1. Some Principles of Plant Breeding Applicable to TACF F. V. Hebard Research Farms Meadowview, VA Fred@acf.org www.acffarms.org

2. Assumptions Leading to Choice of the Backcross Method for Introgressing Blight-Resistance from Chinese into American Chestnut American chestnut lacks blight resistance. Chinese chestnut has the highest level of blight resistance of all species sexually compatible with American chestnut. Blight resistance from Chinese chestnut is inherited fairly simply. American chestnut otherwise is perfectly adapted to our forests.

3. C x A  F 1 x A  B 1 x A  B 2 x A  B 3 x B 3  B 3 -F 2 x B 3 -F 2  B 3 -F 3 Fraction Chinese 1/2 1/4 1/8 1/16 Backcrossing

4. Recovery of American Type The purpose of backcrossing is to dilute out all the traits of the Chinese chestnut tree, except blight resistance, for which we select at each generation. The experience of plant breeders has been that, generally, three backcrosses are enough to recover the recurrent type. The dilution of the fraction of Chinese genes with each step of backcrossing is not exact. Starting with the first backcross, the fractions given are averages over all the progeny for a cross. For instance, first backcross progeny average 1/4 Chinese, 3/4 American genes. However, individual trees can vary from 50% Chinese up to 100% American. In practice the range of variation is less, from about 65% to 85% for 100 B 1 progeny, as assessed by molecular markers.

5. Backcrossing Recovery of American Type We can accelerate the recovery by selecting for American traits and against Chinese traits. Currently, we select for morphological traits. If we were to grow larger numbers of progeny, we could probably accelerate recovery even more using molecular techniques. This would enable us to bypass several generations of backcrossing. Our current use of morphological traits probably balances out the Chinese traits drug along by selection for resistance.

6. Backcrossing Recovery of American Type Chromosomal abnormalities associated with wide crosses, such as inversions and reciprocal translocations, may hinder easy introgression of small segments of Chinese chestnut chromosomes containing resistance genes but little else. Part of the answer to problems such as these is to grow larger numbers of progeny. Molecular techniques also may be very helpful in overcoming such possible obstacles, as well as identifying their occurrence.

7. C x A  F 1 x A  B 1 x A  B 2 x A  B 3 x B 3  B 3 -F 2 x B 3 -F 2  B 3 -F 3 Expected Resistant AllAllAll 1/21/41/8 1/41/161/64 MostMostMost # Genes 123 Backcrossing

8. Number of Genes for Blight Resistance Backcrossing becomes progressively more impractical as the number of genes being introgressed increases. Either so many genes will be lost during backcrossing that the trees will lack adequate levels of blight resistance or so many associated Chinese traits will be retained that the trees will lack adequate American type. Most evidence to date indicates that only a few genes, two or three, confer blight resistance, suggesting that backcrossing will be effective. This assessment could change with further data.

9. Backcrossing Number of Genes for Blight Resistance The evidence that only a few genes are involved with resistance is that: 1)Resistance has not decreased with successive backcrosses. 2)We have recovered F 2 individuals with high levels of resistance at frequencies greater than about 1 in 30, and, correspondingly, 3)the distributions of canker sizes have been less peaked than would be expected if more than 3 genes controlled resistance. 4)Likewise, the tails of the distributions have been less high than would be expected if only 1 gene controlled resistance. 5)The increase in variance at backcross and F 2, compared to homozygous parents, is compatible with 1 to 2 genes controlling resistance. 6)The number of major qtls for resistance on molecular maps has been limited and combinations of any two qtls have predicted high levels of resistance in progeny. 7)However, minor qtls also exist on molecular maps.

10. Backcrossing Why we want 100 nuts per backcross. There is a 99% probability of getting at least one tree heterozygous for three genes with 33 progeny,.01 = (7/8) 33, where 7/8 is the chance of not being homozygous for three genes for resistance. Growing 73 trees per cross gives a 99% chance of getting at least four trees heterozygous for three genes,.01 = 7/8 73 + 73*7/8 72 *1/8 1 + 2628*7/8 71 *1/8 2 + 62196*7/8 70 *1/8 3. About 3/4ths of our planted nuts yield trees, and 73/.75 ~= 100.

