Breeding Cycle Analyser by linzhengnd

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									Breeding Cycle Analyser: deterministic simulator for genetic predictions of gain, diversity loss, cost and time of a breeding cycle
      TroubleSolver                   Get started        Aim             An example




 Genetic parameters within full-sib family                                                               Genetic parameters (large population) Parameter                                Value
                               Variance components Standard deviations        CVASm=        7.07                                           Variance components               CVAm=           10
                          sAS2=           0.5               0.71     sP2=sA2+sD2+sE2=       11.25                                       sAL2=         1                          h2 =       0.08
                                                                                      2
                          sDS 2=         0.750               0.87                  h=        0.04                                       sDL 2=        1                          H2 =       0.17
                           sE 2=           10                3.16                  H2 =      0.11                                        sE 2=       10                      CVAm=          10.0

 General settings and summarizing results                                                                Diversity loss, rotation, BP size                       Final results             Value
                          Choose        Gain (%)                         Cost ("$")       Time (years)                  Rotation age=     60                     Gain/cost                #NAME?
                         alternative       0                                20                3                               BP size     50                     Gain/time                #NAME?
 Progeny test                              3          #NAME?                399             38.9                Group coancestry weight=            100          Group Merit Gain         #NAME?
                                                                                                                              Diversity loss=      0.005         GMG/time                 #NAME?
 Overall                                              #NAME?                419              42           Diversity loss in % of "forest value"=    0.5          Cost/time                    10.00

 Parameters                          Not active     Not active      Progeny test                         Constraint=Annual budget for all BP                           500                  10
   No. of selected individuals=            1              1               1
            Full-sib family size=         90             21              21
  Plant no. per family member=                           20              14                                                                                Apologies
           Selection intensity=         #NAME?         #NAME?             #NAME?
                                                                                          Confusion has been caused as we (Lindgren and Danusevicius) have not made clear in different
            Initiation cost (CIN)=         0              0                 0             papers that within full sib family variance components were used as entries. That is likely to cau
      Cost per test plant (CP)=            1              1                 1             confusion and most readers are led to believe that results apply to other values on variance com
      Cost per genotype (CG)=                             2                 5             they do.
                    Total cost =          90             462               399
                                                                                          The results are still correct, but for different entries than given (if the given entries are interprete
                  Time before =           1               5                 17            large ideal population). It is not likely general conclusions will be seriously misleading. In all c
                  Testing time=           38              34                22            when the results were mentioned (all published papers, all seminars) we have referred to the EX
                     Time after=          0               0                 0             workbooks on the web, and when inserting entries it is always crystal clear that it is within fami
                                                                                          comparison with a large ideal population is made. Thus it is very unlikely that real users of the
                      Duration=          38.9            38.9              38.9           tools for their own purposes will be mislead. However we made this sheet and will make more a
                                                                                          further reduce the risk and control the damage.
                                                                      Confusion has been caused as we (Lindgren and Danusevicius) have not made clear in different
                                                                      papers that within full sib family variance components were used as entries. That is likely to cau
                                                                      confusion and most readers are led to believe that results apply to other values on variance com
                                                                      they do.

                                                                      The results are still correct, but for different entries than given (if the given entries are interprete
                                                                      large ideal population). It is not likely general conclusions will be seriously misleading. In all c
                                                                      when the results were mentioned (all published papers, all seminars) we have referred to the EX
                                                                      workbooks on the web, and when inserting entries it is always crystal clear that it is within fami
                                                                      comparison with a large ideal population is made. Thus it is very unlikely that real users of the
                                                                      tools for their own purposes will be mislead. However we made this sheet and will make more a
                a1=sA=      0.71              0.71         0.35       further reduce the risk and control the damage.
                    a2 =    3.35              1.32         0.96
            rAI=a1/a2=      0.21              0.53         0.37
J-M correlation function     0                 2            2
                           #NAME?            #NAME?      #NAME?
             rY (or R)=    #NAME?            #NAME?      #NAME?
Genetic gain (%)=          #NAME?            #NAME?   Populations can be regarded in different ways
                                                        #NAME?


                                    A population can be regarded as the whole population considered. Or it can be regarded as a
                                    segment of larger population or regarded relative to a reference population. A full sib family can
                                    either be regarded as the whole population or as a part of a larger real or imagined population. In
                                    quantitative genetics it is often stated that there are certain relationships between the variance
                                    components in a full sib as a function in the population it was derived from. These relationships
                                    are however subject to many assumptions. Therefore we will talk about a large ideal population
                                    when the simplest rules apply.

                                    The reason to consider only a single full sib family here is that the EXCEL workbook program
                                    “breeding cycler” considers exclusively within family selection. Thus all these assumptions
                                    connecting the full sib to a large ideal population are circumvented.

                                    Considering the single family as a part of a larger population makes the formula more
                                    complicated and depending on assumptions, which may not be fulfilled. These assumptions
                                    include a genetic balance, which is unlikely to occur in early forest tree breeding, absence of
                                    inbreeding, an infinite population (thus not the “full breeding population”), it matters if it is
                                    parents or progeny etc. Disadvantages is that it is more conventional and traditional to formulate
                                    the variance within a full sib as a part of a large ideal population; that most readers will assume
                                    that the population considered is the large ideal, and that most observations on variance
                                    components reported concerns populations which can be approximated by the large ideal
                                    population.

                                    It may help in constructing formulas relevant for a larger population to start with considering a full sib as
                                    the population and when build the higher hierarchy formulas by considering the relationships between a
                                    full sib and the larger structure in a translation table, and when just insert the translations.


                                    The variances are indexed S for Single Full Sib Family as part of a large ideal population, L.

                                                                                   Designations
                                    Entity                        Population considered
                                                                  General                     One Single full sib           Large ideal
                                                                                              family
                                the variance within a full sib as a part of a large ideal population; that most readers will assume
                                that the population considered is the large ideal, and that most observations on variance
                                components reported concerns populations which can be approximated by the large ideal
                                population.

                                It may help in constructing formulas relevant for a larger population to start with considering a full sib as
                                the population and when build the higher hierarchy formulas by considering the relationships between a
                                full sib and the larger structure in a translation table, and when just insert the translations.


                                The variances are indexed S for Single Full Sib Family as part of a large ideal population, L.

                                                                                             Designations
                                Entity                                     Population considered
                                                                           General                          One Single full sib                   Large ideal
                                                                                                            family
                                Additive variance in
                                goal character (mature                     s Am
                                                                             2
                                                                                                            s Am S
                                                                                                              2
                                                                                                                                                  s Am L
                                                                                                                                                    2

                                character)
                                Coefficient of variance
                                for goal character
                                                                           CV Am                            CV Am S                               CV Am L
                                                                           sA                               s AS                                  s AL
                                Additive variance in                        2                                 2                                     2
                                measured character
                                Dominance variance in
                                measure character                          sD
                                                                            2
                                                                                                            s DS
                                                                                                              2
                                                                                                                                                  s DL
                                                                                                                                                    2

                                Environmental variance
                                in measured character
                                                                           sE
                                                                            2
                                                                                                            sE
                                                                                                             2
                                                                                                                                                  sE
                                                                                                                                                   2




                                                                           Gain formulas

Formulas for genetic gain by selection in a full sib family using different test methods and different reference populations for variance components.
A factor irj m common to all formulas is not shown


Selection     One Single full sib family                                                                                      Full sib family as a part of a large population
criteria
                                                                                Expressed so derivation is transparent                                           Simple expressed
Phenotype                       s AS                                                               0.5s AL                                                                 s Am Ls AL
              s Am S                                                             0.5s Am L
GPhenotype            s   2
                                s   2
                                          s   2
                                                                                                                                                                  2 0.5s   2
                                                                                                                                                                                 0.75s DL  s E
                                                                                                                                                                                        2      2
                           AS        DS        E                                           0.5s AL  0.75s DL  s E
                                                                                                2           2     2
                                                                                                                                                                           AL

Clone                           s AS                                                                     0.5s AL                                                           s Am Ls AL
              s Am S                                                              0.5s ALm
GClone                                    s   2
                                                                                                                                                                                             sE
                                                                                                                                                                                              2
                       s AS  s DS 
                         2      2                                                                                             sE
                                                                                                                               2
                                                                                                                                                                  2 0.5s AL  0.75s DL 
                                                                                                                                                                         2          2
                                                                                             0.5s         0.75s          
                                               E                                                    2                2
                                                                                                    AL               DL
                                            m                                                                                 m                                                               m
Progeny                                   0.5s AS                                                                    0.5 0.5s AL                                                     s Am Ls AL
              s Am S                                                              0.5s ALm
GProgeny                                 0.75s    2
                                                        s   2
                                                                  s   2
                                                                                                                                                                                    0.375s AL  0.7
                                                                                                                                                                                           2
                       0.25s AS 
                             2                                                                                       0.75 * 0.5s AL  0.75s DL  s E
                                                                                                                                 2          2      2
                                                                                                                                                                  4 0.125s AL 
                                                                                                                                                                           2
                                                                                             0.25 * 0.5s         
                                                   AS        DS        E                                    2
                                                                                                            AL
                                                        m                                                                                                                                       m
Formulas for genetic gain by selection in a full sib family using different test methods and different reference populations for variance components.
A factor irj m common to all formulas is not shown


Selection     One Single full sib family                                                                       Full sib family as a part of a large population
criteria
                                                                           Expressed so derivation is transparent                                     Simple expressed
Phenotype                       s AS                                                          0.5s AL                                                          s Am Ls AL
              s Am S                                                        0.5s Am L
GPhenotype            s   2
                                s   2
                                          s   2
                                                                                                                                                      2 0.5s   2
                                                                                                                                                                     0.75s DL  s E
                                                                                                                                                                            2      2
                           AS        DS        E                                      0.5s AL  0.75s DL  s E
                                                                                           2           2     2
                                                                                                                                                               AL

