## Lehninger Principles of Biochemistry 6th Edition

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 $$= 36, 723 mitochondria (C)$$
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 = $$50 molecules of glucose (E)$$