Problem 26 - Basic Principles of Organic Chemistry

Question

A very strong man can lift \(225 \mathrm{~kg}(500 \mathrm{lb}) 2\) meters \((6.5 \mathrm{ft})\). Muscle action gets its energy from the reaction \(\mathrm{ATP}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{ADP}+\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\), a process with a \(\Delta G^{0}\) of \(-7 \mathrm{kcal}\) a. Assuming \(50 \%\) efficiency in the use of the hydrolysis free energy, how many grams of ATP (MW 507 ) would have to be hydrolyzed to achieve this lifting of the weight? (One \(\mathrm{kg}\) raised one meter requires \(2.3\) calof energy.) b. How many grams of glucose would have to be oxidized to \(\mathrm{CO}_{2}\) and water to replenish the ATP used in Part a on the basis of a \(40 \%\) conversion of the energy of combustion to ATP? ( \(\Delta G^{0}\) for combustion of glucose is \(-686 \mathrm{kcal}\) )

Step-by-Step Solution

Answer

a. 149.7 g of ATP; b. 0.679 g of glucose.

Step 1: Calculate the Total Energy Required to Lift the Weight

The energy required to lift a weight is given by the formula \( E = mgh \), where \( m \) is the mass in kg, \( g \) is the acceleration due to gravity in m/s², and \( h \) is the height in meters.We are given that lifting 1 kg by 1 meter requires 2.3 cal of energy. Thus, the energy required to lift a 225 kg weight by 2 meters is:\[ E = 225 \times 2 \times 2.3 \text{ cal} = 1035 \text{ cal} \]

Step 2: Convert Energy from Calories to Kilocalories

Since we often deal with kilocalories in biochemistry, convert the energy from calories to kilocalories.\[ 1035 \text{ cal} = 1.035 \text{ kcal} \]

Step 3: Calculate the Energy from ATP Hydrolysis

Given the reaction \( \text{ATP} + \text{H}_2\text{O} \rightarrow \text{ADP} + \text{H}_2\text{PO}_4^- \) with a \( \Delta G^0 = -7 \text{ kcal} \).With a 50% efficiency, the usable energy from each ATP hydrolysis is:\[ \text{Usable } \Delta G^0 = 0.50 \times 7 \text{ kcal} = 3.5 \text{ kcal per mol} \]

Step 4: Calculate Moles of ATP Required

To achieve 1.035 kcal using 3.5 kcal per mol of ATP, calculate the moles of ATP needed:\[ \text{Moles of ATP} = \frac{1.035 \text{ kcal}}{3.5 \text{ kcal/mol}} \approx 0.295 \text{ mol} \]

Step 5: Convert Moles of ATP to Grams

Using the molar mass of ATP (507 g/mol), convert moles to grams:\[ \text{Mass of ATP} = 0.295 \text{ mol} \times 507 \text{ g/mol} \approx 149.7 \text{ g} \]

Step 6: Calculate the Energy Needed to Replenish ATP Using Glucose

Given \(\Delta G^0\) for glucose combustion as \(-686 \text{ kcal/mol}\) and 40% conversion efficiency to ATP:\[ \text{Usable } \Delta G^0 = 0.40 \times 686 \text{ kcal/mol} = 274.4 \text{ kcal/mol} \]

Step 7: Calculate Moles of Glucose Required

Determine the moles of glucose needed to replenish the ATP:\[ \text{Moles of Glucose} = \frac{1.035 \text{ kcal}}{274.4 \text{ kcal/mol}} \approx 0.00377 \text{ mol} \]

Step 8: Convert Moles of Glucose to Grams

Using the molar mass of glucose (180 g/mol), convert moles of glucose to grams:\[ \text{Mass of Glucose} = 0.00377 \text{ mol} \times 180 \text{ g/mol} \approx 0.679 \text{ g} \]

Key Concepts

ATP Hydrolysis

ATP, or adenosine triphosphate, is considered the energy currency of the cell. It's used extensively in biochemical reactions to power cellular functions. The process of ATP hydrolysis involves the breakdown of ATP into ADP (adenosine diphosphate) and an inorganic phosphate. This reaction releases energy, which is utilized in various biological processes.

The hydrolysis of ATP is accompanied by a standard Gibbs free energy change (\( \Delta G^{0} \)) of approximately \(-7 \text{kcal/mol}\). This reaction's energy can be harnessed to fuel muscular contractions, such as lifting weights or sustaining exercise. However, not all the energy from ATP hydrolysis is converted to work. In many biological systems, the efficiency is around 50%, meaning that only half of the available energy is used to perform useful work. The rest is lost as heat.

In exercises, like the one described, understanding the efficiency of energy usage from ATP is critical. It helps in calculating how much ATP is required to perform a given physical task. This knowledge is crucial for biochemists and physiologists when analyzing energy consumption and metabolism in cells.

Calorimetry in Biochemistry

Calorimetry is the measurement of heat released or absorbed during a chemical reaction, a change of state, or formation of a solution. In biochemistry, calorimetry allows scientists to measure the amount of energy changes that occur during biochemical reactions. It is a powerful technique used to understand the energy dynamics in cellular metabolism.

For the analyzed exercise, energy requirements for lifting weights were calculated using calorimetric data, stating that 1 kg lifted 1 meter requires 2.3 calories of energy. Through calorimetry studies, such values are determined, which are vital for translating physical activities to biochemical energy consumption.

This approach allows researchers and students to draw connections between physical processes and biochemical energy demands. Calorimetry is essential in both educational settings to explain and visualize energy concepts, and in research for precise measurements of energy changes in biological processes.

Glucose Oxidation

Glucose oxidation is a critical metabolic process where glucose is broken down to produce energy. This process occurs through the entirety of cellular respiration, including glycolysis, the Krebs cycle, and the electron transport chain. The ultimate result is the production of ATP from the energy released during the oxidation of glucose.

The complete oxidation of one molecule of glucose releases a theoretical Gibbs free energy change (\( \Delta G^{0} \)) of \(-686 \text{kcal/mol}\). Biological systems can harness about 40% of this energy to synthesize ATP from ADP, with the remaining energy lost as heat. This efficiency of energy conversion is not absolute but varies under different physiological conditions.

The exercise utilized this principle by calculating how much glucose is necessary to replenish ATP used based on the 40% conversion efficiency. By comprehending how glucose oxidation contributes to ATP production, students can appreciate the significance of nutrition and energy metabolism in sustaining biological functions.

Thermodynamics in Biology

Thermodynamics, the study of energy transformations, is foundational in biology. It helps explain how organisms convert chemical energy into useful work and heat. In biochemical systems, the laws of thermodynamics guide the direction and efficiency of energy flow, which is critical in understanding cellular processes.

The first law of thermodynamics, conservation of energy, states that energy can neither be created nor destroyed, only transformed. This principle is evident when ATP hydrolysis converts chemical energy into mechanical work. The second law of thermodynamics, which emphasizes entropy, describes how energy transformations are not 100% efficient. Some energy is always lost as heat, increasing the disorder in the system.

The exercise example reflects these principles by showing a 50% efficiency in ATP energy conversion for muscular work and a 40% efficiency in converting glucose's energy to ATP. This understanding sheds light on the energy economies within living organisms, illustrating how energy is precisely managed and utilized in life's processes.