Download Test 1-2004

Possible answers to the first Plant
Breeding Test
2004
1a. Fill in the blanks in the table below [6 points].
Common name
Scientific name
Region of origin
Glycine max
Northern China
Pea
Syria, Jordan, Israel
Macadamia nut
Macadamia spp.
Sunflower
Helianthus
Potato
Peru
Eleusine coracana
Mid-east Africa
1a. Fill in the blanks in the table below [6 points].
Common name
Scientific name
Region of origin
Soybean
Glycine max
Northern China
Pea
Pisum situva
Syria, Jordan, Israel
Macadamia nut
Macadamia spp.
Australia
Sunflower
Helianthus
South-eastern USA
Potato
Solanum tuberosum
Peru
Millet
Eleusine coracana
Mid-east Africa
1b. List five plant characteristics attributed to
deliberate selection by mankind/plant breeders which
might not have been selected for by natural evolution.
[5 points]
 Less spiny.
 Reduced toxins.
 Larger seeds or fruits.
 Reduced seed dormancy.
 Reduced seed dispersal
mechanism.
2a. Outline (using diagrams as
necessary) a cytoplasmic male sterile
system for producing hybrid seed
involving a CMS genotype, a maintainer
line, and fertility restoration line.
Clearly label the cytoplasm and nuclear
genes of importance in each line and the
resulting hybrid. [5 points].
Start by considering how to increase seed quantities of
the female parent which is cytologically male sterile (Acms-rfrf). This is done by producing an isogenic line
which differs only in that it is male fertile (A’-n-rfrf).
Both A-cms and A’-n have no fertility restoration genes.
After crossing together we have AA'-cms which is male
sterile, with no restoration genes (rfrf). This becomes the
female cross in the hybrid scheme and is crossed to B-n
which is pollen fertile and is homozygous for the
dominant restoration gene (RfRf). This cross results in
hybrid seed of AA'B-cms-RfRf which is pollen fertile as
the restoration allele overrides the cms. The process is
illustrated in diagram form below.
CMS Production System
A-cms - rfrf
male-sterile
x
A’-n - rfrf
male-fertile
AA’-cms - rfrf
male-sterile
x
A’ Maintainer
B-n - RfRf
male-fertile
AA’B-cms - Rfrf
male-fertile
2b. A successful hybrid corn program has
identified 6 parents (labeled A, B, C, D, E,
and F). These lines were crossed in a half
diallel design and the pair-wise hybrids
evaluated in field trials. From these trials
the following yields (t/ha) were obtained.
A x B = 60
A x E = 53
B x D = 61
C x D = 40
D x E = 52
A x C = 65
A x F = 59
B x E = 59
C x E = 42
D x F = 68
A x D = 51
B x C = 63
B x F = 42
C x F = 37
E x F = 33
Single cross hybrid
D x F, yield = 68 t/ha
A x B = 60
A x E = 53
B x D = 61
C x D = 40
D x E = 52
A x C = 65
A x F = 59
B x E = 59
C x E = 42
D x F = 68
A x D = 51
B x C = 63
B x F = 42
C x F = 37
E x F = 33
Three-way cross hybrid
(A x B) x C, yield 62.5 t/ha
A x B = 60
A x E = 53
B x D = 61
C x D = 40
D x E = 52
A x C = 65
A x F = 59
B x E = 59
C x E = 42
D x F = 68
A x D = 51
B x C = 63
B x F = 42
C x F = 37
E x F = 33
Double cross hybrid
(B x F) x (A X D), yield 62.0 t/ha
3a. Complete the table below. [5 points].
Plant
Phenotype
Pest Genotype
A_B_ccD_eeF_
a’a’B’b’c’c’D’D’e’e’f’f
Plant
response
Susceptible
aaB_ccD_E_ff
a’a’B’B’c’cD’D’E’E’f’f
Resistant
’
A_B_C_D_E_ff
aabbccddE_ff
a’a’b’b’c’c’d’d’E’E’f’f’
Resistant
3a. Complete the table below. [5 points].
