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lisov135 [29]
2 years ago
6

What component of enzyme structure is responsible for enzyme specificity

Biology
2 answers:
son4ous [18]2 years ago
4 0
A few enzymes<span> exhibit absolute </span>specificity<span>; that is, they will catalyze only one particular reaction. Other </span>enzymes<span> will be specific for a particular type of chemical bond or functional group. In general, there are four distinct types of </span>specificity<span>: Absolute </span>specificity<span> - the </span>enzyme<span> will catalyze only one reaction.</span>
const2013 [10]2 years ago
3 0
Hey there,

Your question states: <span>What component of enzyme structure is responsible for enzyme specificity?

Based on my understanding of your question, your correct answer would be S</span>ubstrates.<span>Substrates are responsible for enzyme specificity. They do play an important role in the enzyme. There are species that basically observe a chemical reaction.
</span><span>
Your correct answer is </span>\boxed{Substrates}

Hope this helps.
~Jurgen
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The state of health and functioning of the liver is often assessed with dye-tracer techniques. The dye used most frequently is b
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Answer

1.Radioactive or chemical decay

2. X(t) = Xoe^-kt

lnY(t) = -t + lnYo

4. 6.07mg

Explanation:

Let the liver and and blood compartment be represented by the symbol L and B respectively

For the liver

Suppose a first Order removal process started with an amount X, in which amount b disappeared in time t, the process is decay process which can be represented as follows,

∫dX/dt = -K1X

By rearrangement and integration;

∫dX/X = -K1t

ln X = -K1t + C

Since At t= 0, X = Xo then

C = lnXo

The equation becomes:

lnX = -K1t + lnXo

lnX - lnXo = -K1t

ln(X/Xo) = -K1t

X/Xo = e^-kt

X(t) = Xoe^-kt

X(t) = Xoe^-0.5t........(2)

Similarly for B (blood), suppose a first order flow flow of the dye move from the blood to the liver, let Y be the initial concentration, and amount b that has flown to the liver in time t

B---------> B(t)

t=0 Yo 0

t=t. Y-b. b

dY/dt = -K12(Y-b)...........(3)

Let Y-b = Y(t)

∫dY/dt = -∫K12t

By rearrangement and integration;

∫dy/Y(t) = ∫-K12dt

lnY(t) = -K12t + C1

at t= 0, C1 = ln(Yo)

Therefore ln Y(t) = -K12t + lnYo

But K12 =1

ln Y(t) = -t + lnYo.............(4)

(3) The assumptions used here

is that of a decay for the liver . The amount remaining taking as the amount of a substance

(4) using the equation 2,

X(t)= Xo e^-K1t........(2)

For time t = 1hour, and an initial amount X = 10mg, K = 0.5

X(t) = 10× e^-0.5

X(t) = 10 × 0.607

X(t) = 6.07mg

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