Clone                           s AS                                                            0.5s AL                                                        s Am Ls AL
              s Am S                                                         0.5s ALm
GClone                                    s   2
                                                                                                                                                                                 sE
                                                                                                                                                                                  2
                       s AS  s DS 
                         2      2                                                                              s   2
                                                                                                                                                      2 0.5s AL  0.75s DL 
                                                                                                                                                             2          2
                                                                                        0.5s AL  0.75s DL 
                                               E                                             2          2          E
                                            m                                                                   m                                                                 m
Progeny                                   0.5s AS                                                          0.5 0.5s AL                                                   s Am Ls AL
              s Am S                                                         0.5s ALm
GProgeny                                 0.75s    2
                                                        s   2
                                                                  s   2
                                                                                                                                                                        0.375s AL  0.7
                                                                                                                                                                               2
                       0.25s AS 
                             2                                                                             0.75 * 0.5s   2
                                                                                                                               0.75s   2
                                                                                                                                             s   2
                                                                                                                                                      4 0.125s AL 
                                                                                                                                                               2
                                                                                        0.25 * 0.5s AL 
                                                   AS        DS        E                            2                    AL             DL        E
                                                        m                                                                     m                                                     m
me of a breeding cycle
                                  Relation between variance components if a single family or a
                                  large ideal population is regarded as the population


                                  Variance components in a large ideal population
                                  if considered the population or a meta-population
                                  composed of many small populations (full sib
                                  families.
                                  Population = one            Single full sib family
                                  Single full sib family      as part of a Large
                                                              ideal population
                                  2s ASm
                                     2
                                                              s ALm
                                                                2


                                     2CV Am S                 CV Am L

                                  2s AS
                                     2
                                                              s AL
                                                                2


                                  1.33s DS
                                        2
                                                              s DL
                                                                2




ve not made clear in different published
as entries. That is likely to cause               Variance components in a full
other values on variance components than          sib family if considered a
                                                  population or a segment of a
 he given entries are interpreted as for a        large ideal population
 seriously misleading. In all connections
 s) we have referred to the EXCEL                 Population =      Single full sib
stal clear that it is within family and           one Single full family as part
unlikely that real users of the developed
 is sheet and will make more actions to           sib family        of a Large ideal
                                                                    population
                              Additive
                              variance in goal
                                                  s AmS
                                                    2
                                                                     0.5s AmL
                                                                          2

                              character
                              (mature
                              character)
                              Coefficient of
                              variance for goal
                                                  CV Am S            CVAm L / 2
ve not made clear in different published
as entries. That is likely to cause               Variance components in a full
other values on variance components than          sib family if considered a
                                                  population or a segment of a
 he given entries are interpreted as for a        large ideal population
 seriously misleading. In all connections
 s) we have referred to the EXCEL                 Population =      Single full sib
stal clear that it is within family and           one Single full family as part
unlikely that real users of the developed
 is sheet and will make more actions to           sib family        of a Large ideal
                                                                    population
                              Additive
                              variance in goal
                                                  s AmS
                                                    2
                                                                    0.5s AmL
                                                                         2

                              character
                              (mature
                              character)
                              Coefficient of
                              variance for goal
                                                  CV Am S           CVAm L / 2
                              character

                                                  s AS              0.5s AL
                              Additive              2                    2
                              variance in
                              measured
                              character
                              Dominance
                              variance in         s DS
                                                    2
                                                                    0.75s DL
                                                                          2

                              measure
                              character

                                     Why did we choose to look at a single family?
                    The Swedish breeding program was formulated as a mostly within family selection. It is great
                    complications in mixing between family selection and within family selection, but now when
                    breeders focused on within family selection it seemed justified to do a program only for that. We
                    doubt it would have been possible considering both between and within family selection. We
                    thought that this was initiation of more focus on within family breeding. And it seemed simpler
                    not to complicate the formulas.

                    Dag L had develop a “linear” program, GAINPRED, following steps from plus tree selection in the wild
                    forest over crosses to a seed orchard. For long-term breeding that could be transformed into the breeding
                    cycler, there the idea is to mainly focus on what happens in a single family at a certain time. As breeding
                    progress is almost linear under at least the five first generations (Rosvall, Lindgren ???).
                       not to complicate the formulas.

                       Dag L had develop a “linear” program, GAINPRED, following steps from plus tree selection in the wild
                       forest over crosses to a seed orchard. For long-term breeding that could be transformed into the breeding
                       cycler, there the idea is to mainly focus on what happens in a single family at a certain time. As breeding
                       progress is almost linear under at least the five first generations (Rosvall, Lindgren ???).




r variance components.



population

Simple expressed
         s Am Ls AL
2 0.5s   2
         AL    0.75s DL  s E
                      2      2


         s Am Ls AL
                           sE
                            2
2 0.5s AL  0.75s DL 
       2          2

                            m
                   s Am Ls AL
                  0.375s AL  0.75s DL  s E
                         2          2      2
4 0.125s AL 
         2

                              m
r variance components.



population

Simple expressed
         s Am Ls AL
2 0.5s   2
         AL    0.75s DL  s E
                      2      2


         s Am Ls AL
                           sE
                            2
2 0.5s AL  0.75s DL 
       2          2

                            m
                   s Am Ls AL
                  0.375s AL  0.75s DL  s E
                         2          2      2
4 0.125s AL 
         2

                              m
Dag L 000724, Darius 2001.04.24
An excercise

Hi,
I am going to optimise our breeding program for Big Forest Inc. My boss thinks that it is a good ide
optimising the breeding program assuming that it is completely balanced, thus, each parent get ex
The budget allocated to the breeding program is 10$ per year and member of the breeding popula

The alternatives are phenotype test (alt 1); clone test (alt 2) or progeny test (alt 3). The boss believ
coancestry should be penalised as inbreeding, thus the "weight" ("penalty factor" in cell I12) shall b
All calculations (except diversity loss) are within family. The selection should be in two stages,
phenotypic preselection of a shortlist of candidates (stage 1; alternative 1) and then either a polycr
breeding value of the parents (stage 2; alt 3) or cloning and clonal test (stage 2; alt 2). The task is
program so it returns maximum progress in GMG per year. The choices open are propagation met
versus progeny), initial family size to be selected from, number of the individuals to be preselected
number of individuals to be tested per each genetic entry at stage 2, how long the selection trials g

The boss suggests the following assumptions: a test plant costs 1$ for all testing alternatives, 2$ fo
clone) and 2$ (per parent) for initiation: if it is progeny test it costs 1 $ per eximerental plant, 5$ for
produce the test progeny (by polycross or open pollination). Recombination costs 30$ and takes 3
variance is 1, dominance variance shall make 25% of the additve variance in the breeding populat
environmental variance is 17.5 (resulting in the narrow sence heritability of 0.1). The additive stand
of the final forest value (within a full sib) is 10% of the forest value. Time before selection test is es
(T<p) is 1 year for phenotype selection, 4 years for clonal test and 6 years for progeny test. Time a
for all alternatives (i.e. time needed to obtained the improved seed). Specify lacking parameters (if
for the worksheet.

I have heard that you work with a new spreadsheet which may give the answers. Can you please s
something about the exact design close to optimal based on that spreadsheet?
nc. My boss thinks that it is a good idea to start with
ely balanced, thus, each parent get exactly two kids.
ar and member of the breeding population.

or progeny test (alt 3). The boss believes that group
ght" ("penalty factor" in cell I12) shall be set to 100.
 selection should be in two stages, first a
  alternative 1) and then either a polycross test of the
clonal test (stage 2; alt 2). The task is to optimise the
The choices open are propagation method (clones
ber of the individuals to be preselected at stage 1,
 stage 2, how long the selection trials grow.

osts 1$ for all testing alternatives, 2$ for cloning (per
 costs 1 $ per eximerental plant, 5$ for initiation to
 Recombination costs 30$ and takes 3 years. Additive
 ditve variance in the breeding population and
e heritability of 0.1). The additive standard deviation
 value. Time before selection test is established up
st and 6 years for progeny test. Time after is 2 years
d seed). Specify lacking parameters (if any) needed


ay give the answers. Can you please suggest
 that spreadsheet?
Breeding Cycle Analyser: deterministic simulator for genetic predictions of gain, diversity loss, cost and time of a breeding cycle
      TroubleSolver                   Get started        Aim             An example

 Genetic parameters within full-sib family                                                               Genetic parameters (all offspring)
                               Variance components Standard deviations           CVAm=       10                                            Variance components   Parameter               Value
                           sA2=            1                   1.00   sP2=sA2+sD2+sE2=      18.67                                        sA2=        2                           h2 =    0.10
                           sD 2=         0.167                 0.41                h2=       0.05                                        sD 2=      0.5                          H2 =     0.13
                           sE 2=          17.5                 4.18                H2 =      0.06                                        sE 2=      17.5                     CVAm=       14.1

 General settings and summarizing results                                                                Diversity loss, rotation, BP size                       Final results           Value
                          Choose        Gain (%)                         Cost ("$")       Time (years)                  Rotation age=     60                     Gain/cost              #NAME?
                         alternative       0                                20                3                               BP size     50                     Gain/time              #NAME?
 Progeny test                              3          #NAME?                399             38.9                Group coancestry weight=            100          Group Merit Gain       #NAME?
                                                                                                                              Diversity loss=      0.005         GMG/time               #NAME?
 Overall                                              #NAME?                419              42           Diversity loss in % of "forest value"=    0.5          Cost/time                  10.00

 Parameters                          Not active     Not active        Progeny test                       Constraint=Annual budget for all BP                           500                10
   No. of selected individuals=            1              1                 1
            Full-sib family size=         90             21                21
  Plant no. per family member=                           20                14
           Selection intensity=         #NAME?         #NAME?             #NAME?
            Initiation cost (CIN)=         0              0                 0
      Cost per test plant (CP)=            1              1                 1
      Cost per genotype (CG)=                             2                 5
                    Total cost =          90             462               399
                  Time before =           1               5                 17
                  Testing time=           38              34                22
                     Time after=          0               0                 0
                      Duration=          38.9            38.9              38.9
                          a1=sA=          1.00            1.00              0.50
                              a2 =        4.32            1.43              1.25
                     rAI=a1/a2=           0.23            0.70              0.40
       J-M correlation function            0               2                 2
                                        #NAME?         #NAME?             #NAME?
         rY (or R)=   #NAME?   #NAME?   #NAME?
Genetic gain (%)=     #NAME?   #NAME?   #NAME?
me of a breeding cycle
Dag L 000724, Darius 2001.04.24
An excercise

