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klio [65]
1 year ago
11

Phosphofructokinase catalyzes the phosphorylation of fructose 6‑phosphate to fructose 1,6‑bisphosphate in glycolysis. Fructose 1

,6‑bisphosphatase catalyzes the hydrolysis of fructose 1,6‑bisphosphate to fructose 6‑phosphate in gluconeogenesis.
fructose 6- phosphate
phosphofructokinase fructose 1 ,6-bisphosphatase
fructose 1,6 - bisphosphate
How does fructose-2,6-bisphosphate (F26BP) affect the activity of the enzymes phosphofructokinase-1 (PFK) and fructose I ,6-bisphosphatase (FBPase)?
a. increases PFK activity, increases FBPase activity
b. decreases PFK activity, increases FBPase activity
c. decreases PFK activity, decreases FBPase activity
d. increases PFK activity, decreases FBPase activity
Chemistry
1 answer:
umka2103 [35]1 year ago
3 0

Answer:

d. increases PFK activity, decreases FBPase activity

Explanation:

Fructose-2,6-bisphophate is formed by the phosphorylation of fructose-6-phosphate catalyzed by phosphofructokinase-2, PFK-2.

Fructose-2,6-bisphophate functions as an allosteric effector of the enzymes phosphofructokinase-1, PFK-1 and fructose-1,6-bisphosphatase, FBPase.

Fructose-2,6-bisphophate has opposite effects on the enzymes, PFK-1 and FBPase. When it binds to the allosteric site of the enzyme, PFK-1, it increases the enzymes's activity by increasing its affinity for its substrate fructose-6-phosphate and reduces its affinity for its allosteric inhibitors ATP and citrate. However, when it binds to FBPase, it reduces its activity by reducing its affinity for glucose, its substrate

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Compared to micellular Compound 1, Compound 2 is structurally more rigid as a result of what type of interaction?
zimovet [89]

Answer:

D. Intramolecular covalent bond

Explanation:

Compound D is structurally more rigid  as a result of intramolecular covalent bonding. The forces that hold together atoms within a compound are greater as compared to forces holding two molecules  together (intermolecular bonding). On the other hand  Hydrogen bonds are weaker as compared to covalent bonds. Covalent bonds involve the sharing of electrons between two atoms and Hydrogen bonds are formed between a highly electronegative  atom like oxygen, Flourine,Chlorine to hydrogen.  

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2 years ago
Standard Thermodynamic Quantities for Selected Substances at 25 ∘C Reactant or product ΔHf∘(kJ/mol) Al(s) 0.0 MnO2(s) −520.0 Al2
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Answer:

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Explanation:

Starting with standard enthalpies of formation you can calculate the standard enthalpy for the reaction doing this simple calculation:

∑ n *ΔH formation (products) - ∑ n *ΔH formation (reagents)

This is possible because enthalpy is state function meaning it only deppends on the initial and final state of the system (That's why is also possible to "mix" reactions with Hess Law to determine the enthalpy of a new reaction). Also the enthalpy of formation is the heat required to form the compound from pure elements, then products are just atoms of reagents organized in a different form.

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1 year ago
A sample of gas occupies 10 L at STP. What
puteri [66]

Pressure is 5.7 atm

<u>Explanation:</u>

P1 = Standard pressure = 1 atm

P2 = ?  

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T2 = 100°C + 273 K = 373 K

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Plugin the above values as,

$P 2=\frac{1 \text {atm} \times 10 L \times 373 \mathrm{K}}{273 \mathrm{K} \times 2.4 \mathrm{L}}=5.7 \text { atm }

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2 years ago
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