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Biological Chemistry - Cellular Respiration

Grade 9IB

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

🔑Concepts

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Cellular respiration is the process by which cells break down glucose (C6H12O6C_6H_{12}O_6) to release energy in the form of ATPATP (Adenosine Triphosphate).

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Aerobic respiration occurs in the presence of oxygen (O2O_2) and involves three main stages: Glycolysis (in the cytoplasm), the Krebs Cycle, and the Electron Transport Chain (both in the mitochondria).

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Anaerobic respiration occurs when oxygen is absent. In animals, it results in the production of Lactic Acid (C3H6O3C_3H_6O_3), while in plants and yeast, it produces Ethanol (C2H5OHC_2H_5OH) and CO2CO_2.

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The net energy yield of aerobic respiration is approximately 3636 to 3838 molecules of ATPATP per molecule of glucose, whereas anaerobic respiration only yields 22 molecules of ATPATP.

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The mitochondria are the primary organelles responsible for aerobic respiration, featuring a folded inner membrane called cristae to increase surface area for chemical reactions.

📐Formulae

Aerobic Respiration: C6H12O6+6O2→6CO2+6H2O+Energy (ATP)\text{Aerobic Respiration: } C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O + \text{Energy (ATP)}

Anaerobic Respiration (Animals): C6H12O6→2C3H6O3+Energy (ATP)\text{Anaerobic Respiration (Animals): } C_6H_{12}O_6 \rightarrow 2C_3H_6O_3 + \text{Energy (ATP)}

Anaerobic Respiration (Yeast/Plants): C6H12O6→2C2H5OH+2CO2+Energy (ATP)\text{Anaerobic Respiration (Yeast/Plants): } C_6H_{12}O_6 \rightarrow 2C_2H_5OH + 2CO_2 + \text{Energy (ATP)}

💡Examples

Problem 1:

Calculate the number of oxygen (O2O_2) molecules required to fully oxidize 55 molecules of glucose (C6H12O6C_6H_{12}O_6) through aerobic respiration.

Solution:

5×6=30 molecules of O25 \times 6 = 30 \text{ molecules of } O_2

Explanation:

According to the balanced chemical equation for aerobic respiration, 11 molecule of glucose requires 66 molecules of oxygen. Therefore, 55 molecules of glucose require 5×6=305 \times 6 = 30 molecules of O2O_2.

Problem 2:

During heavy exercise, a muscle cell switches to anaerobic respiration. If the cell consumes 1010 molecules of glucose, how many molecules of Lactic Acid (C3H6O3C_3H_6O_3) are produced?

Solution:

10×2=20 molecules of C3H6O310 \times 2 = 20 \text{ molecules of } C_3H_6O_3

Explanation:

In anaerobic respiration in animals, 11 molecule of glucose is broken down into 22 molecules of lactic acid. Thus, 1010 molecules of glucose will yield 2020 molecules of lactic acid.

Problem 3:

Compare the ATPATP efficiency of aerobic vs. anaerobic respiration for 22 molecules of glucose.

Solution:

Aerobic: 2×38=76 ATP\text{Aerobic: } 2 \times 38 = 76 \text{ ATP} Anaerobic: 2×2=4 ATP\text{Anaerobic: } 2 \times 2 = 4 \text{ ATP}

Explanation:

Aerobic respiration is significantly more efficient, producing roughly 1919 times more energy per glucose molecule than anaerobic respiration (3838 vs 22 ATPATP).