Use of DNA information in Genetic Programs .
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Use of DNA information in Genetic Programs.
description
Use of DNA information in Genetic Programs. Outline. DNA Information in Genetic Evaluation: DNA Tests Inclusion in Genetic Evaluations Commercial Ranch Genetic Evaluations Sorting Bulls on DNA Genotyping DNA Parent identification. DNA Test Terminology. - PowerPoint PPT Presentation
Transcript of Use of DNA information in Genetic Programs .
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
• EPD
– Expected Haplotype Effect given sire
genotype
– Polygenic effect
Marker Assisted EPD’s
Progeny Genotype vs. Sire Genotype
Progeny Genotype
Progeny Phenotype
Progeny Genotype
Progeny Phenotype
Sire Haplotype
Sire Genotype
Dam Haplotype
-0.4
-0.3
-0.2
-0.1
0
0.1
0.2
0.3
blank CG AA CG AG CG GG GG AA GG AG GG GG
Observed Sire Genotype Effects (Constructed from Haplotype Effects)
Four Gametes
-1.2
-1.0
-0.8
-0.6
-0.4
-0.2
0.0
0.2
0.4
0.6
0.8
-1.0 -0.8 -0.6 -0.4 -0.2 0.0 0.2 0.4 0.6 0.8
EPD (without marker)
Ma
rke
r A
ssis
ted
EP
D
unit slope CG AA CG AG CG GG GG AA GG AG GG GG
WBSF: EPD vs MA-EPD
-0.4
-0.2
0.0
0.2
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.
The probability of uniquely identifying the sire in a group of “N” bulls is:
( Exclusion rate ) N
Sire Identification
Bulls 0.90 0.95 0.98
2 0.81 0.90 0.96
3 0.73 0.86 0.94
4 0.66 0.81 0.92
5 0.59 0.77 0.90
6 0.53 0.74 0.89
7 0.48 0.70 0.87
8 0.43 0.66 0.85
9 0.39 0.63 0.83
10 0.35 0.60 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
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.
Using probabilities then requires a system for genetic evaluation that “models” sire uncertainty.
Probabilities
Competing sires
Bull 1 = 110/110
Calf = 110/114
Bull 2 = 110/112
If Bull 1: P(110) =1 If Bull 2: P(110) =0.5
Probabilities
Competing sires
Bull 1 = 220/222
Calf = 220/224
Bull 2 = 224/228
Bull 1: P(220)=0.5 Bull 1: P(224)=0.5
Dam genotype224/224
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
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
AID Sire Prob Excl Sire Prob Excl
Example: Bell Ranch Data
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
Experience10 - 15%
chance he still qualifies
The Phenotypic Representation of a Sire Identification Problem
Animal ScoredGenotype
AnimalGenotype
Will use a four allele locus as an example.
The Phenotypic Representation of a Sire Identification Problem
Animal ScoredGenotype
AnimalGenotype
P(A1) = 0.5-EP(A2) = 0.5-EP(A3) = EP(A4) = E
E = simple independent error rate
Bull 1:A1/A2
Population Frequencies
Possible Alleles108 (.4)110 (.3)112 (.2)114 (.1)
Genotyping Errors
Sire Scored Genotype = 108/110
Assume 4% error
Sire Possible Alleles108 (0.48)110 (0.48)112 (0.02)114 (0.02)
Progeny Probabilities
Dam
108 110 112 114
Sire 0.4 0.3 0.2 0.1
108 0.48
110 0.48
112 0.02
114 0.02
Progeny Probabilities
Dam
108 110 112 114
Sire 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 ScoredGenotype
AnimalGenotype
P(A1) = 0.5-EP(A2) = EP(A3) = EP(A4) = 0.5-E
E = simple independent error rate
Calf:A1/A4
Progeny Probabilities
Calf
108 110 112 114
0.985 0.005 0.005 0.005
108 0.005
110 0.005
112 0.005
114 0.985
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 A1017 X1095
1 6 0 5
2 7 6 0
3 7 1 8
4 5 5 6
5 0 1 3
6 0 5 7
7 6 2 7
8 0 0 1
9 6 6 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
