1. Use of DNA information in Genetic Programs.

2. Outline DNA Information in Genetic Evaluation: DNA Tests
Inclusion in Genetic Evaluations
Commercial Ranch Genetic Evaluations
Sorting Bulls on DNA Genotyping
DNA Parent identification

3. Discovery, Validation, Assessment and Application
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

4. DNA Tests for Carcass Merit Traits
Thyroglobulin
Calpain (MARC Discovery)
Calpistatin
Leptin
Three QTL from NCBA Carcass Merit Project (genes unknown)
DGAT1

5. Marker Assisted EPD’s EPD
Expected Haplotype Effect given sire genotype
Polygenic effect

6. Progeny Genotype vs. Sire Genotype
Progeny Phenotype
Sire Haplotype
Sire Genotype
Dam Haplotype
The mixed model analysis is most appropriate as it accounts for differences among sires, e.g., a bull might have a very good/bad Breeding Value for marbling score but have the reverse GeneSTAR marbling genotype
Progeny Genotype
Progeny Phenotype

7. Four Gametes

8. WBSF: EPD vs MA-EPD

9. Commercial Ranch Project and the need for using DNA in sire assignments.

10. Bull Sorting

11. Create genetically diverse groups.
Objective: is to maximize the probability of uniquely identifying one sire to a calf.

12. Outline DNA Information in Genetic Evaluation: DNA Tests
Inclusion in Genetic Evaluations
Commercial Ranch Genetic Evaluations
Sorting Bulls on DNA Genotyping
DNA Parent identification

13. 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.

14. Identification
Identification: Identifying a parent from a group of potential parents (e.g., multiple-sire breeding pastures).

15. 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.

16. 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

17. 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.

18. Sire Identification
The probability of uniquely identifying the sire in a group of “N” bulls is:
( Exclusion rate ) N

19. Two or more qualify 18% of the time
Bulls
0.90
0.95
0.98 2
0.81
0.96 3
0.73
0.86
0.94 4
0.66
0.92 5
0.59
0.77 6
0.53
0.74
0.89 7
0.48
0.70
0.87 8
0.43
0.85 9
0.39
0.63
0.83 10
0.35
0.60
0.82
Two or more qualify 18% of the time

20. 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.

21. 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.

22. 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

23. Probabilities Competing sires Bull 1 = 220/222 Bull 2 = 224/228
Dam genotype
224/224
Calf = 220/224
Bull 1: P(220)=0.5
Bull 1: P(224)=0.5

24. 0.5 of his calves will have the calf genotype in question.
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

25. 0.25 of his calves will have the calf genotype in question.
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

26. 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

27. Example: Bell Ranch Data
3077
106/6 99
3167
A0053 81 11 1
2099J 46 24
3074 87
A8035 12
3057
8101J 70 19
3170
AID Sire Prob Excl Sire Prob Excl

28. Scoring genotypes is NOT a process without error.
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

29. 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

30. chance he still qualifies
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 %
chance he still qualifies

31. The Phenotypic Representation of a Sire Identification Problem
Animal
Genotype
Animal Scored
Genotype
Will use a four allele locus as an example.

32. The Phenotypic Representation of a Sire Identification Problem
Animal
Genotype
Animal Scored
Genotype
P(A1) = 0.5-E
P(A2) = 0.5-E
P(A3) = E
P(A4) = E
E = simple independent error rate
Bull 1:
A1/A2

33. Population Frequencies
Possible Alleles
108 (.4)
110 (.3)
112 (.2)
114 (.1)

34. Sire Scored Genotype = 108/110
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)

35. Progeny Probabilities
Dam
108
110
112
114
Sire
0.4
0.3
0.2
0.1
0.48
0.02

36. Progeny Probabilities
Dam
108
110
112
114
Sire
0.4
0.3
0.2
0.1
0.48
0.192
0.144
0.096
0.048
0.02
0.008
0.006
0.004
0.002

37. The Phenotypic Representation of a Sire Identification Problem
Animal
Genotype
Animal Scored
Genotype
P(A1) = 0.5-E
P(A2) = E
P(A3) = E
P(A4) = 0.5-E
E = simple independent error rate
Calf:
A1/A4

38. Progeny Probabilities
Calf
108
110
112
114
0.985
0.005

39. Progeny Probabilities
Calf
108
110
112
114
0.985
0.005

40. Bell Ranch

41. Progeny Exclusions
X0135
A1017
X1095 1 6 5 2 7 3 8 4 9

42. X0135
A1017
X1095
9999 1 6 5
100 2 7 3 8 4 67 33 9

43. X0135
A1017
X1095
9999 1 6 5
100 2 7 3 8 96 4 99 74 26 65 33 9