Optimize your breeding with Breeding Cycler Efficient longterm
Optimize your breeding with Breeding Cycler Efficient long-term cycling strategy 35 min + 10 min Darius Danusevičius and Dag Lindgren Lithuanian Forest Res. Institute (P 15) Swedish University of Agricultural Sciences
How carry breeding ? More gain Faster More diversity But the budget is so low. . . How to select? phenotype? clones? progeny? ng i d ee br I 1 $ ny oge 3 Pr l No. a tri
Answer is Breeding cycler Deterministic optimizer of one (of many identical) breeding cycles made in Excel Transparent Interactive
Basic feature Complete comparison, as it simultaneously considers: Cost Gain per time Diversity Other things, e. g. to well see the road
It assumes specific longterm strategy Recurrent cycles of mating, testing and balanced selection Adaptive Mating environment Within family selection Breeding population Testing We consider one such breeding population
Key-problem: How to deal with relatedness and gene diversity Solution: Group coancestry (equivalent Status number, Dag Lindgren et al. ) Let's put all homologous genes in a pool Take 2 (at random with replacement). The probability for IBD is group coancestry. f
Components of Long-Term Breeding Gain Initiation Plus trees Selection Mating Long-term breeding Seed orchard Testing
Long term breeding goes for many repeated cycles Mating Long-term breeding Selection Breeding cycler studies Testing what happens in one complete cycle
What happens in one complete cycle? The breeding value increases The gene diversity decreases Long-term breeding How to assign a single value to the increase in BV and the decrease in GD?
Answer is : Group merit weighted average of Breeding Value and Gene Diversity Weighting factor = “Penalty coefficient”; (coancestry of 1 means drop in BV dawn to 0) if coancestry of 1= 100% drop in BV, then coancestry of 0. 005= 0. 5 % drop in BV Lindgren and Mullin 1997
We expressed Group Merit per year to consider 3 key factors: • Genetic gain; • Gene diversity; • Time factor is missing in many BPs in EU Wei and Lindgren 2001
How to consider the cost? Cost of a cycle is depending on number of test plants, mating techniques, testing strategy etc. Mating Selection Long-term breeding Testing
Annual Group Merit progress at a given annual cost for all BP considers four key factors: • Genetic gain; • Gene diversity; • Time; • Cost. Danusevicius and Lindgren 2002
Examples of what Breeding Cycler can do • Which is the best testing strategy • What is optimum breeding population size? • What is the influence of the parameters? • When to select and what numbers to test ? • Where to allocate resources to strengthen your breeding plan? • How to optimize balance among BP members?
How the Cycler works (in principle) Size of breeding population? 1. Input • Genetic parameters Mating • Time components • Cost component 2. Find resource allocation that maximises GM/year? Selection age ? Test method Clone? Progeny? Long-term breeding Testing size ?
Variables - Genetic parameters Additive variance in test Dominance variance in test Environmental variance in test Breeding population size Other…
How the Cycler works (detail) Insert red values. The worksheet will calculate the blue values with the consequences of your choices. Find optimum testing size and testing time to fit into the budget and annual maximize group merit. Or let “SOLVER” find the values which maximise progress in group merit
Time and cost components Cycle cost Under budget constraint • Recombination (cost can be either per BP member or in total) • Cost per tested genotype (it costs to do a clone or a progeny) • Test plant can be economical unit Cycle time • Recombination • Time for e. g. cloning or creation of progeny • Production of test plants • Testing time
Variables - Others Rotation time (for J*M considerations) Annual budget Test method (clonal, progeny or phenotype) J*M development curve Weighting factor for diversity versus gain
In p ut s How the Cycler works Results You do almost nothing – input the parameters and look for result
J-M correlation is important Choice can be made of J-M function including custom, Lambeth and Dill 2001 (genetic) is our favourite.
Constraints and limitations of breeding cycler Sh… in – sh… out= the inputs must be chosen with care The inputs may need adjustment from the most evident for considering factors not considered in the math Breeding heads for an area and get information from a number of sites, this is considered by modification to the variance components Breeding heads for improvement in many characters, we set goal as one character “value forestry”. The “observation” is an index of observations, and J*M has to be adjusted.
Constraints and limitations of breeding cycler - continued Plant cost is seen as independent of age of evaluation. This can cause problems for some type of comparisons. It is rather easy to add many type of considerations to the EXCEL sheet, the problem is that it makes it too complex for the user journal papers.
Cycling will accumulate gain. Where is the limit?
