Molecular Basis of Inheritance PDF

Molecular Basis of Inheritance © 2022, Aakash BYJU'S. All rights reserved Key Takeaways 1 Search for genetic material Griffith’s experiment Nucleic acids 2 Double helix model Hershey and Chase experiment DNA 3 Properties of genetic material (DNA vs RNA) © 2022, Aakash BYJU'S. All rights reserved Ave...

Molecular Basis of Inheritance © 2022, Aakash BYJU'S. All rights reserved Key Takeaways 1 Search for genetic material Griffith’s experiment Nucleic acids 2 Double helix model Hershey and Chase experiment DNA 3 Properties of genetic material (DNA vs RNA) © 2022, Aakash BYJU'S. All rights reserved Avery, MacLeod and McCarty experiment 4 DNA packaging Key Takeaways 5 Semi- conservative replication 6 7 Transcription RNA world and replication Process of replication 8 Prokaryotes Eukaryotes Post-transcriptional modifications 9 tRNA 10 © 2022, Aakash BYJU'S. All rights reserved Genetic code Key Takeaways 11 Regulation of gene expression 12 13 Human genome project Lac operon 14 15 Summary © 2022, Aakash BYJU'S. All rights reserved Translation DNA fingerprinting Search for Genetic Material Mendel’s experiments 1865 Rediscovery of Mendel’s law 1883 1900 Existence of genetic material © 2022, Aakash BYJU'S. All rights reserved Discovery of gene 1902-03 Chromosomal theory of inheritance 1909 Existence of chromosome as an organized structure of DNA was confirmed 1910 Contributions by Morgan 1920s Griffith’s experiment 1928 Quest at molecular level 1944 Hershey and Chase experiment 1952 1926 Avery, MacLeod, McCarty experiment Griffith’s Experiments Streptococcus pneumoniae Is a bacteria that causes pneumonia Smooth colonies S strain Virulent Polysaccharide coat © 2022, Aakash BYJU'S. All rights reserved Rough colonies R strain Non-virulent No Polysaccharide coat Griffith’s Experiments Live S strain Live R strain Heat-killed S strain S strain bacteria isolated from dead mice Mice did not die No living bacteria isolated from live mice © 2022, Aakash BYJU'S. All rights reserved Heat-killed S strain + Live R strain Living S- strain bacteria isolated from dead mice Griffith’s Experiments Conclusion Transformation + Heat killed S strain Live R strain Virulent strain  When S strain and R strain bacteria were mixed, the non-virulent R strain of bacteria got transformed into the virulent S strain bacteria.  This process is called transformation, and through which it happened is called transforming principle.  This ‘transforming principle’ got transferred from the heat-killed S strain.  This had enabled the R strain to synthesise a smooth polysaccharide coat and become virulent.  Griffith concluded that this must be due to the transfer of the genetic material.  The biochemical nature of genetic material was still not defined from his experiments. © 2022, Aakash BYJU'S. All rights reserved Avery, MacLeod and McCarty Experiment  Avery, MacLeod and McCarty purified biochemicals, carbohydrates, proteins, DNA and RNA, from the heat-killed S cells.  They added live R strain bacteria to it.  Sugars/carbohydrates, RNA and proteins showed no transformation. Sugars + RNA + Proteins + DNA + RNase/ Carbohydrase/ Protease + + No Transformation © 2022, Aakash BYJU'S. All rights reserved + +  The one with DNA in it transformed the R strain bacteria into the S strain.  Hence, DNA is the transforming principle.  To confirm, they added either: o Carbohydrase (the enzyme which breaks down carbohydrates in all the solutions) R strain bacteria o RNase (the enzyme which degrades RNA molecules) o Protease (the enzyme that breaks down proteins) S strain bacteria  The solution with DNA caused transformation Transformation of R strain to the virulent S strain Avery, MacLeod and McCarty Experiment Sugars + RNA + Proteins + DNA + DNase DNase DNase DNase + + +  However, when they added DNase in all the solutions, none of the solutions showed transformation.  This proved that DNA is the genetic material. + R strain bacteria No Transformation © 2022, Aakash BYJU'S. All rights reserved Hershey and Chase’s Experiment Phage coat: Protein Bacteriophage: A virus that infects bacteria Phage genome: DNA  They worked with virus (T2 Bacteriophage) which infects E.coli and multiplies inside it. Transduction: Process by which foreign DNA is introduced into a cell by a virus vector. E. coli Medium with radioactive sulfur (S35) Radioactive protein capsid © 2022, Aakash BYJU'S. All rights reserved Medium with radioactive phosphorus (P32) Bacteriophage Radioactive DNA  They grew some viruses on a medium that contained radioactive phosphorus and some others on medium that contained radioactive sulfur. Hershey and Chase’s Experiment Step 1: Infection Both types were allowed to infect normally cultured bacteria separately. Step 2: Blending They were then agitated, to break the contact between virus and bacteria. Step 3: Configuration This separated bacterial cells and viruses into two different levels as bacterial cells are heavier, therefore they settle down. © 2022, Aakash BYJU'S. All rights reserved E. coli E. coli Hershey and Chase’s Experiment Results  The bacteria which were infected with radioactive DNA viruses were radioactive, indicating that DNA was the material that passed from the virus to the bacteria.  