1
|
|
2
|
- Predict young bulls and cows more accurately
- Compare actual DNA inherited
- Use exact relationship matrix G instead of expected values in A
- Trace chromosome segments
- Locate genes with large effects
|
3
|
- Example: Full sibs
- are expected to share 50% of their DNA on average
- may actually share 45% or 55% of their DNA because each inherits a
different mixture of chromosome segments from the two parents.
- Combine genotype and pedigree data to determine exact fractions
|
4
|
- Measures of genetic similarity
- A = Expected % genes identical by descent from pedigree (Wright, 1922)
- G = Actual % of DNA shared (using genotype data)
- T = % genes shared that affect a given trait (using genotype and
phenotype)
- Best measure depends on use
|
5
|
- Models contain markers, not QTLs
- M is markers inherited minus freq
- M M’ / ∑ p(1-p) = G
- List all QTL affecting a trait
- Q is alleles inherited minus freq
- q contains effects of alleles
- u = Q q , var(q) = Vq
- var(u) = E(u u’) = Q Vq Q’ = T
|
6
|
- Three bulls have +50 PTA protein.
- Do they have the same genes?
- Extremely unlikely.
- Bull A could have 10 positive genes.
- Bull B could have 10 positive genes, but on different chromosomes.
- Bull C could have 20 positive and 10 negative genes.
|
7
|
|
8
|
|
9
|
- No known common ancestors
- Many unknown common ancestors born before the known pedigree
- Relationships in base
- 0 ± x.x% due to earlier ancestors
- Called linkage disequilibrium (LD)
- Poor terminology, genes may not be physically linked
|
10
|
|
11
|
|
12
|
|
13
|
|
14
|
|
15
|
|
16
|
|
17
|
|
18
|
- Relationships can be defined as:
- A = expected genes in common
- G = actual DNA in common
- T = QTL alleles in common for a
trait
- Full sibs share 50% ± 3.5% of
DNA.
- “Unrelated” animals share more or fewer unknown ancestors than average.
- Reliability can increase if genomic (G) replace traditional (A)
relationships
|