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LenaWriter [7]
3 years ago
13

Palmitate uniformly labeled with tritium (3H) to a specific activity of 2.48 3 108 counts per minute (cpm) per micromole of palm

itate is added to a mitochondrial preparation that oxidizes it to acetyl-CoA. The acetyl-CoA is isolated and hydrolyzed to acetate. The specific activity of the isolated acetate is 1.00 3 107 cpm/$mol. Is this result consistent with the !-oxidation pathway? Explain. What is the final fate of the removed tritium?
Chemistry
1 answer:
Trava [24]3 years ago
3 0

Palmitate uniformly labeled with tritium (³H) to a specific activity of 2.48×10⁸ counts per minute (cpm) per micromole of palmitate is added to a mitochondrial preparation that oxidizes it to acetyl-CoA. The acetyl-CoA is isolated and hydrolyzed to acetate. The specific activity of the isolated acetate is 1.00 × 10⁷ cpm/μmol. Is this result consistent with the β-oxidation pathway? Explain. What is the final fate of the removed tritium?

Answer:

Explanation:

Lets look at what the  β-oxidation pathway is all about, The   β-oxidation pathway are associates two dehydrogenase enzymes  with itself. These enzymes helps in the removal of hydrogen-hydrogen bond from a fatty acyl-CoA chain. This process occurs at a -CH₂ - CH₂ - and also at -CH₂-CH(OH)- pathway. However , the overall equilibrium outcome of the reaction simultaneous reaction that takes place involves the loss of one of the two hydrogen when the enoyl-CoA intermediate is  formed. What happens afterwards to the  two other hydrogen in the methyl group of acetyl-CoA is that they come with water.

Now;what is  Palmitate ?

Palmitate is a saturated fatty acid organic compound that comprises of 16 carbons and 31 hydrogen; this typical implies that two unit of  carbon attaches itself to about 4/31 of the total present hydrogen.

In addition to the above mentioned, the expected counts per minute for acetyl-CoA with two acetyl hydrogen out of the four acetyl hydrogen labeled is :

\frac{2}{4} * 2.8*10^8 \ cpm/ \mu mol* \frac{4}{31}  \approx   1.6*10^7  \ cpm/ \mu \ mol

Then result is obviously higher than the given. Furthermore, Exchange between  β-ketoacyl-CoA and solvent (water) can be loss of tritium ³H.

The final fate of the removed tritium from Palmitate stem from it how it is seen in water as reduced carriers (FADH₂ ,  NADH) which are therefore re-oxidized by the mitochondria.

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Three kilograms of steam is contained in a horizontal, frictionless piston and the cylinder is heated at a constant pressure of
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Answer:

Final temperature: 659.8ºC

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Internal energy change: 275 kJ

Explanation:

First, considering both initial and final states, write the energy balance:

U_{2}-U_{1}=Q-W

Q is the only variable known. To determine the work, it is possible to consider the reversible process; the work done on a expansion reversible process may be calculated as:

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The pressure is constant, so:  w=P(v_{2}-v_{1} )=0.5*100*1.5=75\frac{kJ}{kg} (There is a multiplication by 100 due to the conversion of bar to kPa)

So, the internal energy change may be calculated from the energy balance (don't forget to multiply by the mass):

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On the other hand, due to the low pressure the ideal gas law may be appropriate. The ideal gas law is written for both states:

P_{1}V_{1}=nRT_{1}

P_{2}V_{2}=nRT_{2}\\V_{2}=2.5V_{1}\\P_{2}=P_{1}\\2.5P_{1}V_{1}=nRT_{2}  

Subtracting the first from the second:

1.5P_{1}V_{1}=nR(T_{2}-T_{1})

Isolating T_{2}:

T_{2}=T_{1}+\frac{1.5P_{1}V_{1}}{nR}

Assuming that it is water steam, n=0.1666 kmol

V_{1}=\frac{nRT_{1}}{P_{1}}=\frac{8.314*0.1666*373.15}{500} =1.034m^{3}

T_{2}=100+\frac{1.5*500*1.034}{0.1666*8.314}=659.76 ºC

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