However, bacteria that were infected with viruses containing radioactive proteins were not radioactive.  This showed that proteins did not enter the bacteria from the viruses.  Hence, it was proved that DNA is the genetic material. Phages Bacteria © 2022, Aakash BYJU'S. All rights reserved Radioactive DNA Nucleic Acids : Discovery Discovered nuclein (DNA) in the nuclei of WBCs 1869 Friedrich Miescher © 2022, Aakash BYJU'S. All rights reserved Erwin Chargaff Albrecht Kossel Late 1800s Determined that DNA contains nitrogenous bases Proposed Tetranucleotide theory 1909 Phoebus Levene 1948 - 1951 Discovered regularity in base ratios of DNA Nucleic Acids Nucleic acids DNA Deoxyribonucleic acid RNA Monomers of nucleic acid Nucleotide Pentose sugar Phosphate group Nitrogenous base Ribonucleic acid Nucleoside Phosphate Nitrogenous base Pentose sugar Nucleotide © 2022, Aakash BYJU'S. All rights reserved Pentose sugar and Phosphate Group RNA has ribose sugar, with -OH group at 2’ position Ribose  Phosphate group links the 3’-carbon of one sugar of one nucleotide to the 5’-carbon of the sugar of the succeeding nucleotide through an ester bond.  Phosphodiester bond is a connecting link between two consecutive nucleotides - DNA has deoxyribose sugar, with -H group at 2’ position Deoxyribose Phosphodiester bond Negative charge © 2022, Aakash BYJU'S. All rights reserved - Nitrogenous Base Heterocyclic Nitrogencontaining compounds Nitrogenous bases Purines Pyrimidines 9 membered double ringed structure 6 membered single ringed structure H O H H N H N C C N C N C C H C N O H O H N N H H Adenine (A) H H C N C C H C N N H N H Guanine (G)  In both DNA and RNA C H C C N C C N H C N C C H C C C H N H O H Cytosine (C) H N C C O H Uracil (U) H N H Thymine (T)  In DNA, cytosine and thymine are found  In RNA, cytosine and uracil are found © 2022, Aakash BYJU'S. All rights reserved O Derivation of DNA structure X-ray crystallography : Chargaff’s rule :  Maurice Wilkins and Rosalind Franklin obtained very fine Xray diffraction pictures of DNA  In DNA, Adenine = Thymine; Cytosine = Guanine  Suggested that structure of DNA was sort of helix with 3.4 Å periodicity  However, did not propose a definitive model for DNA  A+G=C+T  Total number of Purines = Total number of Pyrimidines  Not applicable for single stranded DNA Purines + Adenine © 2022, Aakash BYJU'S. All rights reserved Pyrimidines = Guanine + Cytosine Thymine Double Helix Model Double helix model  James Watson and Francis Crick proposed double helix model  Made up of two polynucleotide chains, existing as a double helix  Two polynucleotide strands are joined together by hydrogen bonds between purines and pyrimidines Sugar – phosphate backbone Nitrogen bases facing inside Sugar phosphate backbone 3’ 5’ Glycosidic linkage 5’ 3’ © 2022, Aakash BYJU'S. All rights reserved Nitrogen bases 5’ Strand 1 3’ Double Helix Model Shallow groove Double helix model 0.34 nm Deep groove Sugar – phosphate backbone 3.4 nm Right-handed coiling Antiparallel strands Helical pitch 3.4 nm Nitrogen bases facing inside Helix diameter 2 nm Helical rise 0.34 nm Strand 2 3’ 5’ 2 nm © 2022, Aakash BYJU'S. All rights reserved One chain runs from 5’→3’ direction and other in 3’ → 5’ Helix pitch can be defined as the height of one complete helix turn Double Helix Model Double helix model Sugar – phosphate backbone Nitrogen bases facing inside Purines Pyrimidines Complementary base pairing A T 2 Hydrogen bonds © 2022, Aakash BYJU'S. All rights reserved C G 3 Hydrogen bonds DNA Forms of DNA B - form  Usual DNA  10 base pairs per turn  Right-handed coiling A - form C - form Z - form  11 base pairs per turn  Like B-form  12 base pairs per turn  Not perpendicular to the  9.33 base pairs per turn  Left-handed coiling axis but slightly tilted  Right-handed coiling  Right-handed coiling  Linear double stranded DNA : found in eukaryotes and PPLO  Repetitive DNA : part of DNA with long sequence of short repetitive DNA called satellite DNA  Palindromic DNA : base sequences which reads the same from either of the strands  Denaturation/ Melting : Separation of two strands of DNA from each other due to breakage of H-bonds when it is exposed to high temperature, acid or alkali  Renaturation/ Annealing : Reassociation of separated DNA by H-bonds formation o DNA with more A = T, low melting areas