Darius Danusevičius and Dag Lindgren Breeding cycler: Efficient long-term cycling strategy

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...

Contents of 40 min. General Breeidng cycler (10 min.) Example of what cycler can do? Testing strategy: optimization and timing (50 min): Single-stage strategies compared, Two-stage strategies compared, Amplified case: Progeny testing versus Pheno/Progeny. Main finding separately for pine and spruce (5 min.) New Cycler and other deterministic worksheets (Seed orchard optimizer)

Answer is Breeding cycler Deterministic optimizer of one (of many identical) breeding cycles made in Excel Transparent Interactive

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).

GD = 1 - group coancestry = the probability that the genes are non-identical, thus diverse. Group coancestry is a measure of gene diversity lost! Gene diversity and coancestry

Components of Tree Breeding Plus trees Long-term breeding Selection Mating Gain Seed orchard Testing Initiation

Long term breeding goes on for many repeated cycles Selection Mating Testing Long-term breeding

Breeding cycler studies what happens in one complete cycle Long-term breeding Selection Mating Testing

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

During one complete cycle Long-term breeding How to consider the cycle time? The cycle takes a number of years, depending on the duration of testing, mating and different waiting times Selection Mating Testing

Answer is: express Group Merit per year to consider 3 key factors: Wei and Lindgren 2001 Genetic gain; Gene diversity; Time.

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 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?

Breeding cycler is based on within family selection Acknowledgement: Large thanks to Swedish breeding for giving us the justification to construct a reasonable simple breeding cycler, that is balanced and where each breeding pop member get exactly one offspring in next generation breeding population. Loss of gene diversity is only a function of Breeding Population Size. It would have been much harder without this simplification! DaDa

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

How the Cycler works Results You do almost nothing – input the parameters and look for result Inputs

Variables - Genetic parameters Additive variance in test Dominance variance in test Environmental variance in test Coefficient of variance for additive “value for forestry” at mature age Breeding population size

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

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, thus the input values must be chosen with care The input values may need an 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 the 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 ~ Optimizing the balance may give 50% more gainOptimizing the balance may give 50% more gain

General M&M

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

Time compnents

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 a present-day budget Budget estimate is taken from pine and spruce breeding plan ~ test size expressed per year and BP member. ~ 10 ’cost units’ for pine, 20- for spruce. Budget constraint

How optimization works under specific scenario? Optimization= optimum combination of testing time and testing size to obtain max GM/Year and to satisfy the budget constraint (use Solver) Optimization= optimum combination of testing time and testing size to obtain max GM/Year and to satisfy the budget constraint (use Solver)

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

Parameters- for reference ParametersMain scenarioAlternative scenarios Additive variance  A 2 ) 1 Dominance variance, % of the additive variance in BP  D 2 ) 250; 100 Narrow-sense heritability (h 2 ) (obtained by changing  E 2 ) ; 0.5 Additive standard deviation at mature age (  Am ), % 105; 20 Diversity loss per cycle, % ;1 Rotation age, years 6010; 120 Time before establishment of the selection test (T BEFORE ), years 1 (phenotype)3; 5 (phenotype) 5 (clone)3; 7 (clone) 17 (progeny)5; 7 (progeny) Recombination cost (C RECOMB ), $ 3015; 50 Cost per genotype (Cg), $ 0.1 (clone),1; 5 (clone), 1 (progeny)0.1; 5 (progeny) Cost per plant (Cp), $ 10.5; 3 Cost per year and parent (constraint) 105; 20 Group Merit Gain per year (GMG/Y) To be maximized

CVa at mature age CV a =14 % is based on pine tests in south Sweden Jansson et al (1998), CV a =14 % is based on pine tests in south Sweden Jansson et al (1998), 1/2 of additive var in pop is within full sib families, 1/2 of additive var in pop is within full sib families, Our program is balanced= gain only from within full-sib selection, Our program is balanced= gain only from within full-sib selection, Thus, CV a within fam= CVa in pop divided by the square root of 2, thus a CV = 10%. Thus, CV a within fam= CVa in pop divided by the square root of 2, thus a CV = 10%. CVa within = sqrt(  2 /2) = sqrt(  2 )/sqrt(2)=  2 /sqrt(2)

