Plant Science 546. Final Examination May 12, Ag.Sci. Room :00am to 12:00 noon

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1 Plant Science 546 Final Examination May 12, 2004 Ag.Sci. Room :00am to 12:00 noon Name : Answer all 16 questions A total of 200 points are available A bonus question is available for an extra 10 points Questions are arranged into two sections Section I Selection/Novel Techniques Practical Applications Section II Basic concepts and Breeding Schemes Genetics/Selection Points available from each part of each question are shown in bold square parenthesis Try to be as brief and concise as possible Please write in a legible form Show any working/calculations Make sure that any additional paper used is attached to the questionnaire

2 2 SECTION I Selection/Novel Techniques Practical Applications Total of 100 points

3 3 1. Describe the major differences in approach to methodologies used in the early, intermediate and advanced stages of selection in a plant breeding program [7 points]. Early generation selection, usually involved 1000's of breeding lines with low quantities of planting material which combined results in ineffective selection on small plots, often single plants. Often the lines to be selected are highly heterozygous genotypes. Intermediate generation selection involves 100's of breeding lines and where more planting material (i.e. seeds) are available. Proper evaluation of quantitatively inherited characters (i.e. yield, quality) can be better assessed. Breeding lines may be more homozygous. At this stage, sufficient phenotypic variation is to allow for some response from selection. Number of sites is somewhat limited. Few (10's) of lines are remaining and large quantities of planting material is available. Testing is conducted throughout the target region. May be site locations. Agronomic trials (i.e. variable nitrogen, pesticides) may be conducted. Breeding lines are confirmed and quality testing assured. On-farm testing begins. Describe which experimental designs might be appropriate for each selection stage [3 points]. EGS - single replicate designs, honeycomb designs, nested progeny designs. IGS - RCB, incomplete block designs. AGS - RCB, incomplete block, split and strip-plot, on farm testing. 2. List three areas where in vitro techniques could be used to enhance a plant breeding program. [3 points] 1. Embryogenesis, anther culture 2. in vitro disease testing 3. plant multiplication, disease-free maintenance stocks List three problems that may be associated with in vitro techniques relating to plant breeding 1. Genotypic specific characteristics 2. Unwanted somatic variation 3. High cost, experience staff, not a "recipe book" procedure [3 points]

4 4 List two methods that could be utilized by plant breeders to increase the mutation events in plant breeding. 1. Chemical (i.e. EMS) 2. Radiation (i.e. gamma or x-rays) [2 Points] Outline two factors that have to be considered when using mutagenesis techniques in a cultivar development scheme. 1. Mutations are random, usually deleterious 2. Human health concerns 3. Takes many cycles to eliminate unwanted variation (backcrossing). [4 points] 3. Why is an appropriate experimental design essential to plant breeding assessment? trials? [3 points]. To obtain a good and accurate estimate of error variance by which significant difference between factors can be estimated. List 5 field factors that need to be considered when setting out plant breeding field trials 1. Previous crop 2. Old river beds 3. Proximity to buildings, trees 4. Proximity to local right of ways 5. Past fertility, chemical application, mainly herbicides 6. Has there been a weed science trial here in the past? [5 points] Outline five difficulties that may be encountered when conducting evaluation trials off-station (i.e. regional testing): 1. Cost of travel. 2. Time involved in visiting sites. 3. Getting good cooperators. 4. Lack of on-site equipment. 5. Lack of on-site storage.

5 5 [5 points] List five uses that plant breeders could use glasshouses or growth chambers for: 1. Crossing. 2. In vitro to in vivo transfer. 3. Generate pre-breeders' seed. 4. Disease or insect testing. 5. Out of season plant growth. [5 points] 4. Progeny means and additive genetic standard deviation of six pea crosses for yield were estimated from a properly controlled experiment. Breeding objectives in this program set a minimum yield target in pea to be 1,950 lb/acre. Considering this objective, which of the six crosses shown below would have most breeding value [7 points]? Cross code Progeny mean Standard deviation 99.SP.36 1, SP.47 1, SP.197 1, SP.199 1, SP.207 1, SP.601 1, Based on (target-mean)/σ conversion and looking up ø values in standardized normal distribution tables we have higher probability of a recombinant inbred line greater than the set target to be 0.45, or 45% of the lines. This would be the progeny of choice. Would you expect to get the same answer with a significantly higher target yield (say 2200 lb/acre)? Explain your answer [3 points]. No, with increasing target values I would expect the standard deviation to become more important and this cross may not result in the highest probability of desirable number of recombinants. 5. A field trial is carried out whereby 600 F 6 pineapple progeny families are evaluated for yield potential at two different locations (Hawaii and the Bahamas, isn't plant breeding a great job?). The narrow-sense heritability for yield between the two sites is h 2 = It is your intent to select the best 35 individual lines with general adaptability to the two environments and you wish to use independent culling to select the adapted cultivars. What percentage should you select at each location to ensure that 35 genotypes are selected at BOTH locations? [10 points].

