Advanced Plant Breeding CSS 650 by 7xS3ef

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									Advanced Plant Breeding CSS 650                               Name            KEY
Take-Home Final Exam, Fall 2011
Due 9:00 am on Friday, December 9, 2011

Part 1 – Recurrent Selection in Radish

        You (and your classmates) have been hired as consultants for a local seed company that
specializes in production of organic vegetable seeds. They would like you to design a population
improvement scheme that they could use to improve a genetically diverse radish population.
Radish seed is an important cash crop in the Willamette Valley, so one of your objectives is to
ensure that the population is well-suited for seed production in the area. The population must
also have desired characteristics for production of organic radishes for the fresh market in the
Willamette Valley, as well as for home gardens in other regions of the US.
        The seed company has the capability to implement the breeding plan using three testing
environments in the Willamette Valley. All of the sites can be irrigated. Growth chambers and
greenhouse facilities can be utilized if needed. A plant pathologist is employed by the company
and is available for assistance on this project. The company is interested in improving the
population on a continuing basis. They understand the need to maintain genetic diversity in the
population and do not expect that each cycle of selection should be suitable for immediate
release as a commercial variety. However, they would like to be able to extract good varieties
from the population in a reasonable time frame, using procedures described in the article
published by the Organic Seed Alliance (2007) entitled “Principles and Practices of Organic
Radish Seed Production in the Pacific Northwest”.
        Additional resources available to you include the lecture presentation on Recurrent
Selection (Module 10), your textbook (Chapter 10), and Chapter 6 from Hallauer and Miranda
(1988). You are welcome to consult other references, but this is not intended as a research
project. If there is information that you think you need that is not in the references provided, you
can make a reasonable assumption and indicate that in your write-up. If any of you have access
to additional information about radish that you think would be relevant for this exercise, please
share that with your classmates on the discussion board in Blackboard. Each of you must submit
your own breeding plan, but you are welcome to exchange ideas with your classmates.

Based on the Terms of Reference for your consultancy, you must respond to the following
questions:

1) What intrapopulation selection schemes can be considered for this radish population? Explain
your reasons for excluding other types of recurrent selection systems.
Possibilities include mass selection, ear-to-row selection, and half-sib selection schemes that do
not require selfing (assuming that radish is highly self-incompatable). It may be possible to
create full-sib families by caging pairs of plants in the greenhouse with pollinators. However, the
benefits of full-sib family selection may not be worth the additional time and expense needed to
create a sufficient number of families in this way.

2) Provide an outline of the selection scheme that you will recommend. Indicate how each stage
of selection will fit into prevailing environments in the Willamette Valley, and the traits that will




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be selected at each stage of the process. Include approximate numbers of families or individuals
that will be evaluated and selected at each stage.

There are various approaches that could be used - I will use a modified ear-to-row selection
scheme. An initial generation will be required to form half-sib families. I could grow about 2000
plants, and evaluate them at the time that they would be ready for market. The best 500 could be
transplanted into a polycross nursery at wide spacing (to permit maximum seed set and avoid
mixing of seed between plants). Additional selection could be practiced at harvest, resulting in
300 half-sib families to begin the recurrent selection scheme.

