Final answer:
In glycolysis, glucose is phosphorylated by hexokinase to form glucose-6-phosphate, consuming ATP and producing ADP. This crucial first step is exergonic, which ensures that the glucose remains trapped within the cell for further metabolic processes.
Step-by-step explanation:
The first stage of glycolysis involves the phosphorylation of glucose to form glucose-6-phosphate (G6P). This reaction is catalyzed by the enzyme hexokinase, which transfers a phosphate group from ATP to glucose, making it more reactive. This phosphorylation process is essential as it not only makes the glucose molecule more chemically active and thus ready for subsequent steps in glycolysis, but it also ensures the molecule remains within the cell due to the negatively charged phosphate group.
Notably, ATP is consumed in this reaction, as it donates a phosphate, resulting in the formation of ADP. This initial step is exergonic, releasing energy with a free energy change (ΔG°) of -17 kJ per the question or -4 Kcal per mole as indicated by the reference material. This difference in numbers might be due to different conditions or sources.
Following this, glucose-6-phosphate is further processed to eventually form two molecules of pyruvate, generating a net yield of two ATPs and two molecules of NADH. These products enter further cellular respiration processes to produce more ATP. The entirety of glycolysis can be summarized by the equation provided where glucose is ultimately transformed into pyruvate, with the generation of ATP and NADH.