Molecular Basis of Inheritance PDF
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This document provides an overview of the molecular basis of inheritance, focusing on nucleic acids (DNA and RNA). It details the structure of nucleotides and the salient features of the DNA double helix. It also discusses the packaging of DNA in prokaryotes and eukaryotes and the central dogma of molecular biology.
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MOLECULAR BASIS OF INHERITANCE NUCLEIC ACIDS 1. DNA (Deoxyribonucleic Acid) and RNA (Ribonucleic Acid) are two types of nucleic acid found in living organisms. 2. DNA acts as genetic material in most of the organisms. 3. RNA also acts as genetic material in...
MOLECULAR BASIS OF INHERITANCE NUCLEIC ACIDS 1. DNA (Deoxyribonucleic Acid) and RNA (Ribonucleic Acid) are two types of nucleic acid found in living organisms. 2. DNA acts as genetic material in most of the organisms. 3. RNA also acts as genetic material in some organisms as in some viruses and acts as messenger. It functions as adapter, structural, and in some cases as a catalytic molecule STRUCTURE OF NUCLEIC ACID NUCLEOTIDE HAS THREE COMPONENTS – 1. Nitrogenous base (A, T/U, C, G) 2. Pentose sugar (Ribose in RNA, Deoxyribose in DNA) 3. Phosphate group TYPES OF NITROGENOUS BASES – a) Purines - Adenine and Guanine b) Pyrimidines - Cytosine, Uracil/Thymine FORMATION OF NUCLEOTIDE N-GLYCOSIDIC LINKAGE Nitrogenous base + Pentose sugar Nucleoside PHOSPHO-ESTER PO4 group at 5’-OH of a nucleoside Nucleotide Two nucleotides 3’-5’ PHOSPHO-DIESTER LINKAGE dinucleotide More nucleotide joins together to form polynucleotide STRUCTURE OF DNA 1. The DNA – it is a long polymer ofdeoxyribonucleotides/deoxyribonucleoside (dNTP). 2. A pair of nucleotide is also known as base pairs. 3. Length of DNA is usually defined as number of nucleotides present in it. 4. Escherichia coli have 4.6 x 106 bp 5. Haploid content of human DNA is 3.3 × 109 bp SALIENT FEATURES OF DOUBLE HELIX STRUCTURE OF DNA 1. Made of two polynucleotide chains. 2. Sugar and phosphate forms the backbone and bases projected to inside. 3. Two chains have anti-parallel polarity. 4. Two strands are held together by hydrogen bond present in between bases. 5. Adenine of one strand pairs with Thymine of another strand by two hydrogen bonds and vice versa. 6. Guanine of one strand pairs with Cytosine of another strand by three hydrogen bonds and vice versa. 7. A purine comes opposite to a pyrimidine. This generates approximately uniform distance between the two strands of the helix. 8. The two chains are coiled in a right – handed fashion. 9. The pitch of the helix is 3.4 nm or 34 A0 10. There are roughly 10 bp in turn. 11. The distance between the bp in a helix is 0.34nm or 3.4 A 0. CENTRAL DOGMA OF MOLECULAR BIOLOGY 1. Proposed by Crick 2. According to which, genetic information flows from DNA to RNA and then to Protein 3. In some viruses like retroviruses, the flow of information is in reverse direction, that is from RNA to DNA, then to mRNA and then to Protein PACKAGING OF DNA HELIX 1. Distance between two conjugative base pairs is 0.34nm, the length of the DNA in a typical mammalian cell will be 6.6 X109 bp X 0.34 X10-9 /bp, it comes about 2.2 meters. 2. The length of DNA is more than the dimension of a typical nucleus (10-6m), how is such a long polymer packaged in a cell? PACKAGING IN PROKARYOTES 1. They do not have definite nucleus. 2. The DNA is not scattered throughout the cell. 3. DNA is held together with some proteins in a region is called ‘nucleoid’. 