Constructing and testing a gene-for gene model
Constructing and testing a gene-for gene model
A crop has two well characterized genes which control resistance to a fungal pathogen with known races 1-3.Line 1 has gene R1 and line 2 has gene R2. After testing a large number of fungal isolates with lots of germplasm, you find that you have two new races and six new sources of resistance.Table 1 shows the reactions of the five fungal races on the different lines that you have identified.Make a model for the genetics of resistance in these lines.Indicate which resistance genes you think lines C through H have and which avirulence genes you think each race has.
The data in Table 2 correspond to segregation ratios (resistant : susceptible) of F2 progeny made from a cross between the two parents given.All of the parental lines are homozygous.The Line U is a universal suscept (i.e. carries no known resistance genes).Use this genetic data to test your model and make any adjustments necessary.
TABLE 1
| Race | |||||
| Plant line | 1 | 2 | 3 | 4 | 5 |
| A (R1) | R | R | S | S | S |
| B (R2) | R | S | R | S | S |
| C | R | R | R | S | S |
| D | R | S | S | R | S |
| E | R | S | R | R | S |
| F | R | R | R | S | S |
| G | R | R | S | R | S |
| H | R | R | R | R | R |
TABLE 2
| Race | |||||
| Plant line | 1 | 2 | 3 | 4 | 5 |
| A x U | 125:36 | 122:38 | 0:161 | 0:148 | 0:170 |
| B x U | 126:44 | 0:150 | 130:45 | 0:169 | 0:155 |
| C x U | 148:8 | 123:43 | 127:40 | 0:160 | 0:148 |
| D x U | 115:44 | 0:148 | 0:157 | 126:39 | 0:160 |
| E x U | 136:7 | 0:140 | 128:39 | 124:43 | 0:167 |
| F x U | 154:12 | 128:42 | 133:39 | 0:171 | 0:150 |
| G x U | 151:13 | 119:47 | 0:159 | 130:36 | 0:164 |
| H x U | 125:41 | 145:9 | 133:41 | 124:42 | 131:42 |
Chi-Square Test
Chi-square test for segregation ratios:
Χ2 = ∑ [(Obs. – Exp.)2 / Exp.]
D.f. = number of classes – 1
Obs. = observed counts in each class
Exp. = expected counts in each class based on Mendelian inheritance (or recombination frequency-adjusted inheritance)
For each of the segregation ratios in Table 2 (only the ones with resistant plants), calculate a chi-square for the null hypothesis that the observed class frequencies equal the expected frequencies for a single R gene vs. two R genes.Adjust your model based on these tests.
Chi-Square Distribution
| df | 0.100 | 0.050 | 0.025 | 0.01 | 0.005 |
| 1 | 2.706 | 3.841 | 5.024 | 6.635 | 7.879 |
| 2 | 4.605 | 5.991 | 7.378 | 9.210 | 10.597 |
| 3 | 6.251 | 7.815 | 9.348 | 11.345 | 12.838 |
| 4 | 7.779 | 9.488 | 11.143 | 13.277 | 14.860 |
| 5 | 9.236 | 11.070 | 12.833 | 15.086 | 16.750 |
