Biochemistry (Looseleaf)
Biochemistry (Looseleaf)
9th Edition
ISBN: 9781319114800
Author: BERG
Publisher: MAC HIGHER
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Chapter 19, Problem 1P
Interpretation Introduction

Interpretation:

The component that does not belong to the chloroplast should be chosen.

Concept introduction:

Chloroplasts produce food for the plants by conversion of the light energy to sugars which can be utilized by the plants. There are two membranes that contain and protect the inner parts of chloroplast, the inner and the outer membrane.

Expert Solution & Answer
Check Mark

Answer to Problem 1P

The correct answer is option (c) Cristae.

Explanation of Solution

Reason for the correct answer:

Option (c) Cristae. It is a fold in the inner membrane of a mitochondrion. The name is from the Latin for crest or plume, and it gives the inner membrane its characteristic wrinkled shape, providing a large amount of surface area for chemical reactions to occur.

Conclusion

Reason for incorrect answer:

Option (a) Thylakoid membrane: The thylakoid system is suspended in the stroma. The thylakoid system is a collection of membranous sacks called thylakoids. The chlorophyll is found in the thylakoids and is the sight for the process of light reactions of photosynthesis to happen. So, this is an incorrect option.

Option (b) Stromal lamellae: The inner membrane of the chloroplast forms a border to the stroma. It regulates passage of materials in and out of the chloroplast. In addition of regulation activity, the fatty acids, lipids and carotenoids are synthesized in the inner chloroplast membrane. So, this is an incorrect option.

Option (d) Granum: The thylakoids are arranged in stacks known as granum. Each granum contains around 10-20 thylakoids. So, this is an incorrect option.

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From data in the table, calculate the AG value for the reactions. Reaction AG' (kJ/mol) Phosphocreatine + H₂O →>> creatine + P -43.0 ADP + Pi → ATP + H₂O +30.5 Fructose +P → fructose 6-phosphate + H₂O +15.9 Phosphocreatine + ADP creatine + ATP AG'O ATP + fructose → ADP + fructose 6-phosphate AG'° kJ/mol kJ/mol
Macmillan Learning The phosphorylation of glucose to glucose 6-phosphate is the initial step in the catabolism of glucose. The direct phosphorylation of glucose by P, is described by the equation Glucose + P ← glucose 6-phosphate + H₂O AG = 13.8 kJ/mol Coupling ATP hydrolysis to glucose phosphorylation makes thermodynamic sense, but consider how the coupling might take place. Given that coupling requires a common intermediate, one conceivable mechanism is to use ATP hydrolysis to raise the intracellular concentration of Pi. The increase in P; concentration would drive the unfavorable phosphorylation of glucose by Pi- Is increasing the P; concentration a reasonable way to couple ATP hydrolysis and glucose phosphorylation? No. The phosphate salts of divalent cations would be present in excess and precipitate out. Yes. Increasing the concentration of P; would decrease K'eq and shift equilibrium to the right. Yes. The extra ATP hydrolysis would provide enough free energy to drive the…
The phosphorylation of glucose to glucose 6-phosphate is the initial step in the catabolism of glucose. The direct phosphorylation of glucose by P, is described by the equation Glucose + P → glucose 6-phosphate + H₂O AG' = 13.8 kJ/mol In principle, at least, one way to increase the concentration of glucose 6-phosphate (G6P) is to drive the equilibrium reaction to the right by increasing the intracellular concentrations of glucose and Pj. The maximum solubility of glucose is less than 1 M, and the normal physiological concentration of G6P is 250 μM. Assume a fixed concentration of P, at 4.8 mM. The calculated value of K'cq is 4.74 × 10-³ M-¹. Calculate the intracellular concentration of glucose when the equilibrium concentration of glucose 6-phosphate is 250 μM, the normal physiological concentration. [glucose] = 10.99 Correct Answer Would increasing the concentration of glucose be a physiologically reasonable way to increase the concentration of G6P? No. Because the concentration of P,…
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