Ask question

Ask question

All categories

kolbaska11 [484]

2 years ago

6

When an acyl group is being transferred from the cytosol to the mitochondria for oxidation, the order of the enzymes it encounte

rs is?

1 answer:

dmitriy555 [2]2 years ago

3 0

CPT-I: Carnitine Translocase: CPT-II

Carnitine palmitoyl transferase II deficiency is an autosomal recessively inherited genetic metabolic disorder characterized by an enzymatic defect that prevents long-chain fatty acids from being transported into the mitochondria for utilization as an energy source.
The disorder presents in one of three clinical forms: lethal neonatal, severe infantile hepatocardiomuscular and myopathic.

What Is Carnitine Palmitoyltransferase II Deficiency?

Carnitine Palmitoyltransferase II (CPT II) deficiency, caused by mutations in the CPT2 gene, is an inherited disease in which the body cannot convert long-chain fatty acids into energy to fuel the body. There are three forms of the disease, and the severity and symptoms vary based on the form. In all three forms, symptoms can be triggered by periods without eating (fasting).

To know more about this follow the link below:

brainly.com/question/16153580

#SPJ4

You might be interested in

Solid NaI is slowly added to a solution that is 0.0079 M Cu+ and 0.0087 M Ag+.Which compound will begin to precipitate first?NaI

NeTakaya

Answer :

AgI should precipitate first.

The concentration of $Ag^+$ when CuI just begins to precipitate is, $6.64\times 10^{-7}M$

Percent of $Ag^+$ remains is, 0.0076 %

Explanation :

$K_{sp}$ for CuI is $1\times 10^{-12}$

$K_{sp}$ for AgI is $8.3\times 10^{-17}$

As we know that these two salts would both dissociate in the same way. So, we can say that as the Ksp value of AgI has a smaller than CuI then AgI should precipitate first.

Now we have to calculate the concentration of iodide ion.

The solubility equilibrium reaction will be:

$CuI\rightleftharpoons Cu^++I^-$

The expression for solubility constant for this reaction will be,

$K_{sp}=[Cu^+][I^-]$

$1\times 10^{-12}=0.0079\times [I^-]$

$[I^-]=1.25\times 10^{-10}M$

Now we have to calculate the concentration of silver ion.

The solubility equilibrium reaction will be:

$AgI\rightleftharpoons Ag^++I^-$

The expression for solubility constant for this reaction will be,

$K_{sp}=[Ag^+][I^-]$

$8.3\times 10^{-17}=[Ag^+]\times 1.25\times 10^{-10}M$

$[Ag^+]=6.64\times 10^{-7}M$

Now we have to calculate the percent of $Ag^+$ remains in solution at this point.

Percent of $Ag^+$ remains = $\frac{6.64\times 10^{-7}}{0.0087}\times 100$

Percent of $Ag^+$ remains = 0.0076 %

8 0

4 years ago

How can you find the charge on po4

Anvisha [2.4K]

Answer:

Answer:

Explanation:

I hope it's helpful!

7 0

3 years ago

Read 2 more answers

Use the reaction 1₂(s) = 12(g), AH = 62.4 kJ/mol, AS = 0.145 kJ/(mol-K), for

Sati [7]

The Reaction is spontaneous when temperature is 430 K. Hence, Option (C) is correct.

<h3></h3><h3>What is Spontaneous reaction ?</h3>

Reactions are favorable when they result in a decrease in enthalpy and an increase in entropy of the system.

When both of these conditions are met, the reaction occurs naturally.

Spontaneous reaction is a reaction that favors the formation of products at the conditions under which the reaction is occurring.

According to Gibb's equation:

ΔG = ΔH - TΔS

ΔG = Gibbs free energy

ΔH = enthalpy change = +62.4 kJ/mol

ΔS = entropy change = +0.145 kJ/molK

T = temperature in Kelvin

ΔG = +ve, reaction is non spontaneous
ΔG = -ve, reaction is spontaneous
ΔG = 0, reaction is in equilibrium

ΔH - TΔS = 0 for reaction to be spontaneous

T = ΔH / ΔS

Here,

T = 500K

Thus the Reaction is spontaneous when temperature is 500 K.

Learn more about Gibbs free energy here ;

https://brainly.in/question/13372282

#SPJ1

3 0

2 years ago

The rock in a lead ore deposit contains 89 % PbS by mass. How many kilograms of the rock must be processed to obtain 1.5 kg of P

Zolol [24]

Answer:

Approximately 1.9 kilograms of this rock.

Explanation:

Relative atomic mass data from a modern periodic table:

Pb: 207.2;
S: 32.06.

To answer this question, start by finding the mass of Pb in each kilogram of this rock.

89% of the rock is $\rm PbS$ . There will be 890 grams of $\rm PbS$ in one kilogram of this rock.

Formula mass of $\rm PbS$ :

$M(\mathrm{PbS}) = 207.2 + 32.06 = 239.26\; g\cdot mol^{-1}$ .

How many moles of $\rm PbS$ formula units in that 890 grams of $\rm PbS$ ?

$\displaystyle n = \frac{m}{M} = \rm \frac{890}{239.26} = 3.71980\; mol$ .

There's one mole of $\rm Pb$ in each mole of $\rm PbS$ . There are thus $\rm 3.71980\; mol$ of $\rm Pb$ in one kilogram of this rock.

What will be the mass of that $\rm 3.71980\; mol$ of $\rm Pb$ ?

$m(\mathrm{Pb}) = n(\mathrm{Pb}) \cdot M(\mathrm{Pb}) = \rm 3.71980 \times 207.2 = 770.743\; g = 0.770743\; kg$ .

In other words, the $\rm PbS$ in 1 kilogram of this rock contains $\rm 0.770743\; kg$ of lead $\rm Pb$ .

How many kilograms of the rock will contain enough $\rm PbS$ to provide 1.5 kilogram of $\rm Pb$ ?

$\displaystyle \frac{1.5}{0.770743} \approx \rm 1.9\; kg$ .

5 0

4 years ago

If 5.100 g of c6h6 is burned and the heat produced from the burning is added to 5691 g of water at 21 °c, what is the final temp

emmasim [6.3K]

<span>C6H12 = 6x12 + 6x1 = 78.
The equation indicates that 2x78 = 156g benzene will produce 6542kJ.
Using proportions you can then calculate that
x/6542kJ = 7.9g / 156g
x = 331.3kJ = 331300J.

heat = mass x ΔT x 4.18J/g°
ΔT = 331300J / (5691g x 4.18J/g°) = 13.9°

final temp = 21 + 14° = 35°C</span>

4 0

4 years ago