also
interested in the correlated response in sugar content (trait y). Given:
• Selection intensity for fruit weight (i) = 1.2
• Heritability of fruit weight (h²_x) = 0.6
• Heritability of sugar content (h²_y) = 0.4
• Genetic correlation between fruit weight and sugar content (r_g) = -0.3
• Phenotypic standard deviation of sugar content (σ_py) = 0.5 °Brix
Calculate the expected correlated response in sugar content when selecting for increased fruit
weight.
Solution:
Step 1: Calculate the square roots of heritabilities h_x = √(0.6) = 0.775 h_y = √(0.4) = 0.632
Step 2: Apply the correlated response equation CR_y = ih_x h_y r_g σ_py CR_y = 1.2 × 0.775
× 0.632 × (-0.3) × 0.5 °Brix CR_y = -0.088 °Brix
Therefore, when selecting for increased fruit weight, the expected correlated response in sugar
content is a decrease of 0.088 °Brix.
1.76 PROBLEM 1: BASIC RESPONSE TO SELECTION
The Breeder’s equation is given by R = h²S, where R is the response to selection, h² is the
narrow-sense heritability, and S is the selection differential. In a population of wheat, the
narrow-sense heritability for grain yield is 0.4, and the selection differential is 500 kg/ha.
Calculate the expected response to selection for grain yield.
Solution:
Given:
• h² = 0.4
• S = 500 kg/ha
Using the Breeder’s equation: R = h²S
R = 0.4 × 500 kg/ha = 200 kg/ha
Therefore, the expected response to selection for grain yield is 200 kg/ha.
1.77 PROBLEM 2: SELECTION INTENSITY
The Breeder’s equation can also be written as R = ih²σ_p, where i is the selection intensity, and
σ_p is the phenotypic standard deviation. In a dairy cattle breeding program, the narrow-sense
heritability for milk yield is 0.3, the phenotypic standard deviation is 1000 kg, and the selection
intensity is 1.4. Calculate the expected response to selection for milk yield.
Solution:
Given:
• h² = 0.3
• σ_p = 1000 kg
• i = 1.4
Using the Breeder’s equation: R = ih²σ_p
R = 1.4 × 0.3 × 1000 kg = 420 kg
Therefore, the expected response to selection for milk yield is 420 kg.
1.78 PROBLEM 3: CALCULATING HERITABILITY
In a selection experiment for plant height in peas, the selection differential was 10 cm, and the
response to selection was 4 cm. Using the Breeder’s equation, calculate the narrow-sense
heritability for plant height in this population.
Solution:
Given:
• S = 10 cm
• R = 4 cm
Using the Breeder’s equation: R = h²S
Rearranging to solve for h²: h² = R / S = 4 cm / 10 cm = 0.4
Therefore, the narrow-sense heritability for plant height in this population is 0.4 or 40
1.79 PROBLEM 4: PREDICTING MULTIPLE GENERATIONS OF SELECTION
Assume you are selecting for increased body weight in a population of chickens. The narrow-
sense heritability for body weight is 0.5, and you can achieve a selection differential of 200 g
per generation. Predict the cumulative response to selection after 3 generations, assuming the
heritability remains constant.
Solution:
Given:
• h² = 0.5
• S = 200 g per generation
• Number of generations = 3
Step 1: Calculate the response to selection per generation R = h²S = 0.5 × 200 g = 100 g
Step 2: Calculate the cumulative response over 3 generations Cumulative R = 3 × 100 g = 300
g
Therefore, the predicted cumulative response to selection after 3 generations is 300 g.
1.80 PROBLEM 5: CORRELATED RESPONSE TO SELECTION
The correlated response (CR_y) in trait y when selecting for trait x is given by the equation:
CR_y = ih_x h_y r_g σ_py
Where i is the selection intensity for trait x, h_x and h_y are the square roots of the
heritabilities for traits x and y respectively, r_g is the genetic correlation between traits x and y,
and σ_py is the phenotypic standard deviation of trait y.
In a tomato breeding program, you are selecting for increased fruit weight (trait x) but are also
interested in the correlated response in sugar content (trait y). Given:
• Selection intensity for fruit weight (i) = 1.2
• Heritability of fruit weight (h²_x) = 0.6
• Heritability of sugar content (h²_y) = 0.4
• Genetic correlation between fruit weight and sugar content (r_g) = -0.3
• Phenotypic standard deviation of sugar content (σ_py) = 0.5 °Brix
Calculate the expected correlated response in sugar content when selecting for increased fruit
weight.
Solution:
Step 1: Calculate the square roots of heritabilities h_x = √(0.6) = 0.775 h_y = √(0.4) = 0.632
Step 2: Apply the correlated response equation CR_y = ih_x h_y r_g σ_py CR_y = 1.2 × 0.775
× 0.632 × (-0.3) × 0.5 °Brix CR_y = -0.088 °Brix
Therefore, when selecting for increased fruit weight, the expected correlated response in sugar
content is a decrease of 0.088 °Brix.
1.81 PROBLEM 1: BASIC RESPONSE TO SELECTION
The Breeder’s equation is given by R = h²S, where R is the response to selection, h² is the
narrow-sense heritability, and S is the selection differential. In a population of wheat, the
narrow-sense heritability for grain yield is 0.4, and the selection differential is 500 kg/ha.
Calculate the expected response to selection for grain yield.
Solution:
Given:
• h² = 0.4
• S = 500 kg/ha
Using the Breeder’s equation: R = h²S
R = 0.4 × 500 kg/ha = 200 kg/ha
Therefore, the expected response to selection for grain yield is 200 kg/ha.
1.82 PROBLEM 2: SELECTION INTENSITY
The Breeder’s equation can also be written as R = ih²σ_p, where i is the selection intensity, and
σ_p is the phenotypic standard deviation. In a dairy cattle breeding program, the narrow-sense
heritability for milk yield is 0.3, the phenotypic standard deviation is 1000 kg, and the selection
intensity is 1.4. Calculate the expected response to selection for milk yield.
Solution:
Given:
• h² = 0.3
• σ_p = 1000 kg
• i = 1.4
Using the Breeder’s equation: R = ih²σ_p
R = 1.4 × 0.3 × 1000 kg = 420 kg
Therefore, the expected response to selection for milk yield is 420 kg.
1.83 PROBLEM 3: CALCULATING HERITABILITY
In a selection experiment for plant height in peas, the selection differential was 10 cm, and the
response to selection was 4 cm. Using the Breeder’s equation, calculate the narrow-sense
heritability for plant height in this population.
Solution:
Given:
• S = 10 cm
• R = 4 cm
Using the Breeder’s equation: R = h²S
Rearranging to solve for h²: h² = R / S = 4 cm / 10 cm = 0.4
Therefore, the narrow-sense heritability for plant height in this population is 0.4 or 40
1.84 PROBLEM 4: PREDICTING MULTIPLE GENERATIONS OF SELECTION
Assume you are selecting for increased body weight in a population of chickens. The narrow-
sense heritability for body weight is 0.5, and you can achieve a selection differential of 200 g
per generation. Predict the cumulative response to selection after 3 generations, assuming the
heritability remains constant.
Solution:
Given:
• h² = 0.5
• S = 200 g per generation
• Number of generations = 3
Step 1: Calculate the response to selection per generation R = h²S = 0.5 × 200 g = 100 g
Step 2: Calculate the cumulative response over 3 generations Cumulative R = 3 × 100 g = 300
g
Therefore, the predicted cumulative response to selection after 3 generations is 300 g.
1.85 PROBLEM 5: CORRELATED RESPONSE TO SELECTION
The correlated response (CR_y) in trait y when selecting for trait x is given by the equation:
CR_y = ih_x h_y r_g σ_py
Where i is the selection intensity for trait x, h_x and h_y are the square roots of the
heritabilities for traits x and y respectively, r_g is the genetic correlation between traits x and y,
and σ_py is the phenotypic standard deviation of trait y.
In a tomato breeding program, you are selecting for increased fruit weight (trait x) but are also
interested in the correlated response in sugar content (trait y). Given:
• Selection intensity for fruit weight (i) = 1.2
• Heritability of fruit weight (h²_x) = 0.6
• Heritability of sugar content (h²_y) = 0.4
• Genetic correlation between fruit weight and sugar content (r_g) = -0.3
• Phenotypic standard deviation of sugar content (σ_py) = 0.5 °Brix
Calculate the expected correlated response in sugar content when selecting for increased fruit
weight.
Solution:
Step 1: Calculate the square roots of heritabilities h_x = √(0.6) = 0.775 h_y = √(0.4) = 0.632
Step 2: Apply the correlated response equation CR_y = ih_x h_y r_g σ_py CR_y = 1.2 × 0.775
× 0.632 × (-0.3) × 0.5 °Brix CR_y = -0.088 °Brix
Therefore, when selecting for increased fruit weight, the expected correlated response in sugar
content is a decrease of 0.088 °Brix.
1.86 PROBLEM 1: BASIC RESPONSE TO SELECTION
The Breeder’s equation is given by R = h²S, where R is the response to selection, h² is the
narrow-sense heritability, and S is the selection differential. In a population of wheat, the
narrow-sense heritability for grain yield is 0.4, and the selection differential is 500 kg/ha.
Calculate the expected response to selection for grain yield.
Solution:
Given:
• h² = 0.4
• S = 500 kg/ha
Using the Breeder’s equation: R = h²S
R = 0.4 × 500 kg/ha = 200 kg/ha
Therefore, the expected response to selection for grain yield is 200 kg/ha.
1.87 PROBLEM 2: SELECTION INTENSITY
The Breeder’s equation can also be written as R = ih²σ_p, where i is the selection intensity, and
σ_p is the phenotypic standard deviation. In a dairy cattle breeding program, the narrow-sense
heritability for milk yield is 0.3, the phenotypic standard deviation is 1000 kg, and the selection
intensity is 1.4. Calculate the expected response to selection for milk yield.
Solution:
Given:
• h² = 0.3
• σ_p = 1000 kg
• i = 1.4
Using the Breeder’s equation: R = ih²σ_p
R = 1.4 × 0.3 × 1000 kg = 420 kg
Therefore, the expected response to selection for milk yield is 420 kg.
1.88 PROBLEM 3: CALCULATING HERITABILITY
In a selection experiment for plant height in peas, the selection differential was 10 cm, and the
response to selection was 4 cm. Using the Breeder’s equation, calculate the narrow-sense
heritability for plant height in this population.
Solution:
Given:
• S = 10 cm
• R = 4 cm
Using the Breeder’s equation: R = h²S
Rearranging to solve for h²: h² = R / S = 4 cm / 10 cm = 0.4
Therefore, the narrow-sense heritability for plant height in this population is 0.4 or 40
1.89 PROBLEM 4: PREDICTING MULTIPLE GENERATIONS OF SELECTION
Assume you are selecting for increased body weight in a population of chickens. The narrow-
sense heritability for body weight is 0.5, and you can achieve a selection differential of 200 g
per generation. Predict the cumulative response to selection after 3 generations, assuming the
heritability remains constant.
Solution:
Given:
• h² = 0.5
• S = 200 g per generation
• Number of generations = 3
Step 1: Calculate the response to selection per generation R = h²S = 0.5 × 200 g = 100 g
Step 2: Calculate the cumulative response over 3 generations Cumulative R = 3 × 100 g = 300
g
Therefore, the predicted cumulative response to selection after 3 generations is 300 g.
1.90 PROBLEM 5: CORRELATED RESPONSE TO SELECTION
The correlated response (CR_y) in trait y when selecting for trait x is given by the equation:
CR_y = ih_x h_y r_g σ_py
Where i is the selection intensity for trait x, h_x and h_y are the square roots of the
heritabilities for traits x and y respectively, r_g is the genetic correlation between traits x and y,
and σ_py is the phenotypic standard deviation of trait y.
In a tomato breeding program, you are selecting for increased fruit weight (trait x) but are also
interested in the correlated response in sugar content (trait y). Given:
• Selection intensity for fruit weight (i) = 1.2
• Heritability of fruit weight (h²_x) = 0.6
• Heritability of sugar content (h²_y) = 0.4
• Genetic correlation between fruit weight and sugar content (r_g) = -0.3
• Phenotypic standard deviation of sugar content (σ_py) = 0.5 °Brix
Calculate the expected correlated response in sugar content when selecting for increased fruit
weight.
Solution:
Step 1: Calculate the square roots of heritabilities h_x = √(0.6) = 0.775 h_y = √(0.4) = 0.632
Step 2: Apply the correlated response equation CR_y = ih_x h_y r_g σ_py CR_y = 1.2 × 0.775
× 0.632 × (-0.3) × 0.5 °Brix CR_y = -0.088 °Brix
Therefore, when selecting for increased fruit weight, the expected correlated response in sugar
content is a decrease of 0.088 °Brix.
1.91 PROBLEM 1: BASIC RESPONSE TO SELECTION
The Breeder’s equation is given by R = h²S, where R is the response to selection, h² is the
narrow-sense heritability, and S is the selection differential. In a population of wheat, the
narrow-sense heritability for grain yield is 0.4, and the selection differential is 500 kg/ha.
Calculate the expected response to selection for grain yield.
Solution:
Given:
• h² = 0.4
• S = 500 kg/ha
Using the Breeder’s equation: R = h²S
R = 0.4 × 500 kg/ha = 200 kg/ha
Therefore, the expected response to selection for grain yield is 200 kg/ha.
1.92 PROBLEM 2: SELECTION INTENSITY
The Breeder’s equation can also be written as R = ih²σ_p, where i is the selection intensity, and
σ_p is the phenotypic standard deviation. In a dairy cattle breeding program, the narrow-sense
heritability for milk yield is 0.3, the phenotypic standard deviation is 1000 kg, and the selection
intensity is 1.4. Calculate the expected response to selection for milk yield.
Solution:
Given:
• h² = 0.3
• σ_p = 1000 kg
• i = 1.4
Using the Breeder’s equation: R = ih²σ_p
R = 1.4 × 0.3 × 1000 kg = 420 kg
Therefore, the expected response to selection for milk yield is 420 kg.
1.93 PROBLEM 3: CALCULATING HERITABILITY
In a selection experiment for plant height in peas, the selection differential was 10 cm, and the
response to selection was 4 cm. Using the Breeder’s equation, calculate the narrow-sense
heritability for plant height in this population.
Solution:
Given:
• S = 10 cm
• R = 4 cm
Using the Breeder’s equation: R = h²S
Rearranging to solve for h²: h² = R / S = 4 cm / 10 cm = 0.4
Therefore, the narrow-sense heritability for plant height in this population is 0.4 or 40
1.94 PROBLEM 4: PREDICTING MULTIPLE GENERATIONS OF SELECTION
Assume you are selecting for increased body weight in a population of chickens. The narrow-
sense heritability for body weight is 0.5, and you can achieve a selection differential of 200 g
per generation. Predict the cumulative response to selection after 3 generations, assuming the
heritability remains constant.
Solution:
Given:
• h² = 0.5
• S = 200 g per generation
• Number of generations = 3
Step 1: Calculate the response to selection per generation R = h²S = 0.5 × 200 g = 100 g
Step 2: Calculate the cumulative response over 3 generations Cumulative R = 3 × 100 g = 300
g
Therefore, the predicted cumulative response to selection after 3 generations is 300 g.
1.95 PROBLEM 5: CORRELATED RESPONSE TO SELECTION
The correlated response (CR_y) in trait y when selecting for trait x is given by the equation:
CR_y = ih_x h_y r_g σ_py
Where i is the selection intensity for trait x, h_x and h_y are the square roots of the
heritabilities for traits x and y respectively, r_g is the genetic correlation between traits x and y,
and σ_py is the phenotypic standard deviation of trait y.
In a tomato breeding program, you are selecting for increased fruit weight (trait x) but are also
interested in the correlated response in sugar content (trait y). Given:
• Selection intensity for fruit weight (i) = 1.2
• Heritability of fruit weight (h²_x) = 0.6
• Heritability of sugar content (h²_y) = 0.4
• Genetic correlation between fruit weight and sugar content (r_g) = -0.3
• Phenotypic standard deviation of sugar content (σ_py) = 0.5 °Brix
Calculate the expected correlated response in sugar content when selecting for increased fruit
weight.
Solution:
Step 1: Calculate the square roots of heritabilities h_x = √(0.6) = 0.775 h_y = √(0.4) = 0.632
Step 2: Apply the correlated response equation CR_y = ih_x h_y r_g σ_py CR_y = 1.2 × 0.775
× 0.632 × (-0.3) × 0.5 °Brix CR_y = -0.088 °Brix
Therefore, when selecting for increased fruit weight, the expected correlated response in sugar
content is a decrease of 0.088 °Brix.
1.96 PROBLEM 1: BASIC RESPONSE TO SELECTION
The Breeder’s equation is given by R = h²S, where R is the response to selection, h² is the
narrow-sense heritability, and S is the selection differential. In a population of wheat, the
narrow-sense heritability for grain yield is 0.4, and the selection differential is 500 kg/ha.