Hi,
I am going to optimise our breeding program for Big Forest Inc. My boss thinks that it is a good ide
optimising the breeding program assuming that it is completely balanced, thus, each parent get ex
The budget allocated to the breeding program is 10$ per year and member of the breeding popula

The alternatives are phenotype test (alt 1); clone test (alt 2) or progeny test (alt 3). The boss believ
coancestry should be penalised as inbreeding, thus the "weight" ("penalty factor" in cell I12) shall b
All calculations (except diversity loss) are within family. The selection should be in two stages,
phenotypic preselection of a shortlist of candidates (stage 1; alternative 1) and then either a polycr
breeding value of the parents (stage 2; alt 3) or cloning and clonal test (stage 2; alt 2). The task is
program so it returns maximum progress in GMG per year. The choices open are propagation met
versus progeny), initial family size to be selected from, number of the individuals to be preselected
number of individuals to be tested per each genetic entry at stage 2, how long the selection trials g

The boss suggests the following assumptions: a test plant costs 1$ for all testing alternatives, 2$ fo
clone) and 2$ (per parent) for initiation: if it is progeny test it costs 1 $ per eximerental plant, 5$ for
produce the test progeny (by polycross or open pollination). Recombination costs 30$ and takes 3
variance is 1, dominance variance shall make 25% of the additve variance in the breeding populat
environmental variance is 17.5 (resulting in the narrow sence heritability of 0.1). The additive stand
of the final forest value (within a full sib) is 10% of the forest value. Time before selection test is es
(T<p) is 1 year for phenotype selection, 4 years for clonal test and 6 years for progeny test. Time a
nc. My boss thinks that it is a good idea to start with
ely balanced, thus, each parent get exactly two kids.
ar and member of the breeding population.

or progeny test (alt 3). The boss believes that group
ght" ("penalty factor" in cell I12) shall be set to 100.
 selection should be in two stages, first a
  alternative 1) and then either a polycross test of the
clonal test (stage 2; alt 2). The task is to optimise the
The choices open are propagation method (clones
ber of the individuals to be preselected at stage 1,
 stage 2, how long the selection trials grow.

osts 1$ for all testing alternatives, 2$ for cloning (per
 costs 1 $ per eximerental plant, 5$ for initiation to
Recombination costs 30$ and takes 3 years. Additive
ditve variance in the breeding population and
e heritability of 0.1). The additive standard deviation
 value. Time before selection test is established up
st and 6 years for progeny test. Time after is 2 years
Breeding Cycle Analyser: deterministic simulator for genetic predictions of gain, diversity loss, cost and time of a breeding cycle
             TroubleSolver                   Get started            Aim            Breeding plan   Assumptions          Initial settings                Results               Costs                Timing

Genetic parameters within full-sib family                                                                        Diversity loss, rotation, BP size
                  Variance components Standard deviations             CVAm=             10                                                                             Parameter               Value
sA2=                  1                               1.00   sP2=sA2+sD2+sE2=         18.67                                                    Weighting factor for group coancestry=              100
sD 2=               0.167                             0.41                 h2=         0.05                                      Rotation age=            60               Diversity loss=            0.005
sE 2=                17.5                             4.18                 H2 =        0.06                                          TheBP size the diversity loss in % of "forestry value"=
                                                                                                                                         value of        50                                              0.5

        General settings and summarizing results                                                             Final results                             Value
                                Choose the    Gain (%)                              Cost ("$")               Gain/cost
                                                                                                   Time (years)                                       #NAME?
        Recombination           alternative      0                                     20              3     Gain/time                                #NAME?
        Stage 1.Phenotype            1        #NAME?                                  100             17     Group Merit Gain                         #NAME?
        Stage 2.Clone                             2             #NAME?                  51             15        GMG/time                             #NAME?
        Overall                                                 #NAME?                 171             35        Cost/time                                4.89

Selection stage 1                                                                                                Selection stage 2
        Parameters                                       C
                                            Phenotype test lone test              Progeny test                   Parameters                                     Clone test
                                                                                                                                                    Phenotype test                             Progeny test
          No. of selected individuals=           3             8                        8                          No. of selected individuals=          1           1                              1
                          Family size=          100           10                        9                                          Family size=          3           3                              3
         Plant no. per family member=                         20                       14                         Plant no. per family member=                       15                            18
                  Selection intensity=         #NAME?             #NAME?             #NAME?                              Selection intensity=          #NAME?               #NAME?                #NAME?
                   Initiation cost (CIN)=         0                 0                   0                                  Initiation cost (CIN)=          0                    0                   0
             Cost per test plant (CP)=            1                 1                   1                             Cost per test plant (CP)=            1                    1                   1
             Cost per genotype (CG)=                                2                   10                            Cost per genotype (CG)=                                   2                   5
                           Total cost =          100               220                 216                                         Total cost =            3                    51                  69
                         Time before =            1                 5                   5                                        Time before =             0                    1                   2
                         Testing time=            16                12                  20                                        Testing time=            2                    12                  10
                            Time after=           0                 2                   0                                           Time after=            2                    2                   0
                             Duration=            17                19                  25                                           Duration=             4                    15                  12
                                 a1=sA=          1.00               1.00               0.50                                                  a1 =         0.82                 0.82                 0.41
                                     a2 =        4.32               1.43               1.25                                                  a2 =         4.28                 1.39                 1.09
                            rAI=a1/a2=           0.23               0.70               0.40                                          rAI=a1/a2=           0.19                 0.59                 0.37
Reduction of initial sA2=   #NAME?   #NAME?   #NAME?
J-M correlation function      0        0        0      J-M correlation function     4        0        0
                            #NAME?   #NAME?   #NAME?                              #NAME?   #NAME?   #NAME?
             rY (or R)=     #NAME?   #NAME?   #NAME?                rY (or R)=    #NAME?   #NAME?   #NAME?
Genetic gain (%)=           #NAME?   #NAME?   #NAME?   Genetic gain (%)=          #NAME?   #NAME?   #NAME?
d time of a breeding cycle                                       Meaning of cell font and colour:                                 Ideas Excel sheets on the
                                                                                                                    TBT-Web. General aboutfor development Tree Breeding Tools WEB site

               How to run   Correlation    Literature   About        Input          Results   Results   Last update of this sheet Darius 04-03-10
                                                                                                                     Changes - recent

                            Genetic parameters in breeding stock                                                    Solver        Scenario to Solver
                                                                    Parameter
                                Variance components Standard deviations              Value
                                   sA2=        2         1.4                 h2 =    0.10
                                   sD 2=     0.5         0.7              H2 =        0.13
                                   sE 2=     17.5        4.2           CVAm=         14.1




Progeny test
Dag L 000724, Darius 2001.04.24
An excercise

Hi,
I am going to optimise our breeding program for Big Forest Inc. My boss thinks that it is a good idea to start with
optimising the breeding program assuming that it is completely balanced, thus, each parent get exactly two kids.
The budget allocated to the breeding program is 10$ per year and member of the breeding population.

The alternatives are phenotype test (alt 1); clone test (alt 2) or progeny test (alt 3). The boss believes that group
coancestry should be penalised as inbreeding, thus the "weight" ("penalty factor" in cell I12) shall be set to 100.
All calculations (except diversity loss) are within family. The selection should be in two stages, first a
phenotypic preselection of a shortlist of candidates (stage 1; alternative 1) and then either a polycross test of the
breeding value of the parents (stage 2; alt 3) or cloning and clonal test (stage 2; alt 2). The task is to optimise the
program so it returns maximum progress in GMG per year. The choices open are propagation method (clones
versus progeny), initial family size to be selected from, number of the individuals to be preselected at stage 1,
number of individuals to be tested per each genetic entry at stage 2, how long the selection trials grow.

The boss suggests the following assumptions: a test plant costs 1$ for all testing alternatives, 2$ for cloning (per
clone) and 2$ (per parent) for initiation: if it is progeny test it costs 1 $ per eximerental plant, 5$ for initiation to
produce the test progeny (by polycross or open pollination). Recombination costs 30$ and takes 3 years. Additive
variance is 1, dominance variance shall make 25% of the additve variance in the breeding population and
environmental variance is 17.5 (resulting in the narrow sence heritability of 0.1). The additive standard deviation
of the final forest value (within a full sib) is 10% of the forest value. Time before selection test is established up
(T<p) is 1 year for phenotype selection, 4 years for clonal test and 6 years for progeny test. Time after is 2 years
for all alternatives (i.e. time needed to obtained the improved seed). Specify lacking parameters (if any) needed
for the worksheet.
Comparison of functions for Juvenile-Mature genetic correla
Try different J-M genetic correlation functions. Use the one you believe best fit your situation
  How to use        Aim       Advice Acknowledgements         This sheet was created by Darius Danusevicius and last edited by

                                                             1.0
Function                rj-m
Lambeth (1980)        #NAME?                                 0.8
Lambeth (2001)        #NAME?




