54. The figure below depicts a hypothetical scheme for synthesizing a target product in plants. A, B, and C are the precursors of a target product D, whereas E is a by-product. The key enzymes of the pathway are indicated as E1—E6. To enhance the levels of target product D, following strategies were tested: (A) Enhancing the activity of the enzyme E5 by over- expression and/or protein engineering. (B) Enhancing the activity of the enzyme E4 by over- expression and/or protein engineering. (C) Enhancing the levels of C. (D) Blocking the activity of E6 by RNA-interference or CRISPR/Cas-mediated knockout. Which of the above mentioned strategies are likely to provide the maximum enhancement of the target product compared to the by-product, if no feedback regulation exists for any of the enzymes in the pathway? (1) A and B (2) B and C (3) C and D (4) A and D
  1. The figure below depicts a hypothetical scheme for synthesizing a target product in plants. A, B, and C are the precursors of a target product D, whereas E is a by-product. The key enzymes of the pathway are indicated as E1—E6. To enhance the levels of target product D, following strategies were tested:

    (A) Enhancing the activity of the enzyme E5 by over- expression and/or protein engineering.
    (B) Enhancing the activity of the enzyme E4 by over- expression and/or protein engineering.
    (C) Enhancing the levels of C.
    (D) Blocking the activity of E6 by RNA-interference or CRISPR/Cas-mediated knockout.
    Which of the above mentioned strategies are likely to provide the maximum enhancement of the target product compared to the by-product, if no feedback regulation exists for any of the enzymes in the pathway?
    (1) A and B (2) B and C
    (3) C and D (4) A and D

    The correct answer is (4) A and D.

    From the figure: A →(E1)→ B →(E2)→ C →(E3)→ D (target).
    From C there is a branch: C →(E5)→ target D and C →(E6)→ by‑product E.
    So C is the common precursor; E5 makes product, E6 makes by‑product.


    Strategy-by-strategy analysis

    (A) Enhancing the activity of enzyme E5 – Helpful

    • E5 converts precursor C directly to the desired product D.

    • Overexpression/protein engineering that increases E5 activity will channel more C into D, increasing target formation at the branch.

    • This clearly raises D relative to E (assuming no feedback).

    (B) Enhancing the activity of enzyme E4 – Not optimal alone

    • E4 acts upstream (e.g., B → C) and simply raises the pool of C.

    • However, C is still partitioned between E5 (to D) and E6 (to E); increasing supply alone will raise flux to both D and E.

    • This does not necessarily maximize D relative to by‑product and is less effective than acting at the branch itself.

    (C) Enhancing the levels of C – Not optimal alone

    • Directly increasing metabolite C again provides more substrate for both E5 and E6.

    • Without changing relative activities of E5 and E6, both D and E rise; the by‑product may still consume a substantial share of C.

    • So C alone is not the best way to favor D over E.

    (D) Blocking activity of E6 (RNAi/CRISPR) – Very helpful

    • E6 converts C to by‑product E.

    • Knocking out or silencing E6 eliminates the competing sink, forcing most or all available C to flow through E5 into D.

    • This both increases D and minimizes E, giving the strongest shift of flux toward the target.


    Why A and D together are best

    • A (↑E5) increases the capacity of the product‑forming branch.

    • D (↓E6) removes the competing by‑product branch.

    • Combining them maximizes flux from C into D and minimizes diversion into E, giving the maximum enhancement of product D compared to by‑product E when no feedback regulation is present.

    Thus, the correct option is (4) A and D.

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