| Literature DB >> 26865846 |
Donghyun Shin1, Kyoung-Do Park2, Sojoeng Ka1, Heebal Kim1, Kwang-Hyeon Cho3.
Abstract
Previous studies in Holstein have shown 35% to 51.8% heritability in milk production traits, such as milk yield, fat, and protein, using pedigree data. Other studies in complex human traits could be captured by common single-nucleotide polymorphisms (SNPs), and their genetic variations, attributed to chromosomes, are in proportion to their length. Using genome-wide estimation and partitioning approaches, we analyzed three quantitative Holstein traits relevant to milk production in Korean Holstein data harvested from 462 individuals genotyped for 54,609 SNPs. For all three traits (milk yield, fat, and protein), we estimated a nominally significant (p = 0.1) proportion of variance explained by all SNPs on the Illumina BovineSNP50 Beadchip (h (2) G ). These common SNPs explained approximately most of the narrow-sense heritability. Longer genomic regions tended to provide more phenotypic variation information, with a correlation of 0.46~0.53 between the estimate of variance explained by individual chromosomes and their physical length. These results suggested that polygenicity was ubiquitous for Holstein milk production traits. These results will expand our knowledge on recent animal breeding, such as genomic selection in Holstein.Entities:
Keywords: Holstein; breeding; genomic selection; heritability; single-nucleotide polymorphism
Year: 2015 PMID: 26865846 PMCID: PMC4742325 DOI: 10.5808/GI.2015.13.4.146
Source DB: PubMed Journal: Genomics Inform ISSN: 1598-866X
Fig. 1Phenotype distribution of the three traits related to milk production.
Estimates of variance explained by all SNPs for the three traits related to milk production
SNP, single-nucleotide polymorphism.
aEstimate of h2 from pedigree analysis in literature [13].
Fig. 2Proportion of variance attributed to each chromosome of three traits of parity 1 against chromosome length. We show proportion of variance of each chromosome, which is significant based on likelihood test (p-value<0.1). (A) Milk yield (slope = 8.05 E-04, S.E = 4.31 E-04, p-value = 8.46 E-02). (B) Milk fat (slope = 5.79 E-04, S.E = 1.84 E-04, p-value = 4.26 E-03). (C) Milk protein (slope = 5.75 E-04, S.E = 2.97 E-04, p-value = 5.24 E-02).
Fig. 3Estimates of the variance explained by all SNPs in genic (intergenic) regions for 3 traits (a, b, and c indicate milk yield, fat, and protein, respectively) against length of genic (intergenic) DNA. Shown in panels (A), (C), and (E) are the results for the genic SNPs, and shown on panels (B), (D), and (F) are the results for intergenic SNPs, under the assumption that genic regions is Wor0 Kb of UTRs. (A) Milk yield, genic region (slope = 7.89 E-04, S.E = 3.30 E-04, p-value = 3.56 E-02). (B) Milk yield, intergenic region (slope = 9.11 E-04, S.E = 6.70 E-04, p-value = 2.07 E-01). (C) Milk fat, genic region (slope = 4.04 E-04, S.E = 2.20 E-04, p-value = 8.01 E-02). (D) Milk fat, intergenic region (slope = 5.23 E-04, S.E = 1.78 E-04, p-value = 7.02 E-03). (E) Milk protein, genic region (slope = 3.66 E-04, S.E = 2.24 E-04, p-value = 1.34 E-01). (F) Milk protein, intergenic region (slope = 7.53 E-04, S.E = 2.24 E-04, p-value = 7.20 E-03).