4. The DNA in nucleoid is organized in large loops held be proteins. PACKAGING IN EUKARYOTES 1. Negatively charged DNA is held with positively charged proteins called HISTONES 2. Histones are positively charged due to rich in basic amino acids like Lysines and arginines 3. Histone is found in 5 iso-forms – H1, H2A, H2B, H3, H4 ---forms a unit of eight molecules by two molecules each of H2A, H2B, H3, H4 -HISTONE OCTAMERE 4. Negatively charged DNA wrapped around positively charged histone octamere to form a structure called NUCLEOSOME – A typical nucleosome contains 200 bp of DNA helix 5. The nucleosomes are seen as ‘BEADS-ON-STRING’ structure when viewed under electron microscope 6. Nucleosome constitutes the repeating unit of a structure in nucleus. 7. Single nucleosome contains about 200 base pairs 8. Chromatin is the repeating unit of nucleosome PACKAGING AT HIGHER LEVEL (Histone + Non-histone protein) EUCHROMATIN 1. Loosely coiled regions of chromatin 2. Stained light 3. Transcriptionally active HETEROCHROMATIN 1. Densely packed 2. Stains dark 3. Transcriptionally inactive THE SEARCH OF GENETIC MATERIAL TRANSFORMING PRINCIPLE 1. Given by Frederick Griffith in 1928. 2. His experiment based on Streptococcus pneumoniae (caused pneumonia). 3. He used two different strains of bacteria – R & S a) Smooth shiny colonies called S strain. b) Rough colonies called R strain. TWO STRAINS OF Streptococcus pneumoniae S → Smooth 1. Secrete a polysaccharide capsule -- Produce smooth colonies on solid media 2. Heat sensitive 3. virulent R → Rough 1. Unable to secrete a capsule 2. Heat resistant 3. Produce colonies with a rough appearance Non-virulent GRIFFITH’S TRANSFORMATION EXPERIMENT FOUR STEPS 1. S strain → Inject into mice → Mice die 2. R strain → Inject into mice → Mice live 3. S strain → Inject into mice → Mice live (Heat-killed) 4. S strain + R strain→ Inject into mice → Mice die (Heat-killed) CONCLUSION OF EXPERIMENT 1. R – Strain bacteria is transformed into S-Strain bacteria due to incorporation of S-strain DNA 2. So that, R-Strains synthesize smooth polysaccharide coat and become virulent (S Strain) 3. However the biochemical nature of genetic material was not defined from his experiment BIOCHEMICAL CHARACTERIZATION OF TRANSFORMING PRINCIPLE 1. Biochemical nature of transforming principle was discovered by Oswald Avery, Colin Macleod and Maclyn McCarty.(1933-44) 2. They purified biomolecules (proteins, DNA and RNA) from the heat killed S cells to see which one could transform live R cells to S cells. 3. Alternatively, They did their experiment in three steps - a) Heat killed S-Strain + Protease + Live R-Strain → Transformation. b) Heat killed S-Strain + RNase + Live R-Strain → Transformation. c) Heat killed S-Strain + DNase + Live R-Strain → NO transformation CONCLUSION OF THE EXPERIMENTS 1. Protein of heat killed S-Strain is not the genetic material 2. RNA of heat killed S-Strain is not the genetic material. 3. DNA of heat killed S-Strain is the genetic material, because DNA digested with DNase mixed with R-strain unable to transform R-Strain to S-Strain. THE GENETIC MATERIAL IS DNA 1. ‘DNA is the genetic material’ is proved by Alfred Hershey and Martha Chase (1952). 2. They worked on the virus that infects bacteria called bacteriophage. 3. During normal infection the bacteriophage first attaches the bacteria cell wall and then inserts its genetic material into the bacterial cell. 4. The viral genetic material became integral part of the bacterial genome and subsequently manufactures more virus particle using host machinery. 5. Hershey and Chase worked to discover whether it was protein or DNA from the viruses that entered the bacteria. EXPERIMENT:(BLENDERS EXPERIMENT) 1. They grew some viruses (bacteriophage) on a medium having radioactive phosphorus as GROUP I and some others on medium having radioactive sulfur as GROUP II. 2. Viruses grown in radioactive Phosphorus have radioactive DNA but not radioactive protein because Phosphorus present in DNA not in protein. 