                                           J-M correlation
Gwaze (2000)          #NAME?
Jansson (2003)        #NAME?
Jansson (2003)        #NAME?
                                                             0.6
Custom                #NAME?
    Choose Q=            0.1
                                                             0.4
                                                                                                            Lambeth (1980)
Custom design box.                                                                                          Lambeth (2001)
         Q               rj-m                                0.2                                            Gwaze (2000)
        0.0              0.10
        0.1              0.40                                                                               Custom
        0.5              0.85                                0.0
        1.0              1.00
                                                                   0.0   0.1   0.2    0.3     0.4     0.5     0.6    0.7
                                                                               Ratio selection/rotation age (Q)
Box for plot (illustrates the difference between the functions)
 Ratio selection J-M correlation coefficient (rj-m)
  over rotation   Lambeth Lambeth           Gwaze                        Jansson   Jansson   Custom
       0.0        #NAME? #NAME? #NAME?                                   #NAME?    #NAME?    #NAME?    Juvenile - mature correlation
       0.1        #NAME? #NAME? #NAME?                                   #NAME?    #NAME?    #NAME?    The common approach is to adjust a fi
       0.2        #NAME? #NAME? #NAME?                                   #NAME?    #NAME?    #NAME?    type rj-m = a + b ln(ageearly/agelate). The
       0.3        #NAME? #NAME? #NAME?                                   #NAME?    #NAME?    #NAME?    (logarithm of age ratio). There is a diff
       0.4        #NAME? #NAME? #NAME?                                   #NAME?    #NAME?    #NAME?    if it is natural or 10-logarithm. That am
       0.5        #NAME? #NAME? #NAME?                                   #NAME?    #NAME?    #NAME?    returns 10log for log in a cell, but the n
       0.6        #NAME? #NAME? #NAME?                                   #NAME?    #NAME?    #NAME?
                                                                                                       LAR always turn large negative at low
       0.7        #NAME? #NAME? #NAME?                                   #NAME?    #NAME?    #NAME?
       0.8        #NAME? #NAME? #NAME?                                   #NAME?    #NAME?    #NAME?
                                                                                                       at low ages (which may concern early
       0.9        #NAME? #NAME? #NAME?                                   #NAME?    #NAME?    #NAME?    Therefore it is suggested to find a bette
       1.0        #NAME? #NAME? #NAME?                                   #NAME?    #NAME?    #NAME?
ature genetic correlations
d by Darius Danusevicius and last edited by Dag Lindgren 04-03-11




     Lambeth (1980)
     Lambeth (2001)
     Gwaze (2000)                               A future perspective
                                             Time machine    You keep measuring,
                                                             while I will jump to
                                                               the future to see
                                                             which trees to select
               0.7     0.8     0.9     1.0
otation age (Q)



  Juvenile - mature correlation
  The common approach is to adjust a fit a function to data. The function is usually of
  type rj-m = a + b ln(ageearly/agelate). The later expression could be called LAR
  (logarithm of age ratio). There is a difficulty with the logarithm because it is not clear
  if it is natural or 10-logarithm. That ambiguity is not improved by EXCEL, which
  returns 10log for log in a cell, but the natural in Visual Basic. A worse problem is that
  LAR always turn large negative at low ages, thus behaves contrary to common sense
  at low ages (which may concern early selection ages actually used in practice).
  Therefore it is suggested to find a better formulation than LAR.
One stage testing strategies

                                                                                          Alt. 2. Selection based on clonal test "C
  Alt 1. Selection based on phenotype "Phenotype strategy" (PH).
                                                                                          among N parents (double-pair mating), f
  Crossings are made among N (for the Swedish Norway spruce program
                                                                                          The best clone in each family is selected
  N=50) parents (double-pair mating), full sib families are tested, one
  individual per family (=1 per parent) is selected
                                                                                         Breeding population member (parent).

Breeding population
member (Parent), N= 50 in
Swedish breeding program
                               P1                    P2                      P3 (…) PN   Produce progeny from each full sib fam.
for N. spruce                                                                            From n family members m ramets are
                                                                                         produced.
Produce n progeny per full
sib fam for testing (n to be                    (…n)                  (…n)               m ramets are produced from each clone
chosen)                                                                                  (N*n) for testing. The best clone in each
                                                                                         family is selected based on performance
                               Select 1 individual     Select 1 individual               of the ramets in the clone test.
                               per full sib family     per full sib family




                               Two-stage (stagewise) testing strategy
                           Two-stage (stagewise) testing strategy
Stage 1. Selection based on phenotype (PH). Crossings are made among 50 parents (double-pair
mating), full sib families are tested, n individuals per family (=1 per parent) are pre-selected
 1. Breeding population member
 (parent), N= 50 in Swedish breeding
                                              P1                     P2                     P3 (…) P50
 program for N. spruce


 2. Produce n candidates per full-sib                            (…n)                     (…n)
 family for testing (n is optional)

                                                     3. Preselect n1        3. Preselect n1
                                                   individuals per full   individuals per full
                                                        sib family             sib family



Stage 2. Re-selection based on clonal test (CL). The n1 individuals preselected in each of the 50 full-
sib families in stage 1 are cloned and tested aiming at selection of one best clone in each family based
on performance of the cuttings.

Produce m cuttings from each
preselected individual (thereby becoming                                                         (…) N=50
an ortet) and reselect 1 best ortet in each
family based on performance of the
ramets.
(…m) (…m) (…m)   (…m) (…m) (…m)
n based on clonal test "Clone strategy" (CL). Crossings are made                Alt. 3. Selection based on progeny testing. "Progeny
ts (double-pair mating), full sib families are cloned and clones are tested.    among N individuals (double-pair mating); full sib offs
 n each family is selected based on performance of the cuttings.                individuals are progeny-tested (polycross or OP). One
                                                                                the next breeding generation cycle based on the perform
                       P1            P2             P3 (…) PN                  Breeding population member (parent), N= 50 in
                                                                               Swedish breeding program for N. spruce          P1
m each full sib fam.
 rs m ramets are                                                               Test n progeny (n is up to you) per full sib
                                  (…n)             (…n)              (…n)      fam by progeny testing (it may be a good idea
                                                                               to apply some early flower stimulation)

d from each clone
 best clone in each                                                            m seedlings (m is up to you) are produced
 d on performance                                                              from each parent for further testing. Select
                                                                               one parent based on performance of the
                       (…m) (…m) (…m) (…m) (…m) (…m) (…m) (…m) (…m)            progeny                                         (…m) (…m) (…m)
he 50 full-
 ly based
ogeny testing. "Progeny strategy" (PR). Crossings are made
pair mating); full sib offspring is produced; in each family n
 (polycross or OP). One parent from each cross is selected for
ycle based on the performance of the half-sib progeny.

     P1                P2                P3 (…) PN

                   (…n)              (…n)              (…n)




      (…m) (…m) (…m)    (…m) (…m) (…m)    (…m) (…m) (…m)
                                                Time and cost components
   Alternative 1. Phenotype test                                                         Established in 1
                                                                                         year after seed           Establishment,
                                        Production in                                    harvest                   maintenance and
                                       nursery (1 year)
                                                                                                                   assessments


         Crossing                       Nursery                     Transportation                              Field trial
   Recombination cost=50,
                                                          Plant dependent cost=1 (per seedling)
        time 4 years


  Alternative 2. Clonal test.                                                                           Established in 5
                                                Cutting of
                              Production of      ramets                                                 years after seed             Establishment,
                                                                                                        harvest
                             ortets (4 years)                   Rooting of ramets                                                    maintenance and
                                                                     (1 year)                                                        assessments


      Crossing                                    Nursery                            Transportation                          Field trial
 Recombination cost=20,
                            Genotype dependent                              Plant dependent cost=1 (per ramet)
      time 4 years
                             cost=2 (per ortet)


   Alternative 3. Progeny test.                                                                                        Established in
                                Production of                                Seed collection,
                                                      Pollination                                                      17 years after        Establishment,
                               female parents                                  extraction
                                                                                                                       seed harvest
                                 (15 years)                      Pollen         (1 year)                                                     maintenance and
                                                                                          Production of F1                                   assessments
                                                                                          seedlings (1 year)



        Crossing                                                 Nursery                                    Transportation              Field trial
   Recombination cost=20,       Genotype dependent cost=5 (per female parent)
                                                                                             Plant dependent cost=1 (per seedling)
        time 4 years

Recombination time                                                        Time before                                        Testing time              Time after
           Time components of complete breeding cyc
          Stage 2. Time after: from selection                    Recombination time:
            of new individuals to start of their            pollen collection, crossing, seed
          flowering (crosses for the next cycle                      maturation…
               can be made). If the selected
           individuals flower, Time after = 0.

                                                   TAfter             TRecomb
                                                                                                  Stage 1. Time be
                                                                                  TBefore          test plant produc


            Stage 2. Testing:                          Breeding                                       Stage 1. Test
          from establishment to
                selection               Ttest           cycle                         Ttest         from establishm
                                                                                                          selection




                                                                                      Stage 1. Time after: from se
                                                        TBefore         TAfter         of candidates to their reprodu
                           Stage 2. Time before:                                         maturity. If the candidates
                            test plant production                                    reproductively mature, Time af
e after
test plant production   reproductively mature, Time af
ing cycle




 tage 1. Time before:
test plant production



  Stage 1. Testing:
from establishment to
      selection




e after: from selection
s to their reproductive
 f the candidates are
  mature, Time after = 0.
Table of contents
Welcome and introduction
Assumptions
Importance of gene diversity
Cost considerations and timing
Group Merit Gain is the target parameter
Gain prediction
Juvenile-mature genetic correlations
Selection intensity
Solver to find maximum Group Merit Gain
Trouble and Training
Acknowledgement and Literature


 Welcome to Breeding Cycle Analyser
 This deterministic simulator shows the effect of different parameters and choices on the
 annual improvement of the breeding population in long-term breeding. The simulator
 considers a complete cycle of within-family (balanced) selection (in one or two stages). The
 simulator can be used as a support to optimise breeding strategy by considering the effect
 of genetic parameters, loss of diversity, cost and time components on benefit per cycle and
 efficiency of different testing alternatives.

 Introduction
 Given the breeding plan based on recurrent cycles, balanced within -family selection and equal parent
 contribution, this simulator is aimed to assess the effect of variable genetic parameters, time and cost
 components as well as to interactively compare the following 3 testing alternatives for stagewise selection of
 one individual within each full-sib family to be used as the parent for the following breeding cycle: (1)
 phenotypic selection, i.e. selection of individual based on its phenotype, (2) clonal test, i.e. selection of
 individual based on performance of vegetativelly propagated cuttings, (3) progeny test, i.e. selection of
 individual based on performance of half-sib progeny (see the sheet "Selection alternatives" in this workbook).
 Thus, the simulator aims (1) at finding the optimum combination of gain, diversity, cost and time (i.e. which of
 the 3 testing alternatives would give the highest value of annual group merit gain (GMG/time) at a given annual
 budget) and (2) to test on whether it is beneficial to have two stages of selection within a breeding cycle.
 Consequently, GMG/time is the parameter to be maximised. In this way many features of the design of
 hypothetical breeding programs can be investigated by considering genetic variation, logistics, cost
 components, time components simultaneously.

