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tigry1 [53]
3 years ago
15

What are some factors that can regulate enzyme activity?

Biology
2 answers:
steposvetlana [31]3 years ago
8 0
Temperature, pH<span>, </span>enzyme concentration<span>, </span>substrate concentration, and the presence of any inhibitors<span> or </span>activators are just some of the factors.
Elena L [17]3 years ago
4 0
Feedback inhibition... where the product of a reaction becomes a competitive inhibitor in order to block more substrate from entering the active site.

Allosteric inhibition... certain enzymes have regions on them away from the active site where special allosteric inhibitor proteins can bind to which change the shape and overall function of the enzyme. This is a form of noncompetitive inhibition.

Other factors could be ph, temperature, amount of substrate/enzyme...
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A random variable X has a mean μ = 10 and a variance σ2 = 4. Using Chebyshev’s theorem, find (a) P(|X −10|≥3); (b) P(|X −10| &lt;
gladu [14]

c) P(5<X<15); (d) the value of the constant c such that P(|X −10|≥c) ≤ 0.04

Answer:

a) 0.4444

b) 0.5556

c) 0.8400

d) 10

Explanation:

Chebyshev’s theorem states that

P(|X - μ| ≥ kσ) = 1/k²

μ = Mean = 10

σ = standard deviation = √variance = √4 = 2

a) P(|X −10|≥3)

Comparing this with P(|X - μ| ≥ kσ)

It is evident that kσ = 3

2k = 3

k = 3/2

k² = 2.25

1/k² = 0.444

P(|X −10|≥3) = 0.444

b) P(|X −10| < 3) = 1 - P(|X −10|≥3)

And P(|X −10|≥3) was found in (a) to be 0.444

So,

P(|X −10| < 3) = 1 - P(|X −10|≥3) = 1 - 0.4444 = 0.5556

c) P(5<X<15) = P(|X-10|<5) = 1 - P(|X-10|≥5)

P(|X-10|≥5) = 1/k²

Comparing with P(|X - μ| ≥ kσ) = 1/k²

where 2k = 5

k = 5/2

k² = 6.25

1/k² = 0.16

P(|X-10|≥5) = 0.16

P(5<X<15) = P(|X-10|<5) = 1 - P(|X-10|≥5) = 1 - 0.16 = 0.8400

d) P(|X −10|≥c) ≤ 0.04

1/k² = 0.04

k = 5

c = kσ = 5×2 = 10

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Answer:

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Explanation:

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Answer: All of the Above

Explanation:

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