Calculate the expected response to selection for grain yield.
Solution:
Given:
• h² = 0.4
• S = 500 kg/ha
Using the Breeder’s equation: R = h²S
R = 0.4 × 500 kg/ha = 200 kg/ha
Therefore, the expected response to selection for grain yield is 200 kg/ha.
1.97 PROBLEM 2: SELECTION INTENSITY
The Breeder’s equation can also be written as R = ih²σ_p, where i is the selection intensity, and
σ_p is the phenotypic standard deviation. In a dairy cattle breeding program, the narrow-sense
heritability for milk yield is 0.3, the phenotypic standard deviation is 1000 kg, and the selection
intensity is 1.4. Calculate the expected response to selection for milk yield.
Solution:
Given:
• h² = 0.3
• σ_p = 1000 kg
• i = 1.4
Using the Breeder’s equation: R = ih²σ_p
R = 1.4 × 0.3 × 1000 kg = 420 kg
Therefore, the expected response to selection for milk yield is 420 kg.
1.98 PROBLEM 3: CALCULATING HERITABILITY
In a selection experiment for plant height in peas, the selection differential was 10 cm, and the
response to selection was 4 cm. Using the Breeder’s equation, calculate the narrow-sense
heritability for plant height in this population.
Solution:
Given:
• S = 10 cm
• R = 4 cm
Using the Breeder’s equation: R = h²S
Rearranging to solve for h²: h² = R / S = 4 cm / 10 cm = 0.4
Therefore, the narrow-sense heritability for plant height in this population is 0.4 or 40
1.99 PROBLEM 4: PREDICTING MULTIPLE GENERATIONS OF SELECTION
Assume you are selecting for increased body weight in a population of chickens. The narrow-
sense heritability for body weight is 0.5, and you can achieve a selection differential of 200 g
per generation. Predict the cumulative response to selection after 3 generations, assuming the
heritability remains constant.
Solution:
Given:
• h² = 0.5
• S = 200 g per generation
• Number of generations = 3
Step 1: Calculate the response to selection per generation R = h²S = 0.5 × 200 g = 100 g
Step 2: Calculate the cumulative response over 3 generations Cumulative R = 3 × 100 g = 300
g
Therefore, the predicted cumulative response to selection after 3 generations is 300 g.
1.100 PROBLEM 5: CORRELATED RESPONSE TO SELECTION
The correlated response (CR_y) in trait y when selecting for trait x is given by the equation:
CR_y = ih_x h_y r_g σ_py
Where i is the selection intensity for trait x, h_x and h_y are the square roots of the
heritabilities for traits x and y respectively, r_g is the genetic correlation between traits x and y,
and σ_py is the phenotypic standard deviation of trait y.
In a tomato breeding program, you are selecting for increased fruit weight (trait x) but are also
interested in the correlated response in sugar content (trait y). Given:
• Selection intensity for fruit weight (i) = 1.2
• Heritability of fruit weight (h²_x) = 0.6
• Heritability of sugar content (h²_y) = 0.4
• Genetic correlation between fruit weight and sugar content (r_g) = -0.3
• Phenotypic standard deviation of sugar content (σ_py) = 0.5 °Brix
Calculate the expected correlated response in sugar content when selecting for increased fruit
weight.
Solution:
Step 1: Calculate the square roots of heritabilities h_x = √(0.6) = 0.775 h_y = √(0.4) = 0.632
Step 2: Apply the correlated response equation CR_y = ih_x h_y r_g σ_py CR_y = 1.2 × 0.775
× 0.632 × (-0.3) × 0.5 °Brix CR_y = -0.088 °Brix
Therefore, when selecting for increased fruit weight, the expected correlated response in sugar
content is a decrease of 0.088 °Brix.
1.101 PROBLEM 1: BASIC RESPONSE TO SELECTION
The Breeder’s equation is given by R = h²S, where R is the response to selection, h² is the
narrow-sense heritability, and S is the selection differential. In a population of wheat, the
narrow-sense heritability for grain yield is 0.4, and the selection differential is 500 kg/ha.
Calculate the expected response to selection for grain yield.
Solution:
Given:
• h² = 0.4
• S = 500 kg/ha
Using the Breeder’s equation: R = h²S
R = 0.4 × 500 kg/ha = 200 kg/ha
Therefore, the expected response to selection for grain yield is 200 kg/ha.
1.102 PROBLEM 2: SELECTION INTENSITY
The Breeder’s equation can also be written as R = ih²σ_p, where i is the selection intensity, and
σ_p is the phenotypic standard deviation. In a dairy cattle breeding program, the narrow-sense
heritability for milk yield is 0.3, the phenotypic standard deviation is 1000 kg, and the selection
intensity is 1.4. Calculate the expected response to selection for milk yield.
Solution:
Given:
• h² = 0.3
• σ_p = 1000 kg
• i = 1.4
Using the Breeder’s equation: R = ih²σ_p
R = 1.4 × 0.3 × 1000 kg = 420 kg
Therefore, the expected response to selection for milk yield is 420 kg.
1.103 PROBLEM 3: CALCULATING HERITABILITY
In a selection experiment for plant height in peas, the selection differential was 10 cm, and the
response to selection was 4 cm. Using the Breeder’s equation, calculate the narrow-sense
heritability for plant height in this population.
Solution:
Given:
• S = 10 cm
• R = 4 cm
Using the Breeder’s equation: R = h²S
Rearranging to solve for h²: h² = R / S = 4 cm / 10 cm = 0.4
Therefore, the narrow-sense heritability for plant height in this population is 0.4 or 40
1.104 PROBLEM 4: PREDICTING MULTIPLE GENERATIONS OF SELECTION
Assume you are selecting for increased body weight in a population of chickens. The narrow-
sense heritability for body weight is 0.5, and you can achieve a selection differential of 200 g
per generation. Predict the cumulative response to selection after 3 generations, assuming the
heritability remains constant.
Solution:
Given:
• h² = 0.5
• S = 200 g per generation
• Number of generations = 3
Step 1: Calculate the response to selection per generation R = h²S = 0.5 × 200 g = 100 g
Step 2: Calculate the cumulative response over 3 generations Cumulative R = 3 × 100 g = 300
g
Therefore, the predicted cumulative response to selection after 3 generations is 300 g.
1.105 PROBLEM 5: CORRELATED RESPONSE TO SELECTION
The correlated response (CR_y) in trait y when selecting for trait x is given by the equation:
CR_y = ih_x h_y r_g σ_py
Where i is the selection intensity for trait x, h_x and h_y are the square roots of the
heritabilities for traits x and y respectively, r_g is the genetic correlation between traits x and y,
and σ_py is the phenotypic standard deviation of trait y.
In a tomato breeding program, you are selecting for increased fruit weight (trait x) but are also
interested in the correlated response in sugar content (trait y). Given:
• Selection intensity for fruit weight (i) = 1.2
• Heritability of fruit weight (h²_x) = 0.6
• Heritability of sugar content (h²_y) = 0.4
• Genetic correlation between fruit weight and sugar content (r_g) = -0.3
• Phenotypic standard deviation of sugar content (σ_py) = 0.5 °Brix
Calculate the expected correlated response in sugar content when selecting for increased fruit
weight.
Solution:
Step 1: Calculate the square roots of heritabilities h_x = √(0.6) = 0.775 h_y = √(0.4) = 0.632
Step 2: Apply the correlated response equation CR_y = ih_x h_y r_g σ_py CR_y = 1.2 × 0.775
× 0.632 × (-0.3) × 0.5 °Brix CR_y = -0.088 °Brix
Therefore, when selecting for increased fruit weight, the expected correlated response in sugar
content is a decrease of 0.088 °Brix.
1.106 PROBLEM 1: BASIC RESPONSE TO SELECTION
The Breeder’s equation is given by R = h²S, where R is the response to selection, h² is the
narrow-sense heritability, and S is the selection differential. In a population of wheat, the
narrow-sense heritability for grain yield is 0.4, and the selection differential is 500 kg/ha.
Calculate the expected response to selection for grain yield.
Solution:
Given:
• h² = 0.4
• S = 500 kg/ha
Using the Breeder’s equation: R = h²S
R = 0.4 × 500 kg/ha = 200 kg/ha
Therefore, the expected response to selection for grain yield is 200 kg/ha.
1.107 PROBLEM 2: SELECTION INTENSITY
The Breeder’s equation can also be written as R = ih²σ_p, where i is the selection intensity, and
σ_p is the phenotypic standard deviation. In a dairy cattle breeding program, the narrow-sense
heritability for milk yield is 0.3, the phenotypic standard deviation is 1000 kg, and the selection
intensity is 1.4. Calculate the expected response to selection for milk yield.
Solution:
Given:
• h² = 0.3
• σ_p = 1000 kg
• i = 1.4
Using the Breeder’s equation: R = ih²σ_p
R = 1.4 × 0.3 × 1000 kg = 420 kg
Therefore, the expected response to selection for milk yield is 420 kg.
1.108 PROBLEM 3: CALCULATING HERITABILITY
In a selection experiment for plant height in peas, the selection differential was 10 cm, and the
response to selection was 4 cm. Using the Breeder’s equation, calculate the narrow-sense
heritability for plant height in this population.
Solution:
Given:
• S = 10 cm
• R = 4 cm
Using the Breeder’s equation: R = h²S
Rearranging to solve for h²: h² = R / S = 4 cm / 10 cm = 0.4
Therefore, the narrow-sense heritability for plant height in this population is 0.4 or 40
1.109 PROBLEM 4: PREDICTING MULTIPLE GENERATIONS OF SELECTION
Assume you are selecting for increased body weight in a population of chickens. The narrow-
sense heritability for body weight is 0.5, and you can achieve a selection differential of 200 g
per generation. Predict the cumulative response to selection after 3 generations, assuming the
heritability remains constant.
Solution:
Given:
• h² = 0.5
• S = 200 g per generation
• Number of generations = 3
Step 1: Calculate the response to selection per generation R = h²S = 0.5 × 200 g = 100 g
Step 2: Calculate the cumulative response over 3 generations Cumulative R = 3 × 100 g = 300
g
Therefore, the predicted cumulative response to selection after 3 generations is 300 g.
1.110 PROBLEM 5: CORRELATED RESPONSE TO SELECTION
The correlated response (CR_y) in trait y when selecting for trait x is given by the equation:
CR_y = ih_x h_y r_g σ_py
Where i is the selection intensity for trait x, h_x and h_y are the square roots of the
heritabilities for traits x and y respectively, r_g is the genetic correlation between traits x and y,
and σ_py is the phenotypic standard deviation of trait y.
In a tomato breeding program, you are selecting for increased fruit weight (trait x) but are also
interested in the correlated response in sugar content (trait y). Given:
• Selection intensity for fruit weight (i) = 1.2
• Heritability of fruit weight (h²_x) = 0.6
• Heritability of sugar content (h²_y) = 0.4
• Genetic correlation between fruit weight and sugar content (r_g) = -0.3
• Phenotypic standard deviation of sugar content (σ_py) = 0.5 °Brix
Calculate the expected correlated response in sugar content when selecting for increased fruit
weight.
Solution:
Step 1: Calculate the square roots of heritabilities h_x = √(0.6) = 0.775 h_y = √(0.4) = 0.632
Step 2: Apply the correlated response equation CR_y = ih_x h_y r_g σ_py CR_y = 1.2 × 0.775
× 0.632 × (-0.3) × 0.5 °Brix CR_y = -0.088 °Brix
Therefore, when selecting for increased fruit weight, the expected correlated response in sugar
content is a decrease of 0.088 °Brix.
1.111 PROBLEM 1: BASIC RESPONSE TO SELECTION
The Breeder’s equation is given by R = h²S, where R is the response to selection, h² is the
narrow-sense heritability, and S is the selection differential. In a population of wheat, the
narrow-sense heritability for grain yield is 0.4, and the selection differential is 500 kg/ha.
Calculate the expected response to selection for grain yield.
Solution:
Given:
• h² = 0.4
• S = 500 kg/ha
Using the Breeder’s equation: R = h²S
R = 0.4 × 500 kg/ha = 200 kg/ha
Therefore, the expected response to selection for grain yield is 200 kg/ha.
1.112 PROBLEM 2: SELECTION INTENSITY
The Breeder’s equation can also be written as R = ih²σ_p, where i is the selection intensity, and
σ_p is the phenotypic standard deviation. In a dairy cattle breeding program, the narrow-sense
heritability for milk yield is 0.3, the phenotypic standard deviation is 1000 kg, and the selection
intensity is 1.4. Calculate the expected response to selection for milk yield.
Solution:
Given:
• h² = 0.3
• σ_p = 1000 kg
• i = 1.4
Using the Breeder’s equation: R = ih²σ_p
R = 1.4 × 0.3 × 1000 kg = 420 kg
Therefore, the expected response to selection for milk yield is 420 kg.
1.113 PROBLEM 3: CALCULATING HERITABILITY
In a selection experiment for plant height in peas, the selection differential was 10 cm, and the
response to selection was 4 cm. Using the Breeder’s equation, calculate the narrow-sense
heritability for plant height in this population.
Solution:
Given:
• S = 10 cm
• R = 4 cm
Using the Breeder’s equation: R = h²S
Rearranging to solve for h²: h² = R / S = 4 cm / 10 cm = 0.4
Therefore, the narrow-sense heritability for plant height in this population is 0.4 or 40
1.114 PROBLEM 4: PREDICTING MULTIPLE GENERATIONS OF SELECTION
Assume you are selecting for increased body weight in a population of chickens. The narrow-
sense heritability for body weight is 0.5, and you can achieve a selection differential of 200 g
per generation. Predict the cumulative response to selection after 3 generations, assuming the
heritability remains constant.
Solution:
Given:
• h² = 0.5
• S = 200 g per generation
• Number of generations = 3
Step 1: Calculate the response to selection per generation R = h²S = 0.5 × 200 g = 100 g
Step 2: Calculate the cumulative response over 3 generations Cumulative R = 3 × 100 g = 300
g
Therefore, the predicted cumulative response to selection after 3 generations is 300 g.
1.115 PROBLEM 5: CORRELATED RESPONSE TO SELECTION
The correlated response (CR_y) in trait y when selecting for trait x is given by the equation:
CR_y = ih_x h_y r_g σ_py
Where i is the selection intensity for trait x, h_x and h_y are the square roots of the
heritabilities for traits x and y respectively, r_g is the genetic correlation between traits x and y,
and σ_py is the phenotypic standard deviation of trait y.
In a tomato breeding program, you are selecting for increased fruit weight (trait x) but are also
interested in the correlated response in sugar content (trait y). Given:
• Selection intensity for fruit weight (i) = 1.2
• Heritability of fruit weight (h²_x) = 0.6
• Heritability of sugar content (h²_y) = 0.4
• Genetic correlation between fruit weight and sugar content (r_g) = -0.3
• Phenotypic standard deviation of sugar content (σ_py) = 0.5 °Brix
Calculate the expected correlated response in sugar content when selecting for increased fruit
weight.
Solution:
Step 1: Calculate the square roots of heritabilities h_x = √(0.6) = 0.775 h_y = √(0.4) = 0.632
Step 2: Apply the correlated response equation CR_y = ih_x h_y r_g σ_py CR_y = 1.2 × 0.775
× 0.632 × (-0.3) × 0.5 °Brix CR_y = -0.088 °Brix
Therefore, when selecting for increased fruit weight, the expected correlated response in sugar
content is a decrease of 0.088 °Brix.
1.116 PROBLEM 1: BASIC RESPONSE TO SELECTION
The Breeder’s equation is given by R = h²S, where R is the response to selection, h² is the
narrow-sense heritability, and S is the selection differential. In a population of wheat, the
narrow-sense heritability for grain yield is 0.4, and the selection differential is 500 kg/ha.
Calculate the expected response to selection for grain yield.
Solution:
Given:
• h² = 0.4
• S = 500 kg/ha
Using the Breeder’s equation: R = h²S
R = 0.4 × 500 kg/ha = 200 kg/ha
Therefore, the expected response to selection for grain yield is 200 kg/ha.
1.117 PROBLEM 2: SELECTION INTENSITY
The Breeder’s equation can also be written as R = ih²σ_p, where i is the selection intensity, and
σ_p is the phenotypic standard deviation. In a dairy cattle breeding program, the narrow-sense
heritability for milk yield is 0.3, the phenotypic standard deviation is 1000 kg, and the selection
intensity is 1.4. Calculate the expected response to selection for milk yield.
Solution:
Given:
• h² = 0.3
• σ_p = 1000 kg
• i = 1.4
Using the Breeder’s equation: R = ih²σ_p
R = 1.4 × 0.3 × 1000 kg = 420 kg
Therefore, the expected response to selection for milk yield is 420 kg.