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, % Narrow-sense heritability Phenotype Clone Progeny

GM/Y digits after comma are important If for Clone GM/Y=0.25%; cycle= 30 years then If for Clone GM/Y=0.25%; cycle= 30 years then Cycle GM=8 % (gain div loss) Cycle GM=8 % (gain div loss) Thus GM/Y reduction by 0.03 (10%) = Cycle gain reduction by 1% Thus GM/Y reduction by 0.03 (10%) = Cycle gain reduction by 1% Loss of Cycle gain by 1% = important loss Loss of Cycle gain by 1% = important loss

(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) Test time

If not enough cuttings, better more clones with less ramets, rather than to reduce ramet number at optimum clone number GM/Y by Phenotype=0, testing time

Higher h 2 = more clones and less ramets Spruce plan 40/15 Ola’s thesis, paper I, Fig. 9= 40 cl with 7 ram at test size 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 Testing time, years Annual Group Merit, % The optimal testing time No effect to test longer than years These years with conservative J-M function (Lambeth 1980) With Lambeth 2001, about years Figure with optimum at main scenario parameters (budget=10) clones/ramets 18/15

How realistic are the optima? Optima depends on budget, h 2, J-M correlation- how realistic are they? Optima depends on budget, h 2, J-M correlation- how realistic are they? 1. Budget is the present-day allocation. Increase will result in more gain. But we test how to optimise the resources we have. 2. h 2 =0,1 seems to be reasonable 3. J-M functions taken from southerly pines, it affects the timing with stand. error of 2 years ( ).

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.

Dominance seems to matter little Annual Group Merit, % Dominance would not markedly affect superior performance of clonal testing Dominance variance (% of additive) Clone Progeny Phenotype

On Genotype cost Tbefore Cost per genotype 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

Mating cost and total budget Clone Progeny Phenotype Budget per year and parent Recombination cost Clone Progeny Phenotype Important factor, we used present-day figures; what happens if it fluctuates If there is need, mating can be expensive Annual Group Merit, %

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).

Let’s 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)

Values- study 2

arrows show main scenario 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 Budget per year and parent (%) Annual Group Merit, % Clone Progeny Phenotype Pheno/Progeny

Budget cuts for Pheno/Progeny Budget  = resources reallocated on cheaper Phenotype test Testing time 10 (stage 1) and 14 (stage 2) little affected by the budget Budget=10Budget=5 Genetic gain, % Stage 1 Phenotype Stage 2 Progeny (44)5(72)

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?

Progeny test with and without phenotypic pre-selection Is there any realistic situation where Progeny testing is superior over Pheno/Progeny (reasonable interactions and scenarios) What and how flat is the optimum age of pre-selection for Pheno/Progeny? (when do we will need flowers?) Phenotype test Pre-selection age? Progeny test

Time and cost components C Per CYCLE = C recomb + n (C G + m C P ), T cycle = T recomb + T MATING + T LAG + T progtest T LAG is crossing lag for progeny test (polycross, seed maturation, seedling production) T MATING age of sufficient flowering capacity to initiate progeny test (for 2-stage strategy it corresponds to the age of phenotypic pre-selection

Parameters study 3

J-M correlation functions Lambeth (1980)= phenotypic fam mean corrs from many trials of 3 temperate conifers Ratio selection/rotation age (Q) J-M genetic correlation coefficient Lambeth (1980) Lambeth & Dill (2001) Gwaze et al. (2000) Gwaze et al. (2000)= genetic correlations from 19 trials with 190 fams of P taeda western USA. Lambeth (2001) Main = genetic corrs in 4 series (15 trials) P taeda (296 fams)

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 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

