plant pathology
BUKET SAHIN PLPTH 732 Fall 2020
Homework #3: Calculating linkage
1) You make a test cross from the following cross: R1 r2 / R1 r2 r1 R2 / r1 R2. You inoculate the progeny with a race that is avirulent on both genes and get 2 susceptible progeny out of 200 test cross progeny. What is the recombination distance? How many plants would be required to detect the same recombination distance with F2 progeny?
R1 r2 / R1 r2 x r1 R2 / r1 R2
F1: R1 r2 / r1 R2
Tester
gamete
Gametes from heterozygous parent
(1-r) / 2 (1-r) / 2
Non-recombinants
r / 2 r / 2
Recombinants
R1r2
r1R2
R1R2
r1r2
r1r2
R1r2
r1r2
r1R2
r1r2
R1R2
r1r2
r1r2
r1r2
Estimation of Recombination Distance
Test cross progeny:
Recombination frequency (distance)
= #recombinants / total progeny
= (2 / 200) x 2
= 2% or 2 map units
R1 r2 / R1 r2 x r1 R2 / r1 R2
F2 Progeny
Frequency
(1-r) / 2
(1-r) / 2
r / 2
r / 2
Gametes
R1r2
r1R2
R1R2
r1r2
(1-r) / 2
R1r2
(1-r)2 /4
(1-r)2 /4
(r(1-r))/4
(r(1-r))/4
(1-r) / 2
r1R2
(1-r)2 /4
(1-r)2 /4
(r(1-r))/4
(r(1-r))/4
r / 2
R1R2
(r(1-r))/4
(r(1-r))/4
r2/4
r2/4
r / 2
r1r2
(r(1-r))/4
(r(1-r))/4
r2/4
r2/4
F2 Progeny:
Expected frequency of r1r2 / r1r2 class
= r2 / 4
= (0.02)2 / 4
= 0.0001 (0.01%)
10,000 plants required to detect 2%.
2) You make a test cross from the following cross: R1 r2 / R1 r2 r1 R2 / r1 R2. You inoculate the progeny with a mixture of two races: Race A is virulent on R1 and avirulent on R2, Race B is avirulent on R1 and virulent on R2. You are fortunate enough to be able to score the races separately. You find 95 test cross progeny that are resistant to race A and susceptible to race B, 101 test cross progeny that are resistant to race B and susceptible to race A, three progeny that are resistant to both races and one that is susceptible to both. What is the recombination distance?
R1 r2 / R1 r2 r1 R2 / r1 R2
F1: R1r2 / r1R2
Tester
gamete
Gametes from heterozygous parent
(1-r) / 2 (1-r) / 2
Non-recombinants
r / 2 r / 2
Recombinants
R1r2
r1R2
R1R2
r1r2
r1r2
R1r2
r1r2
r1R2
r1r2
R1R2
r1r2
r1r2
r1r2
R1r2 / r1r2
r1R2 / r1r2
R1R2 / r1r2
r1r2 / r1r2
Race A
avr1 Avr2
S
R
R
S
Race B
Avr1 avr2
R
S
R
S
Number of progeny
101
95
3
1
Estimation of Recombination Distance
Test cross progeny:
Recombination frequency (distance)
= #recombinants / total progeny
= (3+1) / 200
= 2% or 2 map units
3) You make a test cross from the following cross: R1 R2 / R1 R2 r1 r2 / r1 r2. You inoculate the progeny with a mixture of two races: Race A is virulent on R1 and avirulent on R2, Race B is avirulent on R1 and virulent on R2. You are unable to score the races separately, but you find 99 out of 200 progeny that are resistant to both races. What is the recombination distance?
R1 R2 / R1 R2 r1 r2 / r1 r2
F1: R1 R2 / r1 r2
Tester
gamete
Gametes from heterozygous parent
(1-r) / 2 (1-r) / 2
Non-recombinants
r / 2 r / 2
Recombinants
R1R2
r1r2
R1r2
r1R2
r1r2
R1R2
r1r2
r1r2
r1r2
R1r2
r1r2
r1R2
r1r2
R1R2 / r1r2
r1r2 / r1r2
R1r2 / r1r2
r1R2 / r1r2
Race A
avr1 Avr2
R
S
S
R
Race B
Avr1 avr2
R
S
R
S
Number of progeny
99
–
–
–
Estimation of Recombination Distance
Test cross progeny:
Recombination frequency (distance)
= #recombinants / total progeny
= 1-r = (99/200) x 2
r = 1% or 1 map unit
4) Calculate the expected segregation ratios (R:S) for F2 and test cross progeny given each scenario below. Assume classic gene-for-gene interactions.
Next, suppose you were given the R:S ratios that you just calculated, but not the recombination frequencies. Calculate r from the R:S ratios in scenarios 3 through 6.
R:r2/4 R:r/2
4