o DNA with more G = C than A = T has high melting areas  C- value : Total amount of DNA per genome. Expressed in picogram © 2022, Aakash BYJU'S. All rights reserved DNA  Central dogma of Molecular Biology : Unidirectional flow of information from master copy DNA to working copy RNA and from RNA to building molecule or trait expressing molecule polypeptide; Proposed by Francis Crick Proteins mRNA DNA Transcription Translation Replication  Reverse Central Dogma or Teminism : Reported in 1970 by H. Temin and D. Baltimore o Independently discovered reverse transcription in some viruses o Viruses produce an enzyme reverse transcriptase which synthesizes DNA from RNA template Transcription Reverse transcription © 2022, Aakash BYJU'S. All rights reserved Replication Proteins RNA DNA Translation DNA Packaging : Prokaryotes Nucleoid Nucleoid associated proteins (NAPs)  DNA is not scattered throughout. It does not have a defined nucleus.  DNA is found in cytoplasm in super coiled stage o The coils are maintained by non-histone basic protein polyamines which have positive charge o Packaged structure of DNA is called nucleoid or genophore  Genomic DNA in prokaryotes is organized in large loops held by special proteins called NAPs © 2022, Aakash BYJU'S. All rights reserved DNA Packaging : Eukaryotes Base pairs in a cell = 6.6 x 109 Distance between adjacent base pairs = 0.34 nm Total length of DNA = 6.6 x 109 x 0.34 x 10-9 m = 2.24 m Total length of DNA = 2.2 m  In eukaryotes, the positively charged basic proteins involved in packaging are called histones. Histones are rich in lysine and arginine amino acids.  Histones are organised to form a unit of eight molecules called histone octamer. H1 H2A H2B H3 H4 Histones Histone octamer / Nucleosome core (Positive charge) © 2022, Aakash BYJU'S. All rights reserved Nucleosome Histone octamer binds and wraps DNA approximately 1.7 turns of DNA/ negative 200 base pairs of DNA. charge DNA Packaging : Eukaryotes  DNA present between two adjacent nucleosome is called linker DNA. 10 nm Chromatin  Nucleosomes are seen as beads on thread-like structures in the nucleus under electron microscope.  These structures are known as chromatin because they are seen as coloured bodies when stained. Chromatin fibres are approximately 10 nm in diameter.  The nucleosomes further coils to form solenoid/ chromatin fibre. It has a diameter of 30 nm. Chromosome © 2022, Aakash BYJU'S. All rights reserved  Chromatin fibre further condenses at metaphase stage to form chromosome. o This process requires an additional set of proteins that are collectively called nonhistone chromosomal protein (NHC). DNA Packaging : Eukaryotes  When nucleus is stained some parts of the chromatin are lightly stained whereas others are stained darker. Lightly stained euchromatin  Lightly stained regions are called euchromatin. Dark stained heterochromatin Euchromatin © 2022, Aakash BYJU'S. All rights reserved  Dark stained regions are called heterochromatin.  Loosely packed region  Densely packed region  Stains light  Stains dark  Transcriptionally active  Transcriptionally inactive Heterochromatin Properties of Genetic Material (DNA or RNA) Criteria for genetic material :  Chemical and structural stability  Able to generate its replica (replication)  Provide the scope for slow mutation that is required for evolution  Able to express itself in the form of Mendelian characters DNA being more stable is preferred for storing genetic material, as  Free 2'OH of RNA makes it more labile and easily degradable. Therefore, DNA in comparison is more stable.  Presence of thymine (5-Methyl uracil) at the place of uracil, which provides additional stability to DNA  RNA being unstable, mutates at a faster rate  Viruses having RNA genome can directly code for the synthesis of proteins, hence can easily express the characters © 2022, Aakash BYJU'S. All rights reserved RNA World and Replication  RNA = first genetic material  RNA = adapter, structural molecule and catalytic  Due to stability :  DNA (more stable) - preferred for storage of genetic material o RNA (less stable) - preferred for the transmission of genetic information  Replication : A process of copying and duplicating of the genetic material (DNA)  Watson and Crick - Believed in semi-conservative DNA replication Offspring DNA = Half parent + Half new Parental DNA Offspring DNA = Half parent + Half new  Semi-conservative DNA replication: Two strands of DNA unwind from each other and each act as a template for synthesis of a new, complementary strand © 2022, Aakash BYJU'S. All rights reserved Meselson and Stahl’s Experiment - Setup Grow E. coli in 15NH4Cl Media 15N is a heavier isotope of Nitrogen and not a radioactive isotope. Step 1 15NH 4Cl Step 2 Media + E. coli E. coli Transfer of E.coli with heavier