 It can be seen as a deterministic simulator of breeding strategies. Like a flight simulator responds to rudder
 input, the genetic response of this simulator changes with the design of a breeding operation. The simulator
 considers values of genetic variables, which may not be directly under the breeders control, the cost
 components, as well as the timing of the operation. It also recognises the effect of the breeders intentional
 decisions. The breeder controls the mating system, and can choose if testing is made by vegetative
 propagation or progeny testing or if just phenotypic selection is used. The breeder controls the number of
 parents, the number and size of families, the size of clones and the number of selections (selection intensity
 may be derived from the later). Some things, like the interactive figure, can be moved on the sheet.




Assumptions
No considerations about genotype by environment (GxE) interaction or number of test localities or range to use
the material has been made. The predicting formulas refer to the case where there is no marked diffrence
between the test environment and the range of envoronmets for which the improved material is intended. The

                                                  Page 48
between the test environment and the range of envoronmets for which the improved material is intended. The
effect of GxE interaction if one goes from the parameters of a single site to real forestry over an area could be
considered by setting J-M genetic correlation to a lower value by assigning R (the correction from trials to
practice) a smaller value than 1.
Gain is expressed in a single character (although that character may be a combined index including many
characters). No "after-effects" has been considered, this probably considerably overestimates the gain achieved
by progeny-test based on wind-pollination. Cloning has been assumed not to change the performance or induce
an extra variation, this is probably not true, but it would be difficult to express in formulas or figures.
Note that in a real seed orchard will inbreeding occur. This has not been considered. In a real seed orchard gain
can be boosted by using more ramets of better clones, this is not considered.




Importance of gene diversity
Selection has an effect effects not only on immediate gain, but also on diversity. Gene diversity is required from
seed orchards, but is also useful for long-term breeding where it can be reversed to gain. Gene diversity in the
breeding population allows a harder selection to a seed orchard of given gene diversity, and thus a higher
genetic gain in the production population. Selection tends to make trees more related. These effects are
quantified as group coancestry or status number according to Lindgren et al (1996). It is assumed that the parent
population consists of unrelated trees not affected by inbreeding.


Cost considerations
The cost of the selection alternatives is given as a function of certain cost components. The cost components
can be changed by the user. This makes it possible to compare different options at the same annual budget and
to consider that the testing alternatives may be differently costly. Here costs are seen as dependent mainly on
the number of parents and plants. More complex models can be developed.

Timing
For each alternative, the breeder can input how long time is needed until selection tests can be established and
from measurement of the test until the gain is obtained (e.g. until seed orchard establishment or the first cone
crop in an orchard). When gain is expressed per unit of time (time to get the gain), a fair comparison between the
selection alternatives can be made. The reason that gain increase by time is that the experiments are growing,
and results become more reliable when based on older trials (governed by J-M genetic correlation in gain
prediction formula).


Group Merit advance per year is the target parameter to be maximised
The genetic progress is best measured as Group Merit advance per year over the breeding cycle, which
considers both genetic gain and build-up of relatedness (Wei & Lindgren 2002). Group Merit Gain (GMG) is
estimated as:

GMG = G - cQ,

where, G is additive genetic gain at rotation age, c is a weighting factor between loss of genetic diversity and
genetic gain and also converting gain and diversity to the same scale, Q is raise in group coancestry per
breeding cycle, which, assuming that each parent contributes two offspring to be used as the parents in the next
breeding cycle), was estimated as:

Q = (group coancestry in the selected population) - (group coancestry in the parental population) = (after some
simplification) 0.75/n (selected) - 0.5 /n (parental) = 0.25 /n; where, n is number of parents, which under
balanced selection is the same as number of individuals selected.



Gain prediction
The genetic gain at rotation age from within family selection following each of the selection alternatives is
estimated according to the following formulas (Lindgren and Werner 1989). The breeding value of the founding
parents (the selected plus trees) are usually set to 0. The indici on G refer to the selection alternatives:
"Phenotype"- for selection based on phenotype, "Clone"- for selection based on clonal test and "Progeny"- for
selection based on progeny test. The formulas for gain calculcultation used in this EXCEL book and in the
papers derived usually see the population as a single family when setting the symbols.




                                                  Page 49
                   s Am rj m is A                                            There is a factor which reduces gain, it was considered in the
GPhenotype                                                                   equations by Lindgren and Werner (1989) as a multiplier to
                   s A s D s E
                     2    2    2                                              gain. The sites are not representative for future forestry, they
                                                                              are only a sample and the environments the test environments
                                                                              were drawn from can neither be seen as representing future
                                                                              forestry. The factor can be considered as absorbed in CVAm if
             s     Am   r jm i s                                             that is given a somewhat smaller value than else.
G Clone 
                                          A

                                     sE
                                      2
               s   2
                   A    s   2
                             D   
                                      m
                             s       Am       r j  m i 0 . 5s   A
G Progeny 
                                          0 . 75 s     2
                                                           s        2
                                                                         s   2
                    0 . 25 s     2
                                 A                    A             D        E

                                                           m

Juvenile-mature genetic correlations

You can choose one of the following functions to estimate the J-M genetic correlations:

Function by Lambeth (1980):
rj-m=1.02 + 0.308 * Log(Q), with an adjustment to linearise the function where Q is close to 0 or 1. (Q is ratio of
selection age over rotation age). The VBA code is rg( ). Developed based on phenotypic juvenile-mature
correlation coefficients from numerous trials with temperate and boreal conifer species. It is a most conservative
of the J-M correlation functions available in this simulator (returning lowest correlation coefficients).

Function by Lambeth (2001):
rj-m= 1.02 - 0.098 * (Log(Q))2, with an adjustment to linearise the function where Q is close to 0 and to
constraint it to return no values greater than 1 (Q is ratio of selection age over rotation age). VBA code is rg1( ).
Developed based on juvenile-mature genetic correlation coefficients estimated in 4 series with a total of 15 trials
of Pinus taeda (296 families in total). The function was constructed by pooling "within-series" functions.

Function by Gwaze et al. (2000):
rj-m= 1.03 + 0.215 * LOG(Q). Q is ratio of selection age over rotation age. VBA code is rg2( ). The function was
developed based on juvenile-mature genetic correlation coefficients estimated in 19 trials with a total of 190
families of Pinus taeda from western USA. From the other functions available in Gwaze et al. (2000) study, we
have chosen this function as it provides rj-m values intermediate to Lambeths 1980 and 2001 rj-m functions.

Custom function:
This function is fully dependent on what values of juvenile - mature correlation coefficients you will set for certain
thresholds of ratio selection age over rotation age (based on own experience) in the "Custom box" of "J-M
correlation" worksheet in this workbook. So before referring to it, you should set own J-M genetic correlation
values in "J-M correlation" worksheet.
The age ration (Q) thresholds are: (Q): Q=0, Q=0.1 (10% of rotation age), Q=0.5 (50% of rotation age). Thus,
your setting will affect the trend of rj-m. The following function is used to calculate rj-m (VBA code is rg3 ()):
   If Q < 0.1 And Q >= 0 Then rg1 = a1 + 10 * (a2 - a1) * Q
   If Q < 0.5 And Q >= 0.1 Then rg1 = a2 + 2.5 * (a3 - a2) * (Q - 0.1)
   If Q <= 1 And Q >= 0.5 Then rg1 = a3 + 2 * (1 - a3) * (Q - 0.5)

You can observe the difference between the optional functions in a "J-M correlation" worksheet in this workbook.




Selection intensity
The selection intensity is obtained by calling functions sel(x) (for infinite populations) or SelBurr(j,n) (Burrow's
correction for finite case). These functions exist in Visual Basic Macro modules connected to this workbook and
can be inspected from the Visual Basic Editor. More detail can be found in Lindgren & Nilsson (1985) and
Lindgren & Bondeson (1990). These works could also be consulted when very high accuracy is required.

                                                                 Page 50
can be inspected from the Visual Basic Editor. More detail can be found in Lindgren & Nilsson (1985) and
Lindgren & Bondeson (1990). These works could also be consulted when very high accuracy is required.



You can use "Solver" tool in Excel for optimisation
If you are looking for an optimally efficient solution you can use Solver in the relevant worksheet. Solver can
determine the maximum or minimum value of one specific cell, which can be obtained by changing other specific
cells. In this way, you can maximise Group Merit Gain per year by allowing some values vary while keeping the
other values constant. As a constraint, you may place cost your annual budget. Solver will find the values of all
adjustable cells (e.g. progeny size), which e.g. maximises GMG/Y under your restrictions. To get a better
understanding of this technique you may look at EXCEL Help or try the examples which exist Dag Lindgren's
website. When you run Solver, you can try to place following values into "Options": Precision: 0.00000001;
Tolerance: 0.002; Convergence: 0.0001, but we are uncertain on how to handle this.



Trouble?
Note that it is possible to make invalid input, usually the worksheet will protest by not working properly, which
may cause confusion. You may be in the wrong place of the workbook or worksheet. The formatting may be
wrong. Information about trouble would be gladly received (if your copy is recently collected from the web) by
Dag.Lindgren@genfys.slu.se. There may by mistakes, if you detect one, you make us a big service if you report
it to us.



Training
Some exersizes which may help you understand how this workbook functions can be found in another worksheet
here.



Acknowledgement
We are indebted to a number of people who have commented on former versions of this workbook...