1.118 PROBLEM 3: CALCULATING HERITABILITY
In a selection experiment for plant height in peas, the selection differential was 10 cm, and the
response to selection was 4 cm. Using the Breeder’s equation, calculate the narrow-sense
heritability for plant height in this population.
Solution:
Given:
• S = 10 cm
• R = 4 cm
Using the Breeder’s equation: R = h²S
Rearranging to solve for h²: h² = R / S = 4 cm / 10 cm = 0.4
Therefore, the narrow-sense heritability for plant height in this population is 0.4 or 40
1.119 PROBLEM 4: PREDICTING MULTIPLE GENERATIONS OF SELECTION
Assume you are selecting for increased body weight in a population of chickens. The narrow-
sense heritability for body weight is 0.5, and you can achieve a selection differential of 200 g
per generation. Predict the cumulative response to selection after 3 generations, assuming the
heritability remains constant.
Solution:
Given:
• h² = 0.5
• S = 200 g per generation
• Number of generations = 3
Step 1: Calculate the response to selection per generation R = h²S = 0.5 × 200 g = 100 g
Step 2: Calculate the cumulative response over 3 generations Cumulative R = 3 × 100 g = 300
g
Therefore, the predicted cumulative response to selection after 3 generations is 300 g.
1.120 PROBLEM 5: CORRELATED RESPONSE TO SELECTION
The correlated response (CR_y) in trait y when selecting for trait x is given by the equation:
CR_y = ih_x h_y r_g σ_py
Where i is the selection intensity for trait x, h_x and h_y are the square roots of the
heritabilities for traits x and y respectively, r_g is the genetic correlation between traits x and y,
and σ_py is the phenotypic standard deviation of trait y.
In a tomato breeding program, you are selecting for increased fruit weight (trait x) but are also
interested in the correlated response in sugar content (trait y). Given:
• Selection intensity for fruit weight (i) = 1.2
• Heritability of fruit weight (h²_x) = 0.6
• Heritability of sugar content (h²_y) = 0.4
• Genetic correlation between fruit weight and sugar content (r_g) = -0.3
• Phenotypic standard deviation of sugar content (σ_py) = 0.5 °Brix
Calculate the expected correlated response in sugar content when selecting for increased fruit
weight.
Solution:
Step 1: Calculate the square roots of heritabilities h_x = √(0.6) = 0.775 h_y = √(0.4) = 0.632
Step 2: Apply the correlated response equation CR_y = ih_x h_y r_g σ_py CR_y = 1.2 × 0.775
× 0.632 × (-0.3) × 0.5 °Brix CR_y = -0.088 °Brix
Therefore, when selecting for increased fruit weight, the expected correlated response in sugar
content is a decrease of 0.088 °Brix.
1.121 PROBLEM 1: BASIC RESPONSE TO SELECTION
The Breeder’s equation is given by R = h²S, where R is the response to selection, h² is the
narrow-sense heritability, and S is the selection differential. In a population of wheat, the
narrow-sense heritability for grain yield is 0.4, and the selection differential is 500 kg/ha.
Calculate the expected response to selection for grain yield.
Solution:
Given:
• h² = 0.4
• S = 500 kg/ha
Using the Breeder’s equation: R = h²S
R = 0.4 × 500 kg/ha = 200 kg/ha
Therefore, the expected response to selection for grain yield is 200 kg/ha.
1.122 PROBLEM 2: SELECTION INTENSITY
The Breeder’s equation can also be written as R = ih²σ_p, where i is the selection intensity, and
σ_p is the phenotypic standard deviation. In a dairy cattle breeding program, the narrow-sense
heritability for milk yield is 0.3, the phenotypic standard deviation is 1000 kg, and the selection
intensity is 1.4. Calculate the expected response to selection for milk yield.
Solution:
Given:
• h² = 0.3
• σ_p = 1000 kg
• i = 1.4
Using the Breeder’s equation: R = ih²σ_p
R = 1.4 × 0.3 × 1000 kg = 420 kg
Therefore, the expected response to selection for milk yield is 420 kg.
1.123 PROBLEM 3: CALCULATING HERITABILITY
In a selection experiment for plant height in peas, the selection differential was 10 cm, and the
response to selection was 4 cm. Using the Breeder’s equation, calculate the narrow-sense
heritability for plant height in this population.
Solution:
Given:
• S = 10 cm
• R = 4 cm
Using the Breeder’s equation: R = h²S
Rearranging to solve for h²: h² = R / S = 4 cm / 10 cm = 0.4
Therefore, the narrow-sense heritability for plant height in this population is 0.4 or 40
1.124 PROBLEM 4: PREDICTING MULTIPLE GENERATIONS OF SELECTION
Assume you are selecting for increased body weight in a population of chickens. The narrow-
sense heritability for body weight is 0.5, and you can achieve a selection differential of 200 g
per generation. Predict the cumulative response to selection after 3 generations, assuming the
heritability remains constant.
Solution:
Given:
• h² = 0.5
• S = 200 g per generation
• Number of generations = 3
Step 1: Calculate the response to selection per generation R = h²S = 0.5 × 200 g = 100 g
Step 2: Calculate the cumulative response over 3 generations Cumulative R = 3 × 100 g = 300
g
Therefore, the predicted cumulative response to selection after 3 generations is 300 g.
1.125 PROBLEM 5: CORRELATED RESPONSE TO SELECTION
The correlated response (CR_y) in trait y when selecting for trait x is given by the equation:
CR_y = ih_x h_y r_g σ_py
Where i is the selection intensity for trait x, h_x and h_y are the square roots of the
heritabilities for traits x and y respectively, r_g is the genetic correlation between traits x and y,
and σ_py is the phenotypic standard deviation of trait y.
In a tomato breeding program, you are selecting for increased fruit weight (trait x) but are also
interested in the correlated response in sugar content (trait y). Given:
• Selection intensity for fruit weight (i) = 1.2
• Heritability of fruit weight (h²_x) = 0.6
• Heritability of sugar content (h²_y) = 0.4
• Genetic correlation between fruit weight and sugar content (r_g) = -0.3
• Phenotypic standard deviation of sugar content (σ_py) = 0.5 °Brix
Calculate the expected correlated response in sugar content when selecting for increased fruit
weight.
Solution:
Step 1: Calculate the square roots of heritabilities h_x = √(0.6) = 0.775 h_y = √(0.4) = 0.632
Step 2: Apply the correlated response equation CR_y = ih_x h_y r_g σ_py CR_y = 1.2 × 0.775
× 0.632 × (-0.3) × 0.5 °Brix CR_y = -0.088 °Brix
Therefore, when selecting for increased fruit weight, the expected correlated response in sugar
content is a decrease of 0.088 °Brix.
1.126 PROBLEM 1: BASIC RESPONSE TO SELECTION
The Breeder’s equation is given by R = h²S, where R is the response to selection, h² is the
narrow-sense heritability, and S is the selection differential. In a population of wheat, the
narrow-sense heritability for grain yield is 0.4, and the selection differential is 500 kg/ha.
Calculate the expected response to selection for grain yield.
Solution:
Given:
• h² = 0.4
• S = 500 kg/ha
Using the Breeder’s equation: R = h²S
R = 0.4 × 500 kg/ha = 200 kg/ha
Therefore, the expected response to selection for grain yield is 200 kg/ha.
1.127 PROBLEM 2: SELECTION INTENSITY
The Breeder’s equation can also be written as R = ih²σ_p, where i is the selection intensity, and
σ_p is the phenotypic standard deviation. In a dairy cattle breeding program, the narrow-sense
heritability for milk yield is 0.3, the phenotypic standard deviation is 1000 kg, and the selection
intensity is 1.4. Calculate the expected response to selection for milk yield.
Solution:
Given:
• h² = 0.3
• σ_p = 1000 kg
• i = 1.4
Using the Breeder’s equation: R = ih²σ_p
R = 1.4 × 0.3 × 1000 kg = 420 kg
Therefore, the expected response to selection for milk yield is 420 kg.
1.128 PROBLEM 3: CALCULATING HERITABILITY
In a selection experiment for plant height in peas, the selection differential was 10 cm, and the
response to selection was 4 cm. Using the Breeder’s equation, calculate the narrow-sense
heritability for plant height in this population.
Solution:
Given:
• S = 10 cm
• R = 4 cm
Using the Breeder’s equation: R = h²S
Rearranging to solve for h²: h² = R / S = 4 cm / 10 cm = 0.4
Therefore, the narrow-sense heritability for plant height in this population is 0.4 or 40
1.129 PROBLEM 4: PREDICTING MULTIPLE GENERATIONS OF SELECTION
Assume you are selecting for increased body weight in a population of chickens. The narrow-
sense heritability for body weight is 0.5, and you can achieve a selection differential of 200 g
per generation. Predict the cumulative response to selection after 3 generations, assuming the
heritability remains constant.
Solution:
Given:
• h² = 0.5
• S = 200 g per generation
• Number of generations = 3
Step 1: Calculate the response to selection per generation R = h²S = 0.5 × 200 g = 100 g
Step 2: Calculate the cumulative response over 3 generations Cumulative R = 3 × 100 g = 300
g
Therefore, the predicted cumulative response to selection after 3 generations is 300 g.
1.130 PROBLEM 5: CORRELATED RESPONSE TO SELECTION
The correlated response (CR_y) in trait y when selecting for trait x is given by the equation:
CR_y = ih_x h_y r_g σ_py
Where i is the selection intensity for trait x, h_x and h_y are the square roots of the
heritabilities for traits x and y respectively, r_g is the genetic correlation between traits x and y,
and σ_py is the phenotypic standard deviation of trait y.
In a tomato breeding program, you are selecting for increased fruit weight (trait x) but are also
interested in the correlated response in sugar content (trait y). Given:
• Selection intensity for fruit weight (i) = 1.2
• Heritability of fruit weight (h²_x) = 0.6
• Heritability of sugar content (h²_y) = 0.4
• Genetic correlation between fruit weight and sugar content (r_g) = -0.3
• Phenotypic standard deviation of sugar content (σ_py) = 0.5 °Brix
Calculate the expected correlated response in sugar content when selecting for increased fruit
weight.
Solution:
Step 1: Calculate the square roots of heritabilities h_x = √(0.6) = 0.775 h_y = √(0.4) = 0.632
Step 2: Apply the correlated response equation CR_y = ih_x h_y r_g σ_py CR_y = 1.2 × 0.775
× 0.632 × (-0.3) × 0.5 °Brix CR_y = -0.088 °Brix
Therefore, when selecting for increased fruit weight, the expected correlated response in sugar
content is a decrease of 0.088 °Brix.
1.131 PROBLEM 1: BASIC RESPONSE TO SELECTION
The Breeder’s equation is given by R = h²S, where R is the response to selection, h² is the
narrow-sense heritability, and S is the selection differential. In a population of wheat, the
narrow-sense heritability for grain yield is 0.4, and the selection differential is 500 kg/ha.
Calculate the expected response to selection for grain yield.
Solution:
Given:
• h² = 0.4
• S = 500 kg/ha
Using the Breeder’s equation: R = h²S
R = 0.4 × 500 kg/ha = 200 kg/ha
Therefore, the expected response to selection for grain yield is 200 kg/ha.
1.132 PROBLEM 2: SELECTION INTENSITY
The Breeder’s equation can also be written as R = ih²σ_p, where i is the selection intensity, and
σ_p is the phenotypic standard deviation. In a dairy cattle breeding program, the narrow-sense
heritability for milk yield is 0.3, the phenotypic standard deviation is 1000 kg, and the selection
intensity is 1.4. Calculate the expected response to selection for milk yield.
Solution:
Given:
• h² = 0.3
• σ_p = 1000 kg
• i = 1.4
Using the Breeder’s equation: R = ih²σ_p
R = 1.4 × 0.3 × 1000 kg = 420 kg
Therefore, the expected response to selection for milk yield is 420 kg.
1.133 PROBLEM 3: CALCULATING HERITABILITY
In a selection experiment for plant height in peas, the selection differential was 10 cm, and the
response to selection was 4 cm. Using the Breeder’s equation, calculate the narrow-sense
heritability for plant height in this population.
Solution:
Given:
• S = 10 cm
• R = 4 cm
Using the Breeder’s equation: R = h²S
Rearranging to solve for h²: h² = R / S = 4 cm / 10 cm = 0.4
Therefore, the narrow-sense heritability for plant height in this population is 0.4 or 40
1.134 PROBLEM 4: PREDICTING MULTIPLE GENERATIONS OF SELECTION
Assume you are selecting for increased body weight in a population of chickens. The narrow-
sense heritability for body weight is 0.5, and you can achieve a selection differential of 200 g
per generation. Predict the cumulative response to selection after 3 generations, assuming the
heritability remains constant.
Solution:
Given:
• h² = 0.5
• S = 200 g per generation
• Number of generations = 3
Step 1: Calculate the response to selection per generation R = h²S = 0.5 × 200 g = 100 g
Step 2: Calculate the cumulative response over 3 generations Cumulative R = 3 × 100 g = 300
g
Therefore, the predicted cumulative response to selection after 3 generations is 300 g.
1.135 PROBLEM 5: CORRELATED RESPONSE TO SELECTION
The correlated response (CR_y) in trait y when selecting for trait x is given by the equation:
CR_y = ih_x h_y r_g σ_py
Where i is the selection intensity for trait x, h_x and h_y are the square roots of the
heritabilities for traits x and y respectively, r_g is the genetic correlation between traits x and y,
and σ_py is the phenotypic standard deviation of trait y.
In a tomato breeding program, you are selecting for increased fruit weight (trait x) but are also
interested in the correlated response in sugar content (trait y). Given:
• Selection intensity for fruit weight (i) = 1.2
• Heritability of fruit weight (h²_x) = 0.6
• Heritability of sugar content (h²_y) = 0.4
• Genetic correlation between fruit weight and sugar content (r_g) = -0.3
• Phenotypic standard deviation of sugar content (σ_py) = 0.5 °Brix
Calculate the expected correlated response in sugar content when selecting for increased fruit
weight.
Solution:
Step 1: Calculate the square roots of heritabilities h_x = √(0.6) = 0.775 h_y = √(0.4) = 0.632
Step 2: Apply the correlated response equation CR_y = ih_x h_y r_g σ_py CR_y = 1.2 × 0.775
× 0.632 × (-0.3) × 0.5 °Brix CR_y = -0.088 °Brix
Therefore, when selecting for increased fruit weight, the expected correlated response in sugar
content is a decrease of 0.088 °Brix.
1.136 PROBLEM 1: BASIC RESPONSE TO SELECTION
The Breeder’s equation is given by R = h²S, where R is the response to selection, h² is the
narrow-sense heritability, and S is the selection differential. In a population of wheat, the
narrow-sense heritability for grain yield is 0.4, and the selection differential is 500 kg/ha.
Calculate the expected response to selection for grain yield.
Solution:
Given:
• h² = 0.4
• S = 500 kg/ha
Using the Breeder’s equation: R = h²S
R = 0.4 × 500 kg/ha = 200 kg/ha
Therefore, the expected response to selection for grain yield is 200 kg/ha.
1.137 PROBLEM 2: SELECTION INTENSITY
The Breeder’s equation can also be written as R = ih²σ_p, where i is the selection intensity, and
σ_p is the phenotypic standard deviation. In a dairy cattle breeding program, the narrow-sense
heritability for milk yield is 0.3, the phenotypic standard deviation is 1000 kg, and the selection
intensity is 1.4. Calculate the expected response to selection for milk yield.
Solution:
Given:
• h² = 0.3
• σ_p = 1000 kg
• i = 1.4
Using the Breeder’s equation: R = ih²σ_p
R = 1.4 × 0.3 × 1000 kg = 420 kg
Therefore, the expected response to selection for milk yield is 420 kg.
1.138 PROBLEM 3: CALCULATING HERITABILITY
In a selection experiment for plant height in peas, the selection differential was 10 cm, and the
response to selection was 4 cm. Using the Breeder’s equation, calculate the narrow-sense
heritability for plant height in this population.
Solution:
Given:
• S = 10 cm
• R = 4 cm
Using the Breeder’s equation: R = h²S
Rearranging to solve for h²: h² = R / S = 4 cm / 10 cm = 0.4
Therefore, the narrow-sense heritability for plant height in this population is 0.4 or 40
1.139 PROBLEM 4: PREDICTING MULTIPLE GENERATIONS OF SELECTION
Assume you are selecting for increased body weight in a population of chickens. The narrow-
sense heritability for body weight is 0.5, and you can achieve a selection differential of 200 g
per generation. Predict the cumulative response to selection after 3 generations, assuming the
heritability remains constant.
Solution:
Given:
• h² = 0.5
• S = 200 g per generation
• Number of generations = 3
Step 1: Calculate the response to selection per generation R = h²S = 0.5 × 200 g = 100 g
Step 2: Calculate the cumulative response over 3 generations Cumulative R = 3 × 100 g = 300
g
Therefore, the predicted cumulative response to selection after 3 generations is 300 g.