Site 1: Use to evaluate root characteristics under conditions that represent a typical production
        environment. Plant two replicates of ~300 half-sib families in March. Each replicate
        would consist of a single row of ~25 plants (assuming seed quantities produced on
        individual parent plants are sufficient). Rogue out plants with poor seedling vigor or
        undesirable leaf characteristics. Dig up all of the roots for evaluation at the stage of
        marketable maturity and rate them for “root aspect” (a composite 1-9 score that indicates
        desirable attributes). Several plants from each family could be destructively sampled and
        used to evaluate root quality characterstics such as taste and texture. Transplant the best ~
        10 plants (5 per replicate) from the best 50% of the families to a polycross nursery at
        wide spacing. This nursery must be grown at least one mile from other radish seed
        production fields and from wild radish. Allow the plants to intermate and harvest seed
        from individual plants. These represent the new half-sib families that could potentially be
        evaluated in the next cycle of selection.
Site 2: Use to evaluate disease resistance at the optimum stage for root harvest. Plant two
        replicates a few weeks later than Site 1, inoculate, and irrigate to increase pressure from
        important pathogens. Use the information from this site to remove any families in the
        polycross nursery (Site 1) with unacceptable levels of disease, prior to pollination
        (assume that this would eliminate up to 50 of the 150 families in the nursery).
Site 3: Use to evaluate seed production characteristics of half-sib families. The objective is to
        mimic production of certified radish seed in the valley. The roots will not be transplanted
        because I’m assuming that certified seed is not generally transplanted in the valley (in
        contrast to foundation seed production as described in the handout). I’m also assuming
        that organic growers in the valley will choose to plant seed production fields at relatively
        high density, rather than trying to obtain maximum yields per plant at lower densities.
        Plant two replicates at the time and density characteristic of seed production fields (this
        should be in early spring to permit adequate natural vernalization of roots). Border rows
        may be needed to estimate seed yields in full stands. Data will be collected on seedling
        establishment, as this will be an important trait for good competition against weeds in
        organic fields. From the 100-120 families that were used for recombination and seed
        production at Site 1, select 30 that have good agronomic type and high seed yields. Data
        collected on root aspect and quality and levels of disease resistance from Sites 1 and 2
        could be incorporated in an index to make the final selection, but seed yield (Site 3)
        would receive a relatively high weight because it was not considered when earlier
        selections were made at the other sites. There is no need to save the seed from this site
        after it has been evaluated for yield. However, if you wish to use this field as a seed
        increase you could remove all of the families that were rejected at Sites 1 and 2 before



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       pollination. The seed harvested could be used as breeders seed of a potentially new
       variety. Another generation or two of stringent selection would be needed to attain
       sufficient uniformity of root characteristics for release as a variety, using the protocol
       described in the handout. Improvement of the base population would continue as
       described above. Maintenance of a higher level of genetic variation in the base population
       is essential to maintain high rates of response to selection in the longterm.

As described above, a cycle of selection could be completed each year, utilizing only the main
production season at three sites. Although it takes only 30 days to obtain a marketable root in
radish, seed production requires 150 days, making it unfeasible to obtain two field generations
per year in Oregon. Another option would be to include an extra recombination phase for
production of new half-sib families at a single site in alternate years. Additional mass selection
could be conducted at this time for root attributes. The main benefit of the extra generation for
recombination would be that fewer plants and families could be selected in the first year,
allowing greater selection pressure to be applied, particularly for seed production characterstics
measured at Site 3. For example, a balanced bulk could be created from the best four plants in
the best 15 families at the end of the first year. About 2000 seeds could be planted in a nursery in
the second year, and the best 500 plants could be transplanted at wide spacing. The 300 plants
that produce the most seed could be harvested individually to create half-sib families for
evaluation in the next year.

3) Justify your choice of selection schemes (this should be a fairly comprehensive discussion –
refer to your lecture notes for ideas and examples). Be sure to include a discussion about the
expected gain from selection for your scheme compared to other potential selection strategies.

Given the need to evaluate multiple traits that are expressed at different times and under different
conditions, simple mass selection may not be adequate for improving this population. The
opportunity to select for root characteristics prior to pollination favors an ear-to-row system over
a more conventional half-sib selection, where remnant seed is used for recombination. A cycle of
ear-to-row selection can be completed in a single year, whereas half-sib family selection would
require an additional year to recombine selected families and generate new families for
evaluation.

Expected gain from selection                                                i1 (1/4)2 i2 (3/4)2
For traits measured at Site 1, expected gain from selection would be:
                                                                                     A
                                                                                               A
                                                                                P         Pw
where i1 refers to the standardized selection differential                        hs

among half-sib families, and i2 refers to the selection differential within families.

For traits measured at Site 2, expected gain from selection would be: i(1/4)A Phs
                                                                                 2


This is less than for Site 1 because there is no opportunity for selection within families at Site 2.