3. Viruses grown in radioactive sulfur have radioactive protein not radioactive DNA because sulfur present in protein but not in DNA. 4. Then, he allowed to infect Ecoli with two different bacteriophage (G I & G II) separately and completed his experiment three phases - a) INFECTION: Radioactive phages i.e. radioactive containing bacteriophage were allowed to attach to E.coli bacteria; the phages transfer the genetic material to the bacteria. b) BLENDING: the viral coats were separated from the bacteria surface by agitating them in a blender. c) CENTRIFUGATION: The virus particles were separated from the bacteria by spinning them in a centrifuge machine. OBSERVATION 1. After centrifugation – a) Bacterial cell got settled down at bottom whereas b) Viral coat is found in supernatant 2. Bacteria infected with viruses that had radioactive DNA were radioactive and no radioactivity in the supernatant. 3. Bacteria infected with viruses that had radioactive protein were not radioactive, but radioactivity found in the supernatant. CONCLUSION OF EXPERIMENT DNA is the infecting agent that made the bacteria radioactive hence DNA is the genetic material not the protein. PROPERTIES OF GENETIC MATERIAL (DNA VERSUS RNA) CRITERIA FOR GENETIC MATERIAL 1. It should be able to generate its replica (replication) 2. It should be chemically and structurally stable. 3. It should provide the scope for slow changes (mutation) that required for evolution. 4. It should be able to express itself in the form of ‘Mendelian Character’. Protein dose not fulfill the criteria hence it is not the genetic material – Only RNA and DNA fulfill the criteria RNA IS UNSTABLE 1. RNA is easily Degradable/Changable - due to presence of -OH group at 2’ in ribose sugar, found in RNA --- makes RNA very reactive 2. Acts as catalyst, hence reactive 3. Very unstable and mutates faster --- Consequently the viruses having RNA genome and having shorter life span --- mutate and evolve faster EVIDANCES FOR THE STABILITY OF DNA (BETTER GENETIC MATERIAL - DNA OR RNA) 1. Griffith’s ‘Transforming Principle’ - Heat which killed the bacteria could not destroy DNA 2. Two strands being complementary if separated (by heating) come together, when appropriate conditions are provided 3. Presence of Thymine in place of uracil confers additional stability to DNA 4. Due to the presence of deoxy-ribose, DNA is chemically and structurally more stable than RNA Therefore among the two nucleic acids DNA IS A BETTER GENETIC MATERIAL RNA WORLD 1. RNA is the first genetic material. 2. RNA used to act as a genetic material as well as catalyst. 3. But RNA being catalyst was reactive and hence unstable. 4. Hence DNA has evolved from RNA with chemical modifications that make it more stable. 5. DNA being double stranded and having complementary strand further resists changes by evolving a process of repair. DNA REPLICATION 1. Proposed by Watson and Crick 2. “A process of making DNA copies” 3. Occurs in S-phase of cell cycle before the cell division 4. Two strands get separated and act as template for the synthesis of new complementary strands 5. New DNA molecule has one parental strand and one new strand – Semi - conservative process 6. New strand is always form in 5’ to 3’ direction as per the criteria of DNA polymerase enzyme which forms new strand --- therefore formed on the strand 3’ to 5’ 7. The average rate of polymerization has to be approx. 