1.140 PROBLEM 5: CORRELATED RESPONSE TO SELECTION
The correlated response (CR_y) in trait y when selecting for trait x is given by the equation:
CR_y = ih_x h_y r_g σ_py
Where i is the selection intensity for trait x, h_x and h_y are the square roots of the
heritabilities for traits x and y respectively, r_g is the genetic correlation between traits x and y,
and σ_py is the phenotypic standard deviation of trait y.
In a tomato breeding program, you are selecting for increased fruit weight (trait x) but are also
interested in the correlated response in sugar content (trait y). Given:
• Selection intensity for fruit weight (i) = 1.2
• Heritability of fruit weight (h²_x) = 0.6
• Heritability of sugar content (h²_y) = 0.4
• Genetic correlation between fruit weight and sugar content (r_g) = -0.3
• Phenotypic standard deviation of sugar content (σ_py) = 0.5 °Brix
Calculate the expected correlated response in sugar content when selecting for increased fruit
weight.
Solution:
Step 1: Calculate the square roots of heritabilities h_x = √(0.6) = 0.775 h_y = √(0.4) = 0.632
Step 2: Apply the correlated response equation CR_y = ih_x h_y r_g σ_py CR_y = 1.2 × 0.775
× 0.632 × (-0.3) × 0.5 °Brix CR_y = -0.088 °Brix
Therefore, when selecting for increased fruit weight, the expected correlated response in sugar
content is a decrease of 0.088 °Brix.
1.141 PROBLEM 1: BASIC RESPONSE TO SELECTION
The Breeder’s equation is given by R = h²S, where R is the response to selection, h² is the
narrow-sense heritability, and S is the selection differential. In a population of wheat, the
narrow-sense heritability for grain yield is 0.4, and the selection differential is 500 kg/ha.
Calculate the expected response to selection for grain yield.
Solution:
Given:
• h² = 0.4
• S = 500 kg/ha
Using the Breeder’s equation: R = h²S
R = 0.4 × 500 kg/ha = 200 kg/ha
Therefore, the expected response to selection for grain yield is 200 kg/ha.
1.142 PROBLEM 2: SELECTION INTENSITY
The Breeder’s equation can also be written as R = ih²σ_p, where i is the selection intensity, and
σ_p is the phenotypic standard deviation. In a dairy cattle breeding program, the narrow-sense
heritability for milk yield is 0.3, the phenotypic standard deviation is 1000 kg, and the selection
intensity is 1.4. Calculate the expected response to selection for milk yield.
Solution:
Given:
• h² = 0.3
• σ_p = 1000 kg
• i = 1.4
Using the Breeder’s equation: R = ih²σ_p
R = 1.4 × 0.3 × 1000 kg = 420 kg
Therefore, the expected response to selection for milk yield is 420 kg.
1.143 PROBLEM 3: CALCULATING HERITABILITY
In a selection experiment for plant height in peas, the selection differential was 10 cm, and the
response to selection was 4 cm. Using the Breeder’s equation, calculate the narrow-sense
heritability for plant height in this population.
Solution:
Given:
• S = 10 cm
• R = 4 cm
Using the Breeder’s equation: R = h²S
Rearranging to solve for h²: h² = R / S = 4 cm / 10 cm = 0.4
Therefore, the narrow-sense heritability for plant height in this population is 0.4 or 40
1.144 PROBLEM 4: PREDICTING MULTIPLE GENERATIONS OF SELECTION
Assume you are selecting for increased body weight in a population of chickens. The narrow-
sense heritability for body weight is 0.5, and you can achieve a selection differential of 200 g
per generation. Predict the cumulative response to selection after 3 generations, assuming the
heritability remains constant.
Solution:
Given:
• h² = 0.5
• S = 200 g per generation
• Number of generations = 3
Step 1: Calculate the response to selection per generation R = h²S = 0.5 × 200 g = 100 g
Step 2: Calculate the cumulative response over 3 generations Cumulative R = 3 × 100 g = 300
g
Therefore, the predicted cumulative response to selection after 3 generations is 300 g.
1.145 PROBLEM 5: CORRELATED RESPONSE TO SELECTION
The correlated response (CR_y) in trait y when selecting for trait x is given by the equation:
CR_y = ih_x h_y r_g σ_py
Where i is the selection intensity for trait x, h_x and h_y are the square roots of the
heritabilities for traits x and y respectively, r_g is the genetic correlation between traits x and y,
and σ_py is the phenotypic standard deviation of trait y.
In a tomato breeding program, you are selecting for increased fruit weight (trait x) but are also
interested in the correlated response in sugar content (trait y). Given:
• Selection intensity for fruit weight (i) = 1.2
• Heritability of fruit weight (h²_x) = 0.6
• Heritability of sugar content (h²_y) = 0.4
• Genetic correlation between fruit weight and sugar content (r_g) = -0.3
• Phenotypic standard deviation of sugar content (σ_py) = 0.5 °Brix
Calculate the expected correlated response in sugar content when selecting for increased fruit
weight.
Solution:
Step 1: Calculate the square roots of heritabilities h_x = √(0.6) = 0.775 h_y = √(0.4) = 0.632
Step 2: Apply the correlated response equation CR_y = ih_x h_y r_g σ_py CR_y = 1.2 × 0.775
× 0.632 × (-0.3) × 0.5 °Brix CR_y = -0.088 °Brix
Therefore, when selecting for increased fruit weight, the expected correlated response in sugar
content is a decrease of 0.088 °Brix.
1.146 PROBLEM 1: BASIC RESPONSE TO SELECTION
The Breeder’s equation is given by R = h²S, where R is the response to selection, h² is the
narrow-sense heritability, and S is the selection differential. In a population of wheat, the
narrow-sense heritability for grain yield is 0.4, and the selection differential is 500 kg/ha.
Calculate the expected response to selection for grain yield.
Solution:
Given:
• h² = 0.4
• S = 500 kg/ha
Using the Breeder’s equation: R = h²S
R = 0.4 × 500 kg/ha = 200 kg/ha
Therefore, the expected response to selection for grain yield is 200 kg/ha.
1.147 PROBLEM 2: SELECTION INTENSITY
The Breeder’s equation can also be written as R = ih²σ_p, where i is the selection intensity, and
σ_p is the phenotypic standard deviation. In a dairy cattle breeding program, the narrow-sense
heritability for milk yield is 0.3, the phenotypic standard deviation is 1000 kg, and the selection
intensity is 1.4. Calculate the expected response to selection for milk yield.
Solution:
Given:
• h² = 0.3
• σ_p = 1000 kg
• i = 1.4
Using the Breeder’s equation: R = ih²σ_p
R = 1.4 × 0.3 × 1000 kg = 420 kg
Therefore, the expected response to selection for milk yield is 420 kg.
1.148 PROBLEM 3: CALCULATING HERITABILITY
In a selection experiment for plant height in peas, the selection differential was 10 cm, and the
response to selection was 4 cm. Using the Breeder’s equation, calculate the narrow-sense
heritability for plant height in this population.
Solution:
Given:
• S = 10 cm
• R = 4 cm
Using the Breeder’s equation: R = h²S
Rearranging to solve for h²: h² = R / S = 4 cm / 10 cm = 0.4
Therefore, the narrow-sense heritability for plant height in this population is 0.4 or 40
1.149 PROBLEM 4: PREDICTING MULTIPLE GENERATIONS OF SELECTION
Assume you are selecting for increased body weight in a population of chickens. The narrow-
sense heritability for body weight is 0.5, and you can achieve a selection differential of 200 g
per generation. Predict the cumulative response to selection after 3 generations, assuming the
heritability remains constant.
Solution:
Given:
• h² = 0.5
• S = 200 g per generation
• Number of generations = 3
Step 1: Calculate the response to selection per generation R = h²S = 0.5 × 200 g = 100 g
Step 2: Calculate the cumulative response over 3 generations Cumulative R = 3 × 100 g = 300
g
Therefore, the predicted cumulative response to selection after 3 generations is 300 g.
1.150 PROBLEM 5: CORRELATED RESPONSE TO SELECTION
The correlated response (CR_y) in trait y when selecting for trait x is given by the equation:
CR_y = ih_x h_y r_g σ_py
Where i is the selection intensity for trait x, h_x and h_y are the square roots of the
heritabilities for traits x and y respectively, r_g is the genetic correlation between traits x and y,
and σ_py is the phenotypic standard deviation of trait y.
In a tomato breeding program, you are selecting for increased fruit weight (trait x) but are also
interested in the correlated response in sugar content (trait y). Given:
• Selection intensity for fruit weight (i) = 1.2
• Heritability of fruit weight (h²_x) = 0.6
• Heritability of sugar content (h²_y) = 0.4
• Genetic correlation between fruit weight and sugar content (r_g) = -0.3
• Phenotypic standard deviation of sugar content (σ_py) = 0.5 °Brix
Calculate the expected correlated response in sugar content when selecting for increased fruit
weight.
Solution:
Step 1: Calculate the square roots of heritabilities h_x = √(0.6) = 0.775 h_y = √(0.4) = 0.632
Step 2: Apply the correlated response equation CR_y = ih_x h_y r_g σ_py CR_y = 1.2 × 0.775
× 0.632 × (-0.3) × 0.5 °Brix CR_y = -0.088 °Brix
Therefore, when selecting for increased fruit weight, the expected correlated response in sugar
content is a decrease of 0.088 °Brix.
1.151 PROBLEM 1: BASIC RESPONSE TO SELECTION
The Breeder’s equation is given by R = h²S, where R is the response to selection, h² is the
narrow-sense heritability, and S is the selection differential. In a population of wheat, the
narrow-sense heritability for grain yield is 0.4, and the selection differential is 500 kg/ha.
Calculate the expected response to selection for grain yield.
Solution:
Given:
• h² = 0.4
• S = 500 kg/ha
Using the Breeder’s equation: R = h²S
R = 0.4 × 500 kg/ha = 200 kg/ha
Therefore, the expected response to selection for grain yield is 200 kg/ha.
1.152 PROBLEM 2: SELECTION INTENSITY
The Breeder’s equation can also be written as R = ih²σ_p, where i is the selection intensity, and
σ_p is the phenotypic standard deviation. In a dairy cattle breeding program, the narrow-sense
heritability for milk yield is 0.3, the phenotypic standard deviation is 1000 kg, and the selection
intensity is 1.4. Calculate the expected response to selection for milk yield.
Solution:
Given:
• h² = 0.3
• σ_p = 1000 kg
• i = 1.4
Using the Breeder’s equation: R = ih²σ_p
R = 1.4 × 0.3 × 1000 kg = 420 kg
Therefore, the expected response to selection for milk yield is 420 kg.
1.153 PROBLEM 3: CALCULATING HERITABILITY
In a selection experiment for plant height in peas, the selection differential was 10 cm, and the
response to selection was 4 cm. Using the Breeder’s equation, calculate the narrow-sense
heritability for plant height in this population.
Solution:
Given:
• S = 10 cm
• R = 4 cm
Using the Breeder’s equation: R = h²S
Rearranging to solve for h²: h² = R / S = 4 cm / 10 cm = 0.4
Therefore, the narrow-sense heritability for plant height in this population is 0.4 or 40
1.154 PROBLEM 4: PREDICTING MULTIPLE GENERATIONS OF SELECTION
Assume you are selecting for increased body weight in a population of chickens. The narrow-
sense heritability for body weight is 0.5, and you can achieve a selection differential of 200 g
per generation. Predict the cumulative response to selection after 3 generations, assuming the
heritability remains constant.
Solution:
Given:
• h² = 0.5
• S = 200 g per generation
• Number of generations = 3
Step 1: Calculate the response to selection per generation R = h²S = 0.5 × 200 g = 100 g
Step 2: Calculate the cumulative response over 3 generations Cumulative R = 3 × 100 g = 300
g
Therefore, the predicted cumulative response to selection after 3 generations is 300 g.
1.155 PROBLEM 5: CORRELATED RESPONSE TO SELECTION
The correlated response (CR_y) in trait y when selecting for trait x is given by the equation:
CR_y = ih_x h_y r_g σ_py
Where i is the selection intensity for trait x, h_x and h_y are the square roots of the
heritabilities for traits x and y respectively, r_g is the genetic correlation between traits x and y,
and σ_py is the phenotypic standard deviation of trait y.
In a tomato breeding program, you are selecting for increased fruit weight (trait x) but are also
interested in the correlated response in sugar content (trait y). Given:
• Selection intensity for fruit weight (i) = 1.2
• Heritability of fruit weight (h²_x) = 0.6
• Heritability of sugar content (h²_y) = 0.4
• Genetic correlation between fruit weight and sugar content (r_g) = -0.3
• Phenotypic standard deviation of sugar content (σ_py) = 0.5 °Brix
Calculate the expected correlated response in sugar content when selecting for increased fruit
weight.
Solution:
Step 1: Calculate the square roots of heritabilities h_x = √(0.6) = 0.775 h_y = √(0.4) = 0.632
Step 2: Apply the correlated response equation CR_y = ih_x h_y r_g σ_py CR_y = 1.2 × 0.775
× 0.632 × (-0.3) × 0.5 °Brix CR_y = -0.088 °Brix
Therefore, when selecting for increased fruit weight, the expected correlated response in sugar
content is a decrease of 0.088 °Brix.
1.156 PROBLEM 1: BASIC RESPONSE TO SELECTION
The Breeder’s equation is given by R = h²S, where R is the response to selection, h² is the
narrow-sense heritability, and S is the selection differential. In a population of wheat, the
narrow-sense heritability for grain yield is 0.4, and the selection differential is 500 kg/ha.
Calculate the expected response to selection for grain yield.
Solution:
Given:
• h² = 0.4
• S = 500 kg/ha
Using the Breeder’s equation: R = h²S
R = 0.4 × 500 kg/ha = 200 kg/ha
Therefore, the expected response to selection for grain yield is 200 kg/ha.
1.157 PROBLEM 2: SELECTION INTENSITY
The Breeder’s equation can also be written as R = ih²σ_p, where i is the selection intensity, and
σ_p is the phenotypic standard deviation. In a dairy cattle breeding program, the narrow-sense
heritability for milk yield is 0.3, the phenotypic standard deviation is 1000 kg, and the selection
intensity is 1.4. Calculate the expected response to selection for milk yield.
Solution:
Given:
• h² = 0.3
• σ_p = 1000 kg
• i = 1.4
Using the Breeder’s equation: R = ih²σ_p
R = 1.4 × 0.3 × 1000 kg = 420 kg
Therefore, the expected response to selection for milk yield is 420 kg.
1.158 PROBLEM 3: CALCULATING HERITABILITY
In a selection experiment for plant height in peas, the selection differential was 10 cm, and the
response to selection was 4 cm. Using the Breeder’s equation, calculate the narrow-sense
heritability for plant height in this population.
Solution:
Given:
• S = 10 cm
• R = 4 cm
Using the Breeder’s equation: R = h²S
Rearranging to solve for h²: h² = R / S = 4 cm / 10 cm = 0.4
Therefore, the narrow-sense heritability for plant height in this population is 0.4 or 40
1.159 PROBLEM 4: PREDICTING MULTIPLE GENERATIONS OF SELECTION
Assume you are selecting for increased body weight in a population of chickens. The narrow-
sense heritability for body weight is 0.5, and you can achieve a selection differential of 200 g
per generation. Predict the cumulative response to selection after 3 generations, assuming the
heritability remains constant.
Solution:
Given:
• h² = 0.5
• S = 200 g per generation
• Number of generations = 3
Step 1: Calculate the response to selection per generation R = h²S = 0.5 × 200 g = 100 g
Step 2: Calculate the cumulative response over 3 generations Cumulative R = 3 × 100 g = 300
g
Therefore, the predicted cumulative response to selection after 3 generations is 300 g.
1.160 PROBLEM 5: CORRELATED RESPONSE TO SELECTION
The correlated response (CR_y) in trait y when selecting for trait x is given by the equation:
CR_y = ih_x h_y r_g σ_py
Where i is the selection intensity for trait x, h_x and h_y are the square roots of the
heritabilities for traits x and y respectively, r_g is the genetic correlation between traits x and y,
and σ_py is the phenotypic standard deviation of trait y.
In a tomato breeding program, you are selecting for increased fruit weight (trait x) but are also
interested in the correlated response in sugar content (trait y). Given:
• Selection intensity for fruit weight (i) = 1.2
• Heritability of fruit weight (h²_x) = 0.6
• Heritability of sugar content (h²_y) = 0.4
• Genetic correlation between fruit weight and sugar content (r_g) = -0.3
• Phenotypic standard deviation of sugar content (σ_py) = 0.5 °Brix
Calculate the expected correlated response in sugar content when selecting for increased fruit
weight.
