Some Principles of Plant
Breeding Applicable to TACF
Qui ck Time ™ an d a
TIFF (Unco mp re ssed ) dec ompre ssor
are need ed to s ee th is pi cture.
F. V. Hebard
Research Farms
Meadowview, VA
Fred@acf.org
www.acffarms.org
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.
Backcrossing
Fraction
Chinese
1/2
1/4
1/8
1/16
C x A

F1 x A

B1 x A

B2 x A

B3 x B3
1/16

B3-F2 x B3-F2
1/16

B3-F3
Backcrossing
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 B1 progeny, as assessed by
molecular markers.
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.
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.
Expected
Resistant
Backcrossing
# Genes
1
2
3
C x A

F1 x A

B1 x A

B2 x A

B3 x B3
All
All
All
1/2
1/4
1/8
1/2
1/4
1/8
1/2
1/4
1/8
1/4
1/16 1/64
Most Most Most

B3-F2 x B3-F2

B3-F3
Backcrossing
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.
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 F2 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 F2, 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.
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 = 7/873 + 73*7/872*1/81 + 2628*7/871*1/82 +
62196*7/870*1/83.
• About 3/4ths of our planted nuts yield trees, and 73/.75
~= 100.
Backcrossing
Need more than one American parent
C

F1

B1

B2

B3
x A1
x A2
x A3
x A4
Backcrossing
Need more than one American chestnut line
C

F1 1

B1 1

B2 1

B3 1
x A1
x A5
x A9
x A13
x
C

F1 2

B1 2

B2 2

B3 2
x A2
C

x A6 F13

x A10 B13

x A14 B23

B3 3
x A3
x A7
x A11
x A15
x

B3-F212
B3-F234
C

F1 4

B1 4

B2 4

B3 4
x A4
x A8
x A12
x A16
Intercrossing of Third Backcrosses
With Clapper and Graves, we started from the B1.
C

F1 1

B1 1

B2 1

B3 1
x A0
x A00
x A1
x A5
x
B11 x A2

B22 x A6

B3 2
B11 x A3

B23 x A7

B3 3
x

B3-F212
B3-F234
B11 x A4

B24 x A8

B3 4
Intercrossing of Third Backcrosses
Why we want 20 American lines per source of
resistance for each chapter.
C

F1 1

B1 1

B2 1

B3 1
x A0
x A00
x A1
x A5
x
B11 x A2

B22 x A6

B3 2
B11 x A3

B23 x A7

B3 3
x

B3-F212
B3-F234
B11 x A4

B24 x A8

B3 4
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 B3-F4s, two generations further
than B3-F2.
Intercrossing of Third Backcross Lines
Complete Diallel, No Selfing
B3s 
1
2
3
4
5
6
7
8
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
B3-F2s
Intercrossing of Third Backcross Lines
Half Diallel, No Selfing
B3s 
1
2
3
4
5
6
7
8
12
13
14
15
16
17
18
23
24
25
26
27
28
34
35
36
37
38
45
46
47
48
56
57
58
67
68
B3-F2s
1
2
3
4
5
6
7
8
78
Intercrossing of Third Backcross F2s
Half Diallel, No Selfing
B3-F2s  12
34
56
78
B3-F3s
12
34
56
78
12:34
12:56
12:78
34:56
34:78
56:78
Intercrossing of Third Backcross F3s
Half Diallel, No Selfing
B3-F3s 12:34
12:56
12:78
34:56
34:78
56:78
B3-F4s
12:34
12:56
12:78
34:56
34:78
56:78
12:34-12:56
12:34-12:78
12:34-34:56
12:34-34:78
12:34-56:78
12:56-12:78
12:56-34:56
12:56-34:78
12:56-56:78
12:78-34:56
12:78-34:78
12:78-56:78
34:56-34:78
34:56-56:78
34:78-56:78
Intercrossing of Third Backcross F3s
Half Diallel, No Selfing: Inbreeding Coefficients
B3-F3s 12:34
12:56
12:78
34:56
34:78
56:78
B3-F4s
12:34
12:34-12:56
1/8
12:56
12:34-12:78
1/8
12:56-12:78
1/8
12:78
12:34-34:56
1/8
12:56-34:56
1/8
12:78-34:56
0
34:56
12:34-34:78
1/8
12:56-34:78
0
12:78-34:78
1/8
34:56-34:78
1/8
34:78
12:34-56:78
0
12:56-56:78
1/8
12:78-56:78
1/8
34:56-56:78
1/8
34:78-56:78
1/8
56:78
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 B3 Families
(Lines) on Inbreeding at B3-F4.
Inbreeding Coefficient at B
3 - F4
0.10
0.08
Number of B 3 famil ies per parti al di al lel
24
0.06
0.04
0.02
0.00
0
10
20
30
Number of B 3 Families
40
50
Backcrossing
Why we want nine selected B3-F2
trees per B3 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.
Effect of adding sets of 20 B3-F3 progeny from
our chapter breeding program on inbreeding and
effective population size, for the same source of
blight resistance (inbreeding effective population
size doubles with each additional chapter if
different sources of blight resistance are used).
Number of
Chapters
1
2
3
4
5
Inbreeding
Coefficient
0.0207
0.0115
0.0085
0.0070
0.0060
Inbreeding Effective
Population Size
72
130
176
214
248
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.
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 B3-F2 progeny are needed per B3 line to
capture most alleles.
• If each of those nine to ten sets of B3-F2 progeny is segregated in
a separate orchard, inbreeding at F4 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 Ne).
• Chapter breeding efforts add essential genetic diversity and local
adaptation to the overall program.