2000 bp per sec 8. dNTPs serve dual purposes: a) Provide energy for polymerization b) Acts as substrates for polymerization DNA REPLICATION AND ENZYMES ENZYME FUNCTION TOPOISOMERASE To reduce degree of supercoiling HELICASE – UNZIPPING To unwind the strands by breaking H- ENZYME bonds between bp DNA POLYMERASE III Catalyze the formation of phosphodiester bone to join nucleotides in 5’ to 3’ direction RNA PRIMASE To synthesize primer DNA POLYMERASE I To remove RNA primer DNA LIGASE To join Okazaki fragments SINGLE STRANDED BINDING To bind ssDNA to keep them open PROTEINS after the function of helicase so that new strand can be formed on it PROCESS OF DNA REPLICATION 1. The replication process starts with a small opening in the DNA helix called replication fork. 2. The region where, replication fork formed is called origin of replication. 3. The replication fork is formed by an enzyme called helicase. 4. Two separated strand is called template strands. 5. DNA-dependent DNA polymerase (since it uses a DNA template) catalyzes the polymerization of deoxyribonucleotidesie formation of new strand. 6. DNA polymerase catalyses polymerization only in one direction i.e. 5’→3’. 7. On one strand (template with 3’→5’ polarity. leading strand), formation of new strand occurs as in continuous form whereason another strand (template with 5’→3’ polarity, lagging strand), the polymerization/new strand formation takes place in the form of short fragment calledOkazaki fragment. 8. These okazaki/short fragments are joined by DNA ligase. EXPERIMENTAL PROOF OF SEMICONSERVATIVE NATURE OF REPLICATION 1. Mathew MESSELSON and Franklin STAHL proved - SEMI-CONSERVATIVE NATURE of DNA replication, 1958 2. First in Escherichia coli and in higher organism STEPS OF THE EXPERIMENTS 1. They grew E.coli in 15NH4Cl medium for many generations. (15N is heavy nitrogen not radioactive element) 2. The result was that 15N was incorporated into newly synthesized DNA and other nitrogen containing compound as well. 3. This heavy DNA molecule could be distinguished from normal DNA by centrifugation in a cesium chloride (CsCl) density gradient. 4. Then they transferred the E.coli into a medium with normal 14NH4Cl and let them grow.(E.coli divides in 20 minutes) 5. They took samples at definite time intervals as the cells multiplied, and extracted the DNA that remained as double-stranded helices. 6. Various samples were separated independently on CsCl gradients to measure the densities of DNA. 7. The DNA that was extracted from the culture one generation after the transfer from 15N to 14N medium had a hybrid orintermediate density. 8. DNA extracted from the culture after another generation (after 40 min.) was composed of equal amount of this hybrid DNA and of ‘light ‘DNA. GENE EXPRESSION PROTEIN FORMATION FROM GENE TO PROTEIN ABOUT FUNCTIONAL GENE 1. Cistron: a segment of DNA (structural gene) coding for a polypeptide. 2. Monocistronic: most of eukaryotic structural gene codes for single polypeptide. 3. Polycistronic: Most prokaryotic structural gene code for more than one polypeptides. 4. In eukaryotes the monocistronic structural gens have interrupted coding sequences, the genes are said to be split gene: a) EXON - Coding sequences/expressivesequences, found in appear in mature or processed mRNA b) INTRONS - Non-coding séquences/non-expressive séquence, présent in between exons, never found in mRNA. They are spliced out. TYPES OF RNA Single DNA dependent RNA polymerase catalyzes transcription/synthesis of all three types of RNAs in prokaryotes - 1. mRNA - carries genetic information (in the form of genetic code) from nucleus to the site of protein synthesis i.e. Ribosomes 2. tRNA - brings the amino acids from cytoplasm to the ribosomes acc. To genetic code of mRNA 3. rRNA - is the structural part of the ribosome and also has catalytic role during process of translation STRUCTURE OF tRNA/sRNA AN ADAPTOR MOLECULE 1. Made up of 4 arms and 3 loops 2. tRNA has an anticodon loop that base complementary to the codon 3. It has an amino acid accepter end to which it binds with amino acid 4. Each tRNA bind with specific amino acid i.e 61 types of tRNA found 5. One specific tRNA with anticodon UAC ---- called initiator tRNA 6. There is no tRNA for stop codons. (UAA, UGA, UAG) 7. The secondary structure/two dimensional structure is like CLOVER-LEAF 8. The actual structure of tRNA is compact, looks like inverted ‘L’. GENETIC CODES/CODON/TRIPLETS 1. The codon is triplet. 