DNA (15N) into regular 14NH4Cl media E. coli with heavier DNA Step 3 14NH Cl 4 Media Heavier DNA : settle down as heavier bands Lighter DNA : get suspended in the middle CsCI centrifugation of the DNA samples DNA isolated + CsCl © 2022, Aakash BYJU'S. All rights reserved Bands as result of CsCl centrifugation Centrifuge Meselson and Stahl’s Experiment - Result 0 mins 20 mins 40 mins E. coli divides in 20 minutes 14N15N 15N15N (Hybrid band) 14N14N (Light band) 14N15N (Hybrid band) (Heavy band) Gravitational Force Generation I Result : DNA replicates semi conservatively © 2022, Aakash BYJU'S. All rights reserved Generation II Semi-Conservative Replication Herbert Taylor (on eukaryotes) performed similar experiment as Meselson & Stahl (on prokaryotes) Steps Used radioactive thymidine in root of Vicia faba to detect distribution of newly synthesised DNA in the chromosomes Results Proved DNA replicates semi-conservatively © 2022, Aakash BYJU'S. All rights reserved Faba bean plant Process of Replication Steps 1. Starts at the origin of replication (ori) 2. Activation of deoxyribonucleotides Replication bubble 3. DNA Helicase separates the two strands forming replication fork Helicase Origin of replication: Specific regions of DNA where replication starts Deoxyribonucleotides : dAMP dGMP dTMP dCMP + 2H3PO4 © 2022, Aakash BYJU'S. All rights reserved Phosphorylase Energy ori dATP dGTP dTTP dCTP  DNA Polymerase adds newer dNTPs to 3’ end with the free –OH of primer complementary to the template DNA strand.  dNTPs serve as substrates and provide energy as well. Helicase: Helps unwind DNA Replication fork: A small opening of the DNA helix, a Y- shaped structure Process of Replication Due to unwinding, a supercoiling gets developed on the end of DNA opposite to replicating fork. This tension is released by enzyme topoisomerase. In prokaryotes, DNA gyrase has topoisomerase activity. 4. DNA dependent – DNA Polymerase synthesise the two strands 3’ 5’ RNA primer RNA primer: 5 – 10 nucleotide long RNA fragment that is complementary to the DNA and is synthesized by primase enzyme. DNA polymerase: An enzyme that catalyzes the polymerization of deoxynucleotides.  In prokaryotes, DNA polymerase I, II, III are the enzymes with exonuclease and polymerase are involved in the activity  In eukaryotes, DNA polymerase 𝛂, β, Ɣ,δ, ε are involved © 2022, Aakash BYJU'S. All rights reserved Process of Replication 3’ 5’ 5’ 3’ 5’ 3’ 3’ 5’ Continuous replication 3’ 5’ 3’ 5’ 3’ Discontinuous replication 5’ 3’ 5’ 5’ 3’ 5’ 3’ 5’ 5’ 5. Using leading strand as the template, DNA Polymerase adds nucleotides to the new strands continuously. 6. DNA is added to lagging strands in discontinuous chunks called Okazaki fragments. 7. DNA Ligase fills small gap between the fragments. © 2022, Aakash BYJU'S. All rights reserved 3’ Transcription Copying genetic information from one strand of the DNA into RNA is known as transcription (Heterocatalytic function of DNA). Catalysis Coding strand Noncoding strand 5’ 3’ C T T G A C G C T G A A C T G C G A 3’ Non-template strand 5’ Template strand Gene DNA dependent RNA polymerase :  Uses DNA as template  Catalyses in the direction 5’ 3’  Does not require primer to initiate RNA synthesis  Adds uracil instead of thymine © 2022, Aakash BYJU'S. All rights reserved Transcription Transcription unit Coding strand 5’ 3’ 3’ 5’ Promoter Gene Terminator Gene  Codes for RNA Molecule Template strand Terminator  Located towards 5’ end of the coding strand of the gene  Located towards 3’ end of the coding strand of the gene  RNA polymerase binds here to initiate transcription. Example : TATA box – has sequence TATAAT  Transcription ends at this region Promoter  Recognition sequences : Short and conserved sequences © 2022, Aakash BYJU'S. All rights reserved Why is Only One Strand Transcribed ? If both strands act as a template, they would code for RNA molecule with different sequences 5’ DNA 3’ 5’ C U U G A C RNA 1 C T T G A C G C T G A A C T G C G A G C U 3’ 5’ G A A 3’ 5’ C U G C G RNA 2 Protein 1 Protein 2 One segment of DNA would be coding for two different proteins, which would complicate the genetic information transfer machinery. © 2022, Aakash BYJU'S. All rights reserved A 3’ Why is Only One Strand Transcribed ? 