11. Backcrossing Need more than one American parent C x A1  F 1 x A2  B 1 x A3  B 2 x A4  B 3

12. Backcrossing Need more than one American chestnut line C x A4  F 1 4 x A8  B 1 4 x A12  B 2 4 x A16  B 3 4 C x A3  F 1 3 x A7  B 1 3 x A11  B 2 3 x A15  B 3 3 x C x A2  F 1 2 x A6  B 1 2 x A10  B 2 2 x A14  B 3 2 C x A1  F 1 1 x A5  B 1 1 x A9  B 2 1 x A13  B 3 1 x  B 3 -F 2 12 B 3 -F 2 34

13. Intercrossing of Third Backcrosses With Clapper and Graves, we started from the B 1. B 1 1 x A4  B 2 4 x A8  B 3 4 B 1 1 x A3  B 2 3 x A7  B 3 3 x B 1 1 x A2  B 2 2 x A6  B 3 2 C x A0  F 1 1 x A00  B 1 1 x A1  B 2 1 x A5  B 3 1 x  B 3 -F 2 12 B 3 -F 2 34

14. Intercrossing of Third Backcrosses Why we want 20 American lines per source of resistance for each chapter. B 1 1 x A4  B 2 4 x A8  B 3 4 B 1 1 x A3  B 2 3 x A7  B 3 3 x B 1 1 x A2  B 2 2 x A6  B 3 2 C x A0  F 1 1 x A00  B 1 1 x A1  B 2 1 x A5  B 3 1 x  B 3 -F 2 12 B 3 -F 2 34

15. Intercrossing of Third Backcrosses Why we want 20 American lines per source of resistance for each chapter. Inbreeding is an important parameter in assessing how many lines to use. To look at inbreeding effects, we have to examine B 3 -F 4 s, two generations further than B 3 -F 2.

16. Intercrossing of Third Backcross Lines Complete Diallel, No Selfing B 3 s  1 2 3 4 5 6 7 8  B 3 -F 2 s 1 11 12 13 14 15 16 17 18 2 21 22 23 24 25 26 27 28 3 31 32 33 34 35 36 37 38 4 41 42 43 44 45 46 47 48 5 51 52 53 54 55 56 57 58 6 61 62 63 64 65 66 67 68 7 71 72 73 74 75 76 77 78 8 81 82 83 84 85 86 87 88

17. Intercrossing of Third Backcross Lines Half Diallel, No Selfing B 3 s  1 2 3 4 5 6 7 8  B 3 -F 2 s 1 12 13 14 15 16 17 18 2 23 24 25 26 27 28 3 34 35 36 37 38 4 45 46 47 48 5 56 57 58 6 67 68 7 78 8

18. Intercrossing of Third Backcross F 2 s Half Diallel, No Selfing B 3 -F 2 s   12 34 56 78  B 3 -F 3 s 12 12:34 12:56 12:78 34 34:56 34:78 56 56:78 78

19. Intercrossing of Third Backcross F 3 s Half Diallel, No Selfing B 3 -F 3 s  12:34 12:56 12:78 34:56 34:78 56:78  B 3 -F 4 s 12:34 12:34-12:56 12:34-12:78 12:34-34:56 12:34-34:78 12:34-56:78 12:56 12:56-12:78 12:56-34:56 12:56-34:78 12:56-56:78 12:78 12:78-34:56 12:78-34:78 12:78-56:78 34:56 34:56-34:78 34:56-56:78 34:78 34:78-56:78 56:78

20. Intercrossing of Third Backcross F 3 s Half Diallel, No Selfing: Inbreeding Coefficients B 3 -F 3 s  12:34 12:56 12:78 34:56 34:78 56:78  B 3 -F 4 s 12:34 12:34-12:56 12:34-12:78 12:34-34:56 12:34-34:78 12:34-56:78 1/8 1/8 1/8 1/8 0 12:56 12:56-12:78 12:56-34:56 12:56-34:78 12:56-56:78 1/8 1/8 0 1/8 12:78 12:78-34:56 12:78-34:78 12:78-56:78 0 1/8 1/8 34:56 34:56-34:78 34:56-56:78 1/8 1/8 34:78 34:78-56:78 1/8 56:78

21. Intercrossing of Third Backcrosses Inbreeding decreases rapidly as the number of lines increases to about 20, and then plateaus at a low level. Effect of Number and size of B 3 Families (Lines) on Inbreeding at B 3 -F 4.

22. Backcrossing Why we want nine selected B 3 -F 2 trees per B 3 line. Nine progeny give us a 93% chance of obtaining both alleles at each chromosome locus from a parent, capturing most of the genetic diversity within the parent. The number, 9, comes from the following formula, assuming there are 20 independently assorting chromosome fragments: 0.93 = [1-2 (1-9) ] 20.

24. Population Recovery Thus our breeding program satisfies Franklin’s famous 50/500 rule of conservation genetics, which states that the effective population size has to be at least 50 to avoid immediate collapse from inbreeding depression, and 500 for mutation to offset the long-term erosion of genetic diversity by drift.

25. Conclusions Approximately twenty lines in any breeding unit is enough to avoid immediate collapse of the population due to inbreeding depression. At least nine to ten B 3 -F 2 progeny are needed per B 3 line to capture most alleles. If each of those nine to ten sets of B 3 -F 2 progeny is segregated in a separate orchard, inbreeding at F 4 is reduced. The American Chestnut Foundation’s breeding program will yield adequate effective population sizes (although the projections have a number of assumptions that will not hold true, decreasing the projected N e ). Chapter breeding efforts add essential genetic diversity and local adaptation to the overall program.