Presentation on theme: "Darius Danusevičius and Dag Lindgren Optimize your breeding with Breeding Cycler Efficient long-term cycling strategy 35 min + 10 min Lithuanian Forest."— Presentation transcript:
Darius Danusevičius and Dag Lindgren Optimize your breeding with Breeding Cycler Efficient long-term cycling strategy 35 min + 10 min Lithuanian Forest Res. Institute (P15) Swedish University of Agricultural Sciences
How carry breeding ? How to select? phenotype? clones? progeny? 1$1$ Progeny trial No. 3 I breeding More gain More diversit y Faster But the budget is so low...
Answer is Breeding cycler Deterministic optimizer of one (of many identical) breeding cycles made in Excel Transparent Interactive Communistic
Basic feature Gain per time Cost Diversity Complete comparison, as it simultaneously considers: Other things, e.g. to well see the road
It assumes specific long- term strategy Recurrent cycles of mating, testing and balanced selection Adaptive environment Testing Within family selection Mating We consider one such breeding population Breeding population
Key-problem: How to deal with relatedness and gene diversity Solution: Group coancestry (equivalent Status number, Dag Lindgren et al.) The probability for IBD is group coancestry. f Let's put all homologous genes in a pool Take 2 (at random with replacement).
Components of Long-Term Breeding Plus trees Long-term breeding Selection Mating Gain Seed orchard Testing Initiation
Long term breeding goes for many repeated cycles Selection Mating Testing Long-term breeding Breeding cycler studies what happens in one complete cycle
What happens in one complete cycle? Long-term breeding The breeding value increases The gene diversity decreases How to assign a single value to the increase in BV and the decrease in GD?
weighted average of Breeding Value and Gene Diversity Weighting factor = Penalty coefficient; (coancestry of 1 means drop in BV dawn to 0) Answer is : Group merit Lindgren and Mullin 1997 if coancestry of 1= 100% drop in BV, then coancestry of 0.005= 0.5 % drop in BV
We expressed Group Merit per year to consider 3 key factors: Wei and Lindgren 2001 Genetic gain; Gene diversity; Time. Time factor is missing in many BPs in EU
How to consider the cost? Long-term breeding Cost of a cycle is depending on number of test plants, mating techniques, testing strategy etc. Selection Mating Testing
Annual Group Merit progress at a given annual cost for all BP considers four key factors: Danusevicius and Lindgren 2002 Genetic gain; Gene diversity; Time; Cost.
Examples of what Breeding Cycler can do Which is the best testing strategyWhich is the best testing strategy What is optimum breeding population size?What is optimum breeding population size? What is the influence of the parameters?What is the influence of the parameters? When to select and what numbers to test ?When to select and what numbers to test ? Where to allocate resources to strengthen your breeding plan?Where to allocate resources to strengthen your breeding plan? How to optimize balance among BP members?How to optimize balance among BP members?
How the Cycler works (in principle) Long-term breeding Selection age ? Mating Testing size ? 1. Input Genetic parameters Time components Cost component 2. Find resource allocation that maximises GM/year? Test method Clone? Progeny? Size of breeding population?
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 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 cost Under budget constraint Recombination Time for e.g. cloning or creation of progeny Production of test plants Testing time Cycle 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
How the Cycler works Results You do almost nothing – input the parameters and look for result Inputs
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 for 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
Optimizing the balance 3: Ph/Prog amplified (pine), effect of J-M. Seminar 2009 Best testing strategy Contents Breeding cycler shark
Main findings Clonal test is superior (use for spruce)Clonal test is superior (use for spruce) Progeny testing not efficientProgeny testing not efficient For Pine, use 2 stage Pheno/ProgenyFor Pine, use 2 stage Pheno/Progeny If 2stage is used, pine flowers not needed before age ~ 10-15If 2stage is used, pine flowers not needed before age ~ 10-15 Optimizing the balance may give 50% more gainOptimizing the balance may give 50% more gain
Main inputs and scenarios While testing an alternative parameter value, the other parameters were at main scenario values Low lower reasonable bound Genetic parameters Time components Cost components Main typical for Pine or spruce High higher reasonable bound
Cost per test plant = 1 cost unit, all the other costs expressed as ratio of this 1. 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 Such expression also helped to set the budget constraint corresponding to the present-day budget The time and cost explained Established in 5 years after seed harvest Field trial Establishment, maintenance and assessments Cutting of ramets Rooting of ramets (1 year) Transportation Crossing Recombination cost=20, Time=4 Plant dependent cost=1 (per ramet) Genotype depend. cost=2 (per ortet) Nursery Production of sibs (4 years) Mating time 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. Budget constraint When it is annual, you can optimize time
Single-stage testing strategies
Objective: compare strategies based on phenotype, clone or progeny testing (…n) Phenotype testing N=50 (…n), (…m) and selection age were optimized Clone or progeny testing N=50 (…n) (…m) OBS: Further result on numbers and costs- for one of these families
Results-clonal best, progeny worst At all the scenarios, Clonal was superior, except high h 2. 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, % 0.0 0.1 0.2 0.3 0.4 0.5 0.6 00.10.20.30.40.50.6 Narrow-sense heritability Phenotype Clone Progeny
0.10 0.15 0.20 0.25 0.30 4(59)10(25)15(18)20(14)30(10)40(8) Clone no (ramets per clone) Annual Group Merit, % How flat are the optima (clone)? No marked effect: 12 clones with 22 ramets or 30 clones with 10 ramets. This means: If problems with cloning, better-> clones with clones with < ramets GM/Y by Pheno h 2 =0.1, lower budget, at optimum testing time Optimum 18(15) 17 18 20 22 23 25 Test time
Higher h 2 = more clones and less ramets Spruce plan 40/15 Olas thesis, paper I, Fig. 9= 40 cl with 7 ram at test size 280 0.00 0.10 0.20 0.30 0.40 0.50 00.10.20.30.40.5 Narrow-sense heritability GM/Y, % 13/23 18/15 28/9 46/5 Clone no/ramet no Optimum then is between 18/15 and 30/10 Budget= 10
Clone strategy 0.00 0.05 0.10 0.15 0.20 0.25 0.30 1516171819202122232425 Testing time, years Annual Group Merit, % The optimal testing time No effect to test longer than 18-20 years These 18-20 years with conservative J-M function (Lambeth 1980) With Lambeth 2001, about 15-17 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) 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. Phenotype generates less gain but this is compensated by cheaper and faster cycles.