Example of what Breeding cycler can do studies by Dag and Darius
Contents Optimizing the balance Seminar 2009 3: Ph/Prog amplified (pine), effect of J-M. shark Best testing strategy Breeding cycler
Main findings • Clonal test is superior (use for spruce) • Progeny testing not efficient • For Pine, use 2 stage Pheno/Progeny • If 2 stage is used, pine flowers not needed before age ~ 10 -15 • Optimizing the balance may give 50% more gain
Main inputs and scenarios Low Main High Genetic parameters lower typical for higher Time components reasonable Pine or reasonable bound spruce bound Cost components While testing an alternative parameter value, the other parameters were at main scenario values
Time compnents
The time and cost explained • Cost per test plant = 1 ’cost unit’, all the other costs expressed as ratio of this 1. • Such expression also helped to set the budget constraint corresponding to the present-day budget Production of Cutting of sibs (4 years) ramets Crossing Mating time Rooting of ramets (1 year) Transportation Nursery Recombination Genotype depend. cost=2 (per cost=20, ortet) Time=4 Established in 5 years after seed harvest Establishment, maintenance and assessments Field trial Plant dependent cost=1 (per ramet) Time before Testing time Lag
All these costs should fit to an annual budget Budget estimate is taken from pine and spruce breeding plan ~ test size expressed per year and BP member (can be all BP). ~ 10 ’cost units’ for pine, 20 - for spruce. t e g d Bu t n i a r t cons When it is annual , you can optimize time
Single-stage testing strategies
Objective: compare strategies based on phenotype, clone or progeny testing Clone or progeny testing Phenotype testing N=50 (…n) OBS: Further result on numbers and costs- for one of these families (…n) (…m)(…m)(…m) (…n), (…m) and selection age were optimized
Test 26 clones with 21 ramet (18/15 �budget), select at age 20 Test 182 phenotypes; select at age 15, (�budget: 86, for 17 years) (second best) Test 11 female parents with 47 progeny each; select at age 34 (� budget: 8/34, 40 years) Annual Group Merit, % Results-clonal best, progeny worst 0. 6 0. 5 0. 4 0. 3 0. 2 0. 1 0. 0 Clone Phenotype Progeny 0 0. 1 0. 2 0. 3 0. 4 0. 5 0. 6 Narrow-sense heritability At all the scenarios, Clonal was superior, except high h 2.
How flat are the optima (clone)? h 2=0. 1, lower budget, at optimum testing time This means: If problems with cloning, better-> clones with < ramets 0. 30 Optimum 18(15) Annual Group Merit , % No marked effect: 12 clones with 22 ramets or 30 clones with 10 ramets. 0. 25 0. 20 GM/Y by Pheno 0. 15 0. 10 4(59) 10(25) 15(18) 20(14) 30(10) 40(8) Clone no (ramets per clone) Test time 17 18 20 22 23 25
Higher 2 h = more clones and less ramets Clone no/ramet no 0. 50 Spruce plan 40/15 Ola’s thesis, paper I, Fig. 9= 40 cl with 7 ram at test size 280 0. 40 GM/Y, % Optimum then is between 18/15 and 30/10 46/5 18/15 0. 30 0. 20 28/9 13/23 0. 10 0. 00 0 0. 1 0. 2 0. 3 0. 4 0. 5 Narrow-sense heritability Budget= 10
The optimal testing time • These 18 -20 years with conservative J -M function (Lambeth 1980) • With Lambeth 2001, about 15 -17 years 0. 30 Annual Group Merit, % • No effect to test longer than 18 -20 years Clone strategy 0. 25 0. 20 0. 15 0. 10 0. 05 0. 00 15 16 17 18 19 20 21 22 23 24 25 Testing time, years Figure with optimum at main scenario parameters (budget=10) clones/ramets 18/15
Why Phenotype ≥Progeny ? • Drawbacks of Progeny: long time and high cost (important to consider for improvement) • Phenotype generates less gain but this is compensated by cheaper and faster cycles.
Expensive genotypes are of interest only if it would markedly shorten T before for Progeny or improve cloning So it pays off to make expensive cloning 0. 30 0. 25 Tbefore On Genotype cost Clone 0. 20 0. 15 0. 10 Clone Progeny 0. 15 0. 10 Progeny Phenotype 0. 05 0 1 2 3 4 5 Cost per genotype 6 0 3 6 9 12 15 18 Delay before establishment of selection test (years)
Conclusions • Clonal testing is the best breeding strategy • Phenotype 2 nd best, except very low h 2 • Superiority of the Phenotype over Progeny is minor = additional considerations may be important (idea of a two-stage strategy).