For traits measured at Site 3, selections can only be made after pollination, so the expected gain
from selection is: i(1/8)2 
                           A   Phs




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If the decision is made to include an extra recombination phase in year two, we would essentially
be alternating ear-to-row with mass selection. During the mass selection phase the expected
response to selection would be i 2 P for traits that are measured before pollination, and
                                     A
i (1/2)σ 2 σ P For traits that are selected after pollination.
         A


Although mass selection looks very good in terms of the proportions of additive genetic variance
affected by selection (the numerator), it’s important to remember what the denominators
represent. Phenotypic variance among individual plants includes all of the environmental
variance that acts on individuals. Phenotypic variance among half-sib families is based on plot
averages, and environmental effects are further reduced by replication of plots.




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Advanced Plant Breeding CSS 650 Name                   KEY
Take-Home Final Exam, Fall 2011
Due 9:00 am on Friday, December 9, 2011

Part 2 – Selection for multiple traits in meadowfoam

A breeder wants to improve meadowfoam as an oilseed crop in Oregon. In 2011 she evaluated
270 half-sib families from a breeding population in a yield trial at a single location using a lattice
design with two complete replications (blocks). For the purposes of this exercise, data for seed
weight/plot (seedwt in g), thousand seed weight (TSW in g), and oil content (percent by weight
at ~10% moisture) will be analyzed for a subset of 159 families, ignoring the incomplete
blocking structure in the experiment.


1) To assist in developing a selection index, we will first calculate the genetic variance and
covariance matrix for the three traits (on a family mean basis). Use the following SAS code to
generate univariate analyses for all traits as well as an analysis of covariance among traits. You
will first need to load the datafile ‘mf’ into SAS (the data is in FinalPart2_data.xls).

       proc glm data=mf;
       class rep entry;
       model seedwt TSW oil=rep entry;
       manova h=entry/printh printe;
       random rep entry/test;
       run;


2) You will use the ‘E = Error SSCP Matrix’ and ‘H = Type III SSCP Matrix for Entry’ to do
your calculations.
     Copy the results from your output into Excel.
     Divide each matrix by the appropriate degrees of freedom to obtain the corresponding
       matrices for Mean Squares and Mean Cross Products (you should have one matrix for
       Error and another for Entries).
     Generate a new matrix of genetic variances and covariances for these three traits, using
       the same approach that you used in HW4 to estimate the Genetic Variance among
       families.
     Attach a hard copy of your Excel spreadsheet to your exam (you do not need to submit
       your SAS output).




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3) Use the genetic variance-covariance matrix to calculate genetic correlations among the three
traits. Note that you could use the same process to obtain environmental correlations (from the
Error MCP matrix) and phenotypic correlations (from the MCP matrix for Entries), but you are
not required to do this. The environmental correlations should have been provided in your
computer output (Partial Correlation Coefficients from the Error SSCP Matrix).
          A( XY)
rA 
        2 X)2 Y)
         A(     A(

                         4.07967
rA(seedwt,TSW )                         0.35586
                    549.4144 * 0.239217
                        5.0599
rA(seedwt,oil)                        0.298455
                  549.4144 * 0.523154
                      0.147561
rA(TSW,oil)                          0.41712
                 0.239217 * 0.523154

Are the environmental correlations and genetic correlations similar in magnitude and direction?

No, they are quite different. Environmental correlations with seedwt are close to zero. The
environmental correlation between TSW and oil content is 0.14, but this is not statistically
significant. All of the genetic correlations are positive.

4) Calculation of genetic correlations using MANOVA will give the same results as a mixed
model analysis when the data are balanced. To illustrate the use of mixed models, run the
following SAS code to obtain the genetic correlation between TSW and oil. You will first need
to input the dataset ‘correl’ into SAS (also in FinalPart2_data.xls). This is the same data that you
used for the MANOVA, but the format is different. How do the data sets differ?

For the MANOVA, each trait is considered to be a different variable and each appears in a
different column. For the mixed model analysis, there is a single variable called trait, and the
variable names (seedwt, TSW, and oil) represent different levels of that variable. The traits are
handled as repeated measures on each plot.

