Presentation is loading. Please wait.

Presentation is loading. Please wait.

Predictions of Breed Composition Using Holsteins, Jersey, and Brown Swiss Genotypes K. M. Olson and P. M. VanRaden.

Similar presentations


Presentation on theme: "Predictions of Breed Composition Using Holsteins, Jersey, and Brown Swiss Genotypes K. M. Olson and P. M. VanRaden."— Presentation transcript:

1 Predictions of Breed Composition Using Holsteins, Jersey, and Brown Swiss Genotypes
K. M. Olson and P. M. VanRaden

2 Background 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 SNP and lower genomic relationships and inbreeding Wanted to investigate a more precise way to look at breed composition

3 Objective Use a full and reduced SNP set to predict breed composition with Holstein, Jersey, and Brown Swiss genotypes

4 Materials & Methods Used both females and males
Y- Variable was breed of animal A Holstein would receive a 1 for the Holstein analysis and a 0 for the Jersey and Brown Swiss analysis Animal Breed Holstein Analysis Jersey Analysis BSW Analysis HOL 1 JER BSW

5 Materials & Methods 3 different sizes of SNP sets were used for the genomic evaluation to compute SNP estimates for breed The full 43,385 SNP set The proposed 3 K SNP set The 600 SNP set Each breed has ~ 200 – used for the basic check currently not a genomic evaluation

6 Materials & Methods 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

7 Results Means and standard deviations for given breed of the training (old animals) data set SNP set/ Breed 43 K 3 K 600 Holstein (N = 11,053) 1.000±0.007 0.997±0.041 0.999±0.021 Jersey (N = 2,208) 1.000±0.016 0.990±0.069 0.989±0.047 Brown Swiss (N = 778) 0.994±0.021 0.988±0.035 0.992±0.051

8 Results Means and standard deviations for given breed of the validation (young animals) data set 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 0.992±0.051

9 Results 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

10 Conclusions The 43 K set was the most accurate at prediction of breed composition (SD ranged from 0.8% - 2.8%) The 3 K set could identify individuals that had large amounts (> 13%) of foreign DNA The population of interest would need to be determined

11 Obstacles May not be accurate for animals from different populations
foreign animals older animals There is a patent Located at


Download ppt "Predictions of Breed Composition Using Holsteins, Jersey, and Brown Swiss Genotypes K. M. Olson and P. M. VanRaden."

Similar presentations


Ads by Google