On Genotype cost Tbefore 0.05 0.10 0.15 0.20 0.25 0.30 0123456 Cost per genotype 0.05 0.10 0.15 0.20 0.25 0.30 0369121518 Delay before establishment of selection test (years) 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 Clone Progeny Phenotype Clone Progeny
Conclusions Clonal testing is the best breeding strategy Phenotype 2nd 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).
Lets do it in 2 stages?
Phenotype/Progeny strategy Stage 1 Phenotype select at age 10 (15 only 3% GM lost) Stage 2 Progeny test select at ca 10 (70) (30)
arrows show main scenario 0.10 0.15 0.20 0.25 0.30 1357911131517 Delay before establishment of selection test (years) Phenotype/ProgenyIf Progeny initiated early, may~ Phenotype/Progeny = need for a amplification Phenotype/Progeny is shown with a restriction for Phenotype selection age > 15 Clone = Phenotype/Clone = no need for 2 stages.Clone = Phenotype/Clone = no need for 2 stages. Phenotype/Progeny is 2nd best = best for PinePhenotype/Progeny is 2nd best = best for Pine Clone Progeny Phenotype Pheno/Progeny Results: two-stage 2nd best
Budget cuts = switching to Phenotype tests in Pine If budget is cut by half = simple Phenotype test 0.1 0.2 0.3 05101520 Budget per year and parent (%) Annual Group Merit, % Clone Progeny Phenotype Pheno/Progeny
Why Pheno/Progeny was so good? It generated extra gain by taking advantage of the time before the candidates reach their sexual maturity 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 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? Question for the next study: is there any feasible case where Progeny can be better?
Results: 2 stage is better 2 stage strategy was better under most reasonable values 2 stage strategy was better under most reasonable values Main scenario 0.0 0.3 0.6 0510152025 Annual Group Merit (%) Age of mating for progeny test (years) No marked loss would occur if mating is postponed to age 15No marked loss would occur if mating is postponed to age 15 Pheno/Progeny Progeny
When the loss from optimum is important? Rotation age = 20 0.0 0.3 0.6 0510152025 When early testing is advantageous h 2 is high but then Phenotype alone is better 0.0 0.3 0.6 0510152025 Plant cost= 0.1 0.0 0.3 0.6 0510152025 Rotation is short Plants are cheap h 2 = 0.5 0.0 0.2 0.4 0.6 0.8 0510152025 Pheno/Progeny Progeny Annual Group Merit (%) Age of mating for progeny test (years)
Main findings - spruce (18) (15)(15)(15) (15)(15) (15)(15)(15)(15) (15)(15) Clonal test by far the best Select at age 15 (20) depending on J-M correlation If higher h 2 more clones less ramets Present plans: size 40/15, selection age: 10 years With L(2001), Cycle time~ 21 Gain=8.2 % GM/Y= 0,34%
Main findings- P ine Use 2 stage Pheno/Progeny strategy Stage 1 Phenotype select at age 10 (15 only 3% GM lost) Stage 2 Progeny test select at ca 10 (70) (30) 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
(…) Cycle 1 SPM parental balance (almost current Swedish pine program) Founder selection Mating of founders Select and mate 2 best sibs to create 2 families (…) Cycle 2 Select and mate 2 best sibs to create 2 families (…) Cycle 3 Select and mate 2 best sibs to create 2 families
(…) Cross e.g. 4 best sibs in the 2 best families 2 nd rank family (…) 1 st rank family (…) 3 rd rank family (…) n th rank family (…) Multiple SPMs 2 nd rank family (…) 1 st rank family (…) 3 rd rank family (…) n th rank family (…) Founders Green trees show pedigree
Multiple SPMs Pedigrees BP third generation 2 nd rank family (…) 1 st rank family Trees selected for crossing in the 3 rd generation (…) Pedigree to later breeding population Founders
Note that retrospectively SPM and multiple SPM give identical pedigrees, thus identical increase of coancestry..
10 5 14 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0510152025 Low budget High budget 2 Medium budget Families & parents cost nothing Number of parents per selected family Annual progress (%) 2=phenotypic
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. 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.