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I. The Basic Model
II. Breeding Value
Chapter 7 – The Genetic Model for
Quantitative Traits
III. Gene Combination Value
IV. Producing Ability
Learning Objective: To understand the “Genetic Model” for Quantitative Traits that involves the concepts of Breeding Value and Gene Combination Value. Applications include the use of Expected Progeny Differences (EPD’s) and Producing Ability (PA) for improving traits of economic importance.
Chapter 7 – The Genetic Model for
Quantitative Traits
P = μ + G + E
I. The Basic Model
where
P = the phenotypic value or performance of an individual
animal for a trait,
μ = (the Greek letter mu) the population mean or average
phenotypic value for the trait for all the animals in the
population,
G = the genotypic value of the individual for the trait (i.e., the
effect of the animal’s genes, singly and in combination), and
E = the environmental effect on the individual’s performance
for the trait (i.e., external, non-genetic factors).
II. Breeding Value
Definition: Breeding Value (BV) - The value of an
individual as a genetic parent to its progeny’s performance. The aim is to accurately identify those animals with the best set of genes that constitute BV.
- Each gene imparts a numeric “independent effect” towards the animal’s aggregate breeding value for the trait.
- Each gene imparts a value, which is manifested at the physiological level (“Genetics is the basis of physiological expressions of traits”; a simple example is dwarfism involving a deficiency of growth hormone).
II. Breeding Value
Example using a Single-Locus Model:
B = 10 g; b = -10 g;
complete dominance;
trait is mature body weight in mice.
Genotype BV
BB 20 g
Bb 0 g
bb -20 g
II. Breeding Value
Example using a Single-Locus Model:
B = 10 g; b = -10 g;
complete dominance;
trait is mature body weight in mice.
Genotype BV EPD
BB 20 g +10 g
Bb 0 g 0 g
bb -20 g -10 g
EPD = ½BV
(“Expected Progeny Difference”)
II. Breeding Value
Example using a Single-Locus Model:
B = 10 g; b = -10 g;
complete dominance;
trait is mature body weight in mice.
Genotype BV (G )
BB 20 g 20 g
Bb 0 g 20 g
bb -20 g -20 g
When there is complete dominance, genotypic values (G) are
the same for BB and Bb. Further, G is the value of an
individual’s genes to its own performance.
II. Breeding Value
Definition for EPD: – Half an animal’s estimated breeding value – the expected difference between the mean performance of the individual’s progeny and the mean performance of all progeny (assuming randomly chosen mates).
^ ^
PD = ½BV
^ ^
^ BVS + BVD
BVO = __________
2 (An interim estimate until an animal’s own
records become available [e.g., ET calves].)
III. Gene Combination Value
Definition: The part of an animal’s genotype
(G) that is due to the effects of gene
combination (dominance and epistasis) and
cannot, therefore, be transmitted from
parent to offspring.
G = BV + GCV
BV is the transmissible part of G, whereas GCV
is the non-transmissible part of G.
III. Gene Combination Value
Example using a Single-Locus Model:
B = 10 g; b = -10 g;
complete dominance;
trait is mature body weight in mice.
Genotype BV G GCV
BB 20 g 20 g 0 g
Bb 0 g 20 g 20 g
bb -20 g -20 g 0 g
{G = BV + GCV or GCV = G - BV}
III. Gene Combination Value
Genotype BV G GCV
BB 20 g 20 g 0 g
Bb 0 g 20 g 20 g
bb -20 g -20 g 0 g
Comments: Homozygotes breed as good as or poorly as
they look, whereas Heterozygotes tend to look better
than they breed! Further, seedstock breeders rely on and
actively promote BV’s, whereas commercial producers
market performance based on G (BV is important, while
GCV plays a major role in heterosis expression, a benefit
of crossbreeding).
III. Gene Combination Value
Comments:
_____
GCVO ≠ ½ (GCVS + GCVD)
GCV is also called the “non-additive genetic value”, whereas BV is the “additive genetic value”.
All genes have independent effects (BV), but genes may also contribute to GCV effects.
Review:
P = μ + G + E
G = BV + GCV
G = BV + D + I
BV factors into heritability (Chapter 9)
GCV factors (D (dominance) and I (epistasis)) into heterosis (Chapter 18)
Low
(0-15%)
Most High
Moderate
(15-40%)
Some
Moderate
High
(>40%)
Least
Low
Heritability Heterosis Environment
(BV) (GCV) (E)
Repro/
Health
Production
Yield
Genetic and environmental influences
on livestock traits:
P = μ + BV + GCV + E
Example 1: Consider that egg production in Bantams is affected by only four loci.