2. Three nitrogen base sequences constitute one codon - Made from four N-bases – A, T, G, C 3. Combination of 3 can code all amino acids - 4x4x4 = 64 – more than no. of a/a (20 in no) - 4. NOT AS DOUBLET AS 4X4=16 lesser than the no. of a/a 5. M. Nirenberg, H.Matthaei, S. Ochoa, H. G. Khorana – gave triplet for a/a 6. THERE ARE 64 CODONS– out of which 3 do not code for any a/a – therefore, called STOP CODON - UAA,UAG,UGA 7. AUG – has DUAL function – START CODON, codes for a/a GENETIC CODE Contribution to discovery 1. The process of replication and transcription based on complementarity. 2. The process of translation is the transfer of genetic information form a polymer of nucleotides to a polymer of amino acids. There is no complementarity exist between nucleotides and amino acids. 3. If there is change in the nucleic acid (genetic material) there is change in amino acids in proteins. 4. There must be a genetic code that could direct the sequence of amino acids in proteins during translation. 5. George Gamow proposed the code should be combination of bases, he suggested that in order to code for all the 20 amino acids, the code should be made up of three nucleotides. 6. Har Govind Khorana enables instrumental synthesizing RNA molecules with desired combinations of bases(homopolymer and copolymers). 7. Marshall Nirenberg’s cell – free system for protein synthesis finally helped the discovery of genetic code. 8. Severo Ochoa enzyme (polynucleotide phosphorylase) was also helpful in polymerizing RNA with desired sequences in a template independent manner (enzymatic synthesis of RNA) SALIENT FEATURES OF GENETIC CODE 1. UNAMBIGUOUS – One codon = one a/a 2. DEGENERTE – one a/s , coded by more than codon – called SYNONYMOUS CODON – required to minimize deleterious effects of mutation 3. NON-OVERLAPPING – Each N-base is a part of only one codon 4. COMMALESS – There is no intermediate N-base between codon 5. UNIVERSAL - Same for all organism Initiation codon: AUG is the first codon of all mRNA. And also it codes for methionine (met), hence has dual function. STEP I - TRANSCRIPTION 1. The process of “Copying genetic information from one strand of DNA into RNA” 2. Done by DNA dependent RNA polymerase enzyme 3. PRINCIPLE - Formation of RNA occurs on the basis of complementary sequences with the exception that Uracil is comes in RNA instead of thymine 4. Transcription occurs in TRANSCRIPTION UNIT in DNA 5. TRANSCRIPTION UNIT - a) PROMOTER b) FUNCTIONAL GENE c) TERMINATOR PROMOTER 1. Promoter - Present on one side of structural gene - Towards 5’ end (upstream) of the structural gene/of coding strand 2. It is - a short sequence of DNA that provides binding site for RNA polymerase/TATA binding protein. 3. Also called TATA box - Rich with TATA nts which forma a non-coding sequence 4. If the position of promoter is changed with terminator the definition of coding and template strand will be reversed STRUCTURAL GENE 1. DNA strand having polarity 3’→5’ - TEMPLATE STRAND for transcription 2. Other strand of DNA having polarity 5’→3’ - CODING STRAND TERMINATOR 1. Located on 3’ end (down stream) of the structural gene/of coding strand 2. It is - a short sequence of DNA that recognizes the termination factor (ρ-factor) 3. It terminates the process of transcription WHY STRAND WITH POLARITY 5’→3’ CALLED CODING STRAND? SEQUENCES OF NITROGEN BASE IN hnRNA -formed/transcribed from the template strand - SAME AS THE CODING STRAND OF DNA except having Thymine in place of Uracil. PROCESS OF TRANSCRIPTION PROKARYOTES: Initiation: 1. RNA polymerase binds to the specific site of DNA called promoter. 