Step 1: Understand the origin of lysosomes.
Lysosomes are membrane-bound organelles in the cell that contain digestive enzymes, which help in breaking down excess or worn-out cell parts, and foreign substances.
Step 2: Identify the cellular organelles involved.
The Golgi body (also known as the Golgi apparatus) is responsible for processing and packaging proteins and lipids.
It also forms lysosomes by packaging enzymes into vesicles. Chloroplasts are involved in photosynthesis and do not form lysosomes.
Mitochondria are the energy powerhouses of the cell, not involved in lysosome formation. Ribosomes are responsible for protein synthesis, not for the formation of lysosomes.
Step 3: Conclusion.
Lysosomes are formed in the Golgi body, making it the correct answer.
At 15 atm pressure, $ \text{NH}_3(g) $ is being heated in a closed container from 27°C to 347°C and as a result, it partially dissociates following the equation: $ 2\text{NH}_3(g) \rightleftharpoons \text{N}_2(g) + 3\text{H}_2(g) $ If the volume of the container remains constant and pressure increases to 50 atm, then calculate the percentage dissociation of $ \text{NH}_3(g) $
If equilibrium constant for the equation $ A_2 + B_2 \rightleftharpoons 2AB \quad \text{is} \, K_p, $ then find the equilibrium constant for the equation $ AB \rightleftharpoons \frac{1}{2} A_2 + \frac{1}{2} B_2. $
Consider the following reaction: $ \text{CO}(g) + \frac{1}{2} \text{O}_2(g) \rightarrow \text{CO}_2(g) $ At 27°C, the standard entropy change of the process becomes -0.094 kJ/mol·K. Moreover, standard free energies for the formation of $ \text{CO}_2(g) $ and $ \text{CO}(g) $ are -394.4 and -137.2 kJ/mol, respectively. Predict the nature of the above chemical reaction.