Biochemistry I. Introduction
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1 Biochemistry I Biochemistry is a dichotomous relationship between biology and organic chemistry. Biochemistry is actually the study of chemical reactions within the cell. Types of organisms: Prokaryotes Eukaryotes In addition to molecular structure and function and the relationship between these characteristics, biochemist also think about bioenergetics. 1
2 Biochemist s periodic table Of the hundred plus chemical elements, only about 30 (29%) occur naturally in plants and animals. Elements in red are found in bulk form in living cells and are essential for life. Yellow are trace elements and likely essential. Elements in blue are present in some organisms. 2
3 Elemental composition of the universe. The earth s crust. The human body. 3
4 Biochemistry began about 170 years ago. Classify biomolecules Intermediate metabolism, biochemist interest is in intermediates. C 6 H 12 O 6 intermediate CO 2 + H 2 O Watson & Crick (DNA) Interaction between biomolecules. Super Chemistry Molecular Biology Study of control mechanisms Study of diseases at the molecular level Classifying biomolecules: Approximately 2500 proteins are found in E. coli. Cytochrome C is a protein involved in the metabolic process best known as the electron transport system. General observation: Cytochrome C has been isolated from many different species to compare, contrast and characterize the protein. Proteins are large macromolecules, which are made from individual or single units known as amino acids. DNA & RNA are large macromolecules made from individual or single units known as nucleotides. 4
5 Water is the principle ingredient of the cell. Most abundant chemical compound in the biosphere. General solvent for most all living organisms. Continuous Phase Diffuses through every cell and tissue. Serves in aiding to maintain structure of the macromolecules. Water dictates how proteins fold (hydrophobic/hydrophilic) Amino acid sequence determines shape and shape determines structure. Important in ph and buffers. Cell Size If you were to magnify a cell 10,000 fold, how big would it appear? Assume you are viewing a typical eukaryotic cell with a cellular diameter of 50 microns. The magnified cell would have a diameter of 50 X 10 4 µm = 500 X 10 3 µm = 500 mm. 5
6 Cell Concentration If the cell were muscle cell, how many molecules of actin could it hold assuming there are no other cellular components present? (Actin molecules are spherical with a diameter of 3.6 nm; assume the muscle cell is spherical, the volume of a sphere is 4/3 πr 3 ) The radius of a globular actin molecule is 3.6 nm/2 = 1.8 nm; the volume of the molecule, in cubic meters is 4/3[3.14(1.8 X 10 9 m) 3 ] = 2.44 X m 3 The total number of actin molecules that could fit inside the cell is found by dividing the muscle cell volume by the actin molecule volume. Cell volume = 4/3[3.14(25 X 10 6 m) 3 ] = 6.5 X m 3. Thus the number of actin molecules in the hypothetical muscle cell is: (6.5 X m 3 )(2.44 X m 3 ) = 2.66 X X molecules General information in E. coli DNA The genetic information contained in DNA consists of linear sequences of successive coding units, known as codons. Each codon is a specific sequence of three nucleotides (three nucleotide pairs in double stranded DNA), and each codon code for a single amino acid unit in a protein. The molecular weight of an E. coli DNA molecule is about 3.I X The average molecular weight of a nucleotide pair is 660, and each nucleotide pair contributes 0.34 nm to the length of DNA. 6
7 General information in E. coli DNA Calculate the length of an E. coli DNA molecule. Compare the length of the DNA molecule with the cell dimensions. How does the DNA molecule fit into the cell? Assume that the average protein in E. coli consist of a chain of 400 amino acids. What is the maximum number of proteins that can be coded by an E. coli DNA molecule? The number of nucleotide pairs in the DNA molecule can be calculated by dividing the molecular weight of DNA by that of a single base pair. (3.1 X 10 9 )/(0.66 X 10 3 ) = 4.7 X 10 6 pairs. (4.7 X 10 6 pairs)(0.34 nm/pair) 1.6 X 10 6 nm = 1.6 mm General information in E. coli DNA The length of an E. coli cell is 2.0 µm or mm, which means that the DNA is 1.6 mm/0.002 mm or 800 times longer than the cell. Since the DNA molecule has 4.7 X 10 6 nucleotide pairs, it must have 1/3 this number of triplet codons: (4.7 X 10 6 )/3 = 1.57 X 10 6 codons. If each protein has an average of 400 amino acids, each requiring one codon, the number of proteins that can be coded by E. coli DNA is: (1.57 X 10 6 )/400 = 3,930 proteins per cell 7
8 Viruses Viruses cannot exist independently and are usually not considered a life-form. Viruses are deemed parasites since they are unable to carry out metabolism or reproduction without the help of a host cell. When viruses infect a cell, they take control of the cells metabolic machinery and force it to synthesize nucleic acids and proteins for new virus particle. Viruses are the cause of many plant and animal maladies and their presence in the world has resulted in much human suffering. However, enormous amount of biochemistry has been learned form studies of their actions. 8
9 9
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