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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?
Chemistry
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

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The reaction 2NO2 → 2NO + O2 obeys the rate law: rate = 1.4 x 10-2[NO2]2 at 500 K . What would be the rate constant at 119 K if
Svet_ta [14]

Answer : The value of rate constant at temperature 119 K is 2.46 E-29

Explanation :

As we are the rate law expression as:

Rate=1.4\times 10^{-2}[NO_2]^2  ..........(1)

The general rate law expression will be:

Rate=k[NO_2]^2       ............(2)

By comparing equation 1 and 2 we get:

k=1.4\times 10^{-2}

Now we have to calculate the rate constant at temperature 119 K.

According to the Arrhenius equation,

K=A\times e^{\frac{-Ea}{RT}}

or,

\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_1 = rate constant at T_1  = 1.4\times 10^{-2}

K_2 = rate constant at T_2 = ?

Ea = activation energy for the reaction = 80.0 kJ/mole = 80000 J/mole

R = gas constant = 8.314 J/mole.K

T_1 = initial temperature = 500 K

T_2 = final temperature = 119 K

Now put all the given values in this formula, we get:

\log (\frac{K_2}{1.4\times 10^{-2}})=\frac{80000J/mole}{2.303\times 8.314J/mole.K}[\frac{1}{500}-\frac{1}{119}]

K_2=2.46\times 10^{-29}=2.46E-29

Therefore, the value of rate constant at temperature 119 K is 2.46 E-29

8 0
3 years ago
How much force is needed to accelerate a 1000-kg car at a rate of 3 m/s2? a 1003 N b 0.003 N c 3000 N d 333.3 N
Bogdan [553]

Answer:

Option c.

Explanation:

According to Newton's law, formula to solve this problem is:

F = m . a

Mass → 1000 kg

a → 3 m/s²

F = 1000 kg . 3m/s² →  3000 N

1 N = 1 kg .  m/s²  according to SI, Internation System of units.

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Answer:d

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