Genetic model: P = μ + G + E; {G = BV + GCV}
(μ = 60 eggs)
Complete the table below. Assume the following:
Complete dominance at all loci. No epistasis.
The independent effect of each gene symbol which is capitalized is +3 eggs.
The independent effect of each gene symbol which is not capitalized is -3 eggs.
For homozygous combinations, genotypic values are equal to breeding values.
Genotype BV GCV G E P
AABBCCDD
AABbccDD
aabbCcDd
_________________________________________________
Consider that egg production in Bantams is affected by only four loci.
Genetic model: P = μ + G + E; {G = BV + GCV}
(μ = 60 eggs)
Complete the table below. Assume the following:
Complete dominance at all loci. No epistasis.
The independent effect of each gene symbol which is capitalized is +3 eggs.
The independent effect of each gene symbol which is not capitalized is -3 eggs.
For homozygous combinations, genotypic values are equal to breeding values.
Genotype BV GCV G E P
AABBCCDD +24
AABbccDD +6
aabbCcDd -12
_________________________________________________
Consider that egg production in Bantams is affected by only four loci.
Genetic model: P = μ + G + E; {G = BV + GCV}
(μ = 60 eggs)
Complete the table below. Assume the following:
Complete dominance at all loci. No epistasis.
The independent effect of each gene symbol which is capitalized is +3 eggs.
The independent effect of each gene symbol which is not capitalized is -3 eggs.
For homozygous combinations, genotypic values are equal to breeding values.
Genotype BV GCV G E P
AABBCCDD +24 +24
AABbccDD +6 +12
aabbCcDd -12 0
_________________________________________________
Consider that egg production in Bantams is affected by only four loci.
Genetic model: P = μ + G + E; {G = BV + GCV}
(μ = 60 eggs)
Complete the table below. Assume the following:
Complete dominance at all loci. No epistasis.
The independent effect of each gene symbol which is capitalized is +3 eggs.
The independent effect of each gene symbol which is not capitalized is -3 eggs.
For homozygous combinations, genotypic values are equal to breeding values.
Genotype BV GCV G E P
AABBCCDD +24 0 +24
AABbccDD +6 +6 +12
aabbCcDd -12 +12 0
___________________________________________________
Consider that egg production in Bantams is affected by only four loci.
Genetic model: P = μ + G + E; {G = BV + GCV}
(μ = 60 eggs)
Complete the table below. Assume the following:
Complete dominance at all loci. No epistasis.
The independent effect of each gene symbol which is capitalized is +3 eggs.
The independent effect of each gene symbol which is not capitalized is -3 eggs.
For homozygous combinations, genotypic values are equal to breeding values.
Genotype BV GCV G E P
AABBCCDD +24 0 +24 +4
AABbccDD +6 +6 +12 -10
aabbCcDd -12 +12 0 +12
__________________________________________________
Consider that egg production in Bantams is affected by only four loci.
Genetic model: P = μ + G + E; {G = BV + GCV}
(μ = 60 eggs)
Complete the table below. Assume the following:
Complete dominance at all loci. No epistasis.
The independent effect of each gene symbol which is capitalized is +3 eggs.
The independent effect of each gene symbol which is not capitalized is -3 eggs.
For homozygous combinations, genotypic values are equal to breeding values.
Genotype BV GCV G E P
AABBCCDD +24 0 +24 +4 88
AABbccDD +6 +6 +12 -10 62
aabbCcDd -12 +12 0 +12 72
__________________________________________________
Genotype BV GCV G E P
AABBCCDD +24 0 +24 +4 88
AABbccDD +6 +6 +12 -10 62
aabbCcDd -12 +12 0 +12 72
____________________________________________
1) Which individual lays the most eggs? Explain why.
2) Which individual as a parent would be expected to
produce offspring that lay the most eggs?
3) Give possible reasons why hen #2 has an E value of
-10, while hen #3 has an E value of +12.
Example 2: In mature, male Proghorn antelope, figure that horn length is only
affected by three loci. The horn measurement is the total length of one horn.
Assume symmetry with regards to the genetic model is as follows:
P = μ + BV + GCV + E (μ = 6 inches)
Complete dominance at all loci. No epistasis.
Genotypic values (G) are equal for dominant homozygotes and heterozygotes.
The independent effect of each gene symbol which is capitalized is +2 inches.
The independent effect of each gene symbol which is not capitalized is +1 inches.
Genotype BV GCV G E P
AABBCC +12
AaBbCc +9
AaBBCc +10
Aabbcc +7
_________________________________________
Example 2: In mature, male Proghorn antelope, figure that horn length is only
affected by three loci. The horn measurement is the total length of one horn.
