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Glycogen as Energy Storage: How Long Can a Game Bird Fly? Since ancient times, people have observed that certain game birds, such as grouse, quail, and pheasants, fatigue easily. The Greek historian Xenophon wrote: "The bustards ... can be caught if one is quick in starting them up, for they will fly only a short distance, like partridges, and soon tire; and their flesh is delicious." The flight muscles of game birds rely almost entirely on the use of glucose 1-phosphate to drive ATP synthesis (Chapter 14). The glucose 1-phosphate derives from the breakdown of stored muscle glycogen, catalyzed by the enzyme glycogen phosphorylase. The rate of ATP production is limited by the rate at which glycogen can be broken down. During a "panic flight," the game bird's rate of glycogen breakdown is quite high, approximately \(120 \mu \mathrm{mol} / \mathrm{min}\) of glucose 1-phosphate produced per gram of fresh tissue. Given that the flight muscles usually contain about \(0.35 \%\) glycogen by weight, calculate how long a game bird can fly. (Assume the average molecular weight of a glucose residue in glycogen is \(162 \mathrm{~g} / \mathrm{mol}\). )

Short Answer

Expert verified
The game bird can fly for about 11 seconds before tiring.

Step by step solution

01

Determine Total Glycogen Available

First, find the total amount of glycogen available in the flight muscles. Assume the flight muscle weighs 1 gram. Given that the glycogen content is 0.35% by weight, we calculate the glycogen weight as follows:\[\text{Glycogen available} = 1 \text{ gram} \times 0.0035 = 0.0035 \text{ grams}\]
02

Convert Glycogen to Moles of Glucose

Convert the weight of glycogen to moles of glucose 1-phosphate. Given the molecular weight of glucose is 162 g/mol, use the formula:\[\text{Moles of glucose} = \frac{0.0035 \text{ grams}}{162 \text{ g/mol}} \approx 2.16 \times 10^{-5} \text{ moles}\]
03

Determine the Rate of Glycogen Breakdown

The problem states that the breakdown rate of glycogen is approximately 120 µmol/min per gram of fresh tissue. Convert this to moles per minute:\[\text{Rate of breakdown} = \frac{120 \mu\text{mol}}{1 \times 10^6} = 1.2 \times 10^{-4} \text{ moles/min}\]
04

Calculate Flight Duration

Divide the total moles of glucose by the rate of glycogen breakdown to find out how long the bird can fly:\[\text{Flight duration} = \frac{2.16 \times 10^{-5} \text{ moles}}{1.2 \times 10^{-4} \text{ moles/min}} \approx 0.18 \text{ minutes}\]This result shows that the bird can sustain flight for approximately 0.18 minutes, or about 11 seconds, before exhausting its glycogen reserves.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

ATP synthesis
ATP synthesis is a vital process that fuels many activities in biological organisms. In muscles, ATP (adenosine triphosphate) acts as a primary energy currency. It's crucial for muscle contractions and, therefore, for movements such as flight in birds.
When glucose 1-phosphate is broken down, it drives the synthesis of ATP, ensuring that energy is available quickly when needed. This conversion process is essential, particularly during rapid and intense activities like a bird's panic flight, where the speed of ATP production directly impacts physical performance.
In muscle tissue, the body relies heavily on glucose 1-phosphate derived from glycogen stores, ensuring that energy is always on hand when our muscles start working. This process happens under the enzymatic action of glycogen phosphorylase, showcasing the elegance and efficiency of biological systems."
Glycogen Phosphorylase
Glycogen phosphorylase is a key enzyme in the breakdown of glycogen, a stored form of glucose. Its role is to catalyze the reaction where glycogen is converted into glucose 1-phosphate. This conversion is crucial because it provides the immediate supply of glucose that muscles need under high energy demands, like during a bird's brief bout of flight.
  • It begins by removing glucose residues from the outer branches of glycogen.
  • This produces glucose 1-phosphate, a form that can be readily converted to glucose 6-phosphate and eventually enter glycolysis for energy production.
The activity of glycogen phosphorylase is tightly regulated. For example, during intense physical activity, the enzyme's activity is heightened to meet the increased energy requirements.
Understanding the function and regulation of this enzyme is crucial for comprehending how organisms manage energy stores efficiently.
Muscle Glycogen Breakdown
Muscle glycogen breakdown is a critical metabolic step, especially for rapid and high-intensity actions like bird flight. When carbohydrates are ingested, they are stored as glycogen in muscle tissues for later use. Once a sudden demand for energy arises, such as in a panic flight, muscle cells rapidly convert these glycogen stores into glucose 1-phosphate.
This conversion is facilitated by glycogen phosphorylase and happens very quickly, ensuring a swift energy supply. The glucose 1-phosphate is then channeled into pathways like glycolysis, leading to the production of ATP, which powers muscle contractions.
Key points to remember:
  • Muscle glycogen breakdown is instantaneous and efficient, which is essential for survival behaviors.
  • The process offers a surge of energy that supports short bursts of intense activity.
Overall, the quick breakdown of muscle glycogen to meet energy demands underscores the body's capability to adapt and respond to immediate needs efficiently, which is particularly evident in the flight and movement strategies of game birds.

