Q.200 Zoology ⭐ 360° Option Elimination Active

NEET 2024 Zoology Question 200 Solution

Topic: Structural Organisation in Animals · Official NTA Booklet Code T1

Question

Match List I with List II related to digestive system of cockroach. List I: A. The structures used for storing of food; B. Ring of 6-8 blind tubules at junction of foregut and midgut; C. Ring of 100-150 yellow coloured thin filaments at junction of midgut and hindgut; D. The structures used for grinding the food. List II: I. Gizzard; II. Gastric Caeca; III. Malpighian tubules; IV. Crop. Choose the correct answer from the options given below:

✓

Verified Against NTA Final Answer Key

Option (3) : A-IV, B-II, C-III, D-I

📚 Structural Organisation in Animals Easy

📋 Given Parameters & Problem Setup

List I (Digestive Structures/Descriptions) A: Structures used for storing food; B: Ring of 6-8 blind tubules at junction of foregut and midgut; C: Ring of 100-150 yellow thin filaments at junction of midgut and hindgut; D: Structures used for grinding food

List II (Anatomical Names) I: Gizzard; II: Gastric Caeca; III: Malpighian tubules; IV: Crop

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The 30-Second Exam Hall Shortcut: ⚡ Cockroach Alimentary Canal Landmark Lock

⏱️ Save ~25 seconds

Food storage = Crop (IV); 6-8 blind tubules at foregut/midgut = Hepatic/Gastric caeca (II); 100-150 yellow tubules at midgut/hindgut = Malpighian tubules (III); Grinding teeth = Gizzard / Proventriculus (I)! Match: A-IV, B-II, C-III, D-I!

1 A. Storing of food: The pharynx opens into a narrow tubular oesophagus which opens into a sac-like structure called Crop (IV).

2 B. Ring of 6-8 blind tubules at junction of foregut and midgut: Hepatic or Gastric Caeca, which secrete digestive juice (II).

3 C. Ring of 100-150 yellow coloured thin filaments at junction of midgut and hindgut: Malpighian tubules, which help in excretion (III).

4 D. Grinding the food: Gizzard (proventriculus) has outer circular muscles and thick inner cuticle forming six chitinous teeth (I).

5 Matching order: A-IV, B-II, C-III, D-I -> Option (3) / C.

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Why You Can 100% Trust This Shortcut: NCERT Biology · Structural Organisation in Animals (Cockroach Alimentary Canal)

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Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Review Foregut Organs (Crop and Gizzard)

The oesophagus opens into a sac like structure called crop used for storing of food (A -> IV). The crop is followed by gizzard or proventriculus. It has an outer layer of thick circular muscles and thick inner cuticle forming six highly chitinous plates called teeth. Gizzard helps in grinding the food particles (D -> I).

Step 2

Review Caeca and Excretory Tubules at Gut Junctions

A ring of 6-8 blind tubules called hepatic or gastric caeca is present at the junction of foregut and midgut, which secrete digestive juice (B -> II). At the junction of midgut and hindgut is present another ring of 100-150 yellow coloured thin filamentous Malpighian tubules. They help in removal of excretory products from haemolymph (C -> III).

Step 3

Select Matching Choice

A-IV, B-II, C-III, D-I corresponds to Option (3).

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360° Option Elimination Engine Key Feature

Diagnose every choice: Know exactly why the right option works and why distractors fail.

4-Option Deep Scan

1 (1)

✗ Incorrect

Verdict & Scientific Reason:

Inverts gastric caeca and Malpighian tubules.

⚠️

Student Fallacy (Why candidates pick this trap): Inverts gastric caeca and Malpighian tubules.

2 (2)

✗ Incorrect

Verdict & Scientific Reason:

Mismatches crop with Malpighian tubules.

⚠️

Student Fallacy (Why candidates pick this trap): Mismatches crop with Malpighian tubules.

3 (3)

✓ 100% Correct

Verdict & Scientific Reason:

Accurately pairs all four alimentary canal structures and junctions with their respective physiological names.

💡

Why this is airtight: None.

4 (4)

✗ Incorrect

Verdict & Scientific Reason:

Inverts crop (storing) and gizzard (grinding).

⚠️

Student Fallacy (Why candidates pick this trap): Inverts crop (storing) and gizzard (grinding).

⚠️

NTA Trap Alert: Gastric Caeca vs Malpighian Tubules Position and Count

Confusing the number and junction location of Gastric Caeca (6-8 at foregut/midgut) with Malpighian Tubules (100-150 at midgut/hindgut).

🛡️ How to avoid the -1 mark: Remember: 6-8 blind tubules = Gastric Caeca (secrete enzymes); 100-150 yellow threads = Malpighian Tubules (excretion).

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Core NEET Syllabus Standard

Biology Core Zoology · Structural Organisation in Animals Standard Benchmark

Core Scientific Principle Tested:

"Oesophagus opens into a sac like structure called crop used for storing of food. The crop is followed by gizzard... helps in grinding the food particles. A ring of 6-8 blind tubules called hepatic or gastric caeca is present at junction of foregut and midgut... At junction of midgut and hindgut is present another ring of 100-150 yellow coloured Malpighian tubules."

Syllabus Focus: Alimentary Canal of Cockroach: Crop, Gizzard, Hepatic Caeca, and Malpighian Tubules

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

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NEET 2024 Zoology Question 200 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ Cockroach Alimentary Canal Landmark Lock

Step-by-Step Rigorous Workflow & Mathematical Proof

Review Foregut Organs (Crop and Gizzard)

Review Caeca and Excretory Tubules at Gut Junctions

Select Matching Choice

360° Option Elimination Engine Key Feature

NTA Trap Alert: Gastric Caeca vs Malpighian Tubules Position and Count

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

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