chapter 7 the genetic model for quantitative traitsusers.tamuk.edu/kfsdl00/chapter 7-3335.pdfchapter...

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


B = 10 g; b = -10 g;

complete dominance;


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


B = 10 g; b = -10 g;

complete dominance;


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].)


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.



B = 10 g; b = -10 g;

complete dominance;


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}


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).


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.


(μ = 60 eggs)







AABBCCDD +24

AABbccDD +6

aabbCcDd -12

_________________________________________________



(μ = 60 eggs)







AABBCCDD +24 +24

AABbccDD +6 +12

aabbCcDd -12 0

_________________________________________________



(μ = 60 eggs)







AABBCCDD +24 0 +24

AABbccDD +6 +6 +12

aabbCcDd -12 +12 0

___________________________________________________



(μ = 60 eggs)







AABBCCDD +24 0 +24 +4

AABbccDD +6 +6 +12 -10

aabbCcDd -12 +12 0 +12

__________________________________________________



(μ = 60 eggs)







AABBCCDD +24 0 +24 +4 88

AABbccDD +6 +6 +12 -10 62

aabbCcDd -12 +12 0 +12 72

__________________________________________________


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)


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.


AABBCC +12

AaBbCc +9

AaBBCc +10

Aabbcc +7

_________________________________________





Complete dominance at all loci. No epistasis. Genotypic values (G) are equal for

dominant homozygotes and heterozygotes.




AABBCC +12 +12

AaBbCc +9 +12

AaBBCc +10 +12

Aabbcc +7 +8

__________________________________________










AABBCC +12 0 +12

AaBbCc +9 +3 +12

AaBBCc +10 +2 +12

Aabbcc +7 +1 +8

__________________________________________










AABBCC +12 0 +12 -6 12

AaBbCc +9 +3 +12 +2 20

AaBBCc +10 +2 +12 -8 10

Aabbcc +7 +1 +8 -5 9

___________________________________________






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.


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

____________________________________________________



μ = 600 lbs).







AABBCCDDEE +68 0

AaBbCcDdEe +28 +28

AABbCCDDEE +54 +14

aabbCcddEE -30 0

____________________________________________________



μ = 600 lbs).







AABBCCDDEE +68 0 0

AaBbCcDdEe +28 +28 0

AABbCCDDEE +54 +14 0

aabbCcddEE -30 0 -20

____________________________________________________



μ = 600 lbs).







AABBCCDDEE +68 0 0 +68

AaBbCcDdEe +28 +28 0 +56

AABbCCDDEE +54 +14 0 +68

aabbCcddEE -30 0 -20 -50

____________________________________________________



μ = 600 lbs).







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

____________________________________________________



μ = 600 lbs).


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.


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.


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


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).

http://av.rds.yahoo.com/_ylt=A0Je5XrZ6rpH2hkAUjFvCqMX;_ylu=X3oDMTBvMmFkM29rBHBndANhdl9pbWdfcmVzdWx0BHNlYwNzcg--/SIG=11q63bq9m/EXP=1203518553/**http%3a//myblog.de/noirespoir/page/44328


Genetic model: P = μ + BV + GCV + Ep + Et; PA = BV + GCV + Ep

(μ = 60 eggs)






Genotype BV GCV G Ep Et P PA

AABBCCDD +7 -3

AABbccDD -6 -4

aabbCcDd +2 +10

________________________________________________________



(μ = 60 eggs)







AABBCCDD +24 +7 -3

AABbccDD +6 -6 -4

aabbCcDd -12 +2 +10

________________________________________________________



(μ = 60 eggs)







AABBCCDD +24 +24 +7 -3

AABbccDD +6 +12 -6 -4

aabbCcDd -12 0 +2 +10

________________________________________________________



(μ = 60 eggs)







AABBCCDD +24 0 +24 +7 -3

AABbccDD +6 +6 +12 -6 -4

aabbCcDd -12 +12 0 +2 +10

________________________________________________________



(μ = 60 eggs)







AABBCCDD +24 0 +24 +7 -3 88

AABbccDD +6 +6 +12 -6 -4 62

aabbCcDd -12 +12 0 +2 +10 72

________________________________________________________



(μ = 60 eggs)






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

_______________________________________________________


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


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

chapter 7 the genetic model for quantitative traitsusers.tamuk.edu/kfsdl00/chapter 7-3335.pdfchapter...

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