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Alla [95]
10 months ago
14

The km of an enzyme is 5. 0 mm. Calculate the substrate concentration when this enzyme operates at one‑quarter of its maximum ra

te.
Chemistry
1 answer:
Makovka662 [10]10 months ago
3 0

The substrate concentration of the enzyme operating at one‑quarter of its maximum rate is = 0.333.

Relationship between Km and substrate concentration is -

Km is the concentration of substrate.It allows the enzyme to achieve half Vmax. High Km enzyme requires a higher concentration of substrate to get Vmax. Since, Km is a constant. If the substrate concentration is increased, it has no effect on it.

An enzyme with a high Km has a low affinity for its substrate. The substrate concentration Km corresponds to the substrate concentration.

The substrate concentration at which the reaction rate of the enzyme-catalyzed reaction is half of the maximum reaction rate Vmax.

The equation is:

                <em>V₀ = </em><u><em>Vmax [S] </em></u>

<em>                          [S] + Km</em>

Here,

V₀ is initial rate,

Km is the dissociation constant between the substrate and the enzyme,

Vmax is the maximum rate, and

S is the concentration of substrate.

taking fraction of V₀ and Vmax :

<u><em>    V₀    </em></u><em> = </em><u><em>     [S</em></u><em>]</em><u><em>     </em></u><em>  </em>

<em>Vmax      [S] + Km</em>

<u><em>   </em></u><u>V₀   </u> =   <u>   0.5Km   </u>  = 0.333

Vmax       1.50 + Km

Therefore, the substrate concentration of this enzyme operating at one‑quarter of its maximum rate is = 0.333.

To learn more about substrate concentration,

brainly.com/question/18237939

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15.0 g of cream at 10.0 ℃ are added to an insulated cup containing 150.0 g of coffee at 78.6 °C. Calculate the equilibrium tempe
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Answer:

The equilibrium temperature of the coffee is 72.4 °C

Explanation:

Step 1: Data given

Mass of cream = 15.0 grams

Temperature of the cream = 10.0°C

Mass of the coffee = 150.0 grams

Temperature of the coffee = 78.6 °C

C = respective specific heat of the substances( same as water) = 4.184 J/g°C

Step 2: Calculate the equilibrium temperature

m(cream)*C*(T2-T1) = -m(coffee)*c*(T2-T1)

15.0 g* 4.184 J/g°C *(T2 - 10.0°C) = -150.0g *4.184 J/g°C*(T2-78.6°C)

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3 years ago
How many moles of gas are present in 1.13 L of gas at 2.09 atm and 291 K?
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<h2><u>Answer:</u></h2>

n = 0.0989 moles

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n = PV / RT

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n = (2.09atm)(1.13L) / (0.08206)(291K)

n = 0.0989 moles

3 0
3 years ago
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