2. Promoter of the DNA is recognized by initiation factor or sigma (σ). Elongation: 1. RNA polymerase unzipped the DNA double helix and forms an open loop- TRANSCRIPTION BUBBLE. 2. Using NTPs as substrate, following the rule of complementarity sequences, RNA polymerase catalyzes the formation of hnRNA till the enzyme reaches the terminator gene Termination: 1. RNA polymerase recognizes the terminator gene by a termination- factor called rho (ρ) factor. 2. The RNA polymerase separated from the DNA to stop further elongation of RNA PROCESSING OF hnRNA POST-TRANSCRIPTIONAL CHANGES (occurs inside the nucleus) 1. SPLICING: a) The primary transcript (hn RNA) contain both exons and introns and required to be processed before translationally active (mRNA). b) The introns are removed and exons are joined in a defined order. c) This process is catalyzed by SnRNP, introns removed as spliceosome. 2. CAPPING: an unusual nucleotide called methyl guanosine triphosphate is added to the 5’ end of hnRNA 3. TAILING: Adenylate residues (200-300) are added at 3’ end of hnRNA in a template independent manner. The processed hnRNA is now called mRNA and transported out of the nucleus for translation. Why both strands of DNA not copied during transcription?: 1. If both strand of DNA acts as template - they would form two RNA with different sequences - And if both RNA translated into protein - sequence of amino acids in the protein would be different – which is not possible from one protein 2. Two RNA molecules if produced - would be complementary to each other - hence will form double stranded RNA - would prevent RNA translation into proteinTRANSCRIPTION VS. REPLICATION DIFFERENCE BETWEEN PROKARYOTIC AND EUKARYOTIC TRANCRIPTION PROKARYOTIC TRANSCRIPTION EUKARYOTIC TRANSCRIPTION Occurs in cytoplasm In nucleus No specific period to occur In G1 and G2 phase Coupled to translation Transcription and translation separate Product is effective in situ Products come out of the nucleus for further functioning Only one type of RNA polymerase enzyme Three types of RNA polymerase enzyme required required mRNA is polycystronic mRNA is monocistronic RNA processing like splicing not required Splicing, capping and tailing required STEP II - TRANSLATION 1. The process of “conversion of mRNA into protein” (polymer of a/a which are joined by peptide bonds 2. Translation needs - a) mRNA with genetic codes/codons b) Ribosomes c) Transfer RNA (tRNA) d) a/a 3. Occurs in cytoplasm on ribosomes RIBOSOMES CELLULAR FACTORY FOR PROTEIN SYNTHESIS 1. Made up of 2 ribosomal subunits (small and large) and each is made up of proteins and ribosomal RNA/rRNA 2. Large subunit of ribosome has three sites - a) The A site - Aminoacyl-tRNA b) The P site – Peptidyl-tRNA c) The E site - exit site 3. Ribosomes are of different types – a. 70S (50S+30S) - in prokaryotes and organelles like chloroplast and mitochondria b. 80S (60S+40S) - in eukaryotes PROCESS OF TRANSLATION 1. Before the process of translation, tRNA HAS TO BE CHAGRED 2. CHARGING OF tRNA/AMINOACYLATION OF tRNA Binding of tRNA with specific a/a in the presence of ATP STEPS OF TRANSLATION Initiation: 1. The process of translation or protein synthesis begins with attachment of mRNA with small subunit of ribosome. 2. The ribosome binds to the mRNA at the start codon (AUG). 