Solution:
Step 1: Calculate the square roots of heritabilities h_x = √(0.6) = 0.775 h_y = √(0.4) = 0.632
Step 2: Apply the correlated response equation CR_y = ih_x h_y r_g σ_py CR_y = 1.2 × 0.775
× 0.632 × (-0.3) × 0.5 °Brix CR_y = -0.088 °Brix
Therefore, when selecting for increased fruit weight, the expected correlated response in sugar
content is a decrease of 0.088 °Brix.
1.161 PROBLEM 1: BASIC RESPONSE TO SELECTION
The Breeder’s equation is given by R = h²S, where R is the response to selection, h² is the
narrow-sense heritability, and S is the selection differential. In a population of wheat, the
narrow-sense heritability for grain yield is 0.4, and the selection differential is 500 kg/ha.
Calculate the expected response to selection for grain yield.
Solution:
Given:
• h² = 0.4
• S = 500 kg/ha
Using the Breeder’s equation: R = h²S
R = 0.4 × 500 kg/ha = 200 kg/ha
Therefore, the expected response to selection for grain yield is 200 kg/ha.
1.162 PROBLEM 2: SELECTION INTENSITY
The Breeder’s equation can also be written as R = ih²σ_p, where i is the selection intensity, and
σ_p is the phenotypic standard deviation. In a dairy cattle breeding program, the narrow-sense
heritability for milk yield is 0.3, the phenotypic standard deviation is 1000 kg, and the selection
intensity is 1.4. Calculate the expected response to selection for milk yield.
Solution:
Given:
• h² = 0.3
• σ_p = 1000 kg
• i = 1.4
Using the Breeder’s equation: R = ih²σ_p
R = 1.4 × 0.3 × 1000 kg = 420 kg
Therefore, the expected response to selection for milk yield is 420 kg.
1.163 PROBLEM 3: CALCULATING HERITABILITY
In a selection experiment for plant height in peas, the selection differential was 10 cm, and the
response to selection was 4 cm. Using the Breeder’s equation, calculate the narrow-sense
heritability for plant height in this population.
Solution:
Given:
• S = 10 cm
• R = 4 cm
Using the Breeder’s equation: R = h²S
Rearranging to solve for h²: h² = R / S = 4 cm / 10 cm = 0.4
Therefore, the narrow-sense heritability for plant height in this population is 0.4 or 40
1.164 PROBLEM 4: PREDICTING MULTIPLE GENERATIONS OF SELECTION
Assume you are selecting for increased body weight in a population of chickens. The narrow-
sense heritability for body weight is 0.5, and you can achieve a selection differential of 200 g
per generation. Predict the cumulative response to selection after 3 generations, assuming the
heritability remains constant.
Solution:
Given:
• h² = 0.5
• S = 200 g per generation
• Number of generations = 3
Step 1: Calculate the response to selection per generation R = h²S = 0.5 × 200 g = 100 g
Step 2: Calculate the cumulative response over 3 generations Cumulative R = 3 × 100 g = 300
g
Therefore, the predicted cumulative response to selection after 3 generations is 300 g.
1.165 PROBLEM 5: CORRELATED RESPONSE TO SELECTION
The correlated response (CR_y) in trait y when selecting for trait x is given by the equation:
CR_y = ih_x h_y r_g σ_py
Where i is the selection intensity for trait x, h_x and h_y are the square roots of the
heritabilities for traits x and y respectively, r_g is the genetic correlation between traits x and y,
and σ_py is the phenotypic standard deviation of trait y.
In a tomato breeding program, you are selecting for increased fruit weight (trait x) but are also
interested in the correlated response in sugar content (trait y). Given:
• Selection intensity for fruit weight (i) = 1.2
• Heritability of fruit weight (h²_x) = 0.6
• Heritability of sugar content (h²_y) = 0.4
• Genetic correlation between fruit weight and sugar content (r_g) = -0.3
• Phenotypic standard deviation of sugar content (σ_py) = 0.5 °Brix
Calculate the expected correlated response in sugar content when selecting for increased fruit
weight.
Solution:
Step 1: Calculate the square roots of heritabilities h_x = √(0.6) = 0.775 h_y = √(0.4) = 0.632
Step 2: Apply the correlated response equation CR_y = ih_x h_y r_g σ_py CR_y = 1.2 × 0.775
× 0.632 × (-0.3) × 0.5 °Brix CR_y = -0.088 °Brix
Therefore, when selecting for increased fruit weight, the expected correlated response in sugar
content is a decrease of 0.088 °Brix.
1.166 PROBLEM 1: BASIC RESPONSE TO SELECTION
The Breeder’s equation is given by R = h²S, where R is the response to selection, h² is the
narrow-sense heritability, and S is the selection differential. In a population of wheat, the
narrow-sense heritability for grain yield is 0.4, and the selection differential is 500 kg/ha.
Calculate the expected response to selection for grain yield.
Solution:
Given:
• h² = 0.4
• S = 500 kg/ha
Using the Breeder’s equation: R = h²S
R = 0.4 × 500 kg/ha = 200 kg/ha
Therefore, the expected response to selection for grain yield is 200 kg/ha.
1.167 PROBLEM 2: SELECTION INTENSITY
The Breeder’s equation can also be written as R = ih²σ_p, where i is the selection intensity, and
σ_p is the phenotypic standard deviation. In a dairy cattle breeding program, the narrow-sense
heritability for milk yield is 0.3, the phenotypic standard deviation is 1000 kg, and the selection
intensity is 1.4. Calculate the expected response to selection for milk yield.
Solution:
Given:
• h² = 0.3
• σ_p = 1000 kg
• i = 1.4
Using the Breeder’s equation: R = ih²σ_p
R = 1.4 × 0.3 × 1000 kg = 420 kg
Therefore, the expected response to selection for milk yield is 420 kg.
1.168 PROBLEM 3: CALCULATING HERITABILITY
In a selection experiment for plant height in peas, the selection differential was 10 cm, and the
response to selection was 4 cm. Using the Breeder’s equation, calculate the narrow-sense
heritability for plant height in this population.
Solution:
Given:
• S = 10 cm
• R = 4 cm
Using the Breeder’s equation: R = h²S
Rearranging to solve for h²: h² = R / S = 4 cm / 10 cm = 0.4
Therefore, the narrow-sense heritability for plant height in this population is 0.4 or 40
1.169 PROBLEM 4: PREDICTING MULTIPLE GENERATIONS OF SELECTION
Assume you are selecting for increased body weight in a population of chickens. The narrow-
sense heritability for body weight is 0.5, and you can achieve a selection differential of 200 g
per generation. Predict the cumulative response to selection after 3 generations, assuming the
heritability remains constant.
Solution:
Given:
• h² = 0.5
• S = 200 g per generation
• Number of generations = 3
Step 1: Calculate the response to selection per generation R = h²S = 0.5 × 200 g = 100 g
Step 2: Calculate the cumulative response over 3 generations Cumulative R = 3 × 100 g = 300
g
Therefore, the predicted cumulative response to selection after 3 generations is 300 g.
1.170 PROBLEM 5: CORRELATED RESPONSE TO SELECTION
The correlated response (CR_y) in trait y when selecting for trait x is given by the equation:
CR_y = ih_x h_y r_g σ_py
Where i is the selection intensity for trait x, h_x and h_y are the square roots of the
heritabilities for traits x and y respectively, r_g is the genetic correlation between traits x and y,
and σ_py is the phenotypic standard deviation of trait y.
In a tomato breeding program, you are selecting for increased fruit weight (trait x) but are also
interested in the correlated response in sugar content (trait y). Given:
• Selection intensity for fruit weight (i) = 1.2
• Heritability of fruit weight (h²_x) = 0.6
• Heritability of sugar content (h²_y) = 0.4
• Genetic correlation between fruit weight and sugar content (r_g) = -0.3
• Phenotypic standard deviation of sugar content (σ_py) = 0.5 °Brix
Calculate the expected correlated response in sugar content when selecting for increased fruit
weight.
Solution:
Step 1: Calculate the square roots of heritabilities h_x = √(0.6) = 0.775 h_y = √(0.4) = 0.632
Step 2: Apply the correlated response equation CR_y = ih_x h_y r_g σ_py CR_y = 1.2 × 0.775
× 0.632 × (-0.3) × 0.5 °Brix CR_y = -0.088 °Brix
Therefore, when selecting for increased fruit weight, the expected correlated response in sugar
content is a decrease of 0.088 °Brix.
1.171 PROBLEM 1: BASIC RESPONSE TO SELECTION
The Breeder’s equation is given by R = h²S, where R is the response to selection, h² is the
narrow-sense heritability, and S is the selection differential. In a population of wheat, the
narrow-sense heritability for grain yield is 0.4, and the selection differential is 500 kg/ha.
Calculate the expected response to selection for grain yield.
Solution:
Given:
• h² = 0.4
• S = 500 kg/ha
Using the Breeder’s equation: R = h²S
R = 0.4 × 500 kg/ha = 200 kg/ha
Therefore, the expected response to selection for grain yield is 200 kg/ha.
1.172 PROBLEM 2: SELECTION INTENSITY
The Breeder’s equation can also be written as R = ih²σ_p, where i is the selection intensity, and
σ_p is the phenotypic standard deviation. In a dairy cattle breeding program, the narrow-sense
heritability for milk yield is 0.3, the phenotypic standard deviation is 1000 kg, and the selection
intensity is 1.4. Calculate the expected response to selection for milk yield.
Solution:
Given:
• h² = 0.3
• σ_p = 1000 kg
• i = 1.4
Using the Breeder’s equation: R = ih²σ_p
R = 1.4 × 0.3 × 1000 kg = 420 kg
Therefore, the expected response to selection for milk yield is 420 kg.
1.173 PROBLEM 3: CALCULATING HERITABILITY
In a selection experiment for plant height in peas, the selection differential was 10 cm, and the
response to selection was 4 cm. Using the Breeder’s equation, calculate the narrow-sense
heritability for plant height in this population.
Solution:
Given:
• S = 10 cm
• R = 4 cm
Using the Breeder’s equation: R = h²S
Rearranging to solve for h²: h² = R / S = 4 cm / 10 cm = 0.4
Therefore, the narrow-sense heritability for plant height in this population is 0.4 or 40
1.174 PROBLEM 4: PREDICTING MULTIPLE GENERATIONS OF SELECTION
Assume you are selecting for increased body weight in a population of chickens. The narrow-
sense heritability for body weight is 0.5, and you can achieve a selection differential of 200 g
per generation. Predict the cumulative response to selection after 3 generations, assuming the
heritability remains constant.
Solution:
Given:
• h² = 0.5
• S = 200 g per generation
• Number of generations = 3
Step 1: Calculate the response to selection per generation R = h²S = 0.5 × 200 g = 100 g
Step 2: Calculate the cumulative response over 3 generations Cumulative R = 3 × 100 g = 300
g
Therefore, the predicted cumulative response to selection after 3 generations is 300 g.
1.175 PROBLEM 5: CORRELATED RESPONSE TO SELECTION
The correlated response (CR_y) in trait y when selecting for trait x is given by the equation:
CR_y = ih_x h_y r_g σ_py
Where i is the selection intensity for trait x, h_x and h_y are the square roots of the
heritabilities for traits x and y respectively, r_g is the genetic correlation between traits x and y,
and σ_py is the phenotypic standard deviation of trait y.
In a tomato breeding program, you are selecting for increased fruit weight (trait x) but are also
interested in the correlated response in sugar content (trait y). Given:
• Selection intensity for fruit weight (i) = 1.2
• Heritability of fruit weight (h²_x) = 0.6
• Heritability of sugar content (h²_y) = 0.4
• Genetic correlation between fruit weight and sugar content (r_g) = -0.3
• Phenotypic standard deviation of sugar content (σ_py) = 0.5 °Brix
Calculate the expected correlated response in sugar content when selecting for increased fruit
weight.
Solution:
Step 1: Calculate the square roots of heritabilities h_x = √(0.6) = 0.775 h_y = √(0.4) = 0.632
Step 2: Apply the correlated response equation CR_y = ih_x h_y r_g σ_py CR_y = 1.2 × 0.775
× 0.632 × (-0.3) × 0.5 °Brix CR_y = -0.088 °Brix
Therefore, when selecting for increased fruit weight, the expected correlated response in sugar
content is a decrease of 0.088 °Brix.
1.176 PROBLEM 1: BASIC RESPONSE TO SELECTION
The Breeder’s equation is given by R = h²S, where R is the response to selection, h² is the
narrow-sense heritability, and S is the selection differential. In a population of wheat, the
narrow-sense heritability for grain yield is 0.4, and the selection differential is 500 kg/ha.
Calculate the expected response to selection for grain yield.
Solution:
Given:
• h² = 0.4
• S = 500 kg/ha
Using the Breeder’s equation: R = h²S
R = 0.4 × 500 kg/ha = 200 kg/ha
Therefore, the expected response to selection for grain yield is 200 kg/ha.
1.177 PROBLEM 2: SELECTION INTENSITY
The Breeder’s equation can also be written as R = ih²σ_p, where i is the selection intensity, and
σ_p is the phenotypic standard deviation. In a dairy cattle breeding program, the narrow-sense
heritability for milk yield is 0.3, the phenotypic standard deviation is 1000 kg, and the selection
intensity is 1.4. Calculate the expected response to selection for milk yield.
Solution:
Given:
• h² = 0.3
• σ_p = 1000 kg
• i = 1.4
Using the Breeder’s equation: R = ih²σ_p
R = 1.4 × 0.3 × 1000 kg = 420 kg
Therefore, the expected response to selection for milk yield is 420 kg.
1.178 PROBLEM 3: CALCULATING HERITABILITY
In a selection experiment for plant height in peas, the selection differential was 10 cm, and the
response to selection was 4 cm. Using the Breeder’s equation, calculate the narrow-sense
heritability for plant height in this population.
Solution:
Given:
• S = 10 cm
• R = 4 cm
Using the Breeder’s equation: R = h²S
Rearranging to solve for h²: h² = R / S = 4 cm / 10 cm = 0.4
Therefore, the narrow-sense heritability for plant height in this population is 0.4 or 40
1.179 PROBLEM 4: PREDICTING MULTIPLE GENERATIONS OF SELECTION
Assume you are selecting for increased body weight in a population of chickens. The narrow-
sense heritability for body weight is 0.5, and you can achieve a selection differential of 200 g
per generation. Predict the cumulative response to selection after 3 generations, assuming the
heritability remains constant.
Solution:
Given:
• h² = 0.5
• S = 200 g per generation
• Number of generations = 3
Step 1: Calculate the response to selection per generation R = h²S = 0.5 × 200 g = 100 g
Step 2: Calculate the cumulative response over 3 generations Cumulative R = 3 × 100 g = 300
g
Therefore, the predicted cumulative response to selection after 3 generations is 300 g.
1.180 PROBLEM 5: CORRELATED RESPONSE TO SELECTION
The correlated response (CR_y) in trait y when selecting for trait x is given by the equation:
CR_y = ih_x h_y r_g σ_py
Where i is the selection intensity for trait x, h_x and h_y are the square roots of the
heritabilities for traits x and y respectively, r_g is the genetic correlation between traits x and y,
and σ_py is the phenotypic standard deviation of trait y.
In a tomato breeding program, you are selecting for increased fruit weight (trait x) but are also
interested in the correlated response in sugar content (trait y). Given:
• Selection intensity for fruit weight (i) = 1.2
• Heritability of fruit weight (h²_x) = 0.6
• Heritability of sugar content (h²_y) = 0.4
• Genetic correlation between fruit weight and sugar content (r_g) = -0.3
• Phenotypic standard deviation of sugar content (σ_py) = 0.5 °Brix
Calculate the expected correlated response in sugar content when selecting for increased fruit
weight.
Solution:
Step 1: Calculate the square roots of heritabilities h_x = √(0.6) = 0.775 h_y = √(0.4) = 0.632
Step 2: Apply the correlated response equation CR_y = ih_x h_y r_g σ_py CR_y = 1.2 × 0.775
× 0.632 × (-0.3) × 0.5 °Brix CR_y = -0.088 °Brix
Therefore, when selecting for increased fruit weight, the expected correlated response in sugar
content is a decrease of 0.088 °Brix.
1.181 PROBLEM 1: BASIC RESPONSE TO SELECTION
The Breeder’s equation is given by R = h²S, where R is the response to selection, h² is the
narrow-sense heritability, and S is the selection differential. In a population of wheat, the
narrow-sense heritability for grain yield is 0.4, and the selection differential is 500 kg/ha.
Calculate the expected response to selection for grain yield.
Solution:
Given:
• h² = 0.4
• S = 500 kg/ha
Using the Breeder’s equation: R = h²S
R = 0.4 × 500 kg/ha = 200 kg/ha
Therefore, the expected response to selection for grain yield is 200 kg/ha.
