Lehninger Principles of Biochemistry
Lehninger Principles of Biochemistry
7th Edition
ISBN: 9781464126116
Author: David L. Nelson, Michael M. Cox
Publisher: W. H. Freeman
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Question
Chapter 1, Problem 1P

(a)

Summary Introduction

To determine: The size of a typical eukaryotic cell with cellular diameter of 50μm if the cell is magnified 10,000 times.

Introduction:

The eukaryotic cells are more complex than the prokaryotic cells and they vary based on the structure and functions. The eukaryotes have been evolved from the unicellular organisms in the timeline of evolution.

(a)

Expert Solution
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Explanation of Solution

The diameter (D) of the given eukaryotic cell is 50μm. The power of magnification is 10,000 times. The diameter of the magnified cell is as follows:

D=50μmD(Aftermagnification)=50μm×104=500×103μm=500mm

Conclusion

The size of a typical eukaryotic cell with cellular diameter of 50μm if the cell is magnified 10,000 times is 500mm_.

(b)

Summary Introduction

To determine: The number of molecules of actin in a myocyte of diameter 50μm if the diameter of actin molecules is 3.6nm and cell has no other cellular components.

Introduction:

The eukaryotes have been evolved from the unicellular organisms in the timeline of evolution. The second phase of evolution has been started when the levels of multicellularity diversify the organisms into algal species, fungi, plants, and animals. Myocyte is known as a muscle cell.

(b)

Expert Solution
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Explanation of Solution

The diameter (D) of the given myocyte cell is 50μm. The radius of this cell is half the diameter and is equal to 25μm. The radius of cell with unit in (meter) is 25×106m.

The diameter (D) of the given actin molecule is 3.6nm. The radius of this actin molecule is half the diameter and is equal to 1.8nm. The radius of actin with unit in (meter) is 1.8×109m.

The volume of the given myocyte is as follows:

V(Myocyte)=43πr3=43π(25×106m)3=6.54×1014m3

The volume of the given actin is as follows:

V(Actin)=43πr3=43π(1.8×109m)3=2.44×1026m3

The number of actin molecules in the given myocyte is as follows:

N=V(Myocyte)V(Actin)=6.54×1014m32.44×1026m3=2.7×1012

Conclusion

The number of molecules of actin in a myocyte (if the diameter of actin molecules is 3.6nm_ and cell has no other cellular components) is 2.7×1012_..

(c)

Summary Introduction

To determine: The number of molecules of mitochondria in a liver cell of diameter 50μm if the diameter of action molecules is 1.5μm and cell has no other cellular components is.

Introduction:

The eukaryotic cells are more complex than the prokaryotic cells and they vary based on structure and functions. The eukaryotes have been evolved from the unicellular organisms in the timeline of evolution. Mitochondrion is part of a eukaryotic cells and the energy storage of any cell.

(c)

Expert Solution
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Explanation of Solution

The diameter (D) of the given liver cell is 50μm. The radius of this cell is half the diameter and is equal to 25μm. The radius of cell with unit in (meter) is 25×106m.

The diameter (D) of the given mitochondria is 1.5μm. The radius of this mitochondria is half the diameter and is equal to 0.75nm. The radius of mitochondria with  in meter is 0.75×106m.

The volume of the given liver is as follows:

V(Livercell)=43πr3=43π(25×106m)3=6.54×1014m3

The volume of the given mitochondria is as follows:

V(Mitochondria)=43πr3=43π(0.75×106m)3=1.76×1018m3

The number of mitochondria molecules in the given myocyte is as follows:

N=V(Myocyte)V(Mitochondria)=6.54×1014m31.76×1018m3=36,000

Conclusion

The number of mitochondria in a liver cell (if the diameter of action molecules is 1.5μm_ and cell has no other cellular components) is 36,000_.

(d)

Summary Introduction

To determine: The number of molecules of glucose be present in spherical eukaryotic cell if the cellular concentration is 1mM.

Introduction:

Glucose is known as a simple sugar molecule which possess a molecular formula  C6H12O6. It is referred as the blood sugar.

(d)

Expert Solution
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Explanation of Solution

The volume of the given eukaryotic cell is 6.54×1014m3. The volume of the same cell is as follows:

1m3=(100)3cm36.54×1014m3=6.54×1014m3×(100)3cm3=6.5×108cm3

One cubic centimeter is equal to one millimeter. Therefore, the volume is 6.5×108mL.

The number of molecules in one liter of 1mM solution of glucose is as follows:

N=0.001mol1000mL×6.022×1023molecules1mol=6.022×1017molecules/mL

The total number of glucose in given cell is as follows:

N=Numberofmolecules×Volumeofcell=6.022×1017molecules/mL×6.5×108mL=3.9×1010molecules

Conclusion

The number of molecules of glucose be present in spherical eukaryotic cell if the cellular concentration is 1mM is 3.9×1010_.

(e)

Summary Introduction

To determine: The number of glucose molecules in hexokinase molecule if the hexokinase concentration is 20μm.

Introduction:

Glucose is known as a simple sugar molecule which possess a molecular formula  C6H12O6. It is circulated in an organism’s body and is referred as the blood sugar. Hexokinase refers to an enzyme that helps in phosphorylating the hexoses.

(e)

Expert Solution
Check Mark

Explanation of Solution

The concentration of hexokinase molecule is 20μm and can be written as 0.00002M. The concentration of glucose molecule is 1mM and can be written as 0.001M. The ratio of the concentration of glucose and concentration of hexokinase is as follows:

Ratio=ConcentrationofglucoseConcentrationofhexokinase=0.001M0.00002M=501

One molecule of hexokinase will have 50 molecules of glucose.

Conclusion

The number of glucose molecules in hexokinase molecule (if the hexokinase concentration is 20μm) is 50_.

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