Literature
Danusevicius D & Lindgren D 2002. Efficiency of Selection Based on Phenotype, Clone and Progeny Testing in
Long-term Breeding. Silvae Genetica 51:19-26.
Danusevicius D & Lindgren D 2002. Two-stage selection strategies in tree breeding considering gain, diversity,
time and cost. Forest Genetics. 9:143-157.
Lindgren D and Werner M (1989) Gain generating efficiency of different Norway spruce seed orchard designs.
Includes an appendix by Öje Danell. In Stener L-G and Werner M (editors) Norway spruce: provenances,
breeding and genetic conservation. Institutet for skogsforbättring. Rapport 11:189-206.
Lindgren D, Gea L, & Jefferson P 1996. Loss of genetic diversity monitored by status number. Silvae Genetica,
45:52-59.
Lindgren D, Wei R-P and Lee S. 1997. Optimum family number in the first cycle of a breeding program. For. Sci.
43(2): 206-212.
Lstiburek M 2000. Theoretical analyses of the possible benefit of vegetative propagation for quality birch.
Student thesis at department of forest genetics and plant physiology, SLU.
Wei, R.-P. & Lindgren, D 2001. Optimum breeding generation interval considering build-up of relatedness.
Canadian Journal of Forest Research 31:722-729.
Danusevičius, D. and Lindgren, D. 2003. Clonal testing may be the best approach to long-term breeding of
Eucalyptus. In:Eucalyptus Plantations – Research, Management and Development, R.-P. Wei and D. Xu (eds),.
World Scientific, Singapore, 192-210.
Danusevičius D & Lindgren D 2004. Progeny testing preceded by phenotypic pre-selection - timing
considerations. Submitted.
Danusevičius D & Lindgren D 2004.Optimisation of breeding population size for long-term breeding. submitted
Feb 04
Li H Lindgren D, Danusevicius D, Cui J 2004 Theoretical analyses of long-term poplar breeding. Submitted.




                                                  Page 51
                       Designations                                                 This sheet was last edited by DagL 05-03-15
                       Variance components and genetic parameters
                       sA2 is additive variance
                       sD2 is dominance variance
                       sE2 is environmental variance
                       sP2 is phenotypic variance
                       h2 is individual heritability in narrow sense
                       H2 is individual heritability in wide sense
                       sAm is the "true" additive CV at mature age (in the character we want improved!)
                       rAI is the correlation between selection index and breeding value at the current age
                       rY is juvenile-mature genetic correlation
                       R is a correction from trials to practice, which use different sites, different years, different values and different weights
                       Q is ratio of age of selection over rotation age
                       a1,a2 are intermediate values in the gain formulas given up to the right                      Timing components
                       a2 is sqrt of variance of measured values (phenotypes, index-values)                          T
                       Numbers and cost components                                                                TBefore
                       n is full-sib family size, for a given selection alternative                               Tafter
                       m is number of test plants (seedlings or cuttings) per individual                          Ttest
                       i is selection intensity                                                                   TRecomb
                       Cost components
                       CIN=Cost for initiating the experiment (and for recombination)
                       CP=Cost per added experimental plant (plant-dependent cost)
                       CG=Cost per added genotype (parent) (genotype-dependent cost)


                         Sheets
                         This workbook, EXCEL BREEDINGCYCLE2004.XLS, contains the following worksheets:
                         Simple                     An older version with only one stage, the two stage may be better
                         Two stages                  This is considered the main sheet. Selection is in two stages, e.g. First
                         phenotype when progenyt-test.
                         Explanations (the current) The operations and symbols are explained
                         Time&Cost                   Figure showing the breeding cycle and the three basic strategies
                         TestStrat                   Figure showing successive selection in stages
                         J-M                        Juvenile Mature Correlation Functions Used (alternative functions are
e) at a given annual                  given in this workbook, in earlier workbooks the only alternative was only Lambeth 1981
                         Excersize gives examples
                         PhenotypeProgeny             Two stage specially made for phenotypic preselection followed by progeny
                         testing for Scots pine in Sweden.


                         "Simple" is used for the study Danusevicius D & Lindgren D 2002. Efficiency of Selection Based on
                         Phenotype, Clone and Progeny Testing in Long-term Breeding. Silvae Genetica 51:19-26.
                         "Two stages" is used for the study Danusevicius D & Lindgren D 2002. Two-stage selection strategies in
                         tree breeding considering gain, diversity, time and cost. Forest Genetics. 9:143-157.




                                                                                          Page 52
 the gain achieved
ormance or induce

 seed orchard gain




med that the parent




annual budget and




parison between the




parents in the next




                      Progeny test within family

                                       0.25s Aw  0.5s As  s D  s E
                                              2         2     2     2
                      s 2  0.25s Aw 
                                  2
                        X
                                                     m
                      The criteria used as selection criterium is the means of m approximate halfsibs.


                      The relationship between the progeny and the mother tree is 0.5, the gain formulat will
                      be
                                                          Page 53
                                                                    s Am rj  m i 0.5s A w
                                             GProgeny 
                                                                            2                2   2   2
                                                                 m
                                  The criteria used as selection criterium is the means of m approximate halfsibs.


                                  The relationship between the progeny and the mother tree is 0.5, the gain formulat will
 ain, it was considered in the    be
er (1989) as a multiplier to
ative for future forestry, they
nments the test environments                                                        s Am rj  m i 0.5s A w
seen as representing future                              GProgeny 
                                                                                    0.25s Aw  0.50s As  s D  s E
                                                                                          2          2      2     2
dered as absorbed in CVAm if
                                                                      0.25s A w 
                                                                            2

                                                                                                  m




u will set for certain




                                                                      Page 54
nging other specific




 another worksheet




g program. For. Sci.




ei and D. Xu (eds),.




                       Page 55
                              An alternative way of formulating some relationships
                              According to Danell 1989 (appendix to Lindgren and Werner 1989) gain can be expressed as a product of
                              rTI (correlation between True breeding value and an Index predicting breeding value). Under simple
                              conditions both rTI and the corresponding regression coefficient b and can be expressed in terms of the
                              covariance between the phenotype and the breeding value divided by the variance for the phenotype. Part
                              of the simplification is that if the measurement and the goal characters are the same, i.e. the goal is at the
           a3=sqrt(0.5sA2(1+1/m)+0.25sD2+0.75sD2/m)= the=sqrt(0.5sA2(1+1/m)+0.25sD2+(0.75sD2+sE2/n)/m)= between families predicting breeding
                              measurment age, then a2 symbol rAI may be preferred (correlation between an Index
                              value and Breeding Value for that Index)
                      a3=sqrt[(0.5sA2+0.75sD2)(1-1/m)]=             a2=sqrt((0.5sA2+0.75sD2)(1-1/m)+sE2/n)= within
                              a1 generally corresponds to the covariance between measurment and breeding value, and a2 generally
                              are intermediate values in the the formulas values. to the right
              a1,a2,b1,b2 corresponds to the variance of gainmeasuredgiven up(They can be seen as the root of these expressions).
              a2 is sqrt of variance of measured values (phenotypes, index-values) rAI.
                              Expressions of type a1/a2 generally should correspond to
              a1 is the covariance between observed values (phenotypes) and true values
                                                           s XA
                                                   a1                                bs XA   s XAs XA / s X
                  ir             s                                              sI                            s
                                                                                                           2
           G                 AI          A
                                                           s A             rAI                             XA
                                                   a2     s X                   sA   sA          sA          s Xs A




g. First




                                                      Page 56
Page 57
Page 58
Page 59
ain can be expressed as a product of
breeding value). Under simple
d can be expressed in terms of the
y the variance for the phenotype. Part
 s are the same, i.e. the goal is at the
between an Index predicting breeding


d breeding value, and a2 generally
een as the root of these expressions).




      s XAs XA / s X
                   2
                       s      a
A
                     XA  1
          sA          s X s A a2




                                           Page 60
Try to solve these exercises to learn the programme. Our suggestions on how to gear the programme
exercise A1. The solutions are given at the bottom of this box (to view the solutions, you should mark the text belo
the text colour to black).
----------------------------------------------------------------------------------------------------------------------------- ---
Dag L 000724
Exercise 1
Hi,
I am going to optimise our breeding program for Dream Forest Inc. My boss thinks that it is a good idea to start with optimis
completely balanced, i.e. each parent get exactly two kids (each parent takes part in two crosses, i.e. double
breeding is cycling by double-pair mating and to select one individual within each full-sib family as a parent in next breeding
is 10$ per year and member of the breeding population (founder = cross).

My boss wants me to find out which of the following three alternatives for selection of one genotype per full
Merit Gain per year (GMG/T): (1) phenotypic selection of individuals, (2) clonal test, (3) progeny test
alternatives"), i.e. my choice may be phenotypic selection, versus more accurate breeding value estimates by clonal test or
any harm of rise in coancestry (he thinks that is taken care of by balanced within family selection), thus, the weighting fac
choices open are number of individuals to be tested (family size in phenotypic selection (alternative 1), ortet number and nu
(alternative 2), and number of female parents and seedlings per family in progeny test (alternative 3)).

My boss suggests the following assumptions: one experimental plant costs 1$ for all alternatives, no initiation or other cost
including crossings) costs 30$ per founder and takes 3 years. Within-family variance components are the following: additive
and environmental variance is 3.75. The additive standard deviation of the final forest value (within a full sib) is 10% of t
the selection experiment is 1 year for phenotypic selection (production of seedlings), 4 years for clonal test (basically, pr
(basically, production of female parents). Time after measuring the trial (from selection of individuals to production of the

I have heard that you are working with a new spreadsheet which may give the answers. Can you please suggest which of th
forward of one individual per full sib family (per cross) would be the best as regards group merit gain per unit of time?

----------------------------------------------------------------------------------------------------------------------------- ---
Tips on how to solve the Exercise 1 by using the EXCEL BREEDINGCYCLE tool.
1) Set the variance components in Box 1 (yellow cells): keep additive variance fixed at 1 and set the dominance and environme
changes of heritability in the breeding population in the box to the right), set the other parameters in Box 1 (as indicated in

2) Set the cost and time for recombination in Box 2 and your budget restriction in Box 3 (the programme will automatically ca
components for each of the alternatives in Box 4. Go back to Box 2 and chose the *selection alternative* (cell C18) you will

3) Now you shall find maximum group merit gain per year (GMG/T) for the three alternatives by setting various family size (al
(alt. 3). You may reach your goal by using the following two methods:

a) SLOW method: use the dawnmost box (box to assist): for alt. 1, vary family size in Box 4 until you will get the maximum GM
to copy the values needed), for alt. 2, try to find the combination of number of ramets per ortet and number of ortets which
reached with, say, 12 ramets per ortet; if number of ortets=8 the highest GMG/T was reached with 8 ramets per ortet; and so o
in the dawnmost box, to copy the data. Proceed in this way with alt. 3. The ADVANTAGE of this method is that you will be able
plants in the test.

b) FAST method: use SOLVER to find the family number and size which would give the maximum GMG/T (go to TOOLS+SOLVER). The s
the solver box is the following: Target cell (i.e. cell to maximise) is GMG/T; the cells to be changed (experimented with) ar
(alt.2), and family size (number of female parents) and number of seedlings per family (alt.3). NOTE: when SOLVER finds the m
SOLVER SOLUTION, as it will overwrite the formulas. Rather we would suggest to write dawn the parameters on the paper (GMG/T,
only).
----------------------------------------------------------------------------------------------------------------------------- --------------------------------------------

Solution to Exercise 1 (change colour of the font to see the solution)
Phenotypic selection (alternative 1)
Given the budget of 10 $ per founder and year, the highest group merit gain per year (GMG/T) of 0.3419 may be achieved by tes
after establishment of the selection trial.

