Tiêu đề 5. MOLECULAR BASIS OF INHERITANCE.pdf ✅

Revision Notes Genetic Material 3⁄4 Nucleic Acids DNA and RNA are the two types of nucleic acids. DNA is the genetic material in all organisms except some viruses. RNA is the genetic material in some viruses. RNA mostly functions as messenger. 3⁄4 Structure of Polynucleotide Chain Polynucleotides are the polymers of nucleotides. DNA and RNA are examples of polynucleotides. A nucleotide has 3 components : (i) A nitrogenous base (ii) A pentose sugar (ribose in RNA and deoxyribose in DNA) (iii) A phosphate group Nitrogen bases are of 2 types : (a) Purines : It includes Adenine (A) and Guanine (G). (b) Pyrimidines : It includes Cytosine (C), Thymine (T) and Uracil (U). Thymine (5-methyl Uracil) present only in DNA and Uracil only in RNA ( In place of thymine). A nitrogenous base is linked to the pentose sugar through an N-glycosidic linkage to form nucleoside. Nucleosides in RNA Nucleosides in DNA Adenosine Deoxyadenosine Guanosine Deoxyguanosine Cytidine Deoxycytidine Uridine Deoxythymidine Nitrogen base + sugar + phosphate group = Nucleotide (deoxyribonucleotide). In RNA, every nucleoside residue has an additional –OH group present at 2’-position in the ribose (nucleoside=Ribose sugar+ Base pair) phosphate group is absent in nucleoside. 2 nucleotides are linked through 3’ → 5’ phosphodiester bond to form dinucleotide. When series of nucleotides are linked together, it forms polynucleotide. 3⁄4 Structure of DNA Johann Friedrich Miescher (1869): Identified DNA and named it as ‘Nuclein’. James Watson & Francis Crick proposed the double helix model of DNA. It was based on the X-ray diffraction data produced by Maurice Wilkins & Rosalind Franklin. DNA is made of two polynucleotide chains coiled in a right-handed fashion. Its backbone is formed of sugar and phosphates. The bases project inside. The two chains have anti-parallel polarity i.e., one chain has the polarity 5’ → 3’ and the other has 3’ → 5’. Nitrogen bases of opposite chains are held together by hydrogen bonds forming base pairs (bp). There are two hydrogen bonds between A and T (A = T) and three H-bonds between C and G (C ≡ G). Purine comes opposite to a pyrimidine. This generates a uniform distance between the two strands. 3⁄4 Erwin Chargaff’s Rule Purines and pyrimidines are always in equal amounts i.e., A + G = T + C. In DNA, the proportion of A is equal to T and the proportion of G is equal to C i.e., A = T and G = C. MOLECULAR BASIS OF INHERITANCE 5 CHAPTER List of Topics Topic-1: Nucleic Acid – DNA and RNA Topic-2 : Genetic Code, Translation, Lac Operon, HGP and DNA Fingerprinting Learning Objectives Students will be able to learn about 1. Structure of DNA and its significance. 2. Formation of proteins via DNA through mRNA. 3. Gene Expression and control. 4. Genome projects and its advantages. TOPIC-1 Nucleic Acid – DNA and RNA Concepts Covered  Nucleic acids, packaging of DNA helix,  Nucleosome,  Experiments to show DNA as a genetic material,  RNA, process of protein synthesis.
