A heterozygote of E. coli was produced with the following combination of mutations:​

$$trpR^+ trpO^- trpE^+/trpR^+ trpO^+ trpE^-$$

​​Where R is the repressor, O is the operator, and trpE encodes the first enzyme in the biosynthetic cascade for tryptophan. Assume all other enzymes required are wild type. Which one of the following is the most likely phenotype of this E. coli?​

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Correct option is A

Explanation-​

Allele 1: trpR⁺ trpO⁻ trpE⁺
trpR⁺: Makes functional repressor protein.
trpO⁻: Mutated operator – cannot bind the repressor, so transcription is always ON.
trpE⁺: Produces the functional enzyme needed for tryptophan biosynthesis.
This operon is always ON, even in the presence of tryptophan, because the repressor cannot bind the mutant operator.

Allele 2: trpR⁺ trpO⁺ trpE⁻
trpR⁺: Functional repressor.
trpO⁺: Wild-type operator → can bind the repressor and turn OFF transcription when tryptophan is present.
trpE⁻: Non-functional gene, cannot produce the enzyme.
This copy is regulatable, but it cannot contribute to tryptophan synthesis due to trpE⁻ mutation.

Combined Phenotype (Heterozygote):
The only functional enzyme comes from trpE⁺, which is linked to trpO⁻ (unregulated). So tryptophan is always synthesized, whether tryptophan is present or not. Regulation via trpR and trpO⁺ doesn’t help, because the trpE on that operon is defective.
Therefore, the correct phenotype is:
Option a: Synthesizes tryptophan irrespective of tryptophan status in the medium. (Final Answer)