3. AUG is recognized by the initiator tRNA. Elongation: 1. Done in large unit when it gets attached with the initiation complex 2. Larger subunit has main two site ‘A’ site and ‘P’ site 3. Charged tRNA enter large subunit through A site and then shifted to P site so, that subsequent charged tRNA can enter the ‘A’ site 4. All the charged tRNA enter according to the codon of the mRNA 5. Codon of mRNA and anticodon of tRNA are complementary to each other 6. Formation of peptide bond between two amino acids occur in ‘A’ site-- catalyzed by ribozyme, (23S rRNA in bacteria) 7. Ribosome moves from codon to codon along the mRNA called translocation Termination: 1. Elongation continues until a stop codon arrives at ‘P’ site- as there is no tRNA for stop codon. 2. Further shifting of ribosome leads to separation of polypeptide. 3. An mRNA also has some additional sequences that are not translated called untranslated regions (UTR). REGULATION OF GENE EXPRESSION Regulation of gene expression in eukaryotes takes place in different level: 1. Transcriptional level (formation of primary transcript) 2. Processing level (regulation of splicing) 3. Transport of mRNA from nucleus to the cytoplasm. 4. Translational level. TYPES OF GENES 1. HOUSEKEEPING GENES - are CONSTITUTIVE GENES - Required for the maintenance of basic cellular function 2. TISSUE-SPECIFIC GENES – Gene of a particular tissue/cell type - e.g. Insulin hormone is produced only in pancreas, not in lungs/skin 3. INDUCIBLE GENES – Expressed under specific environmental conditions OPERONS AND TYPES 1. OPERON – ”CLUSTER OF GENES WHICH CODE FOR THE PROTEINS WITH RELATED FUNCTIONS” e.g. gene for different enzymes of the same metabolic pathways. 2. Operon is transcribed as a single mRNA as POLYCISTRONIC mRNA, but translated into individual protein 3. Operon model given by - J Monod, 1961 4. TYPES OF OPERON – a) INDUCIBLE OPERON- function in presence of small effector molecule e.g. lac operon b) REPRESSIBLE OPERON - Function in absence of small effectors molecule and called CO-REPRESSOR e.g. try (trp) operon lac OPERON 1. Lac operon consists of:- a) One regulator gene ( i-gene) - The i-gene codes for repressor of the lac operon b) Three structural genes (z,y,a) c) Operator. (binding site of repressor protein) d) Promoter.(binding site of the RNA polymerase) 2. The structural gene consist of three gene (z, y and a) a) ‘lac z’-gene codes for beta-galactosidase, which hydrolyze lactose into Galactose and glucose. b) ‘lac y’ –gene codes for permease, which increases the permeability of bacterial cell to lactose. c) ‘lac a’-gene codes for transacetylase. 3. All three genes are required for the metabolism of lactose in bacteria. 4. INDUCER: lactose is the substrate for β- galactosidase and it regulates the switching on and off of the lac operon. Hence it is called inducer. MECHANISM OF REGULATION OF LAC OPERON: 1. The repressor protein is synthesized from i-gene (all time constitutively) 2. In the absence of the inducer i.e. lactose the active repressor binds to the operator and prevents RNA polymerase from transcribing the structural gene 3. In the presence of the inducer such as lactose or allolactose, the repressor is inactivated by interaction with inducer. 4. This allows RNA polymerase access to the promoter and transcription proceeds. 5. The regulation of lac operon by repressor is referred to as negative regulation. HUMAN GENOMIC PROJECT 1. The Human Genome Project (HGP), an international research program was started with the goal of “complete mapping and understanding of all the genes of human beings” 2. The Project was coordinated by the National Institutes of Health (NIH) and the U.S. Department of Energy along with the universities across the United States and international partners in the United Kingdom, France, Germany, Japan, and China 3. It began in 1990 and was completed in April 2003, 2 years ahead of its original schedule 4. Apart from sequencing Human Genome, HGP also meant to sequence genome of certain organisms which are used as models in research studies – e.g. mouse, drosophila, yeast, Ecoli, Mycoplasma GOALS OF HGP 1. Identify all the approximately 20.000 – 25000 genes in human DNA. 2. Determine the sequence of all 3 billion chemical base pairs. 