Question:

Identify the nutrient elements in crop plants, capable of getting redistributed under deficiency conditions in descending order:

Show Hint

To remember nutrient mobility:
- Highly Mobile: N, P, K, Mg (deficiency appears first in older leaves).
- Immobile: Ca, B (deficiency appears first in younger growing tips).
- Moderately Mobile: Fe, S, Mn, Cu, Zn.
  • B > S > P
  • Ca > K > Fe
  • Fe > Ca > P
  • N > Fe > B
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The Correct Option is D

Solution and Explanation

Step 1: Understanding the Concept:
When plants experience a nutrient deficiency, some elements can be remobilized from older, mature tissues and transported via the phloem to support new growing tips.
This phloem mobility determines where deficiency symptoms first appear (in older or younger leaves) and varies significantly among different mineral nutrients.

Step 2: Detailed Explanation:

Let us classify the essential plant nutrients based on their phloem mobility and redistribution capacity:
1. Highly Mobile Elements:
These elements are easily remobilized from older leaves and transported to younger tissues.
Deficiency symptoms appear first in older leaves.
Examples include Nitrogen ($N$), Phosphorus ($P$), and Potassium ($K$).
2. Moderately or Weakly Mobile Elements:
These elements show limited remobilization under deficiency conditions.
Examples include Iron ($Fe$), Sulfur ($S$), and Zinc ($Zn$).
3. Immobile Elements:
These elements cannot be remobilized from older leaves because they are bound in cell walls or structural components.
Deficiency symptoms appear first in younger leaves.
Examples include Calcium ($Ca$) and Boron ($B$).
Arranging these categories in descending order of their redistribution capacity (Highly Mobile $>$ Moderately Mobile $>$ Immobile) gives:
\[ N\text{ (Highly Mobile)} > Fe\text{ (Moderately Mobile)} > B\text{ (Immobile)} \]
This sequence represents the correct descending order of redistribution capacity under deficiency conditions.

Step 3: Final Answer:

The correct sequence in descending order of redistribution is N > Fe > B.
This corresponds to Option (D).
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