You will need to include a statement in your data step to remove the variable ‘seedwt’ from the
analysis, because we only wish to obtain a correlation for TSW and oil.

if trait='seedwt' then delete;

Once the data have been input into SAS, run the SAS program (adapted from the article by
Piepho and Möhring, 2011, Crop Sci. 51:1-6):

proc mixed data=correl;
class rep entry plot trait;
model Y=trait trait*rep;
random trait /subject=entry type=unr;
repeated trait / subject=plot type=unr;
run;
quit;



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The analysis is computer intensive, and will take a few minutes to run. A similar analysis could
be done for seedwt and TSW, and seedwt and oil content (but you don’t need to run these
programs). The scale is quite different for seedwt than for the other traits, so I found that I
needed to transform seedwt ( I divided by 10) in order to obtain a solution. The correlations
among traits were not affected by the transformation.

Paste the covariance estimates from SAS below. Explain the meaning of the variance and
correlations that you obtained.

5) Use the output that you have already obtained to calculate heritability for TSW and oil content
(there are several ways to do this - refer to HW4 if necessary).
Cov Parm        Subject   Estimate   Covariance Parameter Estimates
Var(1)          Entry       0.2392   Var(1) is the additive genetic variance for TSW
Var(2)          Entry       0.5232   Var(2) is the additive genetic variance for Oil content
Corr(2,1)       Entry       0.4171   Corr(2,1) is the additive genetic correlation between TSW and
                                     Oil content
Var(1)          Plot        0.2082   This is the error variance for TSW
Var(2)          Plot        0.5384   This is the error variance for Oil content
Corr(2,1)       Plot         0.141   This is the environmental correlation for TSW and Oil content


For TSW
                         2            0.20816 MSF 0.6866
      P  F  2  F  E  0.2392 
        2    2
                  X
                      2
                                                         0.3433
                          2               2     2     2
      F  0.2392
       2



              F 0.2392
                2
         h  2 
            2
                              0.697
              P 0.3433
For Oil content
                               2               0.53839 MSF 1.5847
         P  F  2  F  E  0.52315 
          2     2
                     X
                           2
                                                                            0.7924
                               2                    2          2        2
         F  0.5232
          2


              2 0.5232
         h2  F              0.660
              P 0.7924
                2


6) If the breeder selects the best 10% of the families for oil content, what would be the expected
response to selection after those families are intermated?

        RX  ihX AX  ih2 PX  1.76 * 0.660 * 0.7924  1.03
                          X
        Oil percentage in the population should increase by about one unit.




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7) If the breeder selects the highest 10% of the families for TSW, what change in oil content
would be expected in the next generation?

         CRY  ihXrAAY  1.76 * 0.697 * 0.4171* 0.5232  0.443

        Oil percentage in the population should increase by about 0.44 units.
8) Meadowfoam growers are paid for seed produced by the pound, but processors value
meadowfoam with high oil content. The breeder decides to give equal weight to these two traits
in a selection index. She includes TSW in the index because she knows it is correlated with seed
yield and oil content.

Show how you would set up the matrices for the selection index, using economic weights of +1
for seedwt and oil content, and 0 for TSW.

   b1             σP12     CovP1P2 CovP1P3 -1              σA12     CovA1A2 CovA1A3          a1
   b2      =    CovP1P2      σP22      CovP2P3            CovA1A2    σA22      CovA2A3       a2
   b3           CovP1P3 CovP2P3          σP32             CovA1A3 CovA2A3        σA32        a3


                                                     -1
    b1              1135 4.8111 3.8903                    549.41 4.0797 5.0599                    1
    b2           4.8111 0.3433 0.1712                    4.0797 0.2392 0.1476                    0
    b3            3.8903 0.1712 0.7924                    5.0599 0.1476 0.5232                    1


Using either Excel or R, solve for the values of the coefficients. Refer to the updated handout on
regression using matrices for examples of both methods.

    b1           0.45839
    b2          3.92083
    b3           3.94863

9) For entry 11, average seedwt = 304.5 g, TSW = 10.09 g, and oil content = 26.785%. What
would the index value be for this entry? How would you use this information in your selection?

I = 0.45839*304.5+3.92083*10.09+3.94863*26.785 = 284.905

Select the 16 lines with the highest index scores.




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