# Presentation: 095289

## Methods to explain genomic estimates of breeding value

## Genomic Goals

**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**

## How Related are Relatives?

**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**

## Genomic Relationships

**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**

**Notes:**
Genetic similarity can be defined in several ways using:
A= pedigree data.  Wright, who was a USDA employee btw, used only pedigree data to calculate probabilities that gene pairs are identical by descent.
G = Genomic relationships use genotype data to estimate the fraction of total DNA that individuals actually share.
T = Fractions of alleles that two individuals share for loci that affect a certain trait will be termed QTL or T relationships.  The term QTL often refers only to loci with large effects, but here includes all loci that affect a trait.

## Markers vs QTLs

**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) = V****q**
**var(u) = E(u u’) = Q V****q**** Q’ = ****T**

**Notes:**
Inheritance of DNA can be traced within and across families for hundreds of genes influencing quantitative traits by genotyping many markers across the genome.  Markers may or may not contain loci that affect a trait. Relationship matrix of markers....
Vq is the co-variance matrix of QTL’s
U =  breeding value or sum of genetic effects.
T = QTL relationship matrix.    

Use marker, as not many QTL have been indentified.

## QTL Relationship Matrix (T) 

**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.**

**Notes:**
Unlikely, unless one gene controls the trait.  

## Genes in Common at One Locus

|  | If Full Sib 1 inherits: |  |  |  |  |
| --- | --- | --- | --- | --- | --- |
| If Full Sib 2 inherits: |  | w,y | w,z | x,y | x,z |
|  | w,y | 2 | 1 | 1 | 0 |
|  | w,z | 1 | 2 | 0 | 1 |
|  | x,y | 1 | 0 | 2 | 1 |
|  | x,z | 0 | 1 | 1 | 2 |

**w**** = gene from sire of sire**
**x**** = gene from dam of sire**
**y**** = gene from sire of dam**
**z**** = gene from dam of dam**

**Notes:**
Full sibs, genes in common at one loci.   I have named alleles from grandparents, w x y z.
If wy, and w,y  have 2 alleles in common.   If wz, and w,y    have 1, xy, and w,y also 1.  x,z and w,y  0 in common.

Overall 25% have two, 50% have one, 25% have none, 
Simple Mendelian genetics.    This is for one locus,  what about  multiple loci?????

## Alleles Shared by Sibs

| Indep-endent Loci | Percentage of alleles shared |  |  |  |
| --- | --- | --- | --- | --- |
|  | Full sibs |  | Half sibs |  |
|  | Mean | SD | Mean | SD |
| 1 | 50 | 35.4 | 25 | 17.7 |
| 5 | 50 | 15.8 | 25 | 7.9 |
| 10 | 50 | 11.2 | 25 | 5.6 |
| 50 | 50 | 5.0 | 25 | 2.5 |
| 100 | 50 | 3.5 | 25 | 1.8 |
| Infinite | 50 | 0.0 | 25 | 0.0 |

**Notes:**
Through simulations, Paul , and more recently  Mel, has shown that as the number of independent loci increase, SD decreases.
From 35% for one locus,  16% for 2 loci, 11% for 10  up until we get to 100.    100 is about the maximum independent loci our simulations allow, because after that, loci are NOT independent.
Hypothetically, if we know everything about Sibs,   SD is zero.    But, never get there.....  So how related are individuals?

## Unrelated Individuals?

**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**

**Notes:**
IF pedigrees, stop at 1960,  ancestors are assumed to be unrelated.  But, actually, before the known pedigree,  somewhere they are related......
A better term may be gametic disequilibrium.
Look at these relationships.......  Hopefully, the previous slides will become clear.

## Traditional Pedigree

|  |  |  |
| --- | --- | --- |
|  |  | Sire of Sire |
|  |  Sire |  |
|  |  | Dam of Sire |
|   Animal |  |  |
|  |  | Sire of Dam |
|  |  Dam |  |
|  |  | Dam of Dam |
|  |  |  |