Clonal test (alternative 2)
Given the budget of 10 $ per founder and year, the highest group merit gain per year (GMG/T) of 0.3722 may be achieved by tes
years after establishment of the selection trial.

Progeny test (alternative 3)
Given the budget of 10 $ per founder and year, the highest group merit gain per year (GMG/T) of 0.2654 may be achieved by tes
Given the budget of 10 $ per founder and year, the highest group merit gain per year (GMG/T) of 0.2654 may be achieved by tes
per family in 19.2 years after establishment of the selection trial.

Conclusion (letter to the boss):

My dear boss,
Under the given genetic parameters and cost as well as time scenario, I would recommend to carry out the clonal test by testi
final selection of 1 individual per full-sib family in 19.8 years after establishment of the selection trial. Now, I expect you to rise my salary as much as I have

Results in summary

Alternative/     Selection trial grows/ Number of ortets (or female parents)/ Number of plants per family/                                       GMG/T
Phenotypic (1)      12.6                   -                                     156                                                             0.3419
Clonal test (2)     19.8                   43                                       6                                                            0.3722
Progeny test (3)   19.2                    16                                     17                                                             0.2654

----------------------------------------------------------------------------------------------------------------------------- --------------------------------------------


Exercise 2
Hi,
I am going to optimise our breeding program for Dream Forest Inc. My boss thinks that it is a good idea to start with optimis
completely balanced, i.e. each parent get exactly two kids (each parent takes part in two crosses, i.e. double
breeding is cycling by double-pair mating and to select one individual within each full-sib family as a parent in next breeding
is 10$ per year and member of the breeding population (founder = cross).

My boss wants me to find out which of the following three alternatives for selection of one genotype per full
Merit Gain per year (GMG/T): (1) phenotypic selection of individuals, (2) clonal test, (3) progeny test (see illustration of
alternatives"). My boss believes that group coancestry should be penalised as inbreeding, thus, the weighting factor for grou
this is different from the previous task, and thus the differences in the results indicates what difference it makes if the i
calculations (except diversity loss) are within family. The task is to optimise the program so it returns maximum progress i
are propagation method, how long selection trial grows, number of individuals to be tested.

The boss suggests the following assumptions: An experimental plant costs 1$, no initiation or other costs (e.g. for cloning c
ordinary per plant cost associated to them), cycling (that means the recombination phase including crossings) costs 30$ and
dominance variance 0.25 and environmental 3.75. The standard deviation of the final forest value (within a full sib) is 10% o
experiment is set up is 1 year for phenotypic selection, 4 years for clonal and 6 years for parental selection. Time after ca
lacking parameters (if any) needed for the worksheet.

I have heard that you work with a new spreadsheet which may give the answers. Can you please suggest something about
that spreadsheet?
----------------------------------------------------------------------------------------------------------------------------- ---
Solution to Exercise 2 (change colour of the font to see the solution)
Phenotypic selection (alternative 1)
Given the budget of 10 $ per founder and year, the highest group merit gain per year (GMG/T) of 0.3159 may be achieved by tes
after establishment of the selection trial.

Clonal test (alternative 2)
Given the budget of 10 $ per founder and year, the highest group merit gain per year (GMG/T) of 0.3569 may be achieved by tes
years after establishment of the selection trial.

Progeny test (alternative 3)
Given the budget of 10 $ per founder and year, the highest group merit gain per year (GMG/T) of 0.2495 may be achieved by tes
per family in 22.6 years after establishment of the selection trial.

Conclusion
Under the given genetic parameters and cost as well as time scenario, I would recommend to carry out the clonal test by testi
final selection of 1 individual per full-sib family in 17 years after establishment of the selection trial.

Results in summary

Alternative/     Selection trial grows/ Number of ortets (or female parents)/ Number of plants per family/                                       GMG/T
Phenotypic (1)      13.9                   -                                     169                                                             0.3159
Clonal test (2)     17.0                   46                                       5                                                            0.3565
Progeny test (3)   22.6                    17                                     18                                                             0.2495

----------------------------------------------------------------------------------------------------------------------------- ---
----------------------------------------------------------------------------------------------------------------------------- ---


Task XX+2

Hi,
I am going to optimise our breeding program for Big Forest Inc. My boss thinks it is a good idea to start with optimising it
parent get exactly two kids. The budget allocated to the project is 20$ per year and member of the breeding population.

The alternatives are individual test (alt 1); clone test (alt 2) or parental test (alt 3) . The boss believes that group coancestry s
        how to gear the programme to solve the exercise A1 are given below the
                  you should mark the text below the heading "Solutions" and change

 ------------------------------------------------------------------------




 ss thinks that it is a good idea to start with optimising it by assuming that (1) the program is
akes part in two crosses, i.e. double-pair mating among founders), (2) the idea for long-term
             sib family as a parent in next breeding cycle. The budget allocated to the project


 selection of one genotype per full-sib family (cross) is the best to achieve the highest Group
            (3) progeny test (see illustration of the alternatives in sheet "Selection
ccurate breeding value estimates by clonal test or progeny test. My boss does not believe in
d within family selection), thus, the weighting factor for group coancestry can be set to 0. The
typic selection (alternative 1), ortet number and number of ramets per ortet in clonal test
 progeny test (alternative 3)).

ts 1$ for all alternatives, no initiation or other costs (=0$), cycling (recombination phase
 ly variance components are the following: additive variance is 1, dominance variance is 0.25
e final forest value (within a full sib) is 10% of the forest value. Time before establishment of
seedlings), 4 years for clonal test (basically, production of ortets) and 6 for progeny test
rom selection of individuals to production of their seeds) is 2 for all the alternatives.

e the answers. Can you please suggest which of the following three alternatives for selection
as regards group merit gain per unit of time?

 ------------------------------------------------------------------------


d set the dominance and environmental variance (observe % ratio between dominance and additive variance and
meters in Box 1 (as indicated in the assignment).

 programme will automatically calculate for how long you have to grow the selection trial) as well as cost and time
n alternative* (cell C18) you will test.

s by setting various family size (alt. 1), ortet and ramet number (alt. 2) and number of female parents and progeny


 until you will get the maximum GMG/T (after each input of family size, press the macro button in the dawn most box
tet and number of ortets which would give the highest GMG/T (e.g. if number of ortets=4 the highest GMG/T was
ed with 8 ramets per ortet; and so on until you will find the maximum). Again, after each input, press the macro button
 this method is that you will be able to compare on how GMG/T varies with test age and number of genotypes and


 ximum GMG/T (go to TOOLS+SOLVER). The solver shall be run separately for each of the alternatives. Set up of
hanged (experimented with) are family size (alt.1), family size (i.e. ortet number) and number of ramets per ortet
 ). NOTE: when SOLVER finds the maximum value and asks you how to present the result, DO NOT click on KEEP
wn the parameters on the paper (GMG/T, family size and time until selections made (selection trial grows) or GMG/T

     -----------------------------------------




 /T) of 0.3419 may be achieved by testing 156 seedlings per family and selecting 1 seedling per family in 12.6 years



 /T) of 0.3722 may be achieved by testing 43 ortets with 6 ramets each and selecting 1 ortet (clone) per family in 19.8



 /T) of 0.2654 may be achieved by testing 16 female parents with 17 seedlings each and selecting 1 female parent
/T) of 0.2654 may be achieved by testing 16 female parents with 17 seedlings each and selecting 1 female parent




to carry out the clonal test by testing 43 ortets with 6 ramets each for each of the 50 full-sib families and to make the
tion trial. Now, I expect you to rise my salary as much as I have raised the group merit gain of your forest ...




     -----------------------------------------




 ss thinks that it is a good idea to start with optimising it by assuming that (1) the program is
akes part in two crosses, i.e. double-pair mating among founders), (2) the idea for long-term
             sib family as a parent in next breeding cycle. The budget allocated to the project


 selection of one genotype per full-sib family (cross) is the best to achieve the highest Group
clonal test, (3) progeny test (see illustration of the alternatives in sheet "Selection
ed as inbreeding, thus, the weighting factor for group coancestry shall be set to 100 (note that
ults indicates what difference it makes if the increase in group coancestry is neglected). All
 se the program so it returns maximum progress in Group Merit per year. The chooses open


 s 1$, no initiation or other costs (e.g. for cloning cost, but ramets and progeny have the
mbination phase including crossings) costs 30$ and takes 3 years. Additive variance is 1,
n of the final forest value (within a full sib) is 10% of the forest value. Time before selection
and 6 years for parental selection. Time after can be set to 0 for all alternatives. Specify


nswers. Can you please suggest something about the exact design close to optimal based on

 -------------------------------------------------------------------



/T) of 0.3159 may be achieved by testing 169 seedlings per family and selecting 1 seedling per family in 13.9 years



/T) of 0.3569 may be achieved by testing 46 ortets with 5 ramets each and selecting 1 ortet (clone) per family in 17



/T) of 0.2495 may be achieved by testing 17 female parents with 18 seedlings each and selecting 1 female parent



to carry out the clonal test by testing 46 ortets with 5 ramets each for each of the 50 full-sib families and to make the




 ------------------------------------------------------------------------
------------------------------------------------------------------------




hinks it is a good idea to start with optimising it assuming it is completely balanced, thus each
 year and member of the breeding population.

