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 |


0 comments