Question

Starch (amylose), \(\left(\mathrm{C}_{6} \mathrm{H}_{10} \mathrm{O}_{5}\right) \mathrm{x}\), is a polymer of the sugar glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\), linked by - \(\mathrm{O}\) - bonds; each starch molecule may contain 200 to 1000 glucose units. A mixture of amino acids added to a solution of amylose produces no detectable reaction. One molecule of the enzyme amylase, containing the same amino acids, however breaks (hydrolyzes) \(4000-\mathrm{O}\) - bonds per second, (a) Estimate the mass in picograms of glucose produced by an amylase molecule in 1 day. (b) The heat in calories liberated to a cell by the oxidation of the glucose: \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(\mathrm{aq})+6 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow\) \(6 \mathrm{O}_{2}\) (g) \(+6 \mathrm{H}_{2} \mathrm{O}(\ell) \cdot \Delta \mathrm{H}=-686 \mathrm{Kcal}\). How many calories does the cell obtain from the mass of glucose calculated in part (a)?

Short answer

(a) The mass of glucose produced by an amylase molecule in one day is 103.3 pg. (b) The cell obtains \(3.94 \times 10^{-10}\) calories from the mass of glucose calculated in part (a).

Step-by-step solution

Calculate the number of O-bonds broken in one day

To calculate the number of O-bonds broken by one amylase molecule in a day, we'll multiply the number of bonds it breaks per second (4000) by the total number of seconds in a day (86400). Number of O-bonds broken in one day = 4000 bonds/second * 86400 seconds/day = 345600000 bonds/day

Determine the number of glucose molecules produced

Since one glucose molecule is formed by breaking one O-bond, the number of glucose molecules produced in a day is the same as the number of O-bonds broken. Number of glucose molecules produced = 345600000 molecules/day

Calculate the mass of glucose produced in one day

To calculate the mass of glucose produced in a day, we'll first find the molar mass of glucose (C6H12O6). Then, we'll convert the number of glucose molecules produced into moles and finally into mass (in picograms). Molar mass of glucose = 6(12.01 g/mol C) + 12(1.008 g/mol H) + 6(16.00 g/mol O) = 180.18 g/mol Number of moles of glucose = (Number of glucose molecules produced) / (Avogadro's number) = (345600000 molecules/day) / \(6.022 \times 10^{23}\) molecules/mol = \(5.74 \times 10^{-16}\) mol/day Mass of glucose = (Number of moles of glucose) × (Molar mass of glucose) = \(5.74 \times 10^{-16}\) mol/day × 180.18 g/mol × \(10^{12}\) pg/g = 103.3 pg/day (a) The mass of glucose produced by an amylase molecule in one day is 103.3 pg.

Calculate the energy obtained from the glucose oxidation

To calculate the energy obtained from the glucose oxidation, we'll use the given conversion factor for energy released per mole of glucose. Energy released per mole of glucose = -686 Kcal/mol Energy released per \(5.74 \times 10^{-16}\) mole of glucose = -686 Kcal/mol × \(5.74 \times 10^{-16}\) mol = \(3.94 \times 10^{-13}\) Kcal Since 1 Kcal = 1000 cal, Energy released per \(5.74 \times 10^{-16}\) mole of glucose = \(3.94 \times 10^{-13}\) Kcal × 1000 cal/Kcal = \(3.94 \times 10^{-10}\) cal (b) The cell obtains \(3.94 \times 10^{-10}\) calories from the mass of glucose calculated in part (a).

Key concepts

Understanding Glucose Oxidation

Glucose oxidation is a critical process that occurs in the cells, providing the energy required for various cellular functions. In this context, glucose, with the chemical formula \(C_6H_{12}O_6\), undergoes a biochemical reaction with oxygen that breaks down the glucose molecules and releases energy.
This process can be represented by the chemical equation:

• \(C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O\)

This reaction is exothermic, meaning it releases heat; in fact, it releases approximately 686 kilocalories (Kcal) per mole of glucose oxidized.
Here’s what happens during glucose oxidation:

• Glucose molecules react with oxygen (\(O_2\)) in the body.
• This reaction releases energy, primarily in the form of ATP, the energy currency in the cell.
• The byproducts are carbon dioxide (\(CO_2\)), which is exhaled, and water (\(H_2O\)).

Understanding glucose oxidation is vital because it's the principal way by which the human body generates energy from food.

Mastering Molar Mass Calculation

Calculating the molar mass of a compound like glucose is a foundational aspect of chemistry that allows us to convert between moles and grams. This conversion is essential for quantitatively analyzing chemical reactions.
The formula for glucose is \(C_6H_{12}O_6\), and its molar mass is calculated by summing the atomic masses of all the atoms in the molecule:

• Carbon (C): 12.01 g/mol × 6 = 72.06 g/mol
• Hydrogen (H): 1.008 g/mol × 12 = 12.096 g/mol
• Oxygen (O): 16.00 g/mol × 6 = 96.00 g/mol

Adding these gives us a molar mass of glucose of 180.18 g/mol.
This calculation is crucial because knowing the molar mass allows you to determine how many grams are in a mole of glucose, which is used to find out how much glucose is involved in specific chemical processes, like oxidation.
Once you know the molar mass, you can easily move between moles and grams, which simplifies the calculation of how much energy is released during glucose oxidation.

Explaining Enzyme Catalysis

Enzymes are proteins that serve as catalysts in biological systems, speeding up chemical reactions without being consumed in the process. In our example, the enzyme amylase acts on starch (amylose), breaking it down into glucose molecules.
Here's how enzyme catalysis works:

• Enzymes lower the activation energy required for a reaction, making it easier and faster for the reaction to occur.
• They bind to substrates (in this case, starch) at their active site to form an enzyme-substrate complex.
• This enzyme-substrate interaction facilitates the breaking of -O- bonds in starch, resulting in the production of glucose.

Amylase is particularly efficient, breaking down approximately 4000 -O- bonds every second!
This rate of reaction would be impossible if we relied on the natural breakdown without enzymes, showcasing the incredible efficiency of enzyme catalysis.
Understanding enzymes and their catalytic role helps us appreciate their importance in digestion, metabolism, and various other biological processes.