F 1 TtYy Tall/Resistant

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2 Question A spring barley breeding program has major emphasis in developing cultivars which are short in stature and with yellow stripe resistance. It is known that the inheritance of short plants is controlled by a single completely recessive gene (tt) over tall plants (TT), and that yellow stripe rust resistance is controlled by a single completely dominant gene (YY), over a recessive susceptible gene (yy). The tall gene locus and yellow rust gene locus are located on different chromosomes. Given that a tall resistant plant (TTYY) is crossed to a short susceptible plant (ttyy), both parents being homozygous, what would be the expected proportion of genotypes and phenotypes in the F 1 and F 2 families [12 points].

3 F 1 TtYy Tall/Resistant

4 Gametes from female parent Gametes from male parent TY Ty ty ty TY TTYY TTYy TtYY TtYy Ty TTYy TTyy TtYy Ttyy ty TtYY TtYy ttyy ttyy ty TtYy Ttyy ttyy ttyy 1 TTYY:2 TTYy:1 TTyy:2 TtYY:4 TsYy:2 Ttyy:1 ttyy:2 ttyy:1 ttyy 9 T_Y_ : 3 T_yy : 3 tty_ : 1 ttyy

5 Question How many F 3 plants would need to be assessed to ensure, with 99% certainty, that at least one plant would exist that was short and homozygous yellow rust resistant (i.e. ttyy) [8 points]. ttyy = 9/64 = # = Ln[1-0.99] = Ln(0.01) Ln[ ] Ln(0.8594) = 30.39, need 31 F 3 plants

6 Question A cross is made between two homozygous barley plants. One parent is tall and has single (completely dominant) resistance to yellow rust, but highly susceptible to powdery mildew (TTYYmm), and the other is short (dwarf) with single (completely dominant) resistance to powdery mildew, but is susceptible to yellow rust (ttyymm). Tall plants (T_) are completely dominant to short (tt). If a sample of F 1 plants from this cross were self pollinated, and a population of 12,800 F 2 plants grown. At harvest the short (dwarf) plants are selected and seed from only these plants are planted out in head-rows at F 3. How many of the F 3 head-rows would be resistant to both yellow rust and powdery mildew? How many will be resistant to powdery mildew but susceptible to yellow rust? And, how many will be susceptible to both diseases? [12 points].

7 Question Assume fist that we will select for only the dwarf short these at F 2 (i.e. tt). This being the case we will retain 1/4 of the initial population based on this height selection. From the 12,800 F 2 plants we'll select 3,200 and so we'll have that many F 3 head-rows planted. Now that we have selected for a single gene recessive, we can forget about that height trait as it'll be fixed. We need then only concern ourselves with the two segregating loci being rust and mildew resistance.

8 Question How many will be resistant to powdery mildew but susceptible to yellow rust? And, how many will be susceptible to both diseases? The F 3 head rows will segregate phenotypically as: 25 Y_M_: 15 Y_mm : 15 yy M_: 9 yymm Therefore: Resistant to both = 25/64 x 3,200 = 1,250 Resistant to only rust = 15/64 x 3,200 = 750 Resistant to only mildew = 15/64 x 3,200 = 750 Susceptible to both = 9/64 x 3,200 = 450

9 Question Two homozygous barley parents were crossed to produce an F 1 family. One parent was tall with awns and the other was short and awnless. Tall plants are controlled by a single dominant gene and awned plants are also controlled by a single dominant gene. The F 1 family was crossed to a plant which was short and awnless and the following number of phenotypes observed: Phenotype # observed Tall, awned 954 Short, awned 259 Tall, awnless 221 Short, awnless 966

10 Gametes from female parent Gametes from male parent TA-0.4 Ta-0.1 ta-0.1 ta-0.4 TA 0.4 TTAA 0.16 TTAa 0.04 TtAA 0.04 TtAa 0.16 Ta 0.1 TTAa 0.04 TTaa 0.01 TtAa 0.01 Ttaa 0.04 ta 0.1 TtAA 0.04 TtAa 0.01 ttaa 0.01 ttaa 0.04 ta 0.4 TtAa 0.16 Ttaa 0.04 ttaa 0.04 ttaa TTAA:8 TTAa:1 TTaa:8 TtAA:34 TtAa:8 Ttaa:1 ttaa:8 ttaa:16 ttaa 66 T_A_ : 9 T_aa : 9 tta_ : 16 ttaa

11 Question What is the difference between linkage and pleiotropy? [2 points]. Linkage is when alleles at two loci do not segregate independantly and hence there is linkage disequilibrium in segregating populations. The cause is that the two loci are located on the same chromosome. Pleiotropy is where two characters are controlled by alleles at a single locus.

12 Question Two homozygous squash plants were hybridized and an F 1 family produced. One parent was long and green fruit (LLGG) and the other was round and yellow fruit (llgg) F 2 progeny were examined from selfing the F 1 's and the following number of phenotypes observed: L_G_ L_gg llg_ llgg Explain what may have caused this departuure from a 9:3:3:1 expected frequency of phenotypes [4 points].