1.182 PROBLEM 2: SELECTION INTENSITY
The Breeder’s equation can also be written as R = ih²σ_p, where i is the selection intensity, and
σ_p is the phenotypic standard deviation. In a dairy cattle breeding program, the narrow-sense
heritability for milk yield is 0.3, the phenotypic standard deviation is 1000 kg, and the selection
intensity is 1.4. Calculate the expected response to selection for milk yield.
Solution:
Given:
• h² = 0.3
• σ_p = 1000 kg
• i = 1.4
Using the Breeder’s equation: R = ih²σ_p
R = 1.4 × 0.3 × 1000 kg = 420 kg
Therefore, the expected response to selection for milk yield is 420 kg.
1.183 PROBLEM 3: CALCULATING HERITABILITY
In a selection experiment for plant height in peas, the selection differential was 10 cm, and the
response to selection was 4 cm. Using the Breeder’s equation, calculate the narrow-sense
heritability for plant height in this population.
Solution:
Given:
• S = 10 cm
• R = 4 cm
Using the Breeder’s equation: R = h²S
Rearranging to solve for h²: h² = R / S = 4 cm / 10 cm = 0.4
Therefore, the narrow-sense heritability for plant height in this population is 0.4 or 40
1.184 PROBLEM 4: PREDICTING MULTIPLE GENERATIONS OF SELECTION
Assume you are selecting for increased body weight in a population of chickens. The narrow-
sense heritability for body weight is 0.5, and you can achieve a selection differential of 200 g
per generation. Predict the cumulative response to selection after 3 generations, assuming the
heritability remains constant.
Solution:
Given:
• h² = 0.5
• S = 200 g per generation
• Number of generations = 3
Step 1: Calculate the response to selection per generation R = h²S = 0.5 × 200 g = 100 g
Step 2: Calculate the cumulative response over 3 generations Cumulative R = 3 × 100 g = 300
g
Therefore, the predicted cumulative response to selection after 3 generations is 300 g.
1.185 PROBLEM 5: CORRELATED RESPONSE TO SELECTION
The correlated response (CR_y) in trait y when selecting for trait x is given by the equation:
CR_y = ih_x h_y r_g σ_py
Where i is the selection intensity for trait x, h_x and h_y are the square roots of the
heritabilities for traits x and y respectively, r_g is the genetic correlation between traits x and y,
and σ_py is the phenotypic standard deviation of trait y.
In a tomato breeding program, you are selecting for increased fruit weight (trait x) but are also
interested in the correlated response in sugar content (trait y). Given:
• Selection intensity for fruit weight (i) = 1.2
• Heritability of fruit weight (h²_x) = 0.6
• Heritability of sugar content (h²_y) = 0.4
• Genetic correlation between fruit weight and sugar content (r_g) = -0.3
• Phenotypic standard deviation of sugar content (σ_py) = 0.5 °Brix
Calculate the expected correlated response in sugar content when selecting for increased fruit
weight.
Solution:
Step 1: Calculate the square roots of heritabilities h_x = √(0.6) = 0.775 h_y = √(0.4) = 0.632
Step 2: Apply the correlated response equation CR_y = ih_x h_y r_g σ_py CR_y = 1.2 × 0.775
× 0.632 × (-0.3) × 0.5 °Brix CR_y = -0.088 °Brix
Therefore, when selecting for increased fruit weight, the expected correlated response in sugar
content is a decrease of 0.088 °Brix.
1.186 PROBLEM 1: BASIC RESPONSE TO SELECTION
The Breeder’s equation is given by R = h²S, where R is the response to selection, h² is the
narrow-sense heritability, and S is the selection differential. In a population of wheat, the
narrow-sense heritability for grain yield is 0.4, and the selection differential is 500 kg/ha.
Calculate the expected response to selection for grain yield.
Solution:
Given:
• h² = 0.4
• S = 500 kg/ha
Using the Breeder’s equation: R = h²S
R = 0.4 × 500 kg/ha = 200 kg/ha
Therefore, the expected response to selection for grain yield is 200 kg/ha.
1.187 PROBLEM 2: SELECTION INTENSITY
The Breeder’s equation can also be written as R = ih²σ_p, where i is the selection intensity, and
σ_p is the phenotypic standard deviation. In a dairy cattle breeding program, the narrow-sense
heritability for milk yield is 0.3, the phenotypic standard deviation is 1000 kg, and the selection
intensity is 1.4. Calculate the expected response to selection for milk yield.
Solution:
Given:
• h² = 0.3
• σ_p = 1000 kg
• i = 1.4
Using the Breeder’s equation: R = ih²σ_p
R = 1.4 × 0.3 × 1000 kg = 420 kg
Therefore, the expected response to selection for milk yield is 420 kg.
1.188 PROBLEM 3: CALCULATING HERITABILITY
In a selection experiment for plant height in peas, the selection differential was 10 cm, and the
response to selection was 4 cm. Using the Breeder’s equation, calculate the narrow-sense
heritability for plant height in this population.
Solution:
Given:
• S = 10 cm
• R = 4 cm
Using the Breeder’s equation: R = h²S
Rearranging to solve for h²: h² = R / S = 4 cm / 10 cm = 0.4
Therefore, the narrow-sense heritability for plant height in this population is 0.4 or 40
1.189 PROBLEM 4: PREDICTING MULTIPLE GENERATIONS OF SELECTION
Assume you are selecting for increased body weight in a population of chickens. The narrow-
sense heritability for body weight is 0.5, and you can achieve a selection differential of 200 g
per generation. Predict the cumulative response to selection after 3 generations, assuming the
heritability remains constant.
Solution:
Given:
• h² = 0.5
• S = 200 g per generation
• Number of generations = 3
Step 1: Calculate the response to selection per generation R = h²S = 0.5 × 200 g = 100 g
Step 2: Calculate the cumulative response over 3 generations Cumulative R = 3 × 100 g = 300
g
Therefore, the predicted cumulative response to selection after 3 generations is 300 g.
1.190 PROBLEM 5: CORRELATED RESPONSE TO SELECTION
The correlated response (CR_y) in trait y when selecting for trait x is given by the equation:
CR_y = ih_x h_y r_g σ_py
Where i is the selection intensity for trait x, h_x and h_y are the square roots of the
heritabilities for traits x and y respectively, r_g is the genetic correlation between traits x and y,
and σ_py is the phenotypic standard deviation of trait y.
In a tomato breeding program, you are selecting for increased fruit weight (trait x) but are also
interested in the correlated response in sugar content (trait y). Given:
• Selection intensity for fruit weight (i) = 1.2
• Heritability of fruit weight (h²_x) = 0.6
• Heritability of sugar content (h²_y) = 0.4
• Genetic correlation between fruit weight and sugar content (r_g) = -0.3
• Phenotypic standard deviation of sugar content (σ_py) = 0.5 °Brix
Calculate the expected correlated response in sugar content when selecting for increased fruit
weight.
Solution:
Step 1: Calculate the square roots of heritabilities h_x = √(0.6) = 0.775 h_y = √(0.4) = 0.632
Step 2: Apply the correlated response equation CR_y = ih_x h_y r_g σ_py CR_y = 1.2 × 0.775
× 0.632 × (-0.3) × 0.5 °Brix CR_y = -0.088 °Brix
Therefore, when selecting for increased fruit weight, the expected correlated response in sugar
content is a decrease of 0.088 °Brix.
1.191 PROBLEM 1: BASIC RESPONSE TO SELECTION
The Breeder’s equation is given by R = h²S, where R is the response to selection, h² is the
narrow-sense heritability, and S is the selection differential. In a population of wheat, the
narrow-sense heritability for grain yield is 0.4, and the selection differential is 500 kg/ha.
Calculate the expected response to selection for grain yield.
Solution:
Given:
• h² = 0.4
• S = 500 kg/ha
Using the Breeder’s equation: R = h²S
R = 0.4 × 500 kg/ha = 200 kg/ha
Therefore, the expected response to selection for grain yield is 200 kg/ha.
1.192 PROBLEM 2: SELECTION INTENSITY
The Breeder’s equation can also be written as R = ih²σ_p, where i is the selection intensity, and
σ_p is the phenotypic standard deviation. In a dairy cattle breeding program, the narrow-sense
heritability for milk yield is 0.3, the phenotypic standard deviation is 1000 kg, and the selection
intensity is 1.4. Calculate the expected response to selection for milk yield.
Solution:
Given:
• h² = 0.3
• σ_p = 1000 kg
• i = 1.4
Using the Breeder’s equation: R = ih²σ_p
R = 1.4 × 0.3 × 1000 kg = 420 kg
Therefore, the expected response to selection for milk yield is 420 kg.
1.193 PROBLEM 3: CALCULATING HERITABILITY
In a selection experiment for plant height in peas, the selection differential was 10 cm, and the
response to selection was 4 cm. Using the Breeder’s equation, calculate the narrow-sense
heritability for plant height in this population.
Solution:
Given:
• S = 10 cm
• R = 4 cm
Using the Breeder’s equation: R = h²S
Rearranging to solve for h²: h² = R / S = 4 cm / 10 cm = 0.4
Therefore, the narrow-sense heritability for plant height in this population is 0.4 or 40
1.194 PROBLEM 4: PREDICTING MULTIPLE GENERATIONS OF SELECTION
Assume you are selecting for increased body weight in a population of chickens. The narrow-
sense heritability for body weight is 0.5, and you can achieve a selection differential of 200 g
per generation. Predict the cumulative response to selection after 3 generations, assuming the
heritability remains constant.
Solution:
Given:
• h² = 0.5
• S = 200 g per generation
• Number of generations = 3
Step 1: Calculate the response to selection per generation R = h²S = 0.5 × 200 g = 100 g
Step 2: Calculate the cumulative response over 3 generations Cumulative R = 3 × 100 g = 300
g
Therefore, the predicted cumulative response to selection after 3 generations is 300 g.
1.195 PROBLEM 5: CORRELATED RESPONSE TO SELECTION
The correlated response (CR_y) in trait y when selecting for trait x is given by the equation:
CR_y = ih_x h_y r_g σ_py
Where i is the selection intensity for trait x, h_x and h_y are the square roots of the
heritabilities for traits x and y respectively, r_g is the genetic correlation between traits x and y,
and σ_py is the phenotypic standard deviation of trait y.
In a tomato breeding program, you are selecting for increased fruit weight (trait x) but are also
interested in the correlated response in sugar content (trait y). Given:
• Selection intensity for fruit weight (i) = 1.2
• Heritability of fruit weight (h²_x) = 0.6
• Heritability of sugar content (h²_y) = 0.4
• Genetic correlation between fruit weight and sugar content (r_g) = -0.3
• Phenotypic standard deviation of sugar content (σ_py) = 0.5 °Brix
Calculate the expected correlated response in sugar content when selecting for increased fruit
weight.
Solution:
Step 1: Calculate the square roots of heritabilities h_x = √(0.6) = 0.775 h_y = √(0.4) = 0.632
Step 2: Apply the correlated response equation CR_y = ih_x h_y r_g σ_py CR_y = 1.2 × 0.775
× 0.632 × (-0.3) × 0.5 °Brix CR_y = -0.088 °Brix
Therefore, when selecting for increased fruit weight, the expected correlated response in sugar
content is a decrease of 0.088 °Brix.
1.196 PROBLEM 1: BASIC RESPONSE TO SELECTION
The Breeder’s equation is given by R = h²S, where R is the response to selection, h² is the
narrow-sense heritability, and S is the selection differential. In a population of wheat, the
narrow-sense heritability for grain yield is 0.4, and the selection differential is 500 kg/ha.
Calculate the expected response to selection for grain yield.
Solution:
Given:
• h² = 0.4
• S = 500 kg/ha
Using the Breeder’s equation: R = h²S
R = 0.4 × 500 kg/ha = 200 kg/ha
Therefore, the expected response to selection for grain yield is 200 kg/ha.
1.197 PROBLEM 2: SELECTION INTENSITY
The Breeder’s equation can also be written as R = ih²σ_p, where i is the selection intensity, and
σ_p is the phenotypic standard deviation. In a dairy cattle breeding program, the narrow-sense
heritability for milk yield is 0.3, the phenotypic standard deviation is 1000 kg, and the selection
intensity is 1.4. Calculate the expected response to selection for milk yield.
Solution:
Given:
• h² = 0.3
• σ_p = 1000 kg
• i = 1.4
Using the Breeder’s equation: R = ih²σ_p
R = 1.4 × 0.3 × 1000 kg = 420 kg
Therefore, the expected response to selection for milk yield is 420 kg.
1.198 PROBLEM 3: CALCULATING HERITABILITY
In a selection experiment for plant height in peas, the selection differential was 10 cm, and the
response to selection was 4 cm. Using the Breeder’s equation, calculate the narrow-sense
heritability for plant height in this population.
Solution:
Given:
• S = 10 cm
• R = 4 cm
Using the Breeder’s equation: R = h²S
Rearranging to solve for h²: h² = R / S = 4 cm / 10 cm = 0.4
Therefore, the narrow-sense heritability for plant height in this population is 0.4 or 40
1.199 PROBLEM 4: PREDICTING MULTIPLE GENERATIONS OF SELECTION
Assume you are selecting for increased body weight in a population of chickens. The narrow-
sense heritability for body weight is 0.5, and you can achieve a selection differential of 200 g
per generation. Predict the cumulative response to selection after 3 generations, assuming the
heritability remains constant.
Solution:
Given:
• h² = 0.5
• S = 200 g per generation
• Number of generations = 3
Step 1: Calculate the response to selection per generation R = h²S = 0.5 × 200 g = 100 g
Step 2: Calculate the cumulative response over 3 generations Cumulative R = 3 × 100 g = 300
g
Therefore, the predicted cumulative response to selection after 3 generations is 300 g.
1.200 PROBLEM 5: CORRELATED RESPONSE TO SELECTION
The correlated response (CR_y) in trait y when selecting for trait x is given by the equation:
CR_y = ih_x h_y r_g σ_py
Where i is the selection intensity for trait x, h_x and h_y are the square roots of the
heritabilities for traits x and y respectively, r_g is the genetic correlation between traits x and y,
and σ_py is the phenotypic standard deviation of trait y.
In a tomato breeding program, you are selecting for increased fruit weight (trait x) but are also
interested in the correlated response in sugar content (trait y). Given:
• Selection intensity for fruit weight (i) = 1.2
• Heritability of fruit weight (h²_x) = 0.6
• Heritability of sugar content (h²_y) = 0.4
• Genetic correlation between fruit weight and sugar content (r_g) = -0.3
• Phenotypic standard deviation of sugar content (σ_py) = 0.5 °Brix
Calculate the expected correlated response in sugar content when selecting for increased fruit
weight.
Solution:
Step 1: Calculate the square roots of heritabilities h_x = √(0.6) = 0.775 h_y = √(0.4) = 0.632
Step 2: Apply the correlated response equation CR_y = ih_x h_y r_g σ_py CR_y = 1.2 × 0.775
× 0.632 × (-0.3) × 0.5 °Brix CR_y = -0.088 °Brix
Therefore, when selecting for increased fruit weight, the expected correlated response in sugar
content is a decrease of 0.088 °Brix.
1.201 PROBLEM 1: BASIC RESPONSE TO SELECTION
The Breeder’s equation is given by R = h²S, where R is the response to selection, h² is the
narrow-sense heritability, and S is the selection differential. In a population of wheat, the
narrow-sense heritability for grain yield is 0.4, and the selection differential is 500 kg/ha.
Calculate the expected response to selection for grain yield.
Solution:
Given:
• h² = 0.4
• S = 500 kg/ha
Using the Breeder’s equation: R = h²S
R = 0.4 × 500 kg/ha = 200 kg/ha
Therefore, the expected response to selection for grain yield is 200 kg/ha.
1.202 PROBLEM 2: SELECTION INTENSITY
The Breeder’s equation can also be written as R = ih²σ_p, where i is the selection intensity, and
σ_p is the phenotypic standard deviation. In a dairy cattle breeding program, the narrow-sense
heritability for milk yield is 0.3, the phenotypic standard deviation is 1000 kg, and the selection
intensity is 1.4. Calculate the expected response to selection for milk yield.
Solution:
Given:
• h² = 0.3
• σ_p = 1000 kg
• i = 1.4
Using the Breeder’s equation: R = ih²σ_p
R = 1.4 × 0.3 × 1000 kg = 420 kg
Therefore, the expected response to selection for milk yield is 420 kg.
1.203 PROBLEM 3: CALCULATING HERITABILITY
In a selection experiment for plant height in peas, the selection differential was 10 cm, and the
response to selection was 4 cm. Using the Breeder’s equation, calculate the narrow-sense
heritability for plant height in this population.