        . The boss believes that group coancestry should be penalised as inbreeding, thus the
Breeding Cycle Analyser: deterministic simulator for genetic predictions of gain, diversity loss, cost and time of a breeding cycle
             This sheet is in preparation!of this sheet Dag Lindgren 04-03-14
                                    Last update
Simplified version for phenotype/progeny

   Variance components within full-sib family                   Parameter              Value                  Genetic variation at mature age, rotation age, BP size and diversity loss
                                 Variance components                         CVAm=          7                                                        Parameter             Value
                            sAw2=           0.5          0.71    sPw2=sAw2+sDw2+sE2=      9.33                               Weighting factor for group coancestry=          100
                            sDw 2=         0.083         0.29                   h2=       0.05                  Rotation age=          70                Diversity loss=      0.005
                             sE 2=          8.75         2.96                   H2 =      0.06                     TheBP size the diversity loss in % of "forestry value"=
                                                                                                                       value of        50                                        0.5

   General settings and summarizing results                                            Final results Value
                           Gain (%)       Cost ($)              Time (years)           Gain/cost    #NAME?                          Fixed values and changes compared to the other workshe
                                                                                                                                    Stage 1 will always be Phenotype (1)
   Recombination               0             20                         4              Gain/time    #NAME?
                                                                                                                                    Stage 2 will always be Progeny (3)
   Stage 1.Phenotype       #NAME?           100                        13                           #NAME?
                                                                                       Group Merit Gain                             Number of selected individuals in stage 2 will always be 1
   Stage 2. Progeny                      #NAME?         100             17             GMG/time    #NAME?                           Initiation costs for stage 1 and 2 will be 0
   Overall                               #NAME?         220             34             Cost/time       6.47                         "Time after" for stage 1 and 2 will be 0 (can be absorbed e
                                                                                                                                    Pollen (r) at progeny test is refined
                                                                                                                                    The contraction of genetic variation by preselection is take
Stage 1 Preselection by phenotype                               Stage 2 Selection based on progeny-test among the preselected
   Parameters                          Phenotype test           Progeny test




                                                                                                                                                                               DPM
      No. of selected individuals=          4                          1
                     Family size=          100                         4
    Plant no. per family member=                                      20                                                                                                                 BP N=50
             Selection intensity=         #NAME?                     #NAME?                                   The columns below are used for developing specifics for progeny testing
              Initiation cost (CIN)=        0                           0                                     OP           PC            As it was before March 2004
         Cost per test plant (CP)=          1                           1
         Cost per genotype (CG)=                                        5                                                  2                 5               1
                      Total cost =         100                         100                                                                                                           Field test 50
                    Time before =           1                           5                                                                                                            0      2
                   Selection age=           12                         12
                       Time after=          0                           0
                        Duration=           13                         17                                     OP-selfing       PC                              As it appeared before 04-03-01
                                                                                                                                                 Corrected for maternal or paternal variation, co
                            a1=sA=         0.71                      #NAME?               0.22                     0.22              0.22             0.22        0.22
                                a2 =       3.06                      #NAME?               0.76                     0.82              0.76             0.76        0.74
                       rAI=a1/a2=          0.23                      #NAME?               0.29                     0.26              0.29             0.29
                                                                                                                                                         Field test 50

                                                                                                                                                         0      2




       Remainder of initial sA2=            #NAME?                                           0.1
        J-M correlation function              0                             0
                                            #NAME?                        #NAME?   Progeny test within family
                       rY (or R)=           #NAME?                        #NAME?
        Genetic gain (%)=               #NAME?                            #NAME?                    0.25s Aw  0.5s As  s D  s E
                                                                                                           2         2     2     2
                                                                                   s 2  0.25s Aw 
                                                                                               2
                                                                                     X
                                                                                                                  m
Simplified box                                           Stage
                                                                                   The criteria used as selection criterium within a family is the means of
                                    First                Second                    consider that the mothers are full sibs and the genetic variance between
                            sA2=                     1      #NAME?                 whole offspring population. We assume in this formula that the pollen
                             h2 =                  0.2            0.8              random from the whole offspring (but the effect of inbreeding is neglec
Candidates                                        100                10
Selections                                         10                2
selection intens; i=                        #NAME?          #NAME?                 The truncation preselection reduces the within family genetic variation
selection intens; i=                        #NAME?                                 s Aw '  (1  rAI i (i  t )s Aw
                                                                                     2            2              2

t                                           #NAME?
                                                                                                                    0.5
Var=1-h2i(i-t)                              #NAME?
Gain=ihsA                                   #NAME?          #NAME?                 The relationship between the progeny and the mother tree is 0.5, the ga
Accumulated gain                                           #NAME?                  selection within family will be

                                                                                                                                     s Amwr j  mi 0.5s A w '
                                                                                                             GProgeny 
                                                                                                                                         0.25s Aw '0.5s As  s D  s
                                                                                                                                               2         2      2
                                                                                                                          0.25s A w '
                                                                                                                                2

                                                                                                                                                       m

                                                                                   However, polycross has limited number of fathers in a mix which is as
                                                                                   mothers, thus the variation because of different fathers will be less. Am
                                                                                   variance for the common inheritance for the fathers (which is only half
                                                                                   reduced by (1-1/Y). The one quarter of the dominance variance will als
                                                                                   the formula for the nominator will be
                                                                                                      0.25s Aw  0.25s As  (0.25s As  0.25s D )(1  1 / Y )  0.75
                                                                                                            2          2           2          2
                                                                                   s 2  0.25s Aw 
                                                                                               2
                                                                                     X
                                                                                                                                       m
mothers, thus the variation because of different fathers will be less. Am
variance for the common inheritance for the fathers (which is only half
reduced by (1-1/Y). The one quarter of the dominance variance will als
the formula for the nominator will be
                   0.25s Aw  0.25s As  (0.25s As  0.25s D )(1  1 / Y )  0.75
                         2          2           2          2
s 2  0.25s Aw 
            2
  X
                                                    m
d time of a breeding cycle                              Meaning of cell font and colour:



               Variance components in breeding stock
                                                        Parameter
                    Variance components Standard deviations                  Value
                     sAs2=        1          1.0                  h2 =       0.10
                     sDs 2=     0.25         0.5                 H2 =         0.13
                      sE 2=      8.75        3.0              CVAm=           9.9


nges compared to the other worksheets:    Alternative pine                  strategy (cheap)
  Phenotype (1)
                                         Top grafting Polycross
  Progeny (3)
 dividuals in stage 2 will always be 1
ge 1 and 2 will be 0
1 and 2 will be 0 (can be absorbed elsewhere)Top grafts (4*25)
est is refined




                                                                                                              Next Cycle BP DPM
netic variation by preselection is taken care of.                              Planting

                                                          Planting                Progeny test
                    DPM




                                                                         25*4*40 PC + 25*6*20 OP = 7000

                              BP N=50     Select 4 best                          Grafting
                                                              Select 6 in
                                           in 25 best
                                                               all fam.s     Graft archive (6*25)
ping specifics for progeny testing            fam.s
 efore March 2004
                                                                                                     pollen


                          Field test 5000 F1 (50*100)

                          0      2            12                     18                             30
                                                            Year (approx)

    As it appeared before 04-03-01
                                Field test 5000 F1 (50*100)

                                0      2            12             18         30
                                                              Year (approx)




 2
 D   s E
        2




m within a family is the means of m approximate halfsibs. We
 and the genetic variance between halfsibs is only half that in the
me in this formula that the pollen for the progeny test comes at
t the effect of inbreeding is neglected.


he within family genetic variation available for truncation selection


y and the mother tree is 0.5, the gain formula for truncation


            s Amwr j  mi 0.5s A w '
             0.25s Aw '0.5s As  s D  s E
                   2         2      2     2
25s A w '
    2

                           m

ber of fathers in a mix which is assumed to be equal among
f different fathers will be less. Among effectively Y fathers the
 for the fathers (which is only half the additive variance) will be
of the dominance variance will also be reduced by that factor. So

(0.25s As  0.25s D )(1  1 / Y )  0.75s D  s E
       2          2                       2     2


           m
f different fathers will be less. Among effectively Y fathers the
for the fathers (which is only half the additive variance) will be
of the dominance variance will also be reduced by that factor. So

(0.25s As  0.25s D )(1  1 / Y )  0.75s D  s E
       2          2                       2     2


           m
Dag L 000724, Darius 2001.04.24
An excercise

Hi,
I am going to optimise our breeding program for Big Forest Inc. My boss thinks that it is a good idea to start with
optimising the breeding program assuming that it is completely balanced, thus, each parent get exactly two kids.
The budget allocated to the breeding program is 10$ per year and member of the breeding population.

The alternatives are phenotype test (alt 1); clone test (alt 2) or progeny test (alt 3). The boss believes that group
coancestry should be penalised as inbreeding, thus the "weight" ("penalty factor" in cell I12) shall be set to 100.
All calculations (except diversity loss) are within family. The selection should be in two stages, first a
phenotypic preselection of a shortlist of candidates (stage 1; alternative 1) and then either a polycross test of the
breeding value of the parents (stage 2; alt 3) or cloning and clonal test (stage 2; alt 2). The task is to optimise the
program so it returns maximum progress in GMG per year. The choices open are propagation method (clones
versus progeny), initial family size to be selected from, number of the individuals to be preselected at stage 1,
number of individuals to be tested per each genetic entry at stage 2, how long the selection trials grow.

The boss suggests the following assumptions: a test plant costs 1$ for all testing alternatives, 2$ for cloning (per
clone) and 2$ (per parent) for initiation: if it is progeny test it costs 1 $ per eximerental plant, 5$ for initiation to
produce the test progeny (by polycross or open pollination). Recombination costs 30$ and takes 3 years. Additive
variance is 1, dominance variance shall make 25% of the additve variance in the breeding population and
environmental variance is 17.5 (resulting in the narrow sence heritability of 0.1). The additive standard deviation
of the final forest value (within a full sib) is 10% of the forest value. Time before selection test is established up
(T<p) is 1 year for phenotype selection, 4 years for clonal test and 6 years for progeny test. Time after is 2 years
for all alternatives (i.e. time needed to obtained the improved seed). Specify lacking parameters (if any) needed
for the worksheet.

								
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