13 Question L_G_ L_gg llg_ llgg Explain what may have caused this departuure from a 9:3:3:1 expected frequency of phenotypes [4 points]. This departure from a 9:3:3:1 ratio could be caused by recessive epistasis, where ll is epistatic to G, so llg_ and llgg have the same phenotype.

14 Question Carry out an appropriate test to confirm your hypothesis. L_G_ L_gg llg_ llgg Observed Expected Difference D 2 /exp df = 0.59 ns

15 Question 1a. A cross is made between two homozygous barley plants. One parent is tall and with short leaf margin hairs (TTss), and the other is short with long leaf margin hairs (ttss). Single genes control both plant height and leaf margin hair length. Tall plants (T_) being completely dominant to short (tt), and long leaf margin hairs (S_) dominant to short (ss). If a sample of F 1 plants from this cross were self pollinated, a large population of F 2 plants grown, and their phenotype for height and margin hairs noted, what would be the expected ratio of phenotypes based on the following genetic situations. [8 points].

16 Genetic situation Plant Phenotype T_S_ T_ss tts_ ttss Complete dominance: T dominant to t, S dominant to s Duplicate recessive epistasis: tt epistasis to S and ss; ss epistatic to T and tt Recessive epistasis: tt epistatic to S and ss Duplicate dominant epistasis: T epistasis to S and ss; S epistasis to T and tt

17 Genetic situation Plant Phenotype T_S_ T_ss tts_ ttss Complete dominance: T dominant to t, S dominant to s. Duplicate recessive epistasis: tt epistasis to S and ss; ss epistatic to T and tt. Recessive epistasis: tt epistatic to S and ss. Duplicate dominant epistasis: T epistasis to S and ss; S epistasis to T and tt.

18 Question 2. A spring wheat breeding program aims to develop cultivars that are resistant to foot-rot, controlled by a single completely dominant gene (FF) which confers resistance, and that are resistant to yellow strip rust, which is controlled by a single dominant (YY) gene conferring resistance. A cross is made between two parents where one parent is resistant to foot rot and susceptible to yellow strip rust (FFyy) while the other is resistant to yellow strip rust but susceptible to foot rot (ffyy). A sample of F 1 plants was self-pollinated, without selection, and a large sample of F 2 plants were grown and allowed to self-pollinate. At harvest, only plants that were phenotypically resistant to both diseases (foot-rot and yellow strip rust) are retained and used to plant a large population of F 3 s.

19 Question 2. What proportion of these F 3 plants would be expected to be: Resistant to foot rot and yellow strip rust = Resistant to foot rot but susceptible to yellow strip rust = Susceptible to foot rot but resistant to yellow strip rust = Susceptible to both foot rot and yellow strip rust = [10 points].

20 F 3 F 2 FFYY FFYy FfYY FfYY Total 4/64 8/64 8/64 16/64 FFYY FFYy Ffyy FfYY FfYy Ffyy ffyy ffyy Ffyy

21 Question 2. What proportion of these F 3 plants would be expected to be: Resistant to foot rot and yellow strip rust = 25 Resistant to foot rot but susceptible to yellow strip rust = 5 Susceptible to foot rot but resistant to yellow strip rust = 5 Susceptible to both foot rot and yellow strip rust = 1 [10 points].

22 Question 4a. Potato cyst nematode resistance is controlled by a single completely dominant allele (R). A cross is made between two auto-tetraploid potato cultivars where one parent is resistant to potato cyst nematode and the other is completely susceptible. Progeny from the cross are examined and it is found that 83.3% of the progeny are resistant to potato cyst nematode. What can be determined about the resistant parent in the cross? [4 points]. One parent must be nulliplex (rrrr), and one resistant (Rrrr, RRrr, RRRr, or RRRR). If RRRr or RRRR all progeny are resistant. If Rrrr 50% are resistant (1:1). If RRrr then 83% (5:1) are resistant. Answer parent is duplex (i.e.. RRrr x rrrr).

23 Question 4b. A breeding program aims to produce potato parental lines that are either quadriplex or triplex for the potato cyst nematode resistant gene R. A cross is made between two parents, which are know to be duplex for the resistant gene R. What would be the expected genotype and phenotype of progeny from this diploid x diploid cross? [4 points].

24 Question 4b. 1 RR 4 Rr 1 rr 1 RR 1 RRRR 4 RRRr 1 RRrr 4 Rr 4 RRRr 16 RRrr 4 Rrrr 1 rr 1 RRrr 4 Rrrr 1 rrrr

25 Question 4b. 1 RR 4 Rr 1 rr 1 RR 1 RRRR 4 RRRr 1 RRrr 4 Rr 4 RRRr 16 RRrr 4 Rrrr 1 rr 1 RRrr 4 Rrrr 1 rrrr 1 RRRR: 8 RRRr : 18 RRrr : 8 Rrrr : 1 rrrr

26 Quantitative Genetics