Possible answers to the first Plant Breeding Test 2004

1. Possible answers to the first Plant Breeding Test2004

2. 1a. Fill in the blanks in the table below [6 points].

3. 1a. Fill in the blanks in the table below [6 points].

4. 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.

5. 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].

6. Start by considering how to increase seed quantities of the female parent which is cytologically male sterile (A-cms-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.

8. 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.

9. A x B = 60 A x C = 65 A x D = 51 A x E = 53 A x F = 59 B x C = 63 B x D = 61 B x E = 59 B x F = 42 C x D = 40 C x E = 42 C x F = 37 D x E = 52 D x F = 68 E x F = 33 Single cross hybrid D x F, yield = 68 t/ha

10. A x B = 60 A x C = 65 A x D = 51 A x E = 53 A x F = 59 B x C = 63 B x D = 61 B x E = 59 B x F = 42 C x D = 40 C x E = 42 C x F = 37 D x E = 52 D x F = 68 E x F = 33 Three-way cross hybrid (A x B) x C, yield 62.5 t/ha

11. A x B = 60 A x C = 65 A x D = 51 A x E = 53 A x F = 59 B x C = 63 B x D = 61 B x E = 59 B x F = 42 C x D = 40 C x E = 42 C x F = 37 D x E = 52 D x F = 68 E x F = 33 Double cross hybrid (B x F) x (A X D), yield 62.0 t/ha

12. 3a. Complete the table below. [5 points].

13. 3a. Complete the table below. [5 points].

14. 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].

15. 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.

16. 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.

17. Outline the major features involved in selecting for end-use quality and indicate any problems that breeding for improved quality might cause. [8 points].

18. 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.

19. 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);

20. 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).

21. List three crop species that might be suitable for developing each of: Hybrid cultivars Clonal cultivars Multi-line cultivar

22. 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

23. 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]

24. 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.

25. 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.

26. 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

27. 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].

28. Backcrossing F1 Sselgel x F.Flame 50.0% Self BC1F1 Select for resistance (rr) F1 x F.Flame 75.0% Self BC2F1 Select for resistance (rr) BC1F1 x F.Flame 87.5% Self BC3F1 Select for resistance (rr) BC2F1 x F.Flame 94.8% Self BC4F1 Select for resistance (rr) BC3F1 x F.Flame 96.9% Select Homozygous resistant Self

29. 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].

30. 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.

31. 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).

32. Explain the different seed certification classes of an inbred cultivar [6 points].

33. 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.

34. 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].

35. 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.

36. 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.

37. Bonus question[5 points]I am a hybridist / skeptic (delete as appropriate) about true heterozygous advantage in hybrid crop cultivars because …….

39. The EndThank You