6 6 With h 2 of 0.51 then the correlation coefficient (r) would be ~ 0.7. As we are screening 600 (not 1000) and want 35 then we look up r=0.7 and a target number of [35 * 1000]/600 = 58 selections. Using the inverse tetrachoric correlation tables we obtain this number by (in tables the number is 59) by selecting 10% of lines at one site (say Hawaii) and 15% at the other (say the Bahamas) doubled haploid lines of rice were produced and screened for molecular markers. It was found that these lines were polymorphic for 4 linked molecular markers, on chromosome #1. The 200 doubled haploid lines were evaluated for yield potential over four locations in Describe ALL the processes that would be involved in determining whether any quantitative trait loci (QTL) existed, and describe the process used to map position of these QTL s [10 points]. Find the map distances between the 4 linked molecular markers (loci). Do and ANOVA of the character if interest in the QTL (say yield) and determine whether there are significant differences between the double haploid lines. Do an ANOVA to determine if any single molecular marker is related to the character of interest. Map the QTL. Describe two difficulties that might be encountered in utilizing QTL s in plant breeding selection [4 points]. They are not consistent over sites, years and certainly genotypes examined. Often the molecular information is significantly more detained and accurate than the character of interest. 7. Fully describe the procedure you might use to produce Recommended seed for distribution to farmers from having an F 6 bulk selected population. Describe the procedure you would under take in each year of seed production [6 points]. The breeder here could choose to use a bulk or a pedigree scheme here the major point being that some measure must be taken to ensure a uniform source of Breeders' seed. This is used to plant Foundation Seed, which is used to plant Certified seed, which is used to plant Registered seed which us used by farmers to grow commodity crops. Describe what is meant by: DUS: Distinct for any other cultivar that is available, Uniform over a range of environments, and Stable in performance through reproduction.

7 7 VCU: The cultivar must have value for cultivation and use, so it must have some advantage (yield, quality, etc.) over the cultivars already available. In relationship to plant breeding and cultivar release [4 points]. Describe three methods that could be used to obtain proprietary cultivar protection: [6 points] 1. PVP. 2. Plant Patent. 3. Develop hybrids. 8. Farmer Giles has been growing organic produce ever since he was contracted to supply popcorn at the Janis Joplin concert at Woodstock. In 2002, he was contracted by his royal highness Prince Charles, Duke of Edinburgh, to supply the Royal Canola Oil for the next round of social functions. He did this, and after harvesting and cleaning a bumper crop, he shipped the seed to England for crushing. On arrival Charlie did a standard genetically modified crop (GMO) test and found the organic shipment to be contaminated with Roundup Ready canola seeds. Charlie being an anti-gmo person immediately shipped the seed back to Idaho at great expense. It was discovered that the contamination arose from cross pollination of farmer Giles' crop with a neighbor's Roundup Ready variety. As a result of all this shenanigans, farmer Giles is almost bankrupt. Who should farmer Giles hold responsible, and sue for the damages he incurred: The USDA who deregulated the use of GMO Roundup Ready canola? Monsanto who developed and own the patent on the Roundup Ready gene? The breeding company that licensed the gene from Monsanto and back-crossed it into a locally adapted cultivar? The Seed company that sold the Roundup Ready canola seed to farmer Giles' neighbor? Farmer Giles' neighbor? Prince Charlie and the British Royal family? Donald Rumsfeld? All of the above? or none of the above? All of the above. Actually there is no correct answer but at some time in the near future someone must figure this out, because it's going to happen, and indeed in Canada there are several cases pending. Please explain your answer [10 points].