3. Store this information in data bases. 4. Improve tools for data analysis. 5. Transfer related technologies to other sectors, such as industries. 6. Address the ethical, legal, and social issues (ELSI) that may arise from the project. METHODOLOGY 1. To identify all the genes that expressed as RNA referred as Expressed Sequence Tags (ETSs). 2. Simply sequencing the whole set of genome that contained all the coding and non-coding sequence, and later assigning different regions in the sequence with functions called Sequence Annotation. 3. The commonly used hosts for sequencing were bacteria and yeast and vectors were called as BAC (bacterial artificial chromosome) and YAC(yeast artificial chromosome). SALIENT FEATURES/OUTCOME OF HUMAN GENOME 1. The human genome contains 3164.7 million nucleotide bases. 2. The average gene consists of 3000 bases. 3. The largest known human gene being dystrophin at 2.4 million bases. 4. The total number of gene is estimated at 30000. 5. 99.8 percent nucleotide base sequences are same in all people. 6. The function of 50% genes discovered is unknown. 7. Less than 2 percent of the genome codes for proteins. 8. Repeated sequences make up very large portion of human genome. 9. Chromosome I has most genes (2968) and the Y has the fewest (231). 10. About 1.4 million SNP locations are found where single-base DNA differences (SNPs – single nucleotide polymorphism). APPLICATION OF THE OUTCOME OF HUMAN GENOME PROJECT 1. Identification of DEFECTIVE GENES - Chance of early treatment 2. Identification of GENES WHICH INCREASE THE CHANCES OF SUSCEPTIBLE TO CERTAIN DISEASES - to preventive measures 3. Prediction of PROTEIN/Prediction of DISEASE/DISABILITY due to particular gene - to control these proteins by specially designed drugs DNA FINGERPRINTING 1. To compare DNA of different sources 2. Done using SATELLITE DNA/REPETITIVE SEQUENCE of DNA 3. These sequences DO NOT CODE FOR ANY PROTEINS 4. These sequences SHOW HIGH DEGREE OF POLYMORPHISM and FORM BASIS OF DNA FINGERPRINTING 5. Polymorphism (variation at genetic level) are due to mutations SATELLITE DNA 1. During centrifugation, DNA gets separated by forming two peaks - a) MAJOR PEAK – Bulk DNA b) SMALL/MINOR PEAK - Satellite DNA 2. TYPES OF SATELLITE DNA – a) MINI –SATELLITE b) MICRO – SATELLITE 3. CLASSIFICATION on the basis of - a) Base composition (A:T rich/G:C rich) b) Length of sequence c) Number of repetitive units VNTR 1. It is DNA PROBE- DNA labeled with molecular marker (eg radioactive isotope) 2. VNTR belongs to a class of satellite DNA - Mini-satellite 3. Size of VNTR - 0.1 to 20 kb 4. After hybridization with VNTR probe - autoradiogram gives DNA bands with characteristic pattern for an individual - Differs from individual to individual PROCESS OF DNA FINGERPRINTING 1. DNA fingerprinting - Alec Jeffreys 2. He used VNTR – a form of satellite DNA on high degree of polymorphism 3. PRINCIPLE – Based on HYBRIDIZATION BETWEEN VNTR (DNA PROBE) & ANALYTE DNA 4. Steps of DNA fingerprinting - a) Isolation of DNA b) Digestion of DNA by restriction endonucleases c) Separation of DNA fragments by gel electrophoresis d) Transferring (blotting) of separated DNA fragments to synthetic membranes, such as nitrocellulose or nylon e) Hybridization using radioactive-labeled VNTR probe 5. Detection of hybridized DNA fragments by autoradiography APPLICATIONS 1. Test of paternity. 2. Identify the criminals. 3. Population diversity determination. 4. Determination of genetic diversity.