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Andreas93 [3]
4 years ago
5

In 1906 Harden and Young, in a series of classic studies on the fermentation of glucose to ethanol and CO2 by extracts of brewer

's yeast, made the following observations.
(A) Inorganic phosphate was essential to fermentation; when the supply of phosphate was exhausted, fermentation ceased before all the glucose was used.
(B) During fermentation under these conditions, ethanol, CO2, and a sugar phosphate accumulated.
(C) When arsenate was substituted for phosphate, no sugar phosphate acumulated, but the fermentation proceeded until all the glucose was converted to ethanol and CO2.

Answer the following questions.

1. Which enzyme of glycolysis requires inorganic phosphate and therefore stops when no phosphate is available?
(a) glyceraldehyde 3-phosphate dehydrogenase
(b) phosphoglycerate mutase
(c) phosphofructokinase-1
(d) phosphoglycerate kinase

2. What sugar phosphate accumulates under these conditions?
(a) glucose 1,6-biphosphate
(b) glucose 1-phosphate
(c) fructose 1-phosphate
(d) fructose 1,6-biphosphate

3. Arsenate substitution for phosphate generated an acyl arsenate compound that immediately degraded. What glycolysis intermediate was a product of the spontaneous degradation of this acyl arsenate?
(a) glycerol 3-phosphate
(b) 3-phosphoglycerate
(c) dihydroxyacetone phosphate
(d) glyceraldehyde 3-phosphate
Chemistry
1 answer:
Ludmilka [50]4 years ago
5 0

Answer:

Answers to the (B) Part

1. A

2. D

3. B

Explanation:

Explanation of the (A) Part

a.

G-3-P, also known as Glyceraldehyde-3-Phozphate.

In the dehydrogenase process of G-3-P, phosphate is a compulsory requirement.

During this process, the G-3-P is hereby changed or converted into 1,3-bisphosphoglycerate in presence of NAD+ and Pi. Note that the exhaustion of Pi signal the end of the glycolysis process.

Because there's is glucose in abundance (excess), the glucose is thus, phosphorylated to ATP but no Pi gets released.

b.

During fermentation, CO₂ and Ethanol are produced. But if there's no fermentation of ethanol in the absence of oxygen (anaerobic condition) , the NADP+ will piled up.. So, no new NAD+ would be available for further glycolysis.

Pyruvate is converted to ethanol and CO2 to replenish NAD+ for glycolysis to proceed.

c.

The hexose phosphate is piled up or accumulated to fructose 1, 6 bisphosphate.

The point between the energy input reaction preceding it and the energy forming reaction after it serves as the intermediary of the reaction.

Because of unavailability of P, G-3-P is not broken down. Hence, the reaction will backflow to fructose 1, 6 bisphosphate, which is more stable.

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solniwko [45]

Answer:

2.5L [NaCl] concentrate needs to be 4.8 Molar solution before dilution to prep 10L of 1.2M KNO₃ solution.

Explanation:

Generally, moles of solute in solution before dilution must equal moles of solute after dilution.

By definition Molarity = moles solute/volume of solution in Liters

=> moles solute = Molarity x Volume (L)

Apply moles before dilution = moles after dilution ...

=> (Molarity X Volume)before dilution = (Molarity X Volume)after dilution

=> (M)(2.5L)before = (1.2M)(10.0L)after

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6 0
3 years ago
The valences of metal x,y and z are 1,2 and 3 respectively. What are the formulae of their;a) hydroxides, b) sulphates, c) hydro
Rina8888 [55]

Answer:

See answer below

Explanation:

AS we know that the valence for those metals X, Y, and Z are 1, 2 and 3, we can determine the formula of each compound.

1. Hydroxides.

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Meₐ(OH)ₙ

Where:

a: valence or charge of the hydroxide (Which is -1)

n: valence of the metal.

Following this, the formula for X, Y and Z would be:

XOH

Y(OH)₂

Z(OH)₃

2. Sulphates

Sulphates follow a similar rule of hydroxide in the general molecular formula, but instead of having a charge of -1, it has a charge of -2 so:

Mₐ(SO₄)ₙ

So, following the rule:

X₂SO₄

Y₂(SO₄)₂ ------> YSO₄

Z₂(SO₄)₃

3. Hydrogens

Following the same rule as the previous, hydrogens works with a charge of -1, so:

MₐHₙ

Then:

XH

YH₂

ZH₃

4. Carbonates.

This follows the same rule as sulphates, with the same charge so:

Mₐ(CO₃)ₙ

Then:

X₂CO₃

YCO₃

Z₂(CO₃)₃

5. Nitrates

Follow the same rule as the hydroxides, with the same charge of -1.

Mₐ(NO₃)ₙ

Then:

XNO₃

Y(NO₃)₂

Z(NO₃)₂

6. Phosphates

In the case of phosphates, these have a charge of -3 so:

Mₐ(PO₄)ₙ

Then:

X₃PO₄

Y₃(PO₄)₂

Z₃(PO₄)₃ ----> ZPO₄

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