Answer
Step 1
From the data provided:
The eukaryotic cell has a diameter of =$50µm$
An electron microscope magnifies a cell to 10000 fold=$ 104 μm$
The cell to be magnified has a diameter of =$ 50µm$
Then magnification of the cell can be calculated as follows: $ 50 \times10000$
When micrometer is converted into millimeter
Given that 1 micrometer =$ 10-3 millimeter$
This gives: $50 \times104 \times0.001$
$$=500mm (A)$$
Step 2
Regarding the provided data:
Actin molecule has a diameter of 3.6 nm
Its radius will be = $ (3.6 nm/2) $= $1.8 nm. $
Spheres volume can be calculated from the equation below:
$\frac{4}{3}(πr^3) $
When equated,
r=$1.8 nm$
π = $3.14$
Thus the volume of the actin molecule in mm3 can be calculated by substituting the equation with actual figure as follows:
$\frac{4}{3}[3.14 (1.8 \times10^{-9} m)^3]$
$$=2.44\times 10^{-26} m^3 (B)$$
Step 3
Therefore, the volume of actin molecule = $2.44 \times 10 ^{-26} m^3$
Step 4
The volume of the cell can be calculated as follows
$\frac{4}{3}(πr^3) $…..eq 1
Given:
r= $25\times10^{-6} m$
π = $3.14$
When substituted,
$\frac{4}{3}[3.14(25\times10^{-6} m)^3]$
=$ 6.5 \times10^{-14}m^3$
To find the number of actin molecule in the cell is done as follows
(volume of Muscle cell)/(Volume of actin molecule)
$(6.5\times10^{-14})/(2.44 \times10^{-26} )$
$$= 2.66\times10^{12} molecules$$
Step 5
As per given data:
Mitochondrion have a diameter =$1.5μm $
Then its radius will be =$\frac{1.5μm}{2}= 0.75 μm$
Given that the shape of mitochondrion is spherical, then its volume can be calculated as follows
$\frac{4}{3}(πr^3) $
Step 6
Where:
r=$ 0.75 μm$
π = 3.14
= $\frac{4}{3}[3.14(0.75\times10^{-6} m)^3]$
=$1.77\times1018$
Since the volume of actin is found to be
$6.5\times10^{-14}m^3$
To find the number of mitochondria we can calculate as follows:
(volume of actin molecule)/(volume of mitochondrion)
=$(6.5 \times10^{-14} m^3)/(1.77 \times10^{-18} m^3 )$
=$ 36, 723$
Thus number of mitochondria found in the liver cell $$= 36, 723 mitochondria (C)$$
Work Step by Step
Step 7
Eukaryotic cell has a volume of $6.5 \times10^{-14} m^3$
Avogadro number is given to be = $6.02 \times10^{23}$ molecules/mol
If one liter of 1mM solution has
=$ (0.001mol/100mL)( 6.02 \times10^{23} molecules/mol) $
=$ (0.000001mol/mL)( 6.02 \times10^{23} molecules/mol) $
= $$6.02\times10^{17} molecules/ml (D)$$
Glucose number can be found by multiplying the concentration of glucose by the cell volume product.
Thus:
$ (6.5 \times10^{-8} ml)( 6.02\times10^{17} molecules/ml) $
=$3.91\times10^{10} molecules$
Therefore, glucose molecules that is in eukaryotic cell =$$ 3.91 \times10^{10} molecules (D) $$
Step 8
Hexokinase is an enzyme catalyzing the metabolism of glucose in its initial stage to be converted to glucose-6-phosphate
And from the data given:
Glucose is given concentration = $20µm$
Then, glucose : hexokinase can be found as follows
0.001/0.00002
=$50 molecules$
Therefore, the hexokinase substrate has = $$50 molecules of glucose (E)$$