**Notes:**
We are all familiar with a traditional pedigree chart.   Animal is expected to be an average of his parents.

## Genomic Pedigree

**Notes:**
Here is a simple Genomic pedigree chart.  2 separate chromosome segments are shown for each ancestor.  (During meiosis , they divide-recombination). The earliest generation, are all one color for illustrative purposes.  Dairy cattle have 30 chromosomes, including the   X and Y.   On average each chromosome carries millions of base pairs, and thousands of genes.
For the first segment, the sire has inherited 1⁄2 from each piece.  In the second segment, the majority was inherited from one piece (both of these are an example of a crossover).   The individual has received portions of the first segment that are traced back to the grandparent generation.
2) The dam has inherited most of one segment from one portion of her sire’s piece, and a mixture (double crossover) from her dam’s piece.  The individual has segments that can be traced to all grandparents.
In this example....    Say the light green segment from the paternal grand dam was one that was high milk production;  the offspring, inheriting a large segment from her, could mean that through this recombination, this animal would have a higher PTA for milk that a full sib.    

## Example of a SNP haplotype

**caacg****t****at**** **

**caacg****g****at**** **

- SNP

**atccg****a****at**** **

**atccg****c****at**** **

- ...

- ...

- ...

- ...

- SNP

**tct****a****ggat**** **

**tct****c****ggat**** **

- SNP

- ...

- ...

- Chr1

- Chr2

- Haplotype is a set of single nucleotide polymorphisms (SNPs) associated on a single chromosome. Identification of a few alleles of a haplotype block can identify other polymorphic sites in the region.

- Haplotype 1

**tca**** **

**gac**** **

- Haplotype 2

## SNP Pedigree

- atagatcgatcg

- ctgtagcttagg

- agggcgcgcagt

- cgatctagatcg

- cggtagatcagt

- agagatcgatct

- atggcgcgaacg

- ctatcgctcagg

- ctgtagcgatcg

- agatctagatcg

- agagatcgcagt

- atgtcgctcacg

- ctgtctagatcg

- atgtcgcgcagt

**Notes:**
Very complex, as 100’s of genes, and 1000’s of base pairs....  How to trace these.....

## Haplotype Pedigree

- atagatcgatcg

- ctgtagcttagg

- agggcgcgcagt

- cgatctagatcg

- cggtagatcagt

- agagatcgatct

- atggcgcgaacg

- ctatcgctcagg

- ctgtagcgatcg

- agatctagatcg

- agagatcgcagt

- atgtcgctcacg

- ctgtctagatcg

- atgtcgcgcagt

**Notes:**
A little better.....  Can trace, but still have difficulties......   so move from a haplotye to a genomic pedigree.....

## Translate Haplotype to Genotype

- agatctagatcg

**111211120200**

- ctgtagcgatcg

**Notes:**
One way to do this is to translate haplotype to genotype.  Ongoing discussions on how to best to accomplish this.    In this example aa is always zero, TT would be 2.....

## Genotype PedigreeCount number of second allele

|  |  |  |
| --- | --- | --- |
|  |  | 121101011110 |
|  |  111211120200 |  |
|  |  | 101121101111 |
| 122221121111 |  |  |
|  |   | 101101111102 |
|  |  011111012011 |  |
|  |  | 121120011010 |
|  |  |  |

**0 = ****homozygous for first allele (alphabetically)**
**1 = ****heterozygous**
**2 = ****homozygous for second allele (alphabetically)**

**Notes:**
0 = homozygous for first allele 1 = heterozygous 2 = homozygous for second allele    Many ways to count, this is counting alphabetically....
These numbers could be fed directly into a relationship matrix,  just 0,1,and 2’s.
Major alleles, minor alleles... frequency, what is major in one breed, possibly not in another...   

## Reliability from Full Sibs50,000 markers, 1000 QTLs, sib REL = 99%

|  | Reliability Obtained From: |  |
| --- | --- | --- |
| Full Sibs | A | G |
| 1 | .250 | .261 |
| 10 | .454 | .502 |
| 100 | .495 | .773 |
| 1000 | .499 | .970 |

**A**** = traditional additive relationships,**
**G**** = genomic relationships**

**Notes:**
In simulations, and Paul will go over the math in the next presentation. Lots of markers, 1000 QTL’s, not of major effect, and high rel sibs.
As the number of full sibs increases,  in the A matrix, never rise above 50% reliability.  But with a genomic matix,  with 10 full sibs, already above 50%, with 100 77%....
More reliability for a genomic model than a traditional additive relationship matrix.

## Conclusions

**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**

**Notes:**
Paul follows this talk, and his presentation will hopefully answer any mathematics questions you may have...