Solution:
Given:
• S = 10 cm
• R = 4 cm
Using the Breeder’s equation: R = h²S
Rearranging to solve for h²: h² = R / S = 4 cm / 10 cm = 0.4
Therefore, the narrow-sense heritability for plant height in this population is 0.4 or 40
1.204 PROBLEM 4: PREDICTING MULTIPLE GENERATIONS OF SELECTION
Assume you are selecting for increased body weight in a population of chickens. The narrow-
sense heritability for body weight is 0.5, and you can achieve a selection differential of 200 g
per generation. Predict the cumulative response to selection after 3 generations, assuming the
heritability remains constant.
Solution:
Given:
• h² = 0.5
• S = 200 g per generation
• Number of generations = 3
Step 1: Calculate the response to selection per generation R = h²S = 0.5 × 200 g = 100 g
Step 2: Calculate the cumulative response over 3 generations Cumulative R = 3 × 100 g = 300
g
Therefore, the predicted cumulative response to selection after 3 generations is 300 g.
1.205 PROBLEM 5: CORRELATED RESPONSE TO SELECTION
The correlated response (CR_y) in trait y when selecting for trait x is given by the equation:
CR_y = ih_x h_y r_g σ_py
Where i is the selection intensity for trait x, h_x and h_y are the square roots of the
heritabilities for traits x and y respectively, r_g is the genetic correlation between traits x and y,
and σ_py is the phenotypic standard deviation of trait y.
In a tomato breeding program, you are selecting for increased fruit weight (trait x) but are also
interested in the correlated response in sugar content (trait y). Given:
• Selection intensity for fruit weight (i) = 1.2
• Heritability of fruit weight (h²_x) = 0.6
• Heritability of sugar content (h²_y) = 0.4
• Genetic correlation between fruit weight and sugar content (r_g) = -0.3
• Phenotypic standard deviation of sugar content (σ_py) = 0.5 °Brix
Calculate the expected correlated response in sugar content when selecting for increased fruit
weight.
Solution:
Step 1: Calculate the square roots of heritabilities h_x = √(0.6) = 0.775 h_y = √(0.4) = 0.632
Step 2: Apply the correlated response equation CR_y = ih_x h_y r_g σ_py CR_y = 1.2 × 0.775
× 0.632 × (-0.3) × 0.5 °Brix CR_y = -0.088 °Brix
Therefore, when selecting for increased fruit weight, the expected correlated response in sugar
content is a decrease of 0.088 °Brix.
1.206 PROBLEM 1: BASIC RESPONSE TO SELECTION
The Breeder’s equation is given by R = h²S, where R is the response to selection, h² is the
narrow-sense heritability, and S is the selection differential. In a population of wheat, the
narrow-sense heritability for grain yield is 0.4, and the selection differential is 500 kg/ha.
Calculate the expected response to selection for grain yield.
Solution:
Given:
• h² = 0.4
• S = 500 kg/ha
Using the Breeder’s equation: R = h²S
R = 0.4 × 500 kg/ha = 200 kg/ha
Therefore, the expected response to selection for grain yield is 200 kg/ha.
1.207 PROBLEM 2: SELECTION INTENSITY
The Breeder’s equation can also be written as R = ih²σ_p, where i is the selection intensity, and
σ_p is the phenotypic standard deviation. In a dairy cattle breeding program, the narrow-sense
heritability for milk yield is 0.3, the phenotypic standard deviation is 1000 kg, and the selection
intensity is 1.4. Calculate the expected response to selection for milk yield.
Solution:
Given:
• h² = 0.3
• σ_p = 1000 kg
• i = 1.4
Using the Breeder’s equation: R = ih²σ_p
R = 1.4 × 0.3 × 1000 kg = 420 kg
Therefore, the expected response to selection for milk yield is 420 kg.
1.208 PROBLEM 3: CALCULATING HERITABILITY
In a selection experiment for plant height in peas, the selection differential was 10 cm, and the
response to selection was 4 cm. Using the Breeder’s equation, calculate the narrow-sense
heritability for plant height in this population.
Solution:
Given:
• S = 10 cm
• R = 4 cm
Using the Breeder’s equation: R = h²S
Rearranging to solve for h²: h² = R / S = 4 cm / 10 cm = 0.4
Therefore, the narrow-sense heritability for plant height in this population is 0.4 or 40
1.209 PROBLEM 4: PREDICTING MULTIPLE GENERATIONS OF SELECTION
Assume you are selecting for increased body weight in a population of chickens. The narrow-
sense heritability for body weight is 0.5, and you can achieve a selection differential of 200 g
per generation. Predict the cumulative response to selection after 3 generations, assuming the
heritability remains constant.
Solution:
Given:
• h² = 0.5
• S = 200 g per generation
• Number of generations = 3
Step 1: Calculate the response to selection per generation R = h²S = 0.5 × 200 g = 100 g
Step 2: Calculate the cumulative response over 3 generations Cumulative R = 3 × 100 g = 300
g
Therefore, the predicted cumulative response to selection after 3 generations is 300 g.
1.210 PROBLEM 5: CORRELATED RESPONSE TO SELECTION
The correlated response (CR_y) in trait y when selecting for trait x is given by the equation:
CR_y = ih_x h_y r_g σ_py
Where i is the selection intensity for trait x, h_x and h_y are the square roots of the
heritabilities for traits x and y respectively, r_g is the genetic correlation between traits x and y,
and σ_py is the phenotypic standard deviation of trait y.
In a tomato breeding program, you are selecting for increased fruit weight (trait x) but are also
interested in the correlated response in sugar content (trait y). Given:
• Selection intensity for fruit weight (i) = 1.2
• Heritability of fruit weight (h²_x) = 0.6
• Heritability of sugar content (h²_y) = 0.4
• Genetic correlation between fruit weight and sugar content (r_g) = -0.3
• Phenotypic standard deviation of sugar content (σ_py) = 0.5 °Brix
Calculate the expected correlated response in sugar content when selecting for increased fruit
weight.
Solution:
Step 1: Calculate the square roots of heritabilities h_x = √(0.6) = 0.775 h_y = √(0.4) = 0.632
Step 2: Apply the correlated response equation CR_y = ih_x h_y r_g σ_py CR_y = 1.2 × 0.775
× 0.632 × (-0.3) × 0.5 °Brix CR_y = -0.088 °Brix
Therefore, when selecting for increased fruit weight, the expected correlated response in sugar
content is a decrease of 0.088 °Brix.
1.211 PROBLEM 1: BASIC RESPONSE TO SELECTION
The Breeder’s equation is given by R = h²S, where R is the response to selection, h² is the
narrow-sense heritability, and S is the selection differential. In a population of wheat, the
narrow-sense heritability for grain yield is 0.4, and the selection differential is 500 kg/ha.
Calculate the expected response to selection for grain yield.
Solution:
Given:
• h² = 0.4
• S = 500 kg/ha
Using the Breeder’s equation: R = h²S
R = 0.4 × 500 kg/ha = 200 kg/ha
Therefore, the expected response to selection for grain yield is 200 kg/ha.
1.212 PROBLEM 2: SELECTION INTENSITY
The Breeder’s equation can also be written as R = ih²σ_p, where i is the selection intensity, and
σ_p is the phenotypic standard deviation. In a dairy cattle breeding program, the narrow-sense
heritability for milk yield is 0.3, the phenotypic standard deviation is 1000 kg, and the selection
intensity is 1.4. Calculate the expected response to selection for milk yield.
Solution:
Given:
• h² = 0.3
• σ_p = 1000 kg
• i = 1.4
Using the Breeder’s equation: R = ih²σ_p
R = 1.4 × 0.3 × 1000 kg = 420 kg
Therefore, the expected response to selection for milk yield is 420 kg.
1.213 PROBLEM 3: CALCULATING HERITABILITY
In a selection experiment for plant height in peas, the selection differential was 10 cm, and the
response to selection was 4 cm. Using the Breeder’s equation, calculate the narrow-sense
heritability for plant height in this population.
Solution:
Given:
• S = 10 cm
• R = 4 cm
Using the Breeder’s equation: R = h²S
Rearranging to solve for h²: h² = R / S = 4 cm / 10 cm = 0.4
Therefore, the narrow-sense heritability for plant height in this population is 0.4 or 40
1.214 PROBLEM 4: PREDICTING MULTIPLE GENERATIONS OF SELECTION
Assume you are selecting for increased body weight in a population of chickens. The narrow-
sense heritability for body weight is 0.5, and you can achieve a selection differential of 200 g
per generation. Predict the cumulative response to selection after 3 generations, assuming the
heritability remains constant.
Solution:
Given:
• h² = 0.5
• S = 200 g per generation
• Number of generations = 3
Step 1: Calculate the response to selection per generation R = h²S = 0.5 × 200 g = 100 g
Step 2: Calculate the cumulative response over 3 generations Cumulative R = 3 × 100 g = 300
g
Therefore, the predicted cumulative response to selection after 3 generations is 300 g.
1.215 PROBLEM 5: CORRELATED RESPONSE TO SELECTION
The correlated response (CR_y) in trait y when selecting for trait x is given by the equation:
CR_y = ih_x h_y r_g σ_py
Where i is the selection intensity for trait x, h_x and h_y are the square roots of the
heritabilities for traits x and y respectively, r_g is the genetic correlation between traits x and y,
and σ_py is the phenotypic standard deviation of trait y.
In a tomato breeding program, you are selecting for increased fruit weight (trait x) but are also
interested in the correlated response in sugar content (trait y). Given:
• Selection intensity for fruit weight (i) = 1.2
• Heritability of fruit weight (h²_x) = 0.6
• Heritability of sugar content (h²_y) = 0.4
• Genetic correlation between fruit weight and sugar content (r_g) = -0.3
• Phenotypic standard deviation of sugar content (σ_py) = 0.5 °Brix
Calculate the expected correlated response in sugar content when selecting for increased fruit
weight.
Solution:
Step 1: Calculate the square roots of heritabilities h_x = √(0.6) = 0.775 h_y = √(0.4) = 0.632
Step 2: Apply the correlated response equation CR_y = ih_x h_y r_g σ_py CR_y = 1.2 × 0.775
× 0.632 × (-0.3) × 0.5 °Brix CR_y = -0.088 °Brix
Therefore, when selecting for increased fruit weight, the expected correlated response in sugar
content is a decrease of 0.088 °Brix.
1.216 PROBLEM 1: BASIC RESPONSE TO SELECTION
The Breeder’s equation is given by R = h²S, where R is the response to selection, h² is the
narrow-sense heritability, and S is the selection differential. In a population of wheat, the
narrow-sense heritability for grain yield is 0.4, and the selection differential is 500 kg/ha.
Calculate the expected response to selection for grain yield.
Solution:
Given:
• h² = 0.4
• S = 500 kg/ha
Using the Breeder’s equation: R = h²S
R = 0.4 × 500 kg/ha = 200 kg/ha
Therefore, the expected response to selection for grain yield is 200 kg/ha.
1.217 PROBLEM 2: SELECTION INTENSITY
The Breeder’s equation can also be written as R = ih²σ_p, where i is the selection intensity, and
σ_p is the phenotypic standard deviation. In a dairy cattle breeding program, the narrow-sense
heritability for milk yield is 0.3, the phenotypic standard deviation is 1000 kg, and the selection
intensity is 1.4. Calculate the expected response to selection for milk yield.
Solution:
Given:
• h² = 0.3
• σ_p = 1000 kg
• i = 1.4
Using the Breeder’s equation: R = ih²σ_p
R = 1.4 × 0.3 × 1000 kg = 420 kg
Therefore, the expected response to selection for milk yield is 420 kg.
1.218 PROBLEM 3: CALCULATING HERITABILITY
In a selection experiment for plant height in peas, the selection differential was 10 cm, and the
response to selection was 4 cm. Using the Breeder’s equation, calculate the narrow-sense
heritability for plant height in this population.
Solution:
Given:
• S = 10 cm
• R = 4 cm
Using the Breeder’s equation: R = h²S
Rearranging to solve for h²: h² = R / S = 4 cm / 10 cm = 0.4
Therefore, the narrow-sense heritability for plant height in this population is 0.4 or 40
1.219 PROBLEM 4: PREDICTING MULTIPLE GENERATIONS OF SELECTION
Assume you are selecting for increased body weight in a population of chickens. The narrow-
sense heritability for body weight is 0.5, and you can achieve a selection differential of 200 g
per generation. Predict the cumulative response to selection after 3 generations, assuming the
heritability remains constant.
Solution:
Given:
• h² = 0.5
• S = 200 g per generation
• Number of generations = 3
Step 1: Calculate the response to selection per generation R = h²S = 0.5 × 200 g = 100 g
Step 2: Calculate the cumulative response over 3 generations Cumulative R = 3 × 100 g = 300
g
Therefore, the predicted cumulative response to selection after 3 generations is 300 g.
1.220 PROBLEM 5: CORRELATED RESPONSE TO SELECTION
The correlated response (CR_y) in trait y when selecting for trait x is given by the equation:
CR_y = ih_x h_y r_g σ_py
Where i is the selection intensity for trait x, h_x and h_y are the square roots of the
heritabilities for traits x and y respectively, r_g is the genetic correlation between traits x and y,
and σ_py is the phenotypic standard deviation of trait y.
In a tomato breeding program, you are selecting for increased fruit weight (trait x) but are also
interested in the correlated response in sugar content (trait y). Given:
• Selection intensity for fruit weight (i) = 1.2
• Heritability of fruit weight (h²_x) = 0.6
• Heritability of sugar content (h²_y) = 0.4
• Genetic correlation between fruit weight and sugar content (r_g) = -0.3
• Phenotypic standard deviation of sugar content (σ_py) = 0.5 °Brix
Calculate the expected correlated response in sugar content when selecting for increased fruit
weight.
Solution:
Step 1: Calculate the square roots of heritabilities h_x = √(0.6) = 0.775 h_y = √(0.4) = 0.632
Step 2: Apply the correlated response equation CR_y = ih_x h_y r_g σ_py CR_y = 1.2 × 0.775
× 0.632 × (-0.3) × 0.5 °Brix CR_y = -0.088 °Brix
Therefore, when selecting for increased fruit weight, the expected correlated response in sugar
content is a decrease of 0.088 °Brix.
1.221 PROBLEM 1: BASIC RESPONSE TO SELECTION
The Breeder’s equation is given by R = h²S, where R is the response to selection, h² is the
narrow-sense heritability, and S is the selection differential. In a population of wheat, the
narrow-sense heritability for grain yield is 0.4, and the selection differential is 500 kg/ha.
Calculate the expected response to selection for grain yield.
Solution:
Given:
• h² = 0.4
• S = 500 kg/ha
Using the Breeder’s equation: R = h²S
R = 0.4 × 500 kg/ha = 200 kg/ha
Therefore, the expected response to selection for grain yield is 200 kg/ha.
1.222 PROBLEM 2: SELECTION INTENSITY
The Breeder’s equation can also be written as R = ih²σ_p, where i is the selection intensity, and
σ_p is the phenotypic standard deviation. In a dairy cattle breeding program, the narrow-sense
heritability for milk yield is 0.3, the phenotypic standard deviation is 1000 kg, and the selection
intensity is 1.4. Calculate the expected response to selection for milk yield.
Solution:
Given:
• h² = 0.3
• σ_p = 1000 kg
• i = 1.4
Using the Breeder’s equation: R = ih²σ_p
R = 1.4 × 0.3 × 1000 kg = 420 kg
Therefore, the expected response to selection for milk yield is 420 kg.
1.223 PROBLEM 3: CALCULATING HERITABILITY
In a selection experiment for plant height in peas, the selection differential was 10 cm, and the
response to selection was 4 cm. Using the Breeder’s equation, calculate the narrow-sense
heritability for plant height in this population.
Solution:
Given:
• S = 10 cm
• R = 4 cm
Using the Breeder’s equation: R = h²S
Rearranging to solve for h²: h² = R / S = 4 cm / 10 cm = 0.4
Therefore, the narrow-sense heritability for plant height in this population is 0.4 or 40
1.224 PROBLEM 4: PREDICTING MULTIPLE GENERATIONS OF SELECTION
Assume you are selecting for increased body weight in a population of chickens. The narrow-
sense heritability for body weight is 0.5, and you can achieve a selection differential of 200 g
per generation. Predict the cumulative response to selection after 3 generations, assuming the
heritability remains constant.
Solution:
Given:
• h² = 0.5
• S = 200 g per generation
• Number of generations = 3
Step 1: Calculate the response to selection per generation R = h²S = 0.5 × 200 g = 100 g
Step 2: Calculate the cumulative response over 3 generations Cumulative R = 3 × 100 g = 300
g
Therefore, the predicted cumulative response to selection after 3 generations is 300 g.
1.225 PROBLEM 5: CORRELATED RESPONSE TO SELECTION
The correlated response (CR_y) in trait y when selecting for trait x is given by the equation:
CR_y = ih_x h_y r_g σ_py
Where i is the selection intensity for trait x, h_x and h_y are the square roots of the
heritabilities for traits x and y respectively, r_g is the genetic correlation between traits x and y,
and σ_py is the phenotypic standard deviation of trait y.
