MBG 3060 Lecture Notes - Lecture 6: Epistasis, Total Variation, Standard Deviation
2 May 2018
School
Department
Course
Professor
Phenotype = Genotype + Environment
•
Genotype = Additive + Dominance + Interaction
•
Environment = E permanent + E temporary
•
σ2
•
*see course manual
•
= VA+ VD+ VI+ VEp + VEt
•
P = A + Ep + Et (ignore D and I and let Ep cover all three)
○
Dominance and interaction genetic effects as well as permanent
environment are specific to the animal and re not passed on
•
Genetic Models:
A measure of the sum of effects of individual alleles (genes
working together)
○
Additive genetic component (A)
•
A measure of the combined dominance effects of individual loci
○
Dominance genetic component (D)
•
A measure of the combined interactions between loci (epistasis)
○
Interaction genetic component (I)
•
Additive, Dominance and Interaction Genetic Components
+4" -AA
!
+2" -Aa
!
0" -aa
!
Appetite (A)
○
+4" -GG
!
+3.5" -Gg
□
+ 2"
□
+1.5":
!
0" -gg
!
Growth hormone production (G)
○
0" -nn
!
-1" -Nn (GG)
!
-2" -NN (GG)
!
Nutrient uptake (N)
○
Loci:
•
Height = 5'6" + 0 + 0 + 0
!
Additive = 0 + 0 + 0 = 0"
!
Dominance = 0"
!
Epistasis = 0"
!
'aa gg nn'
○
Height = 5'6" + 2" + (2" + 1.5") + 0" = 5'11.5"
!
Additive = 2" + 2" + 0"
!
Dominance = 1.5"
!
Epistasis = 0"
!
'Aa Gg Nn'
○
Height = 5'6" + 2" + 4" -2" = 5'10"
!
Additive = 2" + 4" + 0
!
Dominance = 0"
!
Epistasis = -2"
!
'Aa GG NN'
○
Additive = 2" + 2" + 0"
!
Dominance = 0"
!
Epistasis = 0"
!
'A G N'
○
*only additive part is passed on in gametes
Genotype
•
Hypothetical Example: Human Height (avg = 5'6")
Proportion of total variance that is genetic
○
Broad sense heritability: H2= σ2/ σ2
•
Proportion of total variance that is additive genetic and 1/2 is
transmitted from parent to offspring
○
Narrow sense heritability: h2= σ2/ σ2
•
Heritability:
'a'ij σ2(A)
○
Based on co-variance among relatives
•
*see course manual --> 2 bO.P. = h2
!
Offspring -midparent (average of 2) regression is preferred
over offspring-parent regression as it includes more
information
!
Regression of offspring (y) on parent (x)1)
ANOVA -phenotypes on specific relatives 2)
Two methods of estimating h2:
•
Estimation of Narrow Sense Heritability
Indicates degree of genetic resemblence between relatives and
generation
○
Helps define how much genetic change would result from
selection
○
Measures the proportion of total variation that is additive genetic
variation
•
Why are we interested in h2?
*see course manual
•
R = XO- XP
○
R = deviation of progeny from original population average
•
S = XS- XP
○
S = average of selected parents deviated from population mean
•
R = S h2
•
Response to Selection:
Average fleece weight = 2.4 kg
•
Select some rams and ewes --> average = 2.6
•
Selection differential: S= 2.6-2.4 = 0.2
•
Expected genetic progress in next generation: R = S h2= (0.2)(0.36) =
0.072 kg
•
Over 10 generations, there will be an increase of 720 g of wool
○
Therefore, selecting sleep for fleece weight results in an increase of 72 g
of wool per sheep in next generation
•
Example: Fleece weight (kg wool/sheep) *h2=0.36
Overall,
R = response to selection (units same as phenotype)
S = selection differential (units same as phenotype)
Fewer individuals -watch for inbreeding
!
Increase S
○
Measure phenotype more precisely or reduce σ2(E)
!
