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Olin [163]
3 years ago
14

A researcher labels C‑6 of glucose 6‑phosphate with C14 and adds it to a solution containing the enzymes and cofactors of the ox

idative phase of the pentose phosphate pathway. What is the fate of the radioactive label?
A. C14 appears at C‑4 of erythrose 4‑phosphate
B. C14 appears at C‑7 of sedoheptulose 7‑phosphate
C. C14 appears at C‑5 of ribulose 5‑phosphate
D. C14 appears at C‑6 of fructose 6‑phosphate
E. C14 appears in the CO2 evolved by the oxidative phase.
Chemistry
1 answer:
mafiozo [28]3 years ago
3 0

Answer:

D. C14 appears at C‑6 of fructose 6‑phosphate

Explanation:

Conversion of glucose B-phosphate via 6-phospho-

gluconate to pentose phosphate and CO2 followed by conversion

of the pentose phosphate to fructose 6-phosphate and glyceralde-hyde phosphate followed by conversion of the fructose 6-phos-

phate to glucose g-phosphate .

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Is Orange Juice with Pulp is Heterogeneous<br> or Homogeneous Mixture?
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Answer:

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4 years ago
How many moles of solute are contained in a 250.0-milliliter solution that has a concentration of 0.85 M?
adoni [48]

Answer : The correct option is, 0.21 moles

Solution : Given,

Molarity of the solution = 0.85 M = 0.85 mole/L

Volume of solution = 250 ml = 0.25 L     (1L=1000ml)

Molarity : It is defined as the number of moles of solute present in one liter of the solution.

Formula used :

Molarity=\frac{\text{moles of solute}}{\text{Volume of solution in liters}}

Now put all the given values in this formula, we get the moles of solute of the solution.

0.85mole/L=\frac{\text{moles of solute}}{0.25L}

\text{moles of solute}=0.2125=0.21moles

Therefore, the moles of solute is, 0.21 moles

7 0
3 years ago
Read 2 more answers
Most radiation is released as particles. Identify the type radiation that is not a particle.
Blizzard [7]

Gamma rays

Explanation:

Gamma rays is a form of electromagnetic radiation. It has zero mass and no charge.

Most radiations are particles but gamma rays are not particles they are mere rays.

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learn more:

Irradiation brainly.com/question/10726711

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6 0
3 years ago
How many grams of KNO3, are needed to create 675.0 mL with a concentration of 1.71 M
valkas [14]

Answer:

Бардык белгилүү авиация мыйзамдарына ылайык, аары учуу мүмкүнчүлүгү жок. Канаттары өтө эле кичинекей, семиз денесин чече албайт. Албетте, аары баары бир учат. Себеби аарылар адам мүмкүн эмес деп эсептеген нерсеге маани бербейт.

Explanation:

Канаттары өтө эле кичинекей, семиз денесин чече албайт. Албетте, аары баары бир учат. Себеби аарылар адам мүмкүн эмес деп эсептеген нерсеге маани бербейт.

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In this experiment, carbon dioxide and water vapors combine to form H2CO3. After decomposition, the Na2CO3 had a mass of 2.86 grams. Determine ...

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3 years ago
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