3’ RNA 2 5’ 5’ G A A C U G C G A C U U G A C G C U Proteins 3’ RNA 1  Two RNA molecules produced would be complementary to each other, hence would form a double stranded RNA  Would prevent RNA from being converted into protein © 2022, Aakash BYJU'S. All rights reserved Types of RNA Types of RNA mRNA (Messenger RNA) rRNA (Ribosomal RNA)  Longest  Carries message from DNA  Template for protein  Smaller  Present in ribosomes  Helps in catalysing protein synthesis  Smallest  Carries correct amino acids to site of protein synthesis synthesis Monocistronic Nucleotide sequence that codes for single protein is called cistron. Polycistronic © 2022, Aakash BYJU'S. All rights reserved tRNA (Transfer RNA)  Contains single cistron  Found in eukaryotes  Contains multiple cistrons  Found in prokaryotes Transcription : Prokaryotes σ Promoter Gene Terminator RNA polymerase Requires DNA as template and RNA polymerase Initiation Sigma factor directs RNA polymerase to bind to promoter and move towards gene © 2022, Aakash BYJU'S. All rights reserved Transcription : Prokaryotes Gene Promoter RNA RNA polymerase unwinds DNA strands and starts adding nucleotides on DNA template Elongation Elongation continues and size of RNA grows © 2022, Aakash BYJU'S. All rights reserved Transcription : Prokaryotes ρ 3’ 5’ Terminator Termination begins once rho factor is attached to RNA Termination Nascent RNA along with RNA polymerase falls off © 2022, Aakash BYJU'S. All rights reserved Transcription : Eukaryotes RNA Polymerase Type of RNA transcribed  Involves 3 RNA polymerases RNA Polymerase I rRNA (28S, 18S and 5.8S)  A bit more complex than the prokaryotic transcription RNA Polymerase II hnRNA (precursor of mRNA) RNA Polymerase III tRNA, scRNA, 5S rRNA, snRNA  Occurs in the nucleus * hnRNA – Heterogeneous RNA mRNA (Messenger RNA) tRNA (Transfer RNA)  Characteristics are same, as of prokaryotes © 2022, Aakash BYJU'S. All rights reserved rRNA (Ribosomal RNA)  Ribosome has large (18S) and small (5S, 5.8S and 28S) subunit snRNA (Small nuclear RNA)  Helps in forming mRNA Post – Transcriptional Modification Splicing  Involves removal of introns (non-functional) and joining of exons in defined order  Mediated by spliceosome (snRNA + proteins) 5’ Exon 1 Intron 1 Intron 2 3’ Exon 2 Spliceosome hnRNA 5’ Exon 1 5’ Intron 1 Exon 1 © 2022, Aakash BYJU'S. All rights reserved Exon 2 Exon 2 Intron 2 3’ 3’  Introns : Non-coding or intervening sequence  Exons : Coding or functional sequence Post – Transcriptional Modification Capping  Addition of unusual nucleotide (mostly methylated guanosine triphosphate) on 5’ end 5’ CAP Exon 1 Exon 2 Tailing  Addition of poly A (200 to 300 residues) tail at 3’ end Poly A 3’ hnRNA (AAAAAAA)n Advantages:  Capping and tailing protects transcript from enzyme attack  Modifications helps in recognition for protein production © 2022, Aakash BYJU'S. All rights reserved Transcription : Prokaryotic vs Eukaryotes Prokaryotic transcription Eukaryotic transcription Occurs in cytoplasm Occurs in nucleus 1 RNA Polymerase 3 RNA Polymerases No transcription factor required Needs transcription factors RNA formed is polycistronic RNA formed is monocistronic No modification required Involves post-transcriptional modification © 2022, Aakash BYJU'S. All rights reserved Genetic Code U C A G U UUU UUC UUA UUG UCU UCC UCA UCG UAU UAC UAA UAG UGU UGC UGA UGG U C A G C CUU CUC CUA CUG CCU CCC CCA CCG CAU CAC CAA CAG CGU CGC CGA CGG U C A AA AUU AUC AUA AUG ACU ACC ACA ACG AAU AAC AAA AAG AGU AGC AGA AGG U C A G G G GUU GUC GUA GUG GCU GCC GCA GCG GAU GAC GAA GAG GGU GGC GGA GGG U C A © 2022, Aakash BYJU'S. All rights reserved G G  George Gamow argued that amino acids must constitute a combination of bases as there are just 4 bases and 20 amino acids.  Three bases would code for 1 amino acid  64 combinations > 20 amino acids  3 letter code (triplets) would be sufficient to code for 20 amino acids Genetic Code mRNA Enzyme is helpful in polymerizing RNA with defined sequences Severo Ochoa de Albornoz Synthesis of artificial mRNA with known sequence Har Gobind Khorana © 2022, Aakash BYJU'S. All rights reserved Cell-free system with required enzymes/ system to produce polypeptides from mRNA outside the cell Marshall Nirenberg Polypeptides Nucleotides decoded Genetic Code Three nitrogenous bases form a codon Codons do not overlap with each other. They are discrete AUG functions as an initiator codon as well as codes for methionine Codon is specific to only one amino acid © 2022, Aakash BYJU'S. All rights reserved Triplet Codon Nonoverlapping Dual Nature Unambiguous Degenerate Some amino acids are coded by more than one codon Stop Signal Three codons do not code for any amino acid and hence function as stop codons Salient features Universal Contiguous Codon codes for the same amino acid across all living organisms and viruses No punctuations between codons in an mRNA tRNA : Structure Amino acid acceptor end Amino acid binding site AA 3’ 5’  Has a 3’ and 5’ end  Is non-linear, clover leaf shaped structure  Actual 3D structure looks like an inverted letter ‘L’ Anticodon loop Anticodons (Complementary to codon) Codons U A C A U G mRNA Diagrammatic representation of structure of t-RNA © 2022, Aakash BYJU'S. All rights reserved  tRNA is called an adapter molecule as it acts as connecting link between amino acids (AAs) and mRNA tRNA : Activation Free amino acids in cytoplasm are inactive Activation Activation happens in presence of aminoacyl synthetase and ATP Active Amino Met Acid Met P i Inactive Amino Acid AMP P i P i P P P i i i ATP Aminoacyl tRNA Synthetase Aminoacyl tRNA Synthetase AA + AMP + Enzyme = Aminoacyl adenylate synthetase complex © 2022, Aakash BYJU'S. All rights reserved tRNA : Charging tRNA without amino acid is called uncharged Charging Addition of amino acid = Charging Active Amino Acid Met P i AMP Uncharged tRNA UA C Aminoacyl tRNA Synthetase © 2022, Aakash BYJU'S. All rights reserved Met P i AMP Charged tRNA UA C Aminoacyl tRNA Synthetase Translation Initiation When small ribosomal subunit binds the mRNA at AUG site, translation starts Met U A 5’ CAP © 2022, Aakash BYJU'S. All rights reserved C 3’ C C A U G A A U G C G C G A U C A G C U G C A A C Translation Next charged tRNA comes to the A site and peptide bond is formed between two amino acid Ribosome moves one whole codon hence A site becomes vacant Elongation tRNA at P site exits through E site and A site accepts next aminoacyl tRNA Polymerisation of amino acid continues Initiator tRNA starts from the P site instead of A site E Met Asn Ala P Leu Met Peptide bond Asn A Ala UU A C G C 3’ A U G C G C G A U C A G C U G C A A C C C A U G A A U G C G U A A U C A G C U G C A A C © 2022, Aakash BYJU'S. All rights reserved Untranslated Regions (UTRs)  Some additional sequences in the mRNA are present that are not translated and are known as UTRs.  They are present at both 5' -end (before the start codon) and at 3' -end (after the stop codon).  They are important for efficient translation. Start codon 5’ CAP 5’ UTR © 2022, Aakash BYJU'S. All rights reserved Stop codon C C A U G A A U G C G C G A U C U A A U G C A A C 3’ UTR 3’ Translation Termination Release factor binds to stop codon, terminating translation Met Asn Ala E Leu Met Asn Ala P A Release Factor 3’ G A U C A G C U G C A A C C C A U G A A U G C G U A A U C A G C U G C A A C © 2022, Aakash BYJU'S. All rights reserved Regulation of Gene Expression  Gene expression is the process by which genetic information stored in the DNA is converted into protein within the cell.  Process of turning gene expression on or off is known as gene regulation.  In eukaryotes, the regulation could be exerted at o Transcriptional level : Formation of primary transcript o Processing level : Regulation of splicing o Transport of mRNA from nucleus to the cytoplasm o Translational level © 2022, Aakash BYJU'S. All rights reserved Lac Operon  Operon is defined as a system where the polycistronic structural gene is regulated by a common promoter and regulatory protein.  E. coli prefers glucose over lactose as an energy source.  However, in absence of glucose, lactose has to be utilized by E. coli as a substitute for energy.  Lactose/ β galactoside is a dimeric sugar (disaccharide) consisting of glucose and galactose. © 2022, Aakash BYJU'S. All rights reserved Lac Operon lac gene  Structural gene which codes for a polycistronic lac mRNA and lactose metabolizing enzymes Pi i P lac gene O lacZ lacY lacA Promoter and operator Pi - Promoter of Inhibitory gene P - Promoter of lac gene i – Inhibitory gene O - Operator of lac gene where the repressor protein binds  Regulatory gene for lac operon  Expressed constitutively  Codes for a repressor protein © 2022, Aakash BYJU'S. All rights reserved T Lac Operon lacZ gene  Lac Z gene codes for β – galactosidase enzyme.  Lactose binds to the active site of β – galactosidase.  Lactose gets digested here into glucose and galactose. Pi lacI P O lacZ lacY lacA T lacY gene  Lac Y gene codes for permease enzyme which is a cell membrane bound enzyme.  It make the cell membrane of E. coli permeable to lactose. © 2022, Aakash BYJU'S. All rights reserved Lac Operon Pi lacI O P lacZ lacY lacA T lacA gene Lactose Digest Glucose and galactose β - galactosidase Lactose E. coli © 2022, Aakash BYJU'S. All rights reserved  Lac A gene codes for transacetylase enzyme.  It helps in trans - acetylation reaction.  Other functions of transacetylase are not known in great detail. Permease Lac Operon Scenario 1 : E. coli does not feed on lactose Glucose Lactose  E. coli does not feed on lactose normally because E. coli prefers glucose over lactose. E. coli RNA pol Pi lacI P Repressor mRNA Translation Active Repressor © 2022, Aakash BYJU'S. All rights reserved O lacZ lacY lacA lac Gene expression - OFF T Lac Operon Scenario 2 : E. coli feeds on lactose  RNA polymerase binds to the promoter of inhibitory gene, and transcribes repressor mRNA which forms active repressor. Lactose  Lactose binds to the active repressor and makes it inactive. RNA pol Pi lacI P O lacZ lacY E. coli lacA Repressor mRNA lac mRNA βPermease galactosidase Active repressor Inactive repressor © 2022, Aakash BYJU'S. All rights reserved T Transacetylase lac gene expression - on Lactose Human Genome Project A thirteen years long project (1990-2003) Aim was to sequence the complete human genome Also known as ‘mega project’ Coordinated by: Partners:  U.S Department of Energy  Wellcome Trust (U.K.)  