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Respiration in Plants - Glycolysis

Grade 11CBSEBiology

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

Glycolysis is the process of partial oxidation of glucose to form two molecules of pyruvic acid (C3H4O3C_3H_4O_3), occurring in the cytoplasm of the cell.

It is also known as the EMP pathway, named after the scientists Embden, Meyerhof, and Parnas.

The process occurs in both aerobic and anaerobic organisms as it does not require O2O_2.

In plants, glucose is derived from sucrose (the product of photosynthesis) or from storage carbohydrates like starch.

The process consists of 10 steps, where the first five steps are the 'Preparatory Phase' (energy investment) and the last five steps are the 'Pay-off Phase' (energy extraction).

Key regulatory step: The conversion of Fructose-6-phosphate to Fructose-1,6-bisphosphate by the enzyme Phosphofructokinase.

ATP is consumed at two steps: (1) Conversion of Glucose to Glucose-6-phosphate and (2) Conversion of Fructose-6-phosphate to Fructose-1,6-bisphosphate.

ATP is produced at two steps (substrate-level phosphorylation): (1) Conversion of 1,3-bisphosphoglyceric acid (BPGA) to 3-phosphoglyceric acid (PGA) and (2) Conversion of Phosphoenolpyruvate (PEP) to Pyruvic acid.

NADH+H+NADH + H^+ is generated during the conversion of Glyceraldehyde-3-phosphate (PGAL) to 1,3-bisphosphoglyceric acid (BPGA).

📐Formulae

C6H12O6+2NAD++2ADP+2Pi2CH3COCOOH+2ATP+2NADH+2H+C_6H_{12}O_6 + 2NAD^+ + 2ADP + 2P_i \rightarrow 2CH_3COCOOH + 2ATP + 2NADH + 2H^+

Net Yield=2 ATP+2 NADH+H+Net\ Yield = 2\ ATP + 2\ NADH + H^+

1 NADH+H+ETS3 ATP1\ NADH + H^+ \xrightarrow{ETS} 3\ ATP

Gross ATP production=4 ATPGross\ ATP\ production = 4\ ATP

Net ATP (Direct)=4 ATP2 ATP=2 ATPNet\ ATP\ (Direct) = 4\ ATP - 2\ ATP = 2\ ATP

💡Examples

Problem 1:

Calculate the total number of ATPATP molecules generated from one molecule of glucose during glycolysis if all NADHNADH produced is processed through the Electron Transport System (ETS).

Solution:

Total ATP=8 ATPATP = 8\ ATP

Explanation:

During glycolysis, there is a net gain of 2 ATP2\ ATP through substrate-level phosphorylation. Additionally, 2 NADH+H+2\ NADH + H^+ are produced. In aerobic conditions, each NADHNADH yields 3 ATP3\ ATP via ETS. Therefore, 2×3=6 ATP2 \times 3 = 6\ ATP (from NADHNADH) + 2 ATP2\ ATP (direct) = 8 ATP8\ ATP.

Problem 2:

Identify the molecule that undergoes cleavage into two triose phosphates during glycolysis.

Solution:

Fructose-1,6-bisphosphate (C6H10O6(PO4)2C_6H_{10}O_6(PO_4)_2)

Explanation:

In the 4th step of glycolysis, the 6-carbon compound Fructose-1,6-bisphosphate is split by the enzyme aldolase into two 3-carbon compounds: Glyceraldehyde-3-phosphate (PGAL) and Dihydroxyacetone phosphate (DHAP).

Glycolysis - Revision Notes & Key Diagrams | CBSE Class 11 Biology