Increase h2
○
To increase R:
•
Selection in Practice
Select best Average S
5457 37
4465 45
3474 54
2485 65
Selection Example: Heifer weight, average = 420 lbs
Standardized selection differential
•
From a standard deviation of 1 (20% of extreme selected)
•
R = S h2= i* σ2(P)* h2
•
I -depends on the proportion of population selected (standardize
distribution by making mean = 0 and s.d = 1) I = S/ var(P)
•
I = Z/P where Z is the height of the curve at the truncation
•
The minimum phenotype in the selected group = t*σ2+
population mean
○
Given t = truncation point,
•
Selection Differential
Fat % -var(P)=0.3
•
Select top 10%, i=1.755 (table 3, p.31)
•
S= 0.53 % fat
•
Therefore, the selected group average is 0.53% fat above population
mean
•
Selection Example: Holstein Cow Milk Production
Let var(P)=2.5, mean X = 10.3, h2=0.5
•
Select best 10% to produce next generation
•
New mean = 10.3 + 2.19 = 12.49
○
R = (1.755)(2.5)(.5) = 2.19
•
i= 2.19/ (2.5)(0.5) = 1.755
○
From table, p = 10%
○
If var(P) = 2.5, h^2 = .5 and R =2.19, how many parents were selected
•
Selection Example:
S = 1/2(Sm + Sf)
○
i= 1/2(im + if)
○
If selection intensity differs between sexes:
•
S = 1/2 (0 + Sf)
○
i= 1/2 (0 + if)
○
R = 1/2 Sf h2
○
Ex. If only cows are selected on the basis of milk production
•
Selection Differential:
Var(P) = 2 eggs, h^2 = 0.3
○
Avg egg production = 160 eggs/yr
•
If = 1.271
○
R = 1/2 (1.271 + 0) (2)(0.2) = 0.3813 eggs
○
Select top 25% females
•
Therefore, new population average after 1 generation = 160.38 eggs
•
For Example, Poultry laying hens
Var (P) = 0.2, h^2 = 0.4
○
Avg height = 16 hands
•
Im = 1.7555, If=1.400
○
I = 1/2(Im + If) = 1.578
○
R = (1.578)(.2)(.4) = 0.126
○
Select tallest 10% stallions, 20% mares
•
New population average after 1 generation = 16.126 hands
•
Selection Example: Horse Height
Rate of response per unit time = R/L
•
L = amount of time for one generation (unit dependent on the genetic
response)
•
Selection over Response Time
Rate of response *see course manual
•
Applied Animal Breeding Connection -The Key Equation
R = S h2--> amount of progress that can be made in one generation
•
Change of R = R/L --> amount of progress that can be made in the time
required for one generation
•
Cattle: 4-7 years
!
Poultry: <1 year
!
Pigs: 1-2 years
!
Humans: 25-35 years
!
Ex.
○
L = generation interval = average age of parents when offspring are born
•
Calculating Genetic Interval:
*G/P
*A/P
*S increases as the
number of parents
selected decreases
6. Heredity and Response to Selection
Tuesday,+ March+ 7,+2017
11:39+AM
Phenotype = Genotype + Environment
•
Genotype = Additive + Dominance + Interaction
•
Environment = E permanent + E temporary
•
σ2
•
*see course manual
•
= VA+ VD+ VI+ VEp + VEt
•
P = A + Ep + Et (ignore D and I and let Ep cover all three)
○
Dominance and interaction genetic effects as well as permanent
environment are specific to the animal and re not passed on
•
Genetic Models:
A measure of the sum of effects of individual alleles (genes
working together)
○
Additive genetic component (A)
•
A measure of the combined dominance effects of individual loci
○
Dominance genetic component (D)
•
A measure of the combined interactions between loci (epistasis)
○
Interaction genetic component (I)
•
Additive, Dominance and Interaction Genetic Components
+4" -AA
!
+2" -Aa
!
0" -aa
!
Appetite (A)
○
+4" -GG
!
+3.5" -Gg
□
+ 2"
□
+1.5":
!
0" -gg
!
Growth hormone production (G)
○
0" -nn
!
-1" -Nn (GG)
!
-2" -NN (GG)
!
Nutrient uptake (N)
○
Loci:
•
Height = 5'6" + 0 + 0 + 0
!
Additive = 0 + 0 + 0 = 0"
!
Dominance = 0"
!
Epistasis = 0"
!
'aa gg nn'
○
Height = 5'6" + 2" + (2" + 1.5") + 0" = 5'11.5"
!
Additive = 2" + 2" + 0"
!
Dominance = 1.5"
!
Epistasis = 0"
!
'Aa Gg Nn'
○
Height = 5'6" + 2" + 4" -2" = 5'10"
!
Additive = 2" + 4" + 0
!
Dominance = 0"
!
Epistasis = -2"
!
'Aa GG NN'
○
Additive = 2" + 2" + 0"
!
Dominance = 0"
!
Epistasis = 0"
!
'A G N'
○
*only additive part is passed on in gametes
Genotype
•
Hypothetical Example: Human Height (avg = 5'6")
Proportion of total variance that is genetic
○
Broad sense heritability: H2= σ2/ σ2
•
Proportion of total variance that is additive genetic and 1/2 is
transmitted from parent to offspring
○
Narrow sense heritability: h2= σ2/ σ2
•
Heritability:
'a'ij σ2(A)
○
Based on co-variance among relatives
•
*see course manual --> 2 bO.P. = h2
!
