April 2010 (1) Prediction of Breed Composition & Multibreed Genomic Evaluations K. M. Olson and P....

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April 2010 (1) Prediction of Breed Composition & Multibreed Genomic Evaluations K. M. Olson and P. M. VanRaden

Transcript of April 2010 (1) Prediction of Breed Composition & Multibreed Genomic Evaluations K. M. Olson and P....

Page 1: April 2010 (1) Prediction of Breed Composition & Multibreed Genomic Evaluations K. M. Olson and P. M. VanRaden.

April 2010 (1)

Prediction of Breed Composition &

Multibreed Genomic Evaluations

K. M. Olson and P. M. VanRaden

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Background - Prediction of Breed 200 Breed specific SNP were used to

verify an animal received the correct breed code in the quality control data step

Several animals had fewer breed-specific SNPs and lower genomic relationships and inbreeding

Wanted to investigate a more precise way to look at breed composition

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Materials & Methods – Prediction of Breed

Y- Variable was breed of animal

Used both females and males

3 different sizes of SNP sets were used for the genomic evaluation

The Full 43,385 SNP set

The proposed 3 K SNP set

The 600 breed specific set

− Each breed has ~ 200 – used for the basic check currently not a genomic evaluation

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Materials & Methods – Prediction of Breed

Training data set – animal reliability set to 99% and parent average reliability set to 50%

Proven as of July 2009

Total of 14,039 animals across all breeds

Validation data set – reliabilities set to 0%

Unproven as of July 2009

15,809 animals across all breeds

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Results – Prediction of Breed

All three tests were able to determine a Holstein that was by pedigree 1/8 (12.5%) Jersey

43 K test predicted her as 85.9% Holstein and 13.3% Jersey

3 K predicted she was 84.4% Holstein and 15.5% Jersey

600 SNP set she was 83.0% Holstein and 16.6% Jersey

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Results – Prediction of Breed

SNP set/ Breed

43 K 3 K 600

Holstein(N = 14,794)

1.000±0.008 1.004±0.031 1.002±0.019

Jersey(N = 919)

0.996±0.028 0.978±0.063 0.989±0.036

Brown Swiss(N = 96)

0.994±0.021 0.989±0.036 0.992±0.051

Means and standard deviations for given breed of the validation data set

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Conclusions – Prediction of Breed The 43 K chip was the most accurate at

prediction of breed composition

The 3 K chip could identify individuals that had large amounts (> 13%) of foreign DNA

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Obstacles – Prediction of Breed

There is a patent

Located at http://www.patentstorm.us/patents/7511127/fulltext.html

May not be accurate for animals from different populations

foreign animals

older animals

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Background - Multibreed

Multibreed methods are currently used in traditional methods

Only within breed methods are used for genomics evaluations

Previous research has shown little improvement in accuracy from using all breeds with the 50K SNP chip however, little research has been done using multi-trait methodology

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Objectives – Multibreed genomic evaluations

To investigate three different methods of multibreed genomic evaluations using Holsteins, Jerseys, and Brown Swiss genotypes

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Materials & Methods – Multibreed (Animals)

The training data set - animals were proven by Nov. 2004 Holsteins – 5,331 Jerseys – 1,361 Brown Swiss – 506

The validation data set - animals were unproven as of Nov. 2004 and proven by June 2009 Holsteins – 2,477 Jerseys – 410 Brown Swiss - 182

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Material & Methods – Multibreed (Methods)

Method 1 estimated SNP effects within breed then applied those effects to the other breeds

Method 2 (across-breed) used a common set of SNP effects from the combined breed genotypes and phenotypes

Method 3 (multi-breed) used a correlated SNP effects using a multitrait method ( as explained by VanRaden and Sullivan, 2010)

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Results – P – Values for Protein Yield

Holstein Jersey Brown Swiss

Traditional

PTA < 0.001 < 0.001 0.061

GPTA < 0.001 < 0.001 0.086

R2adj 0.5045 0.4874 0.1030

Method 1

HOL GPTA < 0.001 0.668 0.344

JER GPTA 0.873 < 0.001 0.844

BSW GPTA 0.813 0.473 0.107

PTA < 0.001 < 0.001 0.054

R2adj 0.5041 0.4854 0.0978

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Results – P-values for protein yield

Holstein Jersey Brown Swiss

Method 2

PTA < 0.001 < 0.001 0.088

GPTA < 0.001 < 0.001 0.316

ABGPTA 0.002 0.290 0.007

R2adj 0.5063 0.4876 0.1337

Method 3

PTA < 0.001 < 0.001 0.080

GPTA 0.742 0.324 0.140

MBGPTA < 0.001 < 0.001 0.060

R2adj 0.5060 0.4916 0.1127

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Results – P-Values for protein yield

Holstein Jersey Brown Swiss

Method 2

PTA <0.001 < 0.001 0.2016

ABGPTA < 0.001 < 0.001 0.0023

R2adj 0.4742 0.4742 0.1336

Method 3

PTA < 0.001 < 0.001 0.055

MBGPTA < 0.001 < 0.001 0.081

R2adj 0.5060 0.4916 0.1067

The traditional GPTA was not included in these analyses

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Conclusions – Multibreed Genomic Evaluation

Method 1 did not help the estimates for genomic evaluations

Method 2 increased the predictive ability, however the traditional GPTA accounted for more variation than the across-breed GPTA

Method 3 increased the predictive ability and the multi-breed GPTA accounted for more variation than the traditional GPTA

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Implications

The multibreed genomic evaluations do slightly increase the accuracy of the evaluations, but may not warrant the increased computational demands

A higher density SNP chip would most likely increase the gains in accuracy for multibreed genomic evaluations

Multibreed would be needed for genomic selection in crossbred herds

Not much demand for that yet