National Institute of Health  Japan  France  Germany  China © 2022, Aakash BYJU'S. All rights reserved Human Genome Project The number of base pairs of the entire human genome is approx 3 X 109 Cost of HGP was $ 9 billion US dollar = 900 crores INR Bioinformatics : Hybrid field that deals with biological data and uses computer science to store, retrieve and analyse them. DNA sequencing – It is a process of identifying the exact sequence of nitrogenous bases in the DNA. © 2022, Aakash BYJU'S. All rights reserved Human Genome Project : Goals Identification of approx. 20,000-25,000 genes in human DNA Determination of 3 billion chemical base pairs of human DNA Storing the information in databases Improvement of tools for data analysis Transfer related technologies to other sectors Address ethical, legal and social issues that may arise from HGP © 2022, Aakash BYJU'S. All rights reserved Human Genome Project : Methodology Sequence annotation : Isolation of DNA Amplification of DNA or creation of copies Sequencing of amplified fragmented DNA Annotation and assigning of DNA © 2022, Aakash BYJU'S. All rights reserved  Fragmented DNA are then cloned in suitable hosts  The commonly used host were yeast and bacteria and vectors were : o BAC (Bacterial Artificial Chromosome) o YAC (Yeast Artificial Chromosome) Human Genome Project : Methodology Expressed sequence tags : Identifying all genes that are expressed as RNA  DNA is isolated from the cell.  mRNA is obtained from this DNA.  Since introns, which are present between two exons, are removed during mRNA synthesis, they are not sequenced.  This way, all the coding genes are isolated and sequenced. Exons Introns Exons 5’ A G T A C C G T A T G T C A T G G C A T A C A T G C T A Exons © 2022, Aakash BYJU'S. All rights reserved A T C G T A A T G T A T G T A C A T A C A T A T A T 3’ T A Exons G C Introns T A 3’ 5’ Human Genome Project : Features Human Genome contains approx. 3164.7 million bp Most of the genome contains repetitive sequences (VNTRs) Average size of gene is 3000 bases Repetitive sequences thought to have no direct coding functions but shed light on chromosome structure, dynamics and evolution Human genome has 30000 genes Function of over 50% genes is unknown Only 2% of genome codes for proteins. © 2022, Aakash BYJU'S. All rights reserved Chromosome 1 has 2968 genes while Y-chromosome has 231 genes SNPs were identified which can be used in disease detection and tracing human history Human Genome Project : Applications Suspect’s DNA extracted Cancer cell Early diagnosis of cancer cells © 2022, Aakash BYJU'S. All rights reserved Suspect C G C G G C C G T A C G In forensic medicine, to match DNA samples of suspects to reach the criminals Normal RBC Defected RBC With genomic sequence, disease like sickle cell anemia can be detected Repetitive Sequences  Repetitive elements that occur multiple times in the nucleic acid sequences(DNA/RNA)  In introns, the sequences can be both repetitive and non-repetitive  Number of these repeats is different in different individuals  Used in the technique of DNA fingerprinting 5’ CTCATGATGATGATGATGTCATCCCGAAATCGTAGCTA 3’ Repetitive sequence 5’ CTTAGGATTCAATCCGATTCATCCCGAAATCGT Non-repetitive sequence © 2022, Aakash BYJU'S. All rights reserved 3’ Repetitive Sequences (Main band is formed by the denser DNA) Bulk DNA Satellite DNA: Highly repetitive DNA sequence that does not (lighter bands) Satellite DNA Amount of DNA measured by absorbance  It is classified on the basis of o length of sequence o number of repetitive units o base composition Density  These repetitive DNA are separated from bulk genomic DNA as different peaks during density gradient centrifugation. © 2022, Aakash BYJU'S. All rights reserved code for proteins and is used for DNA fingerprinting. Satellite DNA - Types Micro satellite Mini satellite 2-6 base pairs repeating units in tandem repeats 10-100 base pairs repeating units in tandem repeats Short Tandem Repeats Variable Number Tandem Repeats Repeat unit size = 2 – 6 base pairs Repeat unit size = 10-100 base pairs Repeated 8 times Repeated 20 times Short Tandem Repeats (STR) © 2022, Aakash BYJU'S. All rights reserved Repeated 4 times Variable Number Tandem Repeats (VNTR) Polymorphism  It is the inheritable mutation observed in a population at a high frequency (Frequency > 0.01).  