8 8 Section II Basic concepts and Breeding Schemes Genetics/Selection Total of 100 points

9 9 9. Three different genes have been identified for yellow stripe rust resistance in winter wheat. Each gene is positioned at different loci on the wheat genome. It is your intent to pyramid these genes (i.e. have all three gene in a single cultivar), describe the procedures you would use, using traditional breeding techniques, to develop a cultivar with all three gene, given that you start with three parents (P 1, P 2, and P 3 ) each with an different resistance gene [9 points]. Cross P 1 x P 2 > F 1, Self F 1 and assess F 2 population for resistance to yellow stripe rust. All progeny should be resistant if P 1 and P 2 have different genes conferring complete resistance. If this is so then choose a number (say 20 F 2 ) lines and cross each to a susceptible genotype, and examine the resistance in the F 2 progeny from this allelism test. If the progeny segregates as a two gene system, keep the original genotype and cross this to P 3. Screen among the progeny, but all should be resistant. You then need to do additional allelism tests to ensure that you do indeed have all three resistance genes pyramided. Lot of work. Describe any advantages that you may have if there were molecular markers that existed for each resistance gene and explain the process you would follow given that you would select plants based on these molecular markers [3 points]. If we have good molecular markers (different) for each resistance gene we do not have to do the complex and time consuming allelism tests. 10. A newly appointed barley breeder was interested in the inheritance and heritability of yield in spring barley. He chose five cultivars from a range that were grown throughout his target region and intercrossed them in all possible cross combinations (excluding reciprocals) including selfs, and he completed a Hayman and Jinks analyses. Some of the results are shown below. Analyses of Variance Source df Sum of Squares Mean Square a *** b ns b *** b Error Based on this analysis of variance we can say that there is significant additive genetic variance, no directional dominance, but significant non-directional dominance. However, we have a complication that we have significant remainder (epitasis) effects.

10 10 V i +W i Analyses of Variance Source df Sum of Squares Mean Square Between arrays *** Within arrays V i -W i Analyses of Variance Source df Sum of Squares Mean Square Between arrays ns Within arrays From these analyses of variance we have (top) significant dominance (again), and from the other (bottom) we have that only dominance is significant. The magnitude of the b 3 SS might suggest that perhaps the error has been underestimated and that epitasis is not significant. Cultivar Mean Yield V i W i Golden Promise Sunrise Reaper Malter Qual Mean We could do a regression of Vi onto Wi and get a slope of 1.03, which is not significantly different from 1 and so the additive/dominance model would be accurate to explain the variability The variance of parents (σ p 2 ) is 92.4, and the variance or array means (σ xr 2 ) is We have that A = 4/7[ ] = 99.61; D = {4[69.14] - A} = h 2 n = 1/2A/[1/2A+1/4D+E] = 0.52, therefore 52% of the total variation is additive in nature. Finally as the high yielding parents have low Vi and Wi values and the low yielding parents have high Vi and Wi value we can state that high yield is associated with an accumulation of dominant alleles. We still don't get it - do we? Complete the analyses, determine whether the additive-dominance model is appropriate, calculate the narrow-sense heritability and describe what can be determined about the inheritance of yield in spring barley [14 points].

11 Why is knowledge of plant evolution important to modern plant breeders? [4 points]. It is necessary to have knowledge of past progress in adapting crop species, if additional advances are to continue into the future. When dealing with a crop species, a plant breeder benefits from knowledge of the time scale of events that have modeled the given crop. For example, the time of domestication, geographic area of origin and prior improvements are all important and will help in setting feasible future objectives. There has always been a question as to if ancient farmers cultivated crops or livestock first (or simultaneously), describe in your own words whether mankind cultivated crop or livestock first, and explain your answer [7 points]. I've always thought that we domesticated livestock first and thereafter decided to cultivate crops in order to feed the critters. Either or both would be correct, as we really don't know for sure. 12. The following data is collected from a properly designed experiment of seed yield on segregating progeny and homozygous parents in a chickpea cross. Whereby the performance of both parents (P 1 and P 2 ), both backcross progeny (B 1 and B 2 ) and the F 3 progeny were evaluated. The data for family mean and within family variance are shown below. Design and complete a suitable test and determine whether a simple additive/dominance model of inheritance is appropriate to explain the inheritance if seed yield in chickpea [12 points]. Family Mean yield Variance for yield P P B B F P 1 = m+a; P 2 = m-a; B 1 = m+ /2a+1/2d ; B 2 = m+1/2a-1/2d; F 3 =m+1/4d 4F 3 -B 1 +B 2-2P 1-2P 2 = 0 4(m+1/4d)-(m+1/2a-1/2d)+(m+1/2a-1/2d)-4(m+a)-4(m-a) = 4m+d-m-1/2a-1/2d+m+1/2a-1/2d-2m+2a-2m-2a = 0 = 4m-m+m-2m-2m-d-1/2d-1/2d -1/2a+1/2a+2a-2a = 0 Test = 4(96.4) (102.6)-2(69.5) = 25.9 Var(Test) = 16(934) (79)+4(39) = 16,004 t oops what's the df = 25.9/126.4 = = ns Therefore the A/D model is adequate

