In a metabolic pathway A → B → C → D → E, enzyme E1 (converting A → B) is allosterically inhibited by product E. A researcher finds that even when [E] is very high, enzyme E2 (converting B → C) remains functional and competent. This feedback regulation mechanism illustrates that:

Question

Biology · Metabolic Pathways and Regulation

Verified Solution4 OptionsUpdated: October 2026👁️1 View

Share:

Correct answer

Feedback inhibition at the first committed step (E1) prevents unnecessary substrate entry while leaving downstream enzymes functional

Save this question and verified answer as a clean study sheet with live clickable link.

👥

Challenge your peer aspirants on WhatsApp or Telegram to solve this question!

Online Test Series

Practice full-length mock tests with real exam timers, instant score analysis, and all-India ranking.

Chapter Practice

Want more practice questions on Metabolic Pathways and Regulation?

Solve complete topic-wise tests with detailed solutions on GkGet.

Recommended practice questions in Biology · Metabolic Pathways and Regulation

The Dandi March, also known as the Salt Satyagraha, was led by which influential leader?

Which act formally granted independence to India, marking the end of British rule?

Who was the key figure in establishing the Indian National Congress?

Answer 1

Non-competitive inhibition of E1 causes conformational change in E2 via subunit interactions

Answer 2

Feedback inhibition at the first committed step (E1) prevents unnecessary substrate entry while leaving downstream enzymes functional

Answer 3

Feed-forward activation of E2 occurs by accumulating substrate B

Answer 4

Competitive feedback inhibition occurs at E2 by end product E

Related Practice Questions