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wel
2 years ago
11

In prolonged fasting conditions acetyl-coa generated from the breakdown of amino acids and fatty acids does not enter the citric

acid cycle in the liver, but acetyl-coa derived from ketone bodies can enter the citric acid cycle in the brain.
Chemistry
1 answer:
Brums [2.3K]2 years ago
7 0

In prolonged fasting conditions acetyl-coa generated from the breakdown of amino acids and fatty acids does not enter the citric acid cycle in the liver, but acetyl-coa derived from ketone bodies can enter the citric acid cycle in the brain. <u>Cholesterol is required in the diet.</u>

<h3>What is amino acids?</h3>

Amino acids are chemical molecules having side chains (R groups) unique to each amino acid as well as amino and carboxylic acid (CO2H) functional groups.

Every amino acid contains the elements carbon (C), hydrogen (H), oxygen (O), and nitrogen (N) (CHON); in addition, the side chains of cysteine and methionine contain sulfur (S), while the less frequent amino acid selenocysteine has selenium (Se). As of 2020, it is known that more than 500 naturally occurring amino acids make up the monomer units of peptides, including proteins.

Despite the fact that there are only 22 proteins, 20 of them have unique specified codons, and another two have unique coding mechanisms: All eukaryotes contain selenocysteine, and pyrrolysine is also present.

To learn more about amino acids from the given link:

brainly.com/question/21327676

#SPJ4

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Evaporation

Explanation:

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3 years ago
What is bias in an experiment?
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3 0
3 years ago
How many are molecules ( or formula) in each sample?
andre [41]

Answer:

  • 4.010 \times 10^{25} \text { molecules of } \mathrm{NaHCO}_{3} \text { present in } 55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}
  • 16.86 \times 10^{26} \text { molecules of } \mathrm{Na}_{3} \mathrm{PO}_{4} \text { present in } 459 \mathrm{kg}\left(4.59 \times 10^{5} \mathrm{gm}\right) \text { of } \mathrm{Na}_{3} \mathrm{PO}_{4}

<u>Explanation</u>:

<u>Number of molecules for 55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}</u>

\text { Firstly molar mass is calculated of } \mathrm{NaHCO}_{3}:

Atomic mass of Na + H + C + 3(O)  = 22.99 + 1.008 + 12.01 + 3 × 16.00 = 84.00 g/mol

\text { Number of molecules of } \mathrm{NaHCO}_{3} \text { in } 55.93 \text { kg are as follows: }

55.93 \times\left(10^{3} \mathrm{gm}\right) \times \frac{1 \mathrm{mol} \mathrm{NaHCO}_{3}}{84.00 \mathrm{gm} \mathrm{NaHCO}_{3}} \times\left(6.022 \times 10^{23} \mathrm{molecules} \text { i.e Avogadro number }\right)

=4.010 \times 10^{26} \text { molecules of } \mathrm{NaHCO}_{3} \text { present in } 55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}

<u>Number of molecules for for \left(4.59 \times 10^{5} \mathrm{gm}\right) \text { of } \mathrm{Na}_{3} \mathrm{PO}_{4}</u>

\text { Firstly molar mass is calculated of } \mathrm{Na}_{3} \mathrm{PO}_{4}

= Atomic mass of 3(Na) + P + 4(O)

= 3(22.99) + 30.97 + 4(16.00) = 163.94 g/mol

459 \times\left(10^{3} \mathrm{gm}\right) \times \frac{1 \mathrm{mol} N a_{3} P O_{4}}{163.94 \mathrm{gm} N a_{3} P O_{4}} \times\left(6.022 \times 10^{23} \mathrm{molecules} \text { i.e Avogadro number) } / 1 \mathrm{mol}\right.

=16.86 \times 10^{26} \text { molecules of } \mathrm{Na}_{3} \mathrm{PO}_{4} \text { present in } 459 \mathrm{kg}\left(4.59 \times 10^{5} \mathrm{gm}\right) \text { of } \mathrm{Na}_{3} \mathrm{PO}_{4}

8 0
3 years ago
Which best explains why the game of economics does not have a single goal?
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3 0
3 years ago
If an ideal gas has a pressure of 2.97 atm, a temperature of 449 K, and has a volume of 58.35 L, how many miles of gas are in th
soldi70 [24.7K]
N = ?

T = 449 K

V = 58.35 L

P =2.97

R = 0.082

Use the clapeyron equation:

P x V = n x R x T

2.97 x 58.35 = n x 0.082 x 449

173.2995 = n x 36.818

n = 173.2995 / 36.818

n = 4.70 moles

hope this helps!



7 0
3 years ago
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