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Elodia [21]
3 years ago
5

An enzyme is discovered that catalyzes the chemical reaction:SAD -------->HAPPY

Chemistry
1 answer:
jekas [21]3 years ago
3 0

Answer: Km = 10μM

Explanation: <u>Michaelis-Menten constant</u> (Km) measures the affinity a enzyme has to its substrate, so it can be known how well an enzyme is suited to the substrate being used. To determine Km another value associated to an eznyme is important: <em>Turnover number (Kcat)</em>, which is the number of time an enzyme site converts substrate into product per unit time.

Enzyme veolcity is calculated as:

V_{0} = \frac{E_{t}.K_{cat}.[substrate]}{K_{m}+[substrate]}

where Et is concentration of enzyme catalitic sites and has to have the same unit as velocity of enzyme, so Et = 20nM = 0.02μM;

To calculate Km:

V_{0}*K_{m} + V_{0}*[substrate] = E_{t}.K_{cat}.[substrate]

K_{m} = \frac{E_{t}.K_{cat}.[substrate]-V_{0}*[substrate]}{V_{0}}

K_{m} = \frac{0.02*600*40-9.6*40}{9.6}

Km = 10μM

<u>The Michaelis-Menten for the substrate SAD is </u><u>10μM</u><u>.</u>

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QveST [7]

Answer:

3CaBr2 + 2LI3PO4 - > Ca3(PO4) 2 + 6LiBr

Explanation:

The first one I did was PO4. There are two on the right side, so I added 2 to Li3PO4 on the other side. That balanced the PO4s and then gave me 6 Lithiums so I balanced that one next on the right side. I added 6 to LiBr which balanced the Li but then gave me 6 Br, so I finished it off by adding 3 in front of CaBr2 which balanced the calcium and bromines.

Here was the process:

CaBr2+2Li3PO4 -> Ca3(PO4)2+LiBr

Balances PO4 (2on both sides)

CaBr2+2Li3PO4 -> Ca3(PO4)2+6LiBr

Balances Lithiums (6 on each side)

3CaBr2+2Li3PO4 -> Ca3(PO4)2+6LiBr

Balances Calciums and Bromines (3 Calciums and 6 Bromines each side)

Hope this helped!

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What would happen to the rate of a reaction with rate law rate = k [NO]^2[H2] if
Ede4ka [16]

The rate of a reaction would be one-fourth.

<h3>Further explanation</h3>

Given

Rate law-r₁ = k [NO]²[H2]

Required

The rate of a reaction

Solution

The reaction rate (v) shows the change in the concentration of the substance (changes in addition to concentrations for reaction products or changes in concentration reduction for reactants) per unit time.  

Can be formulated:  

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\large{\boxed{\boxed{\bold{v~=~-\frac{\Delta A}{\Delta t}}}}

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