It plays a major role in evolution. Polymorphism Single nucleotide Multiple nucleotide  Change in single nucleotide  Change in many nucleotides CTCATGATGATGATGATGTCATCCCGAAATCGT CTCATGATGATGATGAGGTCATCCCGAAATCGT © 2022, Aakash BYJU'S. All rights reserved leading to changes in copy number of repeats CTCATGATGATGATGCGTTCATCCCGAAATCGT DNA Fingerprinting  A technique used to determine the characteristic of an individual’s DNA  Used to compare DNA of two individuals  Was discovered by Sir Alec Jeffreys  Analysing DNA of two different individuals: Alec Jeffreys © 2022, Aakash BYJU'S. All rights reserved o 99.9% genome is similar o Differ by 0.1% (used for DNA fingerprinting) Steps of DNA Fingerprinting DNA isolation: DNA isolation is performed using a biological sample. 2 Molecular scissors DNA isolation Restriction digestion (Cuts DNA into multiple fragments) 1 © 2022, Aakash BYJU'S. All rights reserved Steps of DNA Fingerprinting DNA isolation: DNA isolation is performed using a biological sample 4 3 The samples move under the influence of electric charge Southern blot (Transfer of DNA fragments to synthetic medium) Electrophoresis (Separation of DNA fragments) © 2022, Aakash BYJU'S. All rights reserved Steps of DNA Fingerprinting 6 Radiolabel 5 DNA probes are labelled with radioactive substances Hybridisation (Using labelled VNTR probe) Autoradiography (Detection of hybridised DNA fragments) © 2022, Aakash BYJU'S. All rights reserved As probe binds to the complementary DNA, they send out signal (radiolabels) The banding pattern obtained after exposure to x-ray is analysed. DNA Fingerprinting : Applications Personal identification Paternity – maternity testing Criminal identification and forensics © 2022, Aakash BYJU'S. All rights reserved Summary Griffith concluded that transfer of genetic material takes place due to transforming principle Avery, Macleod and McCarty experiment proved that DNA is the genetic material Hershey and Chase’s experiment also proved that DNA is the genetic material © 2022, Aakash BYJU'S. All rights reserved Summary Double helix model Nitrogen bases facing inside Sugar – phosphate backbone Antiparallel strands Helical pitch 3.4 nm Righthanded coiling Helix diameter 2 nm Helical rise 0.34 nm Purines Complementary base pairing A © 2022, Aakash BYJU'S. All rights reserved Pyrimidines T C G Summary Ideal genetic material  It should be able to replicate.  It should be chemically and structurally stable.  It should provide scope for slow changes (mutation) required for evolution.  It should be able to express itself in the form of 'Mendelian Characters’. © 2022, Aakash BYJU'S. All rights reserved Summary Process of replication Starts at origin of replication DNA helicase separates the two strands forming replication fork DNA ligase fills in the small gaps between the Okazaki fragments of lagging strand Primase binds at the replication fork Primase synthesises RNA primers © 2022, Aakash BYJU'S. All rights reserved DNA polymerase uses the primer to synthesise the two strands (leading and lagging strand) Summary Transcription steps Initiation  RNA polymerase along with sigma factor attaches to the DNA molecule and recognises a promoter sequence.  The DNA double helix unwinds exposing the bases of DNA template strand to form new mRNA. © 2022, Aakash BYJU'S. All rights reserved Elongation Termination  Nucleotides are  RNA polymerase added according to the template strand that enables the growth of mRNA. encounters a terminator sequence, thus causing the release of RNA from RNA polymerase with the help of rho factor. Summary Post-transcriptional modifications Splicing  Removal of introns and joining of exons © 2022, Aakash BYJU'S. All rights reserved Clapping  Addition of an unusual nucleotide at 5’ end Tailing  Adenylate residues (200-300) are added at 3'-end in a template independent manner Summary Steps of translation 1 2 Initiation 3 Elongation Translation Met As n Ala Leu Met E Met U A C 5’ CAP P Met Asn A Ala U U A C G C 5’ As n Al a E P A C G C CAP C C A U G A A U G C  Assembly of mRNA, ribosome and the initiator tRNA © 2022, Aakash BYJU'S. All rights reserved C C A U G A A U G C GC GA U C  Polymerisation of amino acids U GC A A C C CA U G A A U G C G U A A U C A  Release of the polypeptide and disassembly of ribosomes and tRNA Summary Steps of DNA fingerprinting 2 DNA isolation 4 Electrophoresis Restriction digestion 1 © 2022, Aakash BYJU'S. All rights reserved 6 Hybridisation Southern blot 3 Autoradiography 5 Summary Genome has 3164 mbp Gene identification Avg gene size 3000 bp DNA bp determination Genome has 30000 gene Features Goals Function of 50% gene unknown Only 2% code for protein Chromosome 1 has 2968 gene Chromosome Y has 231 gene Information storage Tool improvement Human Genome Project Technology transfer Address concerns Methodology Express sequence tags Sequence annotation © 2022, Aakash BYJU'S. All rights reserved