Let’s do it in 2 stages?
Phenotype/Progeny strategy (70) (30)(30)(30)(30) Stage 1 Phenotype select at age 10 (15 only 3% GM lost) Stage 2 Progeny test select at ca 10
Results: two-stage 2 nd best • Clone = Phenotype/Clone = no need for 2 stages. 0. 30 0. 25 Clone • Phenotype/Progeny is 2 nd 0. 20 best = best for Pine • If Progeny initiated early, may~ Phenotype/Progeny = need for a amplification • Phenotype/Progeny is shown with a restriction for Phenotype selection age > 15 Pheno/Progeny 0. 15 Phenotype 0. 10 1 3 5 7 Progeny 9 11 13 15 17 Delay before establishment of selection test (years) arrows show main scenario
If budget is cut by half = simple Phenotype test Annual Group Merit, % Budget cuts = switching to Phenotype tests in Pine 0. 3 Clone Pheno/Progeny 0. 2 Phenotype Progeny 0. 1 0 5 10 15 20 Budget per year and parent (%)
Why Pheno/Progeny was so good? • It generated extra gain by taking advantage of the time before the candidates reach their sexual maturity • This was more beneficial than single-stage Progeny test at a very early age • Question for the next study: is there any feasible case where Progeny can be better?
• 2 stage strategy was better under most reasonable values • No marked loss would occur if mating is postponed to age 15 Annual Group Merit (%) Results: 2 stage is better Main scenario 0. 6 Pheno/Progeny 0. 3 Progeny 0. 0 0 5 10 15 20 25 Age of mating for progeny test (years)
When the loss from optimum is important? Annual Group Merit (%) When early testing is advantageous Rotation age = 20 0. 6 2 0. 8 h = 0. 5 0. 6 Plant cost= 0. 1 Pheno/Progeny 0. 6 0. 4 0. 3 0. 0 Progeny 0. 2 0. 0 0 5 10 15 20 25 Age of mating for progeny test (years) h 2 is high but then Phenotype alone is better Rotation is short Plants are cheap
Main findings- spruce Clonal test by far the best If higher h 2 more clones less ramets Present plans: size 40/15, selection age: 10 years (18) Select at age 15 (20) depending on J-M correlation (15) (15) With L(2001), Cycle time~ 21 Gain=8. 2 % GM/Y= 0, 34%
Main findings- Pine Use 2 stage Pheno/Progeny strategy (70) (30)(30)(30)(30) Stage 1 Phenotype select at age 10 (15 only 3% GM lost) Stage 2 Progeny test select at ca 10 With L(2001), Cycle time~ 27 Gain=8 % GM/Y= 0, 27%
Conclusions • Under all realistic values, Pheno/Progeny better than Progeny • Sufficient flowering of pine at age 10 is desirable, but the disadvantage to wait until the age of 15 years was minor, • If rotation short, h 2 high, testing cheap, delays from optimum age could be important
Research needs- Faster cloning
Optimizing the Balance: restrict grandparental but relax parental contributions
SPM parental balance (almost current Swedish pine program) Founder selection Mating of founders Select and mate 2 best sibs to create 2 families Cycle 1 (…) Select and mate 2 best sibs to create 2 families Cycle 2 (…) Select and mate 2 best sibs to create 2 families Cycle 3 (…)
Green trees show pedigree Multiple SPMs Founders (…) Cross e. g. 4 best sibs in the 2 best families (…) 1 st rank family (…) nth rank family 3 rd rank family (…) 2 nd rank family (…) 3 rd rank family (…) nth rank family (…) 2 nd rank family
Multiple SPMs Pedigrees BP third generation Founders (…) (…) Pedigree to later breeding population Trees selected for crossing in the 3 rd generation 1 st rank family (…) 2 nd rank family
Note that retrospectively SPM and multiple SPM give identical pedigrees, thus identical increase of coancestry. .
Families & parents cost nothing 0. 6 14 Annual progress (%) 0. 5 10 High budget 0. 4 5 0. 3 0. 2 Medium budget Low budget 0. 1 0. 0 0 2 2=phenotypic 5 10 15 20 Number of parents per selected family 25
Conclusions • Multiple SPM strategy is VERY promising and can beat conventional single SPM strategy with 20 -70 % gain. Variants of strategy 5 which are still more efficient can be constructed.
The end
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