In a tomato breeding program, you are selecting for increased fruit weight (trait x) but are also
interested in the correlated response in sugar content (trait y). Given:
• Selection intensity for fruit weight (i) = 1.2
• Heritability of fruit weight (h²_x) = 0.6
• Heritability of sugar content (h²_y) = 0.4
• Genetic correlation between fruit weight and sugar content (r_g) = -0.3
• Phenotypic standard deviation of sugar content (σ_py) = 0.5 °Brix
Calculate the expected correlated response in sugar content when selecting for increased fruit
weight.
Solution:
Step 1: Calculate the square roots of heritabilities h_x = √(0.6) = 0.775 h_y = √(0.4) = 0.632
Step 2: Apply the correlated response equation CR_y = ih_x h_y r_g σ_py CR_y = 1.2 × 0.775
× 0.632 × (-0.3) × 0.5 °Brix CR_y = -0.088 °Brix
Therefore, when selecting for increased fruit weight, the expected correlated response in sugar
content is a decrease of 0.088 °Brix.
1.226 PROBLEM 1: BASIC RESPONSE TO SELECTION
The Breeder’s equation is given by R = h²S, where R is the response to selection, h² is the
narrow-sense heritability, and S is the selection differential. In a population of wheat, the
narrow-sense heritability for grain yield is 0.4, and the selection differential is 500 kg/ha.
Calculate the expected response to selection for grain yield.
Solution:
Given:
• h² = 0.4
• S = 500 kg/ha
Using the Breeder’s equation: R = h²S
R = 0.4 × 500 kg/ha = 200 kg/ha
Therefore, the expected response to selection for grain yield is 200 kg/ha.
1.227 PROBLEM 2: SELECTION INTENSITY
The Breeder’s equation can also be written as R = ih²σ_p, where i is the selection intensity, and
σ_p is the phenotypic standard deviation. In a dairy cattle breeding program, the narrow-sense
heritability for milk yield is 0.3, the phenotypic standard deviation is 1000 kg, and the selection
intensity is 1.4. Calculate the expected response to selection for milk yield.
Solution:
Given:
• h² = 0.3
• σ_p = 1000 kg
• i = 1.4
Using the Breeder’s equation: R = ih²σ_p
R = 1.4 × 0.3 × 1000 kg = 420 kg
Therefore, the expected response to selection for milk yield is 420 kg.
1.228 PROBLEM 3: CALCULATING HERITABILITY
In a selection experiment for plant height in peas, the selection differential was 10 cm, and the
response to selection was 4 cm. Using the Breeder’s equation, calculate the narrow-sense
heritability for plant height in this population.
Solution:
Given:
• S = 10 cm
• R = 4 cm
Using the Breeder’s equation: R = h²S
Rearranging to solve for h²: h² = R / S = 4 cm / 10 cm = 0.4
Therefore, the narrow-sense heritability for plant height in this population is 0.4 or 40
1.229 PROBLEM 4: PREDICTING MULTIPLE GENERATIONS OF SELECTION
Assume you are selecting for increased body weight in a population of chickens. The narrow-
sense heritability for body weight is 0.5, and you can achieve a selection differential of 200 g
per generation. Predict the cumulative response to selection after 3 generations, assuming the
heritability remains constant.
Solution:
Given:
• h² = 0.5
• S = 200 g per generation
• Number of generations = 3
Step 1: Calculate the response to selection per generation R = h²S = 0.5 × 200 g = 100 g
Step 2: Calculate the cumulative response over 3 generations Cumulative R = 3 × 100 g = 300
g
Therefore, the predicted cumulative response to selection after 3 generations is 300 g.
1.230 PROBLEM 5: CORRELATED RESPONSE TO SELECTION
The correlated response (CR_y) in trait y when selecting for trait x is given by the equation:
CR_y = ih_x h_y r_g σ_py
Where i is the selection intensity for trait x, h_x and h_y are the square roots of the
heritabilities for traits x and y respectively, r_g is the genetic correlation between traits x and y,
and σ_py is the phenotypic standard deviation of trait y.
In a tomato breeding program, you are selecting for increased fruit weight (trait x) but are also
interested in the correlated response in sugar content (trait y). Given:
• Selection intensity for fruit weight (i) = 1.2
• Heritability of fruit weight (h²_x) = 0.6
• Heritability of sugar content (h²_y) = 0.4
• Genetic correlation between fruit weight and sugar content (r_g) = -0.3
• Phenotypic standard deviation of sugar content (σ_py) = 0.5 °Brix
Calculate the expected correlated response in sugar content when selecting for increased fruit
weight.
Solution:
Step 1: Calculate the square roots of heritabilities h_x = √(0.6) = 0.775 h_y = √(0.4) = 0.632
Step 2: Apply the correlated response equation CR_y = ih_x h_y r_g σ_py CR_y = 1.2 × 0.775
× 0.632 × (-0.3) × 0.5 °Brix CR_y = -0.088 °Brix
Therefore, when selecting for increased fruit weight, the expected correlated response in sugar
content is a decrease of 0.088 °Brix.
1.231 PROBLEM 1: BASIC RESPONSE TO SELECTION
The Breeder’s equation is given by R = h²S, where R is the response to selection, h² is the
narrow-sense heritability, and S is the selection differential. In a population of wheat, the
narrow-sense heritability for grain yield is 0.4, and the selection differential is 500 kg/ha.
Calculate the expected response to selection for grain yield.
Solution:
Given:
• h² = 0.4
• S = 500 kg/ha
Using the Breeder’s equation: R = h²S
R = 0.4 × 500 kg/ha = 200 kg/ha
Therefore, the expected response to selection for grain yield is 200 kg/ha.
1.232 PROBLEM 2: SELECTION INTENSITY
The Breeder’s equation can also be written as R = ih²σ_p, where i is the selection intensity, and
σ_p is the phenotypic standard deviation. In a dairy cattle breeding program, the narrow-sense
heritability for milk yield is 0.3, the phenotypic standard deviation is 1000 kg, and the selection
intensity is 1.4. Calculate the expected response to selection for milk yield.
Solution:
Given:
• h² = 0.3
• σ_p = 1000 kg
• i = 1.4
Using the Breeder’s equation: R = ih²σ_p
R = 1.4 × 0.3 × 1000 kg = 420 kg
Therefore, the expected response to selection for milk yield is 420 kg.
1.233 PROBLEM 3: CALCULATING HERITABILITY
In a selection experiment for plant height in peas, the selection differential was 10 cm, and the
response to selection was 4 cm. Using the Breeder’s equation, calculate the narrow-sense
heritability for plant height in this population.
Solution:
Given:
• S = 10 cm
• R = 4 cm
Using the Breeder’s equation: R = h²S
Rearranging to solve for h²: h² = R / S = 4 cm / 10 cm = 0.4
Therefore, the narrow-sense heritability for plant height in this population is 0.4 or 40
1.234 PROBLEM 4: PREDICTING MULTIPLE GENERATIONS OF SELECTION
Assume you are selecting for increased body weight in a population of chickens. The narrow-
sense heritability for body weight is 0.5, and you can achieve a selection differential of 200 g
per generation. Predict the cumulative response to selection after 3 generations, assuming the
heritability remains constant.
Solution:
Given:
• h² = 0.5
• S = 200 g per generation
• Number of generations = 3
Step 1: Calculate the response to selection per generation R = h²S = 0.5 × 200 g = 100 g
Step 2: Calculate the cumulative response over 3 generations Cumulative R = 3 × 100 g = 300
g
Therefore, the predicted cumulative response to selection after 3 generations is 300 g.
1.235 PROBLEM 5: CORRELATED RESPONSE TO SELECTION
The correlated response (CR_y) in trait y when selecting for trait x is given by the equation:
CR_y = ih_x h_y r_g σ_py
Where i is the selection intensity for trait x, h_x and h_y are the square roots of the
heritabilities for traits x and y respectively, r_g is the genetic correlation between traits x and y,
and σ_py is the phenotypic standard deviation of trait y.
In a tomato breeding program, you are selecting for increased fruit weight (trait x) but are also
interested in the correlated response in sugar content (trait y). Given:
• Selection intensity for fruit weight (i) = 1.2
• Heritability of fruit weight (h²_x) = 0.6
• Heritability of sugar content (h²_y) = 0.4
• Genetic correlation between fruit weight and sugar content (r_g) = -0.3
• Phenotypic standard deviation of sugar content (σ_py) = 0.5 °Brix
Calculate the expected correlated response in sugar content when selecting for increased fruit
weight.
Solution:
Step 1: Calculate the square roots of heritabilities h_x = √(0.6) = 0.775 h_y = √(0.4) = 0.632
Step 2: Apply the correlated response equation CR_y = ih_x h_y r_g σ_py CR_y = 1.2 × 0.775
× 0.632 × (-0.3) × 0.5 °Brix CR_y = -0.088 °Brix
Therefore, when selecting for increased fruit weight, the expected correlated response in sugar
content is a decrease of 0.088 °Brix.
1.236 PROBLEM 1: BASIC RESPONSE TO SELECTION
The Breeder’s equation is given by R = h²S, where R is the response to selection, h² is the
narrow-sense heritability, and S is the selection differential. In a population of wheat, the
narrow-sense heritability for grain yield is 0.4, and the selection differential is 500 kg/ha.
Calculate the expected response to selection for grain yield.
Solution:
Given:
• h² = 0.4
• S = 500 kg/ha
Using the Breeder’s equation: R = h²S
R = 0.4 × 500 kg/ha = 200 kg/ha
Therefore, the expected response to selection for grain yield is 200 kg/ha.
1.237 PROBLEM 2: SELECTION INTENSITY
The Breeder’s equation can also be written as R = ih²σ_p, where i is the selection intensity, and
σ_p is the phenotypic standard deviation. In a dairy cattle breeding program, the narrow-sense
heritability for milk yield is 0.3, the phenotypic standard deviation is 1000 kg, and the selection
intensity is 1.4. Calculate the expected response to selection for milk yield.
Solution:
Given:
• h² = 0.3
• σ_p = 1000 kg
• i = 1.4
Using the Breeder’s equation: R = ih²σ_p
R = 1.4 × 0.3 × 1000 kg = 420 kg
Therefore, the expected response to selection for milk yield is 420 kg.
1.238 PROBLEM 3: CALCULATING HERITABILITY
In a selection experiment for plant height in peas, the selection differential was 10 cm, and the
response to selection was 4 cm. Using the Breeder’s equation, calculate the narrow-sense
heritability for plant height in this population.
Solution:
Given:
• S = 10 cm
• R = 4 cm
Using the Breeder’s equation: R = h²S
Rearranging to solve for h²: h² = R / S = 4 cm / 10 cm = 0.4
Therefore, the narrow-sense heritability for plant height in this population is 0.4 or 40
1.239 PROBLEM 4: PREDICTING MULTIPLE GENERATIONS OF SELECTION
Assume you are selecting for increased body weight in a population of chickens. The narrow-
sense heritability for body weight is 0.5, and you can achieve a selection differential of 200 g
per generation. Predict the cumulative response to selection after 3 generations, assuming the
heritability remains constant.
Solution:
Given:
• h² = 0.5
• S = 200 g per generation
• Number of generations = 3
Step 1: Calculate the response to selection per generation R = h²S = 0.5 × 200 g = 100 g
Step 2: Calculate the cumulative response over 3 generations Cumulative R = 3 × 100 g = 300
g
Therefore, the predicted cumulative response to selection after 3 generations is 300 g.
1.240 PROBLEM 5: CORRELATED RESPONSE TO SELECTION
The correlated response (CR_y) in trait y when selecting for trait x is given by the equation:
CR_y = ih_x h_y r_g σ_py
Where i is the selection intensity for trait x, h_x and h_y are the square roots of the
heritabilities for traits x and y respectively, r_g is the genetic correlation between traits x and y,
and σ_py is the phenotypic standard deviation of trait y.
In a tomato breeding program, you are selecting for increased fruit weight (trait x) but are also
interested in the correlated response in sugar content (trait y). Given:
• Selection intensity for fruit weight (i) = 1.2
• Heritability of fruit weight (h²_x) = 0.6
• Heritability of sugar content (h²_y) = 0.4
• Genetic correlation between fruit weight and sugar content (r_g) = -0.3
• Phenotypic standard deviation of sugar content (σ_py) = 0.5 °Brix
Calculate the expected correlated response in sugar content when selecting for increased fruit
weight.
Solution:
Step 1: Calculate the square roots of heritabilities h_x = √(0.6) = 0.775 h_y = √(0.4) = 0.632
Step 2: Apply the correlated response equation CR_y = ih_x h_y r_g σ_py CR_y = 1.2 × 0.775
× 0.632 × (-0.3) × 0.5 °Brix CR_y = -0.088 °Brix
Therefore, when selecting for increased fruit weight, the expected correlated response in sugar
content is a decrease of 0.088 °Brix.
1.241 PROBLEM 1: BASIC RESPONSE TO SELECTION
The Breeder’s equation is given by R = h²S, where R is the response to selection, h² is the
narrow-sense heritability, and S is the selection differential. In a population of wheat, the
narrow-sense heritability for grain yield is 0.4, and the selection differential is 500 kg/ha.
Calculate the expected response to selection for grain yield.
Solution:
Given:
• h² = 0.4
• S = 500 kg/ha
Using the Breeder’s equation: R = h²S
R = 0.4 × 500 kg/ha = 200 kg/ha
Therefore, the expected response to selection for grain yield is 200 kg/ha.
1.242 PROBLEM 2: SELECTION INTENSITY
The Breeder’s equation can also be written as R = ih²σ_p, where i is the selection intensity, and
σ_p is the phenotypic standard deviation. In a dairy cattle breeding program, the narrow-sense
heritability for milk yield is 0.3, the phenotypic standard deviation is 1000 kg, and the selection
intensity is 1.4. Calculate the expected response to selection for milk yield.
Solution:
Given:
• h² = 0.3
• σ_p = 1000 kg
• i = 1.4
Using the Breeder’s equation: R = ih²σ_p
R = 1.4 × 0.3 × 1000 kg = 420 kg
Therefore, the expected response to selection for milk yield is 420 kg.
1.243 PROBLEM 3: CALCULATING HERITABILITY
In a selection experiment for plant height in peas, the selection differential was 10 cm, and the
response to selection was 4 cm. Using the Breeder’s equation, calculate the narrow-sense
heritability for plant height in this population.
Solution:
Given:
• S = 10 cm
• R = 4 cm
Using the Breeder’s equation: R = h²S
Rearranging to solve for h²: h² = R / S = 4 cm / 10 cm = 0.4
Therefore, the narrow-sense heritability for plant height in this population is 0.4 or 40
1.244 PROBLEM 4: PREDICTING MULTIPLE GENERATIONS OF SELECTION
Assume you are selecting for increased body weight in a population of chickens. The narrow-
sense heritability for body weight is 0.5, and you can achieve a selection differential of 200 g
per generation. Predict the cumulative response to selection after 3 generations, assuming the
heritability remains constant.
Solution:
Given:
• h² = 0.5
• S = 200 g per generation
• Number of generations = 3
Step 1: Calculate the response to selection per generation R = h²S = 0.5 × 200 g = 100 g
Step 2: Calculate the cumulative response over 3 generations Cumulative R = 3 × 100 g = 300
g
Therefore, the predicted cumulative response to selection after 3 generations is 300 g.
1.245 PROBLEM 5: CORRELATED RESPONSE TO SELECTION
The correlated response (CR_y) in trait y when selecting for trait x is given by the equation:
CR_y = ih_x h_y r_g σ_py
Where i is the selection intensity for trait x, h_x and h_y are the square roots of the
heritabilities for traits x and y respectively, r_g is the genetic correlation between traits x and y,
and σ_py is the phenotypic standard deviation of trait y.
In a tomato breeding program, you are selecting for increased fruit weight (trait x) but are also
interested in the correlated response in sugar content (trait y). Given:
• Selection intensity for fruit weight (i) = 1.2
• Heritability of fruit weight (h²_x) = 0.6
• Heritability of sugar content (h²_y) = 0.4
• Genetic correlation between fruit weight and sugar content (r_g) = -0.3
• Phenotypic standard deviation of sugar content (σ_py) = 0.5 °Brix
Calculate the expected correlated response in sugar content when selecting for increased fruit
weight.
Solution:
Step 1: Calculate the square roots of heritabilities h_x = √(0.6) = 0.775 h_y = √(0.4) = 0.632
Step 2: Apply the correlated response equation CR_y = ih_x h_y r_g σ_py CR_y = 1.2 × 0.775
× 0.632 × (-0.3) × 0.5 °Brix CR_y = -0.088 °Brix
Therefore, when selecting for increased fruit weight, the expected correlated response in sugar
content is a decrease of 0.088 °Brix.
