Given the following: You have a [hormone] plasma = 1 µM, which is 60 % bound by

Question

Given the following: You have a [hormone]plasma = 1 µM, which is 60 % bound by plasma proteins. The liver enzyme responsible for metabolizing this hormone does so at a constant rate per [hormone]; i.e., the enzyme follows first-order kinetics. The decay constant, kel, is 0.578. Calculate the t1/2 and then determine the [hormone]plasma 6 half-lives later. Fill in the following table below.

Number of half-lives

Time passed

[Hormone]plasma

1

2

3

4

5

6

Number of half-lives

Time passed

[Hormone]plasma

1

2

3

4

5

6

Explanation / Answer

t1/2 = (0.693/ k) = (0.693/0.578) = 1.2 sec (since unit is not given, I have assumed it to be second).

N0 = initial concentration of the hormone is given = 1micro mole. One enzyme can bind only one reactant at a time. It is also given that 60% of the hormone is bound to plasma proteins. So, only 40% of this hormone i.e. 0.40 micromole will be able to react with the enzyme. So, N0 = 0.40

After one half life, the final concentration Nt1= 0.20 micromole. Amount of hormone present in plasma will be = 0.20 + 0.60 = 0.80 micromole.

After second half life, the final concentration will be half of Nt1 , i.e. Nt2 = 0.10 micromole. Amount of hormone present in plasma will be = 0.10 + 0.60 = 0.70 micromole. ,

So, the table will be:-

Number of half-lives

Time passed (in sec)

[Hormone]plasma (in micromole)

1

1.2

0.20 +0.60 = 0.80

2

2.4

0.10 +0.60 = 0.70

3

3.6

0.05+0.60

4

4.8

0.025+0.60

5

6.0

0.0125 +0.60

6

7.2

0.00625 +0.60

Number of half-lives

Time passed (in sec)

[Hormone]plasma (in micromole)

1

1.2

0.20 +0.60 = 0.80

2

2.4

0.10 +0.60 = 0.70

3

3.6

0.05+0.60

4

4.8

0.025+0.60

5

6.0

0.0125 +0.60

6

7.2

0.00625 +0.60

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