12 We're at it again and imagining that you will get a real breeding job after you graduate. In this job you have been given the task of selecting for only disease resistance that is to be combined in a barley cultivar with one of three dwarfing genes. All disease resistance in barley is controlled by single dominant alleles, while dwarfism is controlled by a single recessive allele. A fellow worker in the company breeds barley only yield and quality, both of which are quantitatively inherited. Do you think that you and your fellow co-worker will adopt the same breeding scheme, and why? Explain the major differences between the schemes. [12 points]. No. The two schemes would likely be highly different from each other as one is dealing entirely with single gene traits and that other is dealing with polygenic traits. In the first a simple pedigree breeding scheme may be highly effective whereby plants are continually selected based on single plant evaluation of qualitative traits (which can often be successful). In addition, this first scheme, there may be test crossing involved to differentiate homozygous from heterozygous disease resistance. In the yield and quality situation, it is unlikely that any single plant selection will be conducted and it is more possible that replicated yield and quality trials will be needed. Molecular markers may be used in these schemes, in the first they may be highly successful as we are dealing with single genes of interest. In the latter QTL's will need to be considered along with all the difficulties that QTL's have shown. 14. In your job as barley breeder breeding for disease resistance combined with a recessive dwarfing gene, you have made a cross between two homozygous barley parents. One parent was resistant to yellow strip rust but susceptible to mildew, with no dwarfing gene (YYmmTT), while the other is resistant to mildew but susceptible to yellow strip rust, and has the erectoides dwarfing gene (yymmtt). It is known that the dwarfing gene and mildew resistance gene are on the same chromosome (with 20% recombination) and that the yellow rust resistance gene is on a different chromosome. After producing F 1 seed in a glasshouse you grow out the F 2 population in the field and select all plants that are dwarf and resistant to mildew. These plants are trashed separately and F 3 head rows planted the following year. Starting with 4,000 F 2 plants, how many of the F 3 head rows are: This is actually a whole lot easier than anyone thought. Or so I must assume as no-one came real close to getting the correct answer. First consider the first year of selection, where we started out (i.e. planted) 4,000 F 2 plants, and we are selecting based on dwarfism and resistance to mildew (ignore yellow rust for now as you know it segregates independently from these two). You know that dwarfism and mildew resistance are linked (20% recombination). So we have.

13 13 MT -0.1 Mt -0.4 mt -0.4 mt -0.1 MT-0.1 Mt-0.4 MMtt 0.16 Mmtt 0.4 mt-.4 mt-0.1 Mmtt 0.4 So the only F 2 phenotypes we will select are mildew resistant and dwarf which can only come from two types so we have MMtt=16/100 and mmtt=8/100 or a total of 24% evaluated so we select 960 F 2 plants and grow out 960 F 3 head rows. These F3's will all be dwarf as we have fixed dwarf (tt) and 1/3 (320) will segregate for mildew resistance the other 2/3 (640) being homozygous. The 320 that are Mmtt will segregate as 80 MMtt : 160 Mmtt : 80 mmtt. The 640 will remain MMTT. So at F3 we have 720 MMtt : 160 Mmtt and 80mmtt. Within each of these we will have the yellow rust resistance segregating in a 1:2:1 ratio in each class. So 180 MMttYY : 360 MMttYy : 180 MMttyy and 40 MMttYY : 80 MmttYy : 40 mmttyy and 20 mmttyy : 40 mmttyy : 20 mmttyy. Short in stature? All of these selected = 960. Segregating for mildew resistance? 160. Segregating for yellow rust resistance? 1/2 (i.e. 1:2:1 segregation) = 480. Short in stature, and homozygous resistant to both diseases? = 220. [16 points]. 15. Describe four ways that can be used to estimate heritability in plant breeding: 1. P 1, P 2, F 1, F 2, B 1, and B 2, or simply P 1, P 2, and F 2, etc. 2. Hayman and Jinks diallel cross. 3. Parent v off spring regress, or correlation. 4. Response from selection. [8 points] List three limitations of heritability in plant breeding: 1. Often technical assumption inherent in the theory. 2. Tend to change over time and over locations. 3. Use second order statistics which are usually less precise. [3 points]