Offspring -midparent (average of 2) regression is preferred
over offspring-parent regression as it includes more
information
!
Regression of offspring (y) on parent (x)
1)
ANOVA -phenotypes on specific relatives
2)
Two methods of estimating h2:
•
Estimation of Narrow Sense Heritability
Indicates degree of genetic resemblence between relatives and
generation
○
Helps define how much genetic change would result from
selection
○
Measures the proportion of total variation that is additive genetic
variation
•
Why are we interested in h2?
*see course manual
•
R = XO- XP
○
R = deviation of progeny from original population average
•
S = XS- XP
○
S = average of selected parents deviated from population mean
•
R = S h2
•
Response to Selection:
Average fleece weight = 2.4 kg
•
Select some rams and ewes --> average = 2.6
•
Selection differential: S= 2.6-2.4 = 0.2
•
Expected genetic progress in next generation: R = S h2= (0.2)(0.36) =
0.072 kg
•
Over 10 generations, there will be an increase of 720 g of wool
○
Therefore, selecting sleep for fleece weight results in an increase of 72 g
of wool per sheep in next generation
•
Example: Fleece weight (kg wool/sheep) *h2=0.36
Overall,
R = response to selection (units same as phenotype)
S = selection differential (units same as phenotype)
Fewer individuals -watch for inbreeding
!
Increase S
○
Measure phenotype more precisely or reduce σ2(E)
!
Increase h2
○
To increase R:
•
Selection in Practice
Select best Average S
5457 37
4465 45
3474 54
2485 65
Selection Example: Heifer weight, average = 420 lbs
Standardized selection differential
•
From a standard deviation of 1 (20% of extreme selected)
•
R = S h2= i* σ2(P)* h2
•
I -depends on the proportion of population selected (standardize
distribution by making mean = 0 and s.d = 1) I = S/ var(P)
•
I = Z/P where Z is the height of the curve at the truncation
•
The minimum phenotype in the selected group = t*σ2+
population mean
○
Given t = truncation point,
•
Selection Differential
Fat % -var(P)=0.3
•
Select top 10%, i=1.755 (table 3, p.31)
•
S= 0.53 % fat
•
Therefore, the selected group average is 0.53% fat above population
mean
•
Selection Example: Holstein Cow Milk Production
Let var(P)=2.5, mean X = 10.3, h2=0.5
•
Select best 10% to produce next generation
•
New mean = 10.3 + 2.19 = 12.49
○
R = (1.755)(2.5)(.5) = 2.19
•
i= 2.19/ (2.5)(0.5) = 1.755
○
From table, p = 10%
○
If var(P) = 2.5, h^2 = .5 and R =2.19, how many parents were selected
•
Selection Example:
S = 1/2(Sm + Sf)
○
i= 1/2(im + if)
○
If selection intensity differs between sexes:
•
S = 1/2 (0 + Sf)
○
i= 1/2 (0 + if)
○
R = 1/2 Sf h2
○
Ex. If only cows are selected on the basis of milk production
•
Selection Differential:
Var(P) = 2 eggs, h^2 = 0.3
○
Avg egg production = 160 eggs/yr
•
If = 1.271
○
R = 1/2 (1.271 + 0) (2)(0.2) = 0.3813 eggs
○
Select top 25% females
•
Therefore, new population average after 1 generation = 160.38 eggs
•
For Example, Poultry laying hens
Var (P) = 0.2, h^2 = 0.4
○
Avg height = 16 hands
•
Im = 1.7555, If=1.400
○
I = 1/2(Im + If) = 1.578
○
R = (1.578)(.2)(.4) = 0.126
○
Select tallest 10% stallions, 20% mares
•
New population average after 1 generation = 16.126 hands
•
Selection Example: Horse Height
Rate of response per unit time = R/L
•
L = amount of time for one generation (unit dependent on the genetic
response)
•
Selection over Response Time
Rate of response *see course manual
•
Applied Animal Breeding Connection -The Key Equation
R = S h2--> amount of progress that can be made in one generation
•
Change of R = R/L --> amount of progress that can be made in the time
required for one generation
•
Cattle: 4-7 years
!
Poultry: <1 year
!
Pigs: 1-2 years
!
Humans: 25-35 years
!
Ex.
○
L = generation interval = average age of parents when offspring are born
•
Calculating Genetic Interval:
*G/P
*A/P
*S increases as the
number of parents
selected decreases
6. Heredity and Response to Selection
Tuesday,+ March+ 7,+2017 11:39+AM
