Advanced Plant Breeding PBG 650 Midterm 2, Fall 2015 Name KEY Please show your work. 1) A plant breeder is evaluating a random-mating population of Russian dandelion as a potential source of natural rubber. She conducts progeny trials and estimates the breeding values of genotypes segregating at a single locus that affects rubber production. Frequency A 1A 1 A 1A 2 A 2A 2 8 pts 0.36 0.48 0.16 Genotypic Breeding Value Value 25 5.28 20 -1.32 11 -7.92 a) What is the additive genetic variance for this trait at this locus in this population? There are many ways to solve this problem with the information provided. Method 1: Use the formula VA=2pq2 p=0.36 + 0.5*0.48 = 0.36 + 0.24 = 0.60 q = 1 – 0.6 = 0.4 To calculate you could work with known formulas for the individual genotypes, or use a, d, and –a and the formula = a + d(q - p). For example, the breeding value of A1A1 is 2q = 5.28, so =5.28/(2*0.4)=6.6 Alternatively, with MP = 18, a=7, and d=2, = a + d(q - p)=7 + 2(0.4-0.6)=7-0.4=6.6 VA=2pq2=2*0.6*0.4*6.62 = 20.9088 Method 2: Calculate the variance among breeding values V (Y) f i Yi2 μY2 0.36 *5.282 0.48* ( 1.32) 2 0.16 * ( 7.92) 2 0 2 10.036 0.836 10.036 20.9088 8 pts b) She decides to cross an A1A1 genotype with an A1A2 genotype. What is the expected breeding value of the offspring from this cross? The expected breeding value of the progeny will be the average breeding value of the parents. In this case the average of 5.28 and -1.32 is 3.96/2 = 1.98 1 2) An experiment is conducted to estimate additive genetic variance from 200 half-sib families, obtained from noninbred parents. Each plot represents a family, and the families are replicated in four randomized blocks (r=4) and evaluated in two years (y=2). Source 6 pts df MS Expected Mean Square Years y-1 Blocks(Years) r-1 Families f-1 381 2 e2 r FY ry F2 Families x Years (f-1)(y-1) 183 2 e2 r FY Error y(r-1)(f-1) 55 e2 a) Calculate the genetic variance among families. F2 (MS MS ) / ry (381 183) / 8 24.75 F FY b) Calculate additive genetic variance for the reference population. 6 pts 6 pts 2A 42F 4 * 24.75 99 c) What is the heritability for this trait, assuming that selection would be based on half-sib family means averaged across blocks and years? (Hint: 𝜎𝑃2 = 𝑀𝑆(𝑓𝑎𝑚𝑖𝑙𝑖𝑒𝑠)/𝑟𝑦) F2 F2 F2 24.75 24.75 24.75 h 2 2 0.5197 381 183 P F ry y F2 X2 24.75 47.625 8 8 2 2 e 2 FY 2 10 pts 3) In the equation below, fill in the appropriate coefficients for the covariance of full-sibs when two-locus epistasis is present. CovFS 8 pts 1 2 2 2 A2 14 D2 14 AA 81 AD 1 16 2 DD 4) The mixed model equation that is used to estimate Best Linear Unbiased Predictors is Y = X + Z + Briefly explain what X, , Z and in the model represent. X is the design matrix for fixed effects Beta is the matrix of fixed effects that we wish to estimate Z is the design matrix for random effects Mu is the matrix of random effect 8 pts 5) The benefits of using BLUP values for selection rather than the average performance of an individual or family would be greatest when….(circle the best answer for each of the four criteria) Criteria Heritability Low High Parents Related Unrelated Data Balanced (genotypes are evaluated to the same extent) Unbalanced Lifespan of the species Short Long 3 6) Heritability can be estimated from parent-offspring regression. Which statement about parent-offspring regression is true? 6 pts a) Heritability is expected to be higher when it is estimated from regression on midparent values than it is for regression on a single parent. b) Heritability is expected to be higher when we evaluate more offspring from each parent. c) The slope for regression of offspring on midparent is expected to be twice as large as the slope for regression on one parent. d) The standard error of the heritability will be the same for regression on midparent values as it is for regression on a single parent. 6 pts 7) A breeder selfs individual plants from a noninbred meadowfoam population in the greenhouse. She harvests the seed from each plant and grows a row of 10 plants from each parent plant in an outdoor nursery. She hires some honeybees to intermate all of the plants in the nursery. Each row of 10 plants that came from a common parent is harvested in bulk to create a half-sib family for progeny tests in the next cropping season. Can she assume 1 that the covariance for half-sib families is 4 𝜎𝐴2 , or will she need to make an adjustment for inbreeding of the parents? Explain your answer. 1 Yes, the covariance should be 4 𝜎𝐴2 and no adjustment for inbreeding is needed. Each S1 family represents the genotype of a noninbred parent plant. 6) 6 pts 10 pts A plant breeder has been hired to initiate a quinoa breeding program in Montana. He has obtained some germplasm and plans to make diallel crosses using Griffing’s Method 4. He is not sure how he should interpret his results. a) List at least three questions that you would ask him to help him decide how to interpret his data. How were the parents chosen and what is their breeding history? How many parents are there? Is his main interest in this particular set of parents and how they perform in crosses, or does he want to understand more about the inheritance of specific traits in quinoa? b) Describe (in general) how his answers would determine the inferences that can be made from his experiment. The main issue is to determine if genotypes are fixed or random and if the assumption that the parents are a representative sample from of a random-mating reference population can be met. If he is using a random model and the necessary assumptions can be met, then he can estimate additive genetic variance and dominance variance (Model II). If the genotypes 4 are fixed and/or the assumptions are not met, then he can only estimate GCA and SCA (Model I). 7) Using 10 barley plants (A, B, C, D, E, F, G, H, I, and J), describe how you would make 1) Design I crosses (Nested) 12 pts 2) Design II crosses (Factorial) For each mating design, indicate how the crosses would form 1) A half-sib family Give an example for each mating design: You don’t have to designate all possible half-sib and full-sib families. 2) A full-sib family Nested Design Cross A as a male to females B, C, D, and E Cross F as a male to females G, H, I, and J An example of one half-sib family would include AxB, AxC, AxD, and AxE (all have A as a common parent). AxB represents a full-sib family Design II A B C D E F X X X X X G X X X X X H X X X X X I X X X X X J X X X X X An example of one half-sib family would include AxF, AxG, AxH, AxI, and AxJ (all have A as a common parent). AxF represents a full-sib family 5