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Respiration in Plants - Glycolysis

Grade 11CBSEBiology

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

🔑Concepts

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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.

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

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

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

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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).

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

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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.

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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.

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

📐Formulae

C6H12O6+2NAD++2ADP+2Pi→2CH3COCOOH+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 ATP−2 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).