Assume symmetry with regards to the genetic model is as follows:
P = μ + BV + GCV + E (μ = 6 inches)
Complete dominance at all loci. No epistasis. Genotypic values (G) are equal for
dominant homozygotes and heterozygotes.
The independent effect of each gene symbol which is capitalized is +2 inches.
The independent effect of each gene symbol which is not capitalized is +1 inches.
Genotype BV GCV G E P
AABBCC +12 +12
AaBbCc +9 +12
AaBBCc +10 +12
Aabbcc +7 +8
__________________________________________
Example 2: In mature, male Proghorn antelope, figure that horn length is only
affected by three loci. The horn measurement is the total length of one horn.
Assume symmetry with regards to the genetic model is as follows:
P = μ + BV + GCV + E (μ = 6 inches)
Complete dominance at all loci. No epistasis. Genotypic values (G) are equal for
dominant homozygotes and heterozygotes.
The independent effect of each gene symbol which is capitalized is +2 inches.
The independent effect of each gene symbol which is not capitalized is +1 inches.
Genotype BV GCV G E P
AABBCC +12 0 +12
AaBbCc +9 +3 +12
AaBBCc +10 +2 +12
Aabbcc +7 +1 +8
__________________________________________
Example 2: In mature, male Proghorn antelope, figure that horn length is only
affected by three loci. The horn measurement is the total length of one horn.
Assume symmetry with regards to the genetic model is as follows:
P = μ + BV + GCV + E (μ = 6 inches)
Complete dominance at all loci. No epistasis. Genotypic values (G) are equal for
dominant homozygotes and heterozygotes.
The independent effect of each gene symbol which is capitalized is +2 inches.
The independent effect of each gene symbol which is not capitalized is +1 inches.
Genotype BV GCV G E P
AABBCC +12 0 +12 -6 12
AaBbCc +9 +3 +12 +2 20
AaBBCc +10 +2 +12 -8 10
Aabbcc +7 +1 +8 -5 9
___________________________________________
Example 2: In mature, male Proghorn antelope, figure that horn length is only
affected by three loci. The horn measurement is the total length of one horn.
Assume symmetry with regards to the genetic model is as follows:
P = μ + BV + GCV + E (μ = 6 inches)
Genotype BV GCV G E P
AABBCC +12 0 +12 -6 12
AaBbCc +9 +3 +12 +2 20
AaBBCc +10 +2 +12 -8 10
Aabbcc +7 +1 +8 -5 9
___________________________________________
1) Which individual has the longest horns? Explain why.
2) Which individual as a parent would be expected to
produce offspring that has the longest horns?
3) Give possible reasons why animal #2 has an E value of
+2, while animal #3 has an E value of -8 inches.
Example 3: Consider a hypothetical quantitative trait (e.g., calf weaning
weight) affected by five loci. (Genetic model: P = μ + BV + D + I + E;
μ = 600 lbs).
Assume the following: The independent effect of each gene symbol which is capitalized is +10 lbs
for loci A through D, but also for each e gene at the E locus;
The independent effect of each gene symbol which is not capitalized is -4 lbs for loci A through D, except for each E gene at the E locus which is -6 lbs;
Complete dominance exists only at loci B and D; and
Epistasis exists only between genotypes aa and EE, which is -20 lbs.
Genotype BV D I G E P__
AABBCCDDEE +68
AaBbCcDdEe +28
AABbCCDDEE +54
aabbCcddEE -30
____________________________________________________
Example 3: Consider a hypothetical quantitative trait (e.g., calf weaning
weight) affected by five loci. (Genetic model: P = μ + BV + D + I + E;
μ = 600 lbs).
Assume the following: The independent effect of each gene symbol which is capitalized is +10 lbs
for loci A through D, but also for each e gene at the E locus;
The independent effect of each gene symbol which is not capitalized is -4 lbs for loci A through D, except for each E gene at the E locus which is -6 lbs;
Complete dominance exists only at loci B and D; and
Epistasis exists only between genotypes aa and EE, which is -20 lbs.
Genotype BV D I G E P__
AABBCCDDEE +68 0
AaBbCcDdEe +28 +28
AABbCCDDEE +54 +14
aabbCcddEE -30 0
____________________________________________________
Example 3: Consider a hypothetical quantitative trait (e.g., calf weaning
weight) affected by five loci. (Genetic model: P = μ + BV + D + I + E;
μ = 600 lbs).
Assume the following: The independent effect of each gene symbol which is capitalized is +10 lbs
for loci A through D, but also for each e gene at the E locus;
The independent effect of each gene symbol which is not capitalized is -4 lbs for loci A through D, except for each E gene at the E locus which is -6 lbs;
Complete dominance exists only at loci B and D; and
Epistasis exists only between genotypes aa and EE, which is -20 lbs.
