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Darya [45]
3 years ago
5

g In animal tissues the rate of conversion of pyruvate to acetyl-CoA is regulated by the ratio of phosphorylated and dephosphory

lated PDH complex. Describe what happens to this reaction when a preparation of rabbit muscle mitochondria containing PDH complex is treated with:
Chemistry
1 answer:
Colt1911 [192]3 years ago
5 0

Answer: seen below.

Explanation: since the different means use in treating rabbit muscle mitochondria containing PDH complex. I will give different reactions that occurs.

The mitochondria preparation responds as follow;

Active pyruvate dehydrogenase (dephosphorylated) is converted to inactive pyruvate dehydrogenase (phosphorylated) and the rate of conversion of pyruvate to acetyl-CoA decreases.

The phosphoryl group on pyruvate dehydrogenase (dephosphorylated) phosphate is removed enzymatically to give active pyruvate dephosphorylated which increases the rate of conversion of pyruvate to acetyl-CoA.

Malonate inhibit succinate dehydrogenase, and citrate accumulates. Accumulation of this citrate inhibits citrate synthase, and acetyl-CoA accumulates. Increased level of this acetyl-CoA inhibits pyruvate dephosphorylated and the rate of conversion of pyruvate to acetyl-CoA decreases.

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What mass of F2 is needed to produce 180 g of PF3 if the reaction has a 78.1% yield?
nexus9112 [7]

Answer:

91.26 g

Explanation:

Given data:

Mass of PF₃ = 180 g

Mass of F₂ required = ?

Solution:

Chemical equation:

P₄ + 6F₂   → 4PF₃

Moles of PF₃:

Number of moles = mass/ molar mass

Number of moles = 180 g/ 88 g/mol

Number of moles = 2.05 mol

Now we will compare the moles of PF₃ with F₂.

                        PF₃            :           F₂

                          4               :           6

                          2.05         :           6/4×2.05 = 3.075

Mass of  F₂:

Mass of F₂ = moles × molar mass

Mass of F₂ = 3.075 mol × 38 g/mol

Mass of F₂ =  116.85 g

If reaction yield is 78.1%:

116.85 /100 ×78.1 = 91.26 g

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Explanation:

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