J-M correlation affects pre-selection age Optimum selection age depends on efficiency of Phenotype to generate enough gain to motivate prolongation of testing for an unit of time.Optimum selection age depends on efficiency of Phenotype to generate enough gain to motivate prolongation of testing for an unit of time Ratio selection/rotation age (Q) J-M genetic correlation coefficient Do we have J-M estimates for spruce and pine? Gwaze et al. (2000) 7 Lambeth & Dill (2001) 10 Lambeth (1980) 12 Gain increases fast by time Gain would increase faster if switching to progeny test The gain generating efficiency mainly depends on slope of J-M correlation function.The gain generating efficiency mainly depends on slope of J-M correlation function.

When the loss from optimum is important? Rotation age = When early testing is advantageous h 2 is high but then Phenotype alone is better Plant cost= Rotation is short Plants are cheap h 2 = Pheno/Progeny Progeny Annual Group Merit (%) Age of mating for progeny test (years)

Better crossings are motivated Crossing lag and genotype costs had no marked effect = the crosses can be made over a longer time to simultaneously test all pre-selected individuals and their flowering may be induced at a higher cost. Crossing lag= ; 0.26 Crossing lag= ; 0.25 Pheno/Progeny Progeny Annual Group Merit (%) Age of mating for progeny test (years)

These are as for our interactive scenario: low heritability (0,01), long rotation (80 y)= less J-M at pre-selection, weak J-M correlation (L1980) Progeny is motivated when conditions disfavour Phenotype But the optima flat and scenario unrealistic Pheno/Progeny Progeny Interactive scenario Annual Group Merit (%) Age of mating for progeny test (years)

Optimum test time and size for pine (for one of the 50 full sib fams) Long-term breeding Stage 2. Progeny- test each of those 5 with 30 offspring Stage 1: Test 70 full-sibs Mating 2-4 years, at a high cost if feasible Lag- 3-4 years Cycle time~ 27 Gain=8 % GM/Y= 0,27% Select back the best of 5 when progeny- test age is 10 Select 5 at age 10

What if no pine flowers until age 25? Pheno/Progeny is still leading Phenotype with selection age of 25 is better Progeny is the last Budget cuts, high h 2 will favour Phenotype This means, singe stage Phenotype cycle time > 25 years and For the two-stage, pre-selection not at its optimum age (10 years) Main (h=0.1, budget=10), Flowers at age Progeny Phenotype Pheno/ Progeny Annual Group Merit, %

May be 2 cycles of Phenotype instead of Pheno/Progeny? Cycle, years GM/year, % GM/cycle 2 cycle s of Pheno Phenotype200,1523,046,08 Pheno/Prog 400,1817,26 Answer is No: 7,26 is > 6,08

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

Main findings: cloning is the best strategy

Our main findings

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%

Research needs- Faster cloning

Research needs (a PhD thesis) Faster, better cloning: embryogenesis, rooting, C-effects (especially for pine)Faster, better cloning: embryogenesis, rooting, C-effects (especially for pine) Sufficient flowering at age 10 (15) for pineSufficient flowering at age 10 (15) for pine Documentation of flowering in breeding stockDocumentation of flowering in breeding stock How sexual maturation, flowering abundance are related to breeding value?How sexual maturation, flowering abundance are related to breeding value?

Optimizing the Balance: restrict grandparental but relax parental contributions

Balanced long-term breeding Some unbalance has advantages in long term breeding The inoptimality loss seem to be small; it is tricky to utilize inbalance, and the balance is unlikely to affect most of the recommendations much.

Unbalance at initiation of breeding At initiation of breeding there is no “balance”. Truncating tested plus trees to long term breeding has sometimes been done with inoptimal unbalance in the Swedish breeding program. I believe it is more optimal to sacrify the gene diversity in the initial selection a bit slower. This has been discussed i förädlingsrådet 1999, and one argument is in Routsalainen’s thesis (2002), so I will not discuss this more here.

(…) 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..

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 % 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 % gain. Variants of strategy 5 which are still more efficient can be constructed.

The end