Genotype BV D I G E P__
AABBCCDDEE +68 0 0
AaBbCcDdEe +28 +28 0
AABbCCDDEE +54 +14 0
aabbCcddEE -30 0 -20
____________________________________________________
Example 3: Consider a hypothetical quantitative trait (e.g., calf weaning
weight) affected by five loci. (Genetic model: P = μ + BV + D + I + E;
μ = 600 lbs).
Assume the following: The independent effect of each gene symbol which is capitalized is +10 lbs
for loci A through D, but also for each e gene at the E locus;
The independent effect of each gene symbol which is not capitalized is -4 lbs for loci A through D, except for each E gene at the E locus which is -6 lbs;
Complete dominance exists only at loci B and D; and
Epistasis exists only between genotypes aa and EE, which is -20 lbs.
Genotype BV D I G E P__
AABBCCDDEE +68 0 0 +68
AaBbCcDdEe +28 +28 0 +56
AABbCCDDEE +54 +14 0 +68
aabbCcddEE -30 0 -20 -50
____________________________________________________
Example 3: Consider a hypothetical quantitative trait (e.g., calf weaning
weight) affected by five loci. (Genetic model: P = μ + BV + D + I + E;
μ = 600 lbs).
Assume the following: The independent effect of each gene symbol which is capitalized is +10 lbs
for loci A through D, but also for each e gene at the E locus;
The independent effect of each gene symbol which is not capitalized is -4 lbs for loci A through D, except for each E gene at the E locus which is -6 lbs;
Complete dominance exists only at loci B and D; and
Epistasis exists only between genotypes aa and EE, which is -20 lbs.
Genotype BV D I G E P__
AABBCCDDEE +68 0 0 +68 +21 689
AaBbCcDdEe +28 +28 0 +56 -15 641
AABbCCDDEE +54 +14 0 +68 +22 690
aabbCcddEE -30 0 -20 -50 0 550
____________________________________________________
Example 3: Consider a hypothetical quantitative trait (e.g., calf weaning
weight) affected by five loci. (Genetic model: P = μ + BV + D + I + E;
μ = 600 lbs).
Genotype BV D I G E P__
AABBCCDDEE +68 0 0 +68 +21 689
AaBbCcDdEe +28 +28 0 +56 -15 641
AABbCCDDEE +54 +14 0 +68 +22 690
aabbCcddEE -30 0 -20 -50 0 550
____________________________________________________
1) Which individual has the heaviest weight? Explain why.
2) Which individual as a parent would be expected to
produce offspring that has the heaviest weight?
3) Give possible reasons why animal #3 has an E value of
+22 lbs, while animal #2 has an E value of -15 lbs.
IV. Producing Ability
Definition: Producing Ability (PA) - The performance potential of an individual for a repeated trait.
Repeated trait – A trait for which individuals commonly have more than one performance record (e.g., antler score, ease of birthing, egg production, fertility, litter size, milk production, race performance, and wool yield).
P = μ + BV + GCV + Ep + Et {E = Ep + Et}
^
PA = G + Ep
IV. Producing Ability
where
Ep = An environmental effect that permanently
influences an individual’s performance for a
repeated trait (e.g., an animal born in a severe
drought year or in an ample rainfall year), and
Et = An environmental effect that only influences a
single performance record (e.g., a muddy race
track or a track in exceptionally ideal condition).
IV. Producing Ability - Problem 7.7
Consider the Thoroughbred stallions: Raise-A-
Ruckus and Presidium. Raise-A-Ruckus’s BV for
racing time is -8 seconds. He was particularly well
trained, having a permanent environmental effect of
-6 seconds. Presidium’s BV for racing time is -12
seconds, but his permanent environmental effect is
+2 seconds. Assume both horses have GCV’s of 0.
Answer the following:
a. Calculate progeny difference (EPD) for each horse {EPD = ½BV}.
R’R: ½(-8) = -4 sec Presidium: ½(-12) = -6 sec
b. Calculate producing ability (PA) for each horse {PA = BV + GCV + Ep}.
R’R: PA = -8 + 0 + (-6) = -14 sec Presidium PA = -12 + 0 +2 = -10 sec
c. Which horse would you bet on in a race? Why?
R’R because he has the better PA (-14 vs. -10 seconds) (i.e., 4 sec faster).
d. Which horse would you breed mares to? Why?
Presidium because he has the better BV (-6 vs. -4 seconds) (i.e., his progeny
are predicted to be faster by 2 sec).