Plant
Phenotype
Plant
response
Pest Genotype
A_B_ccD_eeF_
a’a’B’b’c’c’D’D’e’e’f’f
Resistant
’
aaB_ccD_E_ff
-‘-‘b’b’-‘-‘d’d’e’e’-‘-‘
Susceptible
a’a’B’B’c’cD’D’E’E’f’f
A_- - C_ - - - -F_
Resistant
’
A_B_C_D_E_ff
a’a’b’b’c’c’d’d’E’E’f’f’ Resistant
aabbccddE_ff
-‘-‘-‘-‘-‘-‘-‘-‘e’e’-‘-‘
Resistant
Explain the difference between vertical and
horizontal disease resistance and outline any
advantages or disadvantages of each form of
resistance as it relates to cultivar
development. [5 points].
Resistance that is controlled by many genes
shows a continually variable degree of
resistance and is referred to as horizontal
resistance. Vertical resistance is associated
with the ability of single genes to control
specific races of a disease or pest.
Horizontal resistance is more durable as the
pest must overcome all the resistance genes.
The advantage of vertical resistance is that
individual alleles can be readily identified
and transferred from one genotype to
another.
Horizontal resistance is more durable as the
pest must overcome all the resistance genes.
Conversely vertical resistance is less durable,
particularly to diseases with a rapid spread,
like air-borne fungi. However, single gene
resistance to many pests has remained
functional and effective over many years (i.e.
potato cyst nematode). Indeed developing
horizontal resistance can be slow as it is
difficult to accumulate multiple resistance
genes into a single genotype.
Outline the major features involved in
selecting for end-use quality and indicate any
problems that breeding for improved quality
might cause. [8 points].
The major difficulty in breeding for improved quality
characteristics is being able to test many (often
several thousand) lines quickly, cheaply, and in an
effective manner which simulates actual large-scale
quality of say a bread making plant. Quality in many
crops is a subjective matter where by individual users
(say people) have specific preferences on what
constitutes “good quality”. Indeed for many process
foods, processors want raw products to have no taste
as it is easier for them to produce a uniform product
by adding flavor, or indeed color. Finally it is often
difficult to obtain what constitutes good quality from
industry as they are very coveted about their
processing process.
List, and briefly describe four plant parts that can
be responsible for asexual reproduction.
[8 points].
 a bulb is a large bud with a stem at its lower
end (e.g. an onion);
 a corm is very much like a bulb in size and
form but has a different internal structure
(e.g. crocus);
 a rhizome is a horizontal stem that grows at
or below the soil surface (e.g. iris);
List, and briefly describe four plant parts that can
be responsible for asexual reproduction.
[8 points].
 a stolon is a stem that grows horizontally on
the soil surface (e.g.strawberry);
 a tuber is a swollen stem that enlarges
beneath the soil surface. The eyes of a tuber
are the sites for buds from which stems and
roots can develop (e.g. potato).
List three crop species that might be suitable
for developing each of:
Hybrid cultivars
Clonal cultivars
Multi-line cultivar
List three crop species that might be suitable
for developing each of:
Hybrid cultivars
Corn; Tomato; Canola
Clonal cultivars
Potato; Strawberry; Apple
Multi-line cultivar
Wheat; Barley; Pea
Explain the difference between a pedigree
breeding scheme and a bulk breeding scheme
indicating any advantages or disadvantages
of each scheme when breeding new barley
cultivars. [8 points]
In a bulk breeding scheme the major advantage is that
conscious selection is not attempted until plants have
been selfed for a number of generations and hence plants
are nearly homozygous. This avoids the difficulty of
selection among segregating populations where
phenotypic expression will be greatly affected by levels
of dominance in the heterozygotes. This method is also
one of the least expensive methods of producing
populations of inbred lines. The disadvantage of this
scheme is the time from initial crossing until yield trials
are grown. In addition, it has often been found that the
natural selection, which occurs through bulk population
growth, is not always that which is favorable for growth
in agricultural practice.
In a pedigree breeding scheme, single plant
selection is carried out at every generation, even
amongst segregation lines. If selection is effective
on a single plant basis, pedigree breeding schemes
allow inferior genotypes to be discarded early in
the breeding scheme, without the need of tested in
more extensive, and costly, yield trials. However,
pedigree breeding schemes offer little opportunity
to select for quantitatively inherited characters
until much later in the breeding scheme.
List three techniques that can be used to
accelerate breeding material towards
homozygosity in inbred line cultivar
development [6 points].
 Single seed descent.
 Off-season sites.
 Doubles haploidy
A world famous Scottish plant breeder, Angus McTavish,
has identifies a new single recessive gene resistance to
cereal root nematode, a major pest in your target region.