Molecular Basis of Inheritance The base ratio A + T/G + C may vary from species to species but constant for a given species. Length of DNA = number of base pairs × distance between two adjacent base pairs. f 174 (a bacteriophage) has 5386 nucleotides. Bacteriophage lambda has 48502 base pairs (bp). E. coli has 4.6 × 106 bp. Haploid content of human DNA = 3.3 × 109 bp. Number of base pairs in human = 6.6 × 109 Length of DNA in humans = 6.6 × 109 bp × 0.34 × 10–9 m/bp = 2.2 m Length of DNA in E. coli =1.36 mm (1.36 × 10–3 m). \ The number of base pairs = 1.36 × 10–3 m/0.34 × 10–9 m/bp = 4 × 106 bp. 3⁄4 Packaging of DNA Helix In prokaryotes (e.g., E. coli), the DNA molecule is held with some positively charged non-histone basic proteins like negatively charged polyamines and form ‘nucleoid’. In eukaryotes, it involves a number of molecules. His- tones, Histone octamer, Nucleosome, Chromatin. Two types of chromatin are: (a) Euchromatin: Loosely packed and transcriptionally active chromatin and is light-stained. (b) Heterochromatin: Densely packed and inactive region of chromatin and stains dark. In eukaryotes, there is a set of positively charged basic proteins called histones. Histone proteins are rich in positively charged basic amino acid residues lysine and arginine. There are five types of histone proteins-H1, H2A, H2B, H3 and H4. Two molecules each of H2A, H2B, H3 and H4 organize to form a unit of eight molecules called as histone octamer. Negatively charged DNA is wrapped around positively charged histone octamer to form a structure called a nucleosome. Nucleosomes are connected with the help of linker DNA on which H1 Histone is present. 3⁄4 Nucleosome A typical nucleosome contains 200 bp of DNA helix. Therefore, the total number of nucleosomes in human = 6.6 × 109 bp/200 bp = 3.3 × 107 . Nucleosomes constitute the repeated unit to form chromatin. Chromatin is the thread-like stained bodies. Nucleosomes in chromatin appears as “beads- on-string” when it is viewed under the electron microscope. Chromatin is packaged to form a solenoid or a zig-zag structure. Further supercoiling constitute a looped structure called chromatin fibre. These chromatin fibres further coil and condense at the metaphase stage of cell division to form chromosomes. Chromatin is packaged → solenoid → chromatin fibres → coiled and condensed at metaphase stage → chromosomes. Higher level packaging of chromatin requires non- histone chromosomal (NHC) proteins. Example-1 Q. Calculate the number of beaded structures (nucleosomes) present in the nucleus of diploid eukaryotic cell which possess 2.2 × 106 bp. Sol. One nucleosome has 200 bp. The number of beaded structures (nucleosomes) present in the nucleus of diploid eukaryotic cell which possess 2.2 × 106 bp. \ 2.2 1 ́ 0 200 6 = 1.1 × 104 or 11 × 103 nucleosomes 3⁄4 The Search for Genetic Material Griffith’s Experiment - Transforming Principle Griffith (1928) used mice and a bacterial strain, Streptococcus pneumoniae. Streptococcus pneumoniae has two strains : (a) Smooth (S) strain (Virulent): Has polysaccharide mucous coat. Causes pneumonia. (b) Rough (R) strain (Non-virulent): No mucous coat. Does not cause pneumonia. 3⁄4 Experiment S-strain → Inject into mice → Mice die R-strain → Inject into mice → Mice live S-strain (Hk) → Inject into mice → Mice live S-strain (Hk) + R-strain (live) → Inject into mice → Mice die He concluded that there exists some ‘transforming principle’, that is transferred from heat-killed S-strain to R-strain. It enabled R-strain to synthesise smooth polysaccharide coat and become virulent. This must be due to the transfer of genetic material. 3⁄4 Biochemical Characterisation of Transforming Principle Oswald Avery, Colin MacLeod & Maclyn McCarty in 1944 worked to determine the biochemical nature of ‘transforming principle’ in Griffith’s experiment. They purified biochemicals (proteins, DNA, RNA, etc.) from heat-killed S cells using suitable enzymes. They discovered that: (a) Digestion of protein and RNA (using Proteases and RNases) did not affect transformation. So, the transforming substance was not a protein or RNA. (b) Digestion of DNA with DNase inhibited transformation. It means that DNA caused the transformation of R cells to S cells i.e., DNA was the transforming substance. 3⁄4 The Genetic Material is DNA The fact that DNA is the genetic material also came from the experiments of Alfred Hershey and Martha Chase (1952). They worked with viruses that infect bacteria and are called bacteriophages. 3⁄4 Hershey-Chase Experiment—Blender Experiment Hershey and Chase made two preparations of bacteriophage – In one, proteins were labelled with 35S by putting in a medium containing radioactive sulphur (35S). In the second, DNA was labelled with 32P by putting in a medium containing radioactive Phosphorous (32P).