14 In 2003, a budding oat breeder conducted an experiment whereby he grew both parents (P 1 and P 2 ) and the F 1, F 2, B 1, and B 2 families from a typical cross in his breeding program, and evaluated them for plant yield (kg/plot) in a properly designed experiment. The following variance from each family was found: σ 2 P1 = σ 2 P2 = σ 2 F1 = σ 2 F2 = σ 2 B1 = σ 2 B2 = On inspection of the population he had a hunch that it would have potential for breeding and selection. As a result he took a random sample of F 2 single plants and grew them out as head rows at F 3 on harvesting and weighing the F 3 head rows individually, he found the average yield to be 96 Mt ha -1 with variance (σ 2 F3) of he decided to select based on F 3 yield at the 10% level (Hint: i = 10% selection), what would his expected yield be if he grew the selected lines out at F 4 [8 points]. E=[ ]/4=11.6 D = 4[σ 2 B1 + σ 2 B2 - σ 2 F2 - E]; A = 2[σ 2 F2-1/4D]; σ F2 = 1/4A/σ 2 F2 D = 27.88; A= 62.1; h 2 n = 0.63 Base yield = 96 Mt ha -1 ; R = iσ F2 h 2 n = * 7.04 * 0.63 = 7.78 Yield of selected population = If this breeders were not as smart and only grew out both parents (P 1 and P 2 ) and the F 1, F 2 would this make his estimate more or less accurate, and why [4 points]. No - it would not as he is dealing with an inbreeding crop and with narrow-sense heritability. In this case he/she would have to use the broad-sense heritability where dominant genetic variance (D) changes from generation to generation with greater homozygosity.

15 15 Bonus Questions [10 points] What will be the biggest challenges facing plant breeders in the next 20 years? Virtually anything would have done here. Including get a better handle on QTL's and how they can be used. How to transform polygenic traits. How to handle GxE. How to handle changes in food prospective in the western world and starvation on other regions. Coping with multinational seed/breeding/chemical organizations. Finally, if you're in a University based program - how to avoid lay-off, furlough and administration! Bonus Bonus Question [No extra points] A cross is made between two homozygous barley parents. One of the parents is resistant to mildew, susceptible to rust and tall, AAbbTT, where A is a single doiminant gene conferring resistance to mildew, b is a recessive gene for susceptibility to rust and T is a single dominant tall gene. The other parent is susceptible to mildew, resistant to rust, and short (dwarf), aabbtt, where a is a single recessive gene for susceptibility to mildew, B is a single dominant gene for resistance to rust and t is a single recessive gene for dwarfism. The F 1 progeny are self pollinated and the F 2 progeny are screened for resistace to mildew and rust and for dwarfism. What proportion of the F 2 population will be phenotypically resistant to both diseases and be dwarf? Start with the dwarf gene, ¼ (say 16/64) of the F 2 will be dwarf (all others are therefore discarded). From these: 16/64 tt AA Aa aa 4/64 8/64 4/64 Discard the aa for Mildew susc. BB Bb bb BB Bb bb Discard the bb for 1/64 2/64 1/64 2/64 4/64 2/64 rust susc. So you have 1/64 + 2/64 + 2/64 + 4/64 = A_B_tt, which is resistant to both mildew and rust and dwarf. Now given that only the plants resistant to mildew and rust and that are dwarf are selected at F 2 and these grown on as F 3 plants, what proportion of these F 3 s will be

16 16 homozygous resistant to both diseases and dwarf? What proportion will be phenotypically resistant to both diseases and dwarf? There are 4 types of possible F 2 selections: AABBtt = 1 = 4 AABbtt = 2 = 8 AaBBtt = 2 = 8 AaBbtt = 4 = 16 Which will segregate at F 3 to be: AABBtt = 1 = 4 = All AABBtt = 4 AABBtt = 2 AABbtt = 2 = 8 AABbtt = 4 AAbbtt = 2 = Susceptible to rust AABBtt = 2 AaBBtt = 2 = 8 AaBBtt = 4 aabbtt = 2 = Susceptible to mildew AaBbtt = 4 = 1AABBtt : 2AABbtt : 1AAbbtt : 2AaBBtt : 4AaBbtt : 2AaBBtt : 1aaBBtt : 2aaBbtt : 1aaBbtt = genotypically, or 9A_B_tt : 3 A_bbtt : 3 aab_tt : 1 aabbtt. Homozygous resistant to mildew and rust and dwarf = /36 = 9/36 = ¼ Phenotypically resistant to both and dwarf = = 25/36