Angus, being a Scot and the salt of the earth every one of
them, kindly provides you with seed from his malting
quality cultivar ‘Sselgel’ which is homozygous for the
resistance gene. However, this cultivar is not adapted to
your growing condition around Budville, and only
produces half the yield of your newly released nematode
susceptible brewing cultivar called ‘Fire Flame’. Outline
all the stages of a scheme which will allow you to
develop a new cultivar (called ‘Pmihcknurd’) for the
Budweiser beer company which will be 95% ‘Fire Flame’
with nematode resistance. [8 points].
Backcrossing
Sselgel x F.Flame
F1 x F.Flame
BC1F1 x F.Flame
BC2F1 x F.Flame
BC3F1 x F.Flame
F1
BC1F1
BC2F1
BC3F1
BC4F1
Select Homozygous
resistant
50.0%
Self
Self
Self
Self
Select for
resistance (rr)
75.0%
Select for
resistance (rr)
87.5%
Select for
resistance (rr)
94.8%
Select for
resistance (rr)
96.9%
Self
Describe, (using a suitable diagram if necessary) a breeding
scheme used to develop a superior clonal strawberry
cultivar. Outline any advantages or disadvantages of the
clonal breeding scheme. [9 points].
Year 1
Crosses
Year 2
Screen 10,000
seedlings in glasshouse
Year 3-4
Unreplicated small
plot field trial
Year 5-6
Replicated field trial
100 entries x 4 reps
Disease free
increase 100 entries
Year 7-8
Replicated field trial
30 entries x 3 sites
Disease free
increase 30 entries
Replicated field trial
Year 9-10 10 entries x 12 sites
Disease free
increase 10 entries
Each year 100 different parental cross combinations
will be made in a glasshouse and 100 seedlings from
each cross will be grown in 4 inch pots in a glasshouse.
Individual plants will be selected according to berry
size, shape and color, and the best 1000 lines will be
selected and grown clonally in the field at one site.
Based only on visual appraisal, clones will be selected
and the best 100 will be increased clonally and grown
in a replicated yield trial in the 5th-6th year. At this
stage these 100 lines will also be maintained and
increased at a location where disease risks are
minimal.
In the 7th and 8th year the best 30 lines are tested in
yield trials at three locations. Clonal material for
these trials will come from the disease free location in
year 4. After the 10th year regional trials the most
superior cultivar will be released. Advantage of clonal
breeding is that the genotype of selected lines is fixed
genetically as grown of subsequent generations are
cloned. Care must be taken to maintain disease free
stocks. Clonal reproduction limits the plot sizes used
in the yield trials. Yield trials take two years for get
data (the transplant year and the year following).
Explain the different seed certification classes of an
inbred cultivar [6 points].
The different certification classes are Foundation
seed, Certified seed and Registered seed. A fourth
class of seed is Breeders’ seed but this class of seed
does not require State certification and inspection by
State inspectors. Foundation seed must be produced
by planting either previous Foundation seed or
Breeder’s seed. Certified seed can only be produced
by planting Foundation seed, and Registered seed can
only be produced by planting either Foundation seed
of Certified seed. If Foundation seed is planted and
the criteria set for Certified seed is not met but the
requirements for Registered seed are achieved then
the seed can be certified as Registered, skipping the
Certified stage.
You have identified one of your advanced breeding lines
which you intend to release as a new cultivar. Outline a
scheme you will use to develop Breeder’s Seed. [5 points].
Either a bulk or pedigreed scheme could be use. In
a bulk scheme the breeder would collect seed from
a number of individual plants (usually as single
heads in cereals or whole plants in say pea). Plants
are chosen to be highly homogeneous. Seed from
the selected plants will be trashed as a bulk and the
Breeders’ seed planted using this bulked seed.
In a pedigree scheme a breeder sill select a
number (usually 200-400) which show
homogeneity. Seed is threshed separately from
each plant and planted as head-row plots the
following year. The head row-plots are inspected
for uniformity and off-types discarded. Quality is
often checked to ensure homogeneity. At harvest
only the most homogeneous head row-plots are
harvested and combined and used to plant the
Breeder’s seed.
Bonus question
[5 points]
I am a hybridist / skeptic
(delete as appropriate)
about true heterozygous advantage in
hybrid crop cultivars because …….
The End
Thank You