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juin [17]
3 years ago
12

Consider the biochemical pathway: A + E1 –> B + E2 –> C (Note that A, B, and C are molecules in the pathway, E1 is enzyme

1, and E2 is enzyme 2.) If enzyme 2 is inactive, which of the following compounds will accumulate?
Chemistry
1 answer:
MAVERICK [17]3 years ago
5 0

Answer:

The compound that will accumulate is B.

Explanation:

Enzymes in any biochemical process are the protein molecules that speed up the rate of the biological reaction. It lowers down the activation energy of the reactant molecules.

Since from the question:

<u>A + E1 –> B + E2 –> C</u>

Given that A, B and C are the molecules in the pathway and E1 and E2 are the enzymes.

<u>This means that A undergoes some biological reaction in which E1 is acting a catalyst to give B and which further undergoing another biological reaction to give C in the presence of enzyme E2.</u>

Since, E2 is inactive,

<u>A will form B but B will not be able to undergo further reaction and thus, C can not be formed.</u>

<u>Thus A is consumed and C is not formed. The compound that will accumulate is B.</u>

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9. What is the name of the molecule?
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3–methyl–2–butanol

Explanation:

To name the compound, we must:

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8 0
3 years ago
What is the solubility of La(IO₃)₃ in a solution that contains 0.100 M IO₃⁻ ions? (Ksp of La(IO₃)₃ is 7.5 × 10⁻¹²)
love history [14]

<u>Solution and Explanation:</u>

[La3+] = 0.1 M

<u>At the equilibrium: </u>

La(IO3)3 <---->     La3+     +         3 IO3-  

                    0.1 +s             3s      

\mathrm{Ksp}=[\mathrm{La} 3+][\mathrm{IO} 3-]^{\wedge} 3

7.5 * 10^{\wedge}-12=(0.1+s) *(3 s)^{\wedge} 3

Since Ksp is small, s can be ignored as compared to the 0.1

The above shown expression thus becomes:

7.5 * 10 \wedge-12=(0.1) *(3 s) \wedge 3

7.5 * 10^{\wedge}-12=0.1 * 27(\mathrm{s})^{\wedge} 3

s=1.406 * 10^{\wedge}-4 \mathrm{M}

<u>Answer: 1.4*10^-4 M </u>

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If you have 1 proton and 2 electrons, what is the overall charge of the ion?
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The overall charge will be negative 1
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