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Most popular questions from this chapter

13\. Optimal Glycogen Structure Muscle cells need rapid access to large amounts of glucose during heavy exercise. This glucose is stored in liver and skeletal muscle in polymeric form as particles of glycogen. The typical glycogen \(\beta\)-particle contains about 55,000 glucose residues (see Eig_ 15-2). Meléndez-Hevia, Waddell, and Shelton (1993), explored some theoretical aspects of the structure of glycogen, as described in this problem. a. The cellular concentration of glycogen in liver is about \(0.01 \mu \mathrm{M}\). What cellular concentration of free glucose would be required to store an equivalent amount of glucose? Why would this concentration of free glucose present a problem for the cell? Glucose is released from glycogen by glycogen phosphorylase, an enzyme that can remove glucose molecules, one at a time, from one end of a glycogen chain (see Eig. 15-3). Glycogen chains are branched (see Eig.15-2), and the degree of branching - the number of branches per chain - has a powerful influence on the rate at which glycogen phosphorylase can release glucose. b. Why would a degree of branching that was too low (i.e., below an optimum level) reduce the rate of glucose release? (Hint: Consider the extreme case of no branches in a chain of 55,000 glucose residues.) c. Why would a degree of branching that was too high also reduce the rate of glucose release? (Hint: Think of the physical constraints.) Meléndez-Hevia and colleagues did a series of calculations and found that two branches per chain (see Eig_15-2) was optimal for the constraints described above. This is what is found in glycogen stored in muscle and liver. To determine the optimum number of glucose residues per chain, Meléndez-Hevia and coauthors considered two key parameters that define the structure of a glycogen particle: \(t=\) the number of tiers of glucose chains in a particle (the mole-cule in Eig.15-2 has five tiers); \(g_{c}=\) the number of glucose residues in each chain. The \(y\) set out to find the values of \(t\) and \(g_{c}\) that would maximize three quantities: (1) the amount of glucose stored in the particle \(\left(G_{\mathrm{T}}\right)\) per unit volume; (2) the number of unbranched glucose chains \(\left(C_{A}\right)\) per unit volume (i.e., number of A chains in the outermost tier, readily accessible to glycogen phosphorylase); and (3) the amount of glucose available to phosphorylase in these unbranched chains \(\left(G_{\mathrm{PT}}\right)\). d. Show that \(C_{A}=2^{t-1}\). This is the number of chains available to glycogen phosphorylase before the action of the debranching enzyme. e. Show that \(C_{\mathrm{T}}\), the total number of chains in the particle, is given by \(C_{\mathrm{T}}=2^{t}-1\). For purposes of this calculation, consider the primers to be a single chain. Thus \(G_{\mathrm{T}}=g_{\mathrm{c}}\left(C_{\mathrm{T}}\right)=g_{c}\left(2^{t}-1\right)\), the total number of glucose residues in the particle. f. Glycogen phosphorylase cannot remove glucose from glycogen chains that are shorter than five glucose residues. Show that \(G_{\mathrm{PT}}=\left(g_{e}-4\right)\left(2^{t-1}\right)\). This is the amount of glucose readily available to glycogen phosphorylase.g. Based on the size of a glucose residue and the location of branches, the thickness of one tier of glycogen is \(0.12 g_{\mathrm{c}} \mathrm{nm}+0.35 \mathrm{~nm}\). Show that the volume of a particle, \(V_{5}\), is given by the equation $$ V_{\mathrm{s}}=4 / 3 \pi t^{3}\left(0.12 g_{\mathrm{c}}+0.35\right)^{3} \mathrm{~nm}^{3} $$ Meléndez-Hevia and coauthors then determined the optimum values of \(t\) and \(g_{c}\) - those that gave the maximum value of a quality function, \(f\), that maximizes \(G_{\mathrm{T}}, C_{A}\), and \(G_{P T}\), while minimizing \(V_{8}: f=\frac{G_{\mathrm{T}} C_{\mathrm{A}} G \mathrm{PT}}{V_{8}}\). They found that the optimum value of \(g_{c}\) is independent of \(t .\) h. Choose a value of \(t\) between 5 and 15 and find the optimum value of \(g_{\mathrm{c}}\). How does this compare with the \(g_{e}\) found in liver glycogen (see Egg.15-2)? (Hint: You may find it useful to use a spreadsheet program.)

Enzyme Activity and Physiological Function The \(V_{\max }\) of the glycogen phosphorylase from skeletal muscle is much greater than the \(V_{\max }\) of the same enzyme from liver tissue. a. What is the physiological function of glycogen phosphorylase in skeletal muscle? In liver tissue? b. Why does the \(V_{\max }\) of the muscle enzyme need to be greater than that of the liver enzyme?

Glycogen Breakdown in Migrating Birds Unlike a rabbit, running all-out for a few moments to escape a predator, migratory birds require energy for extended periods of time. For example, ducks generally fly several thousand miles during their annual migration. The flight muscles of migratory birds have a high oxidative capacity and obtain the necessary ATP through the oxidation of acetyl-CoA (obtained from fats) via the citric acid cycle. Compare the regulation of muscle glycolysis during short-term intense activity, as in a fleeing rabbit, and during extended activity, as in a migrating duck. Why must the regulation in these two settings be different?

Hormonal Control of Metabolic Fuel Between your evening meal and breakfast, your blood glucose drops and your liver becomes a net producer rather than consumer of glucose. Describe the hormonal basis for this switch, and explain how the hormonal change triggers glucose production by the liver.

Metabolic Effects of Mutant Enzymes Predict and explain the effect on glycogen metabolism of each of the listed defects caused by mutation: (a) Loss of the cAMPbinding site on the regulatory subunit of protein kinase A (PKA) (b) Loss of the protein phosphatase inhibitor (inhibitor 1 in Fig. 15-16) (c) Overexpression of phosphorylase \(b\) kinase in liver (d) Defective glucagon receptors in liver.

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