Principles of Biochemistry (Loose Leaf) (6th Edition) Edit editionThis problem has been solved:
Looking for the textbook?- CH1
- CH2
- CH3
- CH4
- CH5
- CH6
- CH7
- CH8
- CH9
- CH10
- CH11
- CH12
- CH13
- CH14
- CH15
- CH16
- CH17
- CH18
- CH19
- CH20
- CH21
- CH22
- CH23
- CH24
- CH25
- CH26
- CH27
- CH28
- 1P
- 2P
- 3P
- 4P
- 5P
- 6P
- 7P
- 8P
- 9P
- 10P
- 11P
- 12P
- 13P
- 14P
- 15P
- 16DAP
(a) According to the given data:
The typical eukaryotic cell has a cellular diameter = 50 µm.
An electron microscope is used to magnify a cell of = 10,000 fold.
The diameter of the magnified cell is = 50 µm.
Electron microscope is used to magnify a cell to 10,000 fold = 104 µm.
The magnified cell can be calculated by the following equation:
(b) According to the given data:
The diameter of the actin molecule = 3.6 nm (D = r/2).
The radius of a globular actin molecule is = 
The volume of the sphere is calculated by the following equation:
Where,
r = 
= 3.14.
The volume of one actin molecule (in cu.mm) is given by:
Hence, the volume of one actin molecule is
.
The cell volume can be calculated as follows:
…… (1)
Where,
Substitute the values in equation (1)
The number of actin molecules accommodated inside the cell is calculated as follows:
= 2.66 x 1012 or 2.7 x 1012 molecules.
(c) According to the given data:
The diameter of the mitochondrion is given by = 
The radius of the mitochondria is given by =
.
The shape of the mitochondria is spherical.
Therefore the volume of a spherical mitochondrion is calculated by the equation:
= 4/3pr3
Where,
= 0.75 µm. 
= 3.14.
=
= 1.77 x 10-.
The value obtained from volume of the actin molecule is found to be:
Therefore, the number of mitochondria that is present in the liver cell is 3600 mitochondria.
(d) The volume of the eukaryotic cell is = 6.5 x 10-14 m3 (or)
The volume of the eukaryotic cell is = 6.5 x 10-8 cm3 or 6.5 x 10-8 ml.
Avogadro number = 6.02 x 1023molecules/mol.
One Liter of 1 mM solution consists of:
The number of glucose molecules is calculated by: Multiplying the concentration of glucose and the product of the cell volume.
Therefore, the number of glucose molecules present in eukaryotic cell is 
(e) The enzyme hexokinase catalyzes the metabolism of glucose in the first step, during the conversion of glucose into glucose-6-Phosphate.
According to the given data:
The concentration of glucose is given by = 20 µm.
The concentration of glucose/hexokinase is:
Thus, 50 molecules of glucose are available as substrate in hexokinase.
Corresponding textbook
Plus, we regularly update and improve textbook solutions based on student ratings and feedback, so you can be sure you're getting the latest information available.
The best part? As a Chegg Study subscriber, you can view available interactive solutions manuals for each of your classes for one low monthly price. Why buy extra books when you can get all the homework help you need in one place?
