Use of DNA information in Genetic
Programs.
Outline
1. DNA Information in Genetic Evaluation:
• DNA Tests
• Inclusion in Genetic Evaluations
2. Commercial Ranch Genetic Evaluations
• Sorting Bulls on DNA Genotyping
• DNA Parent identification
DNA Test Terminology
Discovery, Validation, Assessment and Application
Discovery: Process of identifying QTL
Validation: Process of replicating results in independent data
through blind testing
Assessment: Process of evaluating the effect of the QTL in a broader
context (other traits and environments)
Application: Process of using the DNA information in genetic
decisions
DNA Tests for Carcass Merit
Traits
•Thyroglobulin
•Calpain (MARC Discovery)
•Calpistatin
•Leptin
•Three QTL from NCBA Carcass Merit Project
(genes unknown)
•DGAT1
Marker Assisted EPD’s
• EPD
– Expected Haplotype Effect given sire
genotype
– Polygenic effect
Progeny Genotype vs. Sire Genotype
Sire Genotype
Dam Haplotype
Sire Haplotype
Progeny Genotype
Progeny Genotype
Progeny Phenotype
Progeny Phenotype
Observed Sire Genotype Effects
(Constructed from Haplotype Effects)
0.3
0.2
0.1
0
-0.1
-0.2
-0.3
-0.4
blank
CG AA CG AG CG GG GG AA GG AG GG GG
Four Gametes
WBSF: EPD vs MA-EPD
unit slope
CG AA
CG AG
CG GG
GG AA
GG AG
GG GG
0.8
0.6
Marker Assisted EPD
0.4
0.2
0.0
-0.2
0.2
-0.4
0.0
-0.6
-0.8
-0.2
-1.0
-1.2
-1.0
-0.4
-0.8
-0.6
-0.4
-0.2
0.0
EPD (without marker)
0.2
0.4
0.6
0.8
Commercial Ranch Project and the
need for using DNA in sire
assignments.
Bull Sorting
Create genetically diverse groups.
Objective: is to maximize the probability of uniquely identifying one
sire to a calf.
Outline
1. DNA Information in Genetic Evaluation:
• DNA Tests
• Inclusion in Genetic Evaluations
2. Commercial Ranch Genetic Evaluations
• Sorting Bulls on DNA Genotyping
• DNA Parent identification
Verification
Verification: Verifying that the putative
parent is the real parent.
In the seedstock industry, pedigree integrity
is the primary reason for DNA testing for
parent verification
AI sires, ET cows and calves, random checks.
Identification
Identification: Identifying a parent from a group
of potential parents (e.g., multiple-sire breeding
pastures).
Practical Application
We are currently developing a program for genetic
evaluation for the commercial sector.
A problem is that the large commercial ranches use
multiple-sire pastures so DNA testing for
identification becomes necessary.
Perfect World
Begin by assuming that genotypes are scored
without error. Process of excluding bulls.
A mismatch between the genotype of the putative
sire and the calf in question.
Sire = 110/110
Calf = 112/114
Panel Exclusion Rate
Measure of the effectiveness of a DNA panel
to exclude an animal as a parent.
Probability of excluding as the parent any
animal drawn at random from the
population.
Sire Identification
The probability of uniquely identifying the
sire in a group of “N” bulls is:
( Exclusion rate ) N
Bulls
0.90
0.95
0.98
2
3
4
0.81
0.73
0.66
0.90
0.86
0.81
0.96
0.94
0.92
5
6
7
8
0.59
0.53
0.48
0.43
0.77
0.74
0.70
0.66
0.90
0.89
0.87
0.85
9
10
0.39
0.35
0.63
0.60
0.83
0.82
Two or more qualify 18% of the time
Multiple Qualifying Sires
Could run more markers (a second panel). If
this was a seedstock problem probably would.
In the commercial program however this is
not cost effective, so we compute the
probability that each qualifying sire is the
true sire.
Commercial Genetic Evaluation
Using probabilities then requires a system for
genetic evaluation that “models” sire uncertainty.
Under a sire uncertainty model
do not need to uniquely identify the sire.
We will use the probability associated with each
bull of being the sire.
Probabilities
Competing sires
Bull 1 = 110/110
Bull 2 = 110/112
Calf = 110/114
If Bull 1: P(110) =1
If Bull 2: P(110) =0.5
Probabilities
Competing sires
Bull 1 = 220/222
Dam genotype
224/224
Bull 1: P(220)=0.5
Bull 2 = 224/228
Calf = 220/224
Bull 1: P(224)=0.5
Two Qualifying Bulls
Bull 1:
P(locus one) = 1.0
P(locus two) = 0.5
0.5 of his calves will have the calf genotype in
question.
