a) Parental generation:
Heavy, green, long (Hh Gg Ll) x Light, yellow, ovate (hh gg ll)
F1 Test cross:
F1 (Hh Gg Ll) x Light, yellow, ovate (hh gg ll)
b) To construct a genetic map, we first need to determine the recombination frequencies between the traits.
251 + 129 + 1 + 112 = 493 (total offspring in the F2)
(251+129)/493 = 0.742 (recombination frequency between heavy bloom and oblate fruit)
(129+1)/493 = 0.264 (recombination frequency between oblate fruit and green skin)
(1+112)/493 = 0.236 (recombination frequency between green skin and heavy bloom)
Based on these recombination frequencies, we can order the genes as follows:
Heavy bloom - oblate fruit - green skin - long fruit
The distances between the genes can be calculated by multiplying the recombination frequencies by 100:
Heavy bloom and oblate fruit: (0.742) x 100 = 74.2 map units
Oblate fruit and green skin: (0.264) x 100 = 26.4 map units
Green skin and long fruit: (0.236) x 100 = 23.6 map units
Therefore, the genetic map for these three linked genes is:
Heavy bloom --- 74.2 map units --- oblate fruit --- 26.4 map units --- green skin --- 23.6 map units --- long fruit
In an East Indian mango, three genes are linked on the same chromosome. Heavy bloom is dominant to light bloom, skin that is green is dominant to skin that is yellow, and fruit with a long shape is dominant to ovate shape. A plant that is true-breeding for the dominant traits was crossed to a light, yellow and ovate fruit. The F1 plants were then testcrossed to light, yellow and ovate fruit. The following results were obtained:
251 heavy, green, long
111 heavy, green, ovate
133 heavy, yellow, ovate
21 heavy, yellow, long
255 light, yellow, oblate
129 light, yellow, long
112 light, green, long
1 light, green, ovate
a) Diagram the crosses for the parental generation and F1 test cross.
Construct a genetic map that shows the order of these three genes and the distances between them.
1 answer
1 answer