Plant Science 446/546

Class Test March 25, 1996

Ag.Sci. Room 339

9.00pm to 10.30pm

Name :

Answer all 7 questions

A total of 100 points are available

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 1. A cross is made between two homozygous barley parents. One parent is tall and resistant to mildew (i.e., TTRR) and the other parent is short and susceptible to mildew (i.e., ttrr). Height and mildew resistance are both qualitatively inherited with tall is dominant to short and resistant dominant to susceptible.

1a. If the F1 family from this cross were self pollinated how many F2 plants would you need to grow to ensure a 99% certainty of having at least one plant which was resistant to mildew and short [4 points].

1b. Consider now the same cross (TTRR x ttrr) and in a breeding program 6400 F2 plants were evaluated. At harvest only the short mildew resistant plants were selected and grown as F3 head rows (i.e. only these selected types were evaluated at the F3 stage). Determine the expected number of genotypes and phenotypes you would have when you harvest the F3 rows [6 points]. 2a. Given values for the variance of the F2, and variance of both parents (P1 and P(2) and 2 the F1, estimate h b and explain what this value means in genetic terms [5 points].

V(F2) = 436.72 V(F1) = 111.72 V(P1) = 164.13 V(P2) = 109.33

2b. Given the variance from two parents (P1 and P2), the F1, F2, B1 and B2 families, 2 estimate the narrow-sense heritability (h n) and explain what the value means in genetic terms [10 points].

V(P1) = 9.5 V(P2) = 7.4 V(F1) = 8.6 V(F2) = 17.7 V(B1) = 14.3 V(B2) = 15.2

3a. Past researchers have shown that the inheritance of plant height in Brassica napus can be explained by an additive/dominance model. Below are shown family means, the standard errors (s.e.), (NOTE these are not the standard errors of the mean), and the number of plants that these data are estimated from the P1, P2, F1 and F2 families.

Number Family Mean s.e. Plants P1 40.2 0.142 31 P2 19.3 0.151 31 F1 35.4 0.099 31 F2 22.7 0.462 31

Using an appropriate statistical test determine if the result from past researchers is true based on these data (i.e. that the additive/dominance model is adequate to describe the observed variation) [12 points].

3b. If the additive/dominance model does not explain the variation found for plant height, list three reasons what might be the cause? [3 points].

1. 2. 3.

4a. Explain the meaning of a "Test cross" as applied to testing linkage disequilibrium in qualitative genetics. [1 point].

4b. Two homozygous barley genotypes are chosen for a linkage study. One of the parents had a long awn and short in stature. The other parent has a short awn and is tall in stature. Long awn is controlled by a single dominant gene (AA = Long awn = dominant over aa) and plant height is controlled by a single dominant gene (TT = Tall = completely dominant over tt). The two lines are crossed (i.e ttAA x TTaa) and the resulting F1 is test crossed to a homozygous genotype with short awn and short stature (aatt). 3000 plants from the test cross are grown and the following phenotypic frequencies were observed:

Tall and Long awn (TA) 308 Short and Long awn (tA) 1076 Tall and Short awn (Ta) 1324 Short and Short awn (ta) 292

Determine the percentage recombination [2 point].

4c. A sample of seed from the original F1 (i.e., TtAa) was increased to the F2 and the following are the numbers of plant types observed: Tall and Long awn (TA) 2650 Short and Long awn (tA) 1133 Tall and Short (Ta) 1163 Short and Short awn (ta) 54

Based on the recombination percentage from question 4b (above), determine the expected number of each phenotype in the F2. Use a Chi-square test to determine if your expected numbers are the same are those observed in the F2, write answer on next page [12 points].

5. Consider that the additive/dominance model of inheritance is adequate to explain the variation in the cross between P1 and P2, two diploid homozygous parents.

5a. In terms of m, [a] and [d], what would be the expected performance of each parent (P1 and P2), the F1 and F2 families and both backcross families (B1 and B2) [6 point].

P1 = P2 = F1 = F2 = B1 = B2 =

5b. In a single cross between two homozygous barley cultivars (P1 and P2) the yield of P1 was 3694 lbs/acre, the yield of P2 was found to be 2744 lbs/acre and the one of the backcross families (B1) showed a yield of 3622 lbs/acre. Given that the additive/dominance model is adequate to explain genetic variation for yield in barley, what would the expected yield of the F1, F2 and B2 families be? [4 points].

6a. Explain (using A as a dominant allele and a as a recessive allele) the difference between genotypes that are nulliplex, simplex, duplex, triplex and quadriplex for a single dominant gene [2 points].

6b. Assuming no complications such as double reduction, what would be the expected ratio of nulliplex, simplex, duplex, triplex and quadriplex resulting from a cross between two autotetraploid lines that are duplex for a single gene (i.e. AAaa x AAaa) [8 points].

7. A new oil crop (Brassica gasolinous) has been discovered which may have potential as a renewable biological fuel oil substitute. This diploid species is tolerant to inbreeding and is self compatible. A preliminary genetic experiment was designed to examine the inheritance of seed yield (YIELD) and percentage oil content (%OIL). This experiment involved a 4 x 4 half diallel (including selfs). The four homozygous parental lines are represented by the codes AAA, BBB, CCC and DDD. The half diallel array values (averaged over two replicates), array means, general combing ability (GCA) values, mean squares from the analyses of variance (Griffin style), Vi and Wi values (Hayman & Jinks analysis), variance of array means (Vxr) and parental variances (Vp), and the one-way analyses of variance for Vi+Wi and Vi- Wi are shown below for each character.

Array Means

YIELD OIL

AAA 40.5 20.5 BBB 38.5 29.5 20.5 25.0 CCC 37.0 28.0 19.5 23.0 26.0 30.5 DDD 32.5 20.5 18.5 10.0 24.5 27.5 31.0 36.0

AAA BBB CCC DDD AAA BBB CCC DDD

Array means 37.1 29.1 25.8 20.4 22.1 24.7 27.6 29.8

GCA values +9.0 +1.0 -2.3 -7.7 -3.9 -1.3 +1.6 +3.7

Analyses of Variance

Source df YIELD %OIL

G.C.A. 3 796.5 180.6 S.C.A. 6 90.5 30.1 Replicate Blocks 1 0.4 0.1 Replicate error 9 51.48 5.72

Vi and Wi values

YIELD %OIL Vi Wi Vi Wi

AAA 46.2 171.3 15.6 50.7 BBB 218.2 376.7 36.3 75.0 CCC 297.7 436.7 58.3 98.0 DDD 344.3 472.3 97.3 132.3

Vxr and Vp values YIELD %OIL

Vxr 196.9 44.4 Vp 687.6 180.7

One-way analyses of Variance

YIELD %OIL Source df Vi+Wi Vi-Wi Vi+Wi Vi-Wi

Between 3 8760 28.0 628 7.68 Within 4 133 17.0 115 4.77

2 7a. Without using regression, estimate the narrow-sense heritability (h n) for seed yield, and explain this values in genetic variance terms [5 points].

7b. Explain the analyses for YIELD and outline any conclusions which can be drawn for these data [8 points].

7c. Explain the analysis for %OIL and outline any conclusions which can be drawn from these data [8 points].

7d. Which one of these four genotypes would you choose as a parent in your breeding program and explain your choice [4 points].

BONUS QUESTION

Describe any difficulties suggested from these analyses in a breeding program designed at selecting lines with high yield and high percentage of oil [5 bonus points].