Step-by-Step Solution
Auxotrophs are organisms which cannot synthesize a specific metabolite at its own and require that nutrient to be supplemented for their growth. Most of the auxotrophs bear a mutation that makes them unable to synthesize an essential compound. In the given case, the compound E does not support the growth of any mutant. So, it must be the initial product of the biochemical pathway.
Followed by the compound E, the compound A supports growth of only mutant 5, so the second compound in the biochemical pathway should be “A.” Followed by the compound A, the compound C supports growth of mutants 4 and 5, so the third compound in the biochemical pathway should be “C.”
The given data is as follows:
Followed by the compound C, the compound B supports growth of mutants 2, 4 and 5, so the fourth compound in the biochemical pathway should be “B.” Followed by the compound B, the compound G supports growth of mutants 1, 2, 4 and 5, so the fifth compound in the biochemical pathway should be “D.” The compound G supports the growth of mutants from 1 to 5. So, it must be the end product of the biochemical pathway.
Thus, the order of Biochemical pathway is as follows: 
The mutant-1 is blocked between compound B and D, the mutant-2 is blocked between compound C and B, the mutant-3 is blocked between compound D and G, the mutant-4 is blocked between compound A and C, and the mutant-5 is blocked between compound E and A. Thus, the pathway at which each mutant blocked is as follows:
Mutants are selected by either positive (direct) or negative (indirect) selection. Indirect selection takes place when the mutants are cultured in conditions, where the growth of the mutant is different from the growth of the wild type. A biochemical mutant can grow on complete medium but not on minimal medium, because it cannot synthesize a particular essential compound for its growth.
In the given case, a heterokaryon composed of double mutants 1, 3 and 2, 4 can grow on a minimal medium. This is because heterokaryon 1, 3 is functional with 2 and 4, whereas the heterokaryon 2, 4 is functional with 1 and 3. Therefore, the whole pathway would be operational and the heterokaryon survives.
A heterokaryon composed of double mutants 1, 3 and 3, 4 cannot grow on a minimal medium. This is because both heterokaryons 1, 3 and 3, 4 are mutants for “3” and this cannot produce the compound G, and thus the heterokaryon dies.
A heterokaryon composed of double mutants 1, 2 and 2, 4 and 1, 4 can grow on a minimal medium. This is because heterokaryon 1, 2 is functional with 4, the heterokaryon 2, 4 is functional with 1 and the heterokaryon 1, 4 is functional with 2. Therefore, the whole pathway would be operational and the heterokaryon survives.
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