Question:
ग्लूकोज के 206 अणुओं से ग्लाइकोलिसिस के अंत में पायरुविक अम्ल के कितने अणु उत्पन्न होते हैं?
ग्लूकोज के 206 अणुओं से ग्लाइकोलिसिस के अंत में पायरुविक अम्ल के कितने अणु उत्पन्न होते हैं?
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1 glucose → 2 pyruvate glycolysis का basic fact है।
Updated On: Jun 22, 2026
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The Correct Option is B
Solution and Explanation
पद 1: प्रश्न को समझना
हमें glycolysis के अंत में pyruvate molecules की संख्या निकालनी है।
पद 2: मुख्य सूत्र या दृष्टिकोण
एक glucose molecule से: \[ 2 \text{ pyruvate} \] पद 3: विस्तृत व्याख्या
Given: \[ 206 \text{ glucose} \] So: \[ 206 \times 2 = 412 \] पद 4: अंतिम उत्तर
अतः सही विकल्प (B) है।
हमें glycolysis के अंत में pyruvate molecules की संख्या निकालनी है।
पद 2: मुख्य सूत्र या दृष्टिकोण
एक glucose molecule से: \[ 2 \text{ pyruvate} \] पद 3: विस्तृत व्याख्या
Given: \[ 206 \text{ glucose} \] So: \[ 206 \times 2 = 412 \] पद 4: अंतिम उत्तर
अतः सही विकल्प (B) है।
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