Example 1: Consider that egg production in Bantams is affected by only four loci.
Genetic model: P = μ + BV + GCV + Ep + Et; PA = BV + GCV + Ep
(μ = 60 eggs)
Complete the table below. Assume the following:
Complete dominance at all loci. No epistasis.
The independent effect of each gene symbol which is capitalized is +3 eggs.
The independent effect of each gene symbol which is not capitalized is -3 eggs.
For homozygous combinations, genotypic values are equal to breeding values.
Genotype BV GCV G Ep Et P PA
AABBCCDD +7 -3
AABbccDD -6 -4
aabbCcDd +2 +10
________________________________________________________
Example 1: Consider that egg production in Bantams is affected by only four loci.
Genetic model: P = μ + BV + GCV + Ep + Et; PA = BV + GCV + Ep
(μ = 60 eggs)
Complete the table below. Assume the following:
Complete dominance at all loci. No epistasis.
The independent effect of each gene symbol which is capitalized is +3 eggs.
The independent effect of each gene symbol which is not capitalized is -3 eggs.
For homozygous combinations, genotypic values are equal to breeding values.
Genotype BV GCV G Ep Et P PA
AABBCCDD +24 +7 -3
AABbccDD +6 -6 -4
aabbCcDd -12 +2 +10
________________________________________________________
Example 1: Consider that egg production in Bantams is affected by only four loci.
Genetic model: P = μ + BV + GCV + Ep + Et; PA = BV + GCV + Ep
(μ = 60 eggs)
Complete the table below. Assume the following:
Complete dominance at all loci. No epistasis.
The independent effect of each gene symbol which is capitalized is +3 eggs.
The independent effect of each gene symbol which is not capitalized is -3 eggs.
For homozygous combinations, genotypic values are equal to breeding values.
Genotype BV GCV G Ep Et P PA
AABBCCDD +24 +24 +7 -3
AABbccDD +6 +12 -6 -4
aabbCcDd -12 0 +2 +10
________________________________________________________
Example 1: Consider that egg production in Bantams is affected by only four loci.
Genetic model: P = μ + BV + GCV + Ep + Et; PA = BV + GCV + Ep
(μ = 60 eggs)
Complete the table below. Assume the following:
Complete dominance at all loci. No epistasis.
The independent effect of each gene symbol which is capitalized is +3 eggs.
The independent effect of each gene symbol which is not capitalized is -3 eggs.
For homozygous combinations, genotypic values are equal to breeding values.
Genotype BV GCV G Ep Et P PA
AABBCCDD +24 0 +24 +7 -3
AABbccDD +6 +6 +12 -6 -4
aabbCcDd -12 +12 0 +2 +10
________________________________________________________
Example 1: Consider that egg production in Bantams is affected by only four loci.
Genetic model: P = μ + BV + GCV + Ep + Et; PA = BV + GCV + Ep
(μ = 60 eggs)
Complete the table below. Assume the following:
Complete dominance at all loci. No epistasis.
The independent effect of each gene symbol which is capitalized is +3 eggs.
The independent effect of each gene symbol which is not capitalized is -3 eggs.
For homozygous combinations, genotypic values are equal to breeding values.
Genotype BV GCV G Ep Et P PA
AABBCCDD +24 0 +24 +7 -3 88
AABbccDD +6 +6 +12 -6 -4 62
aabbCcDd -12 +12 0 +2 +10 72
________________________________________________________
Example 1: Consider that egg production in Bantams is affected by only four loci.
Genetic model: P = μ + BV + GCV + Ep + Et; PA = BV + GCV + Ep
(μ = 60 eggs)
Complete the table below. Assume the following:
Complete dominance at all loci. No epistasis.
The independent effect of each gene symbol which is capitalized is +3 eggs.
The independent effect of each gene symbol which is not capitalized is -3 eggs.
For homozygous combinations, genotypic values are equal to breeding values.
Genotype BV GCV G Ep Et E P PA
AABBCCDD +24 0 +24 +7 -3 +4 88 +31
AABbccDD +6 +6 +12 -6 -4 -10 62 +6
aabbCcDd -12 +12 0 +2 +10 +12 72 +2
_______________________________________________________
Genotype BV GCV G Ep Et P PA
AABBCCDD +24 0 +24 +7 -3 88 +31
AABbccDD +6 +6 +12 -6 -4 62 +6
aabbCcDd -12 +12 0 +2 +10 72 +2
_______________________________________________________
1) Which individual lays the most eggs?
P
2) Which individual as a parent would be expected to
produce offspring that lay the most eggs?
BV
3) Which hen is predicted to lay more eggs in the future
with regards to her producing ability?
PA