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

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

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 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

Four Gametes

WBSF: EPD vs MA-EPD

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 DNA Information in Genetic Evaluation: DNA Tests Inclusion in Genetic Evaluations 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

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

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 Bull 2 = 224/228 Dam genotype 224/224 Calf = 220/224 Bull 1: P(220)=0.5 Bull 1: P(224)=0.5

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

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

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 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

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

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

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 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 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

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

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)

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

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

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

Progeny Probabilities Calf   108 110 112 114 0.985 0.005

Progeny Probabilities Calf   108 110 112 114 0.985 0.005 0.004925 0.000025 0.970225

Bell Ranch

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

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

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