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BaLLatris [955]
3 years ago
10

The second messenger cyclic AMP (cAMP) is synthesized from ATP by the activity of the enzyme adenylyl cyclase. Cyclic AMP, in tu

rn, activates protein kinase A (PKA), also called cAMP‑dependent protein kinase, which is responsible for most of the effects of cAMP within the cell. Determine the correct steps in the activation of PKA, and then place them in the correct order, starting after the adenylyl cyclase reaction.
Chemistry
1 answer:
scoundrel [369]3 years ago
8 0

Answer: 1. Cyclic AMP(cAMP) concentration increases;

2. cAMP then binds to the two regulatory subunits;

3. A change in the conformation of enzyme results. The autoinhibitory sequences move away from the active site of the catalytic subunits;

4. The catalytic sites are now free to catalyse the various reactions of PKA.

(Note: The question did not list the steps to be placed in the correct order)

Explanation:

The inactive form of Protein Kinase A (PKA) contains two identical sub-units (C) and two identical regulatory sub-units (R). Two auto-inhibitory sequences of the regulatory subunits occupy the catalytic sites of the catalytic subunits. After the reaction of adenylyl cyclase, this inactive form of PKA is activated in the following steps:

1.<em> </em><em>Cyclic AMP</em><em>(</em><em>cAMP</em><em>) concentration increases;</em>

<em>2. </em><em>cAMP</em><em> then binds to the two regulatory subunits;</em>

<em>3. A change in the conformation of enzyme results. The autoinhibitory sequences move away from the active site of the catalytic subunits;</em>

<em>4. The catalytic sites are now free to catalyse the reactions various of </em><em>PKA</em><em> which includes the regulation of glycogen, sugar, and lipid metabolism. </em>

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3 years ago
A solution is prepared by mixing 93.0 mL of 5.00 M HCl and 37.0 mL of 8.00 M HNO3. Water is then added until the final volume is
Charra [1.4K]

Answer:

[H^{+}] = 0.761 \frac{mol}{L}

[OH^{-}]=1.33X10^{-14}\frac{mol}{L}

pH = 0.119

Explanation:

HCl and HNO₃ both dissociate completely in water. A simple method is to determine the number of moles of proton from both these acids and dividing it by the total volume of solution.

n_{H^{+} } from HCl = [HCl](\frac{mol}{L}). V_{HCl}(L)  \\ n_{H^{+} } from HNO_{3}  = [HNO_{3}](\frac{mol}{L}). V_{HNO_{3}}(L)

Here, n is the number of moles and V is the volume. From the given data moles can be calculated as follows

n_{H^{+} } from HCl = (5.00)(0.093)

n_{H^{+} } from HCl = 0.465 mol

n_{H^{+} } from HNO_{3}  = (8.00)(0.037)

n_{H^{+} } from HNO_{3}  = 0.296 mol

n_{H^{+}(total) } = 0.296 + 0.465

n_{H^{+}(total) } = 0.761 mol

For molar concentration of hydrogen ions:

[H^{+}]  = \frac{n_{H^{+}}(mol)}{V(L)}

[H^{+}] = \frac{0.761}{1.00}

[H^{+}] = 0.761 \frac{mol}{L}

From dissociation of water (Kw = 1.01 X 10⁻¹⁴ at 25°C) [OH⁻] can be determined as follows

K_{w} = [H^{+} ][OH^{-} ]

[OH^{-}]=\frac{Kw}{[H^{+}] }

[OH^{-}]=\frac{1.01X10-^{-14}}{0.761 }

[OH^{-}]=1.33X10^{-14}\frac{mol}{L}

The pH of the solution can be measured by the following formula:

pH = -log[H^{+} ]

pH = -log(0.761)

pH = 0.119

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