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Dafna1 [17]
4 years ago
12

It was found that a reversible inhibitor and a substrate bind an enzyme, but at different sites. The inhibitor was quite potent

at reducing enzyme activity, but a Scatchard analysis indicated it could not actually bind the enzyme unless the substrate was bound first. These results led to what conclusion? A. It was a competitive inhibitor. [B [U) B. It was a mixed inhibitor. C. It was an uncompetitive inhibitor. D. It was an allosteric modulator. E. It was a mechanism-based inactivator.
Chemistry
1 answer:
natita [175]4 years ago
8 0

Answer:

The correct option is C (Uncompetitive inhibitor)

Explanation:

  • Uncompetitive inhibitor takes place when an inhibitor binds at different site (allosteric site) of the enzyme after the substrate forms a complex with the enzyme.
  • The Uncompetitive inhibitors bind to a different site on the enzyme while the substrate binds to the active site on the enzyme.
  • It differs from non-competitive inhibition because non-competitive inhibition can occur with or without the presence of the substrate.
  • Uncompetitive inhibition is also referred to as Anti-competitive inhibition.
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Most scientific theories involving microscopic and macroscopic phenomenon have taken several years to be developed; however, this theories are still under revision till date.

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Explain why a strong acid must have a weak conjugate base
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Explanation:

The specie produced when a strong acid is dissociated in water along with the proton is the conjugate base of that acid.

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3 years ago
A 12.0 g sample of a metal is heated to 90.0 ◦C. It is then dropped into 25.0 g of water. The temperature of the water rises fro
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Answer:

The specific heat of the metal is 0.34 J/g°C

Explanation:

Step 1: Data given

Mass of the metal = 12.0 grams

Mass of the water = 25.0 grams

Initial temperature of the metal = 90.0 °C

Initial temperature of the water = 22.5 °C

Final temperature = 25 °C

Specific heat of water = 4.184 J/g°C

Step 2: Calculate the specific heat of the metal

Qlost = Qgained

Q = m*c*ΔT

Qmetal = -Qwater

m(metal) *c(metal)* ΔT(metal) = -m(water) * c(water) *ΔT(water)

⇒ with mass of metal = 12.0 grams

⇒ with c(metal) = TO BE DETERMINED

⇒ with ΔT(metal) = T2 - T1  = 25.0°C - 90.0 °C = -65.0 °C

⇒ with mass of water = 25.0 grams

⇒ with c(water) = 4.184 J/g°C

⇒ with ΔT(water) = T2 - T1 = 25.0 - 22.5 °C = 2.5 °C

12.0 * c(metal) * -65.0 °C = -25.0g * 4.184 J/g°C * 2.5°C

-780.0 * c(metal) = -2615  ( 2.6*10^3 with sig figs)

c(metal) = 0.335 (=0.34 with sig figs)

The specific heat of the metal is 0.34 J/g°C

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Answer:

85

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