Locus 1: 110/114
Locus 2: 220/224
Two Qualifying Bulls
Bull 2:
P(locus one) = 0.5
P(locus two) = 0.5
0.25 of his calves will have the calf genotype in
question.
Locus 1: 110/114
Locus 2: 220/224
Two Qualifying Bulls
Bull 1 = 0.50
Bull 2 = 0.25
Bull 1 is twice as likely as bull 2 to be the sire so the
probability of each bull is then:
Bull 1 = 2/3
Bull 2 = 1/3
Example: Bell Ranch Data
AID
Sire
Prob
Excl
Sire
Prob
Excl
3077
106/6
99
0
3167
A0053
81
0
106/6
11
1
3077
2099J
46
0
106/6
24
0
3074
106/6
87
0
A8035
12
0
3057
8101J
70
0
A0053
19
0
3170
8101J
70
0
A0053
19
0
Real World
Scoring genotypes is NOT a process without error.
A mismatch between the genotype of the sire and
calf in question does not exclude the bull.
Sire = 110/110
Calf = 112/114
Types of Scoring Errors
Independent of genotype (2-base pair repeats):
Base pair mis-reads (usually two bases off)
More likely in large DNA repeat segments
Dependent of genotype (2-base pair repeats):
Heterozygotes for alleles differing by two
bases are read as a homozygote for the smaller
allele: genotype 110/112 => scored as 110/110
Real World
A mismatch between the genotype of the sire and
calf in question does not exclude the bull.
Sire = 110/110
Calf = 112/114
Experience
10 - 15%
chance he still qualifies
The Phenotypic Representation of a
Sire Identification Problem
Animal
Genotype
Animal Scored
Genotype
Will use a four allele locus as an example.
The Phenotypic Representation of a
Sire Identification Problem
Animal
Genotype
Animal Scored
Genotype
Bull 1:
A1/A2
P(A1) = 0.5-E
P(A2) = 0.5-E
P(A3) = E
P(A4) = E
E = simple independent error rate
Population Frequencies
Possible Alleles
108 (.4)
110 (.3)
112 (.2)
114 (.1)
Genotyping Errors
Sire Scored Genotype = 108/110
Assume 4% error
Sire Possible Alleles
108 (0.48)
110 (0.48)
112 (0.02)
114 (0.02)
Progeny Probabilities
Dam
Sire
108
0.48
110
0.48
112
0.02
114
0.02
108
110
112
114
0.4
0.3
0.2
0.1
Progeny Probabilities
Dam
Sire
108
110
112
114
0.4
0.3
0.2
0.1
108
0.48
0.192
0.144
0.096
0.048
110
0.48
0.192
0.144
0.096
0.048
112
0.02
0.008
0.006
0.004
0.002
114
0.02
0.008
0.006
0.004
0.002
The Phenotypic Representation of a
Sire Identification Problem
Animal
Genotype
Animal Scored
Genotype
Calf:
A1/A4
P(A1) = 0.5-E
P(A2) = E
P(A3) = E
P(A4) = 0.5-E
E = simple independent error rate
Progeny Probabilities
Calf
108
0.005
110
0.005
112
0.005
114
0.985
108
110
112
114
0.985
0.005
0.005
0.005
Progeny Probabilities
Calf
108
110
112
114
0.985
0.005
0.005
0.005
108
0.005
0.004925
0.000025
0.000025
0.000025
110
0.005
0.004925
0.000025
0.000025
0.000025
112
0.005
0.004925
0.000025
0.000025
0.000025
114
0.985
0.970225
0.004925
0.004925
0.004925
Bell Ranch
Progeny Exclusions
X0135
1
2
3
4
5
6
7
8
9
A1017
X1095
6
7
0
6
5
0
7
5
0
0
1
5
1
5
8
6
3
7
6
0
6
2
0
6
7
1
7
X0135
A1017 X1095
X0135
A1017
X1095
9999
1
6
0
5
0
100
0
0
2
7
6
0
0
0
100
0
3
7
1
8
0
0
0
100
4
5
5
6
0
0
0
100
5
0
1
3
100
0
0
0
6
0
5
7
100
0
0
0
7
6
2
7
0
0
0
100
8
0
0
1
67
33
0
0
9
6
6
7
0
0
0
100
X0135 A1017 X1095
X0135
A1017
X1095
9999
1
6
0
5
0
100
0
0
2
7
6
0
0
0
100
0
3
7
1
8
0
96
0
4
4
5
5
6
0
1
0
99
5
0
1
3
100
0
0
0
6
0
5
7
100
0
0
0
7
6
2
7
0
74
0
26
8
0
0
1
65
33
2
0
9
6
6
7
0
0
0
100