Molecular Basis of Inheritance PDF

6. MOLECULAR BASIS OF INHERITANCE What is Genetic material? The substance which stores biological information in a coded form, transfers it to the next generation and causes its expression in the offspring is called Genetic material. Characteristics of Genetic Material:  A Molecule that...

6. MOLECULAR BASIS OF INHERITANCE What is Genetic material? The substance which stores biological information in a coded form, transfers it to the next generation and causes its expression in the offspring is called Genetic material. Characteristics of Genetic Material:  A Molecule that can act as genetic material must have the following properties: (i) It should be present in every cell. (ii) Its amount should be same in all somatic cells. (iii) It should show diversity corresponding to the variety existing in the organisms. (iv) It should be able to precisely duplicate itself, forming carbon copies. (v) It should be chemically and structurally stable. (vi) It should be able to pass its copies into the next progeny. (vii) It should occasionally develop inheritable changes (mutations) that are necessary to allow adaptations and scope for evolution. (viii) It should be able to store the information in the coded form for the control of biological functions of the cell. (ix) It should be able to express itself in the form of Mendelian characters.  Sutton and Boveri in 1902, proposed that the genetic information is passed between generations by way of Chromosomes not proteins.  The nucleic acids i.e. DNA and RNA can replicate, but not protein.  The predominant genetic material is DNA, while few viruses like tobacco mosaic virus have RNA as the genetic material.  Transformation and the Transforming Principle:  Frederick Griffith (1928) conducted a series of experiments with Streptococcus pneumonia (Pneumococcus), the bacterium causing pneumonia.  He observed two strains of this bacterium, one forming smooth colonies, virulent strain (S- strain) which is protected by a polysaccharide capsule and causes Pneumonia.  The other one forming rough colonies, non virulent (R-strain) without capsule and it did not cause Pneumonia.  When live S-type cells were injected into the mice, they produced pneumonia i.e they were (pathogenic/virulent), and the mice dead.  When R-type cells were injected into the mice, the disease was not produced (non pathogenic/ non-virulent) and the mice did not die.  When heat-killed S-type cells were injected into the mice, the disease did not appear.  When heat-killed S-strain cells were mixed with live R-strain cells and injected into the mice, the mice died and he could isolate live S-strain cells from the body of the mice. 89 DELHI PUBLIC SCHOOL, HYDERABAD BIOLOGY  He concluded that the R-strain bacteria had somehow been transformed by the heat-killed S-strain bacteria, which must be due to the transfer of the genetic material (transforming principle). Sl No Form of Pneumococcus injected Effect on mice 1. Live rough non- capsulated survived 2. Live smooth capsulated died 3. Heat-killed smooth survived 4. Heat killed smooth + live rough died  Later Oswald Avery, Colin Mac Leod and Maclyn Mc Carty (1933-44) discovered that DNA from the heat-killed S-strain bacteria caused the transformation of R-strain bacteria.  They also discovered that Proteases and RNase did not affect transformation, while DNAse inhibited the process, this proves that DNA is the transforming principle.  O.T. Avery, C. Macleod and Mc. Carty separated the extract of smooth, virulent bacteria into Protein, Carbohydrates, DNA fractions.  Each fraction was separately added to the culture medium containing live rough bacteria and observed the following. Mixture Result R- Type Bacteria + Carbohydrates of S- Type R- Type Bacteria Bacteria R - type Bacteria + Protein of S-type Bacteria R – Type Bacteria R – Type Bacteria + DNA of S-type Bacteria S – Type Bacteria R – Type Bacteria + DNA of S-type Bacteria + R – Type Bacteria Deoxy-Ribonuclease Results of Avery’s Experiments  The 2’-OH group in the nucleotides of RNA is a reactive group and makes RNA liable and easily degradable, RNA as a catalyst is also more reactive and hence DNA has the property to be the genetic material. DNA is the Genetic Material:  The proof for DNA as the genetic material came from the experiments of Alfred Hershey and Martha Chase (1952), who worked with Bacteriophages (virus that infect bacteria).  The Bacteriophage on infection injects only the DNA into the bacterial cell and not the protein coat, the bacterial cell treats the viral DNA as its own and subsequently manufactures more virus particles.  They made two different preparations of the phage, in one, the DNA was made radioactive with 32P and in other, the protein coat was made radioactive with 35S.  These two phage preparations were allowed to infect the bacterial cells separately.  Soon after infection, the cultures were gently agitated in a blender to separate the adhering protein coat of the virus from the bacterial cells. 90 DELHI PUBLIC SCHOOL, HYDERABAD BIOLOGY  The culture was also centrifuged to separate the viral coat and the bacterial cells.  It was found that when the phage containing radioactive DNA was used to infect the bacteria, its radioactivity was found in the bacterial cells (in the sediment) indicating that the DNA has been injected into the bacterial cell.  So, DNA is the genetic material and not proteins. Deoxyribonucleic Acid (DNA)  DNA is along polymer of Deoxy-ribonucleotides, whose length is defined as the number of nucleotides or base pairs (bp).  The number of base pairs is characteristic of every organism/species e.g. 174 phage - 5386 bp Lambda phage - 48502 bp Escherichia coli - 4.6 x 10 bp6 Human beings - 3.3 x 109 bp (haploid number)  DNA was discovered b y Fredrich Meischer (1869) as an acidic substance in the nucleus, he called it ‘Nuclein’(a) Structure of a Polynucleotide Chain of DNA.  Each nucleotide has three components, a Nitrogenous base, a Deoxyribose (pentose) sugar and a Phosphate group.  Nitrogenous bases are of two types : i) Purines (Adenine and Guanine) and ii) Pyrimidines (Cytosine and Thymine)  A Nitrogenous base is linked to the (pentose ) Deoxyribose sugar through a N- glycosidic linkage to form a Nucleoside.  When a phosphate group is attached to 5’-OH of a nucleoside through a Phospho-diester linkage a corresponding nucleotide is formed.  Two nucleotides are linked through 3’-5’ phospho-diester linkage to form a Dinucleotide and when many nucleotides are linked in this manner, a polynucleotide is formed. 91 DELHI PUBLIC SCHOOL, HYDERABAD BIOLOGY  The polynucleotide chain has at the 5’ –end of the sugar a free phosphate moiety (it is called 5’-end) and at the 3’-end a OH group (it is called 3’ end).  The back bone of the polynucleotide is formed by the sugar and phosphates, while the nitrogen bases project from the backbone. (b) Double Helix – Model of DNA..  Watson and Francis Crick (1953) proposed a double-helix mode of DNA, based on the X-ray diffraction data, produced by Maurice Wilkins and Rosalind Franklin.  One of the important features of this model is the complementary base-pairing.  Later Erwin Chargaff observed that in a double-stranded DNA, the ratios between adenine and thymine and that between guanine and cytosine are constant, i.e A : T = 1 and G : C = 1.  The salient features of the double helical model are given below: (i) DNA is made of two polynucleotide chains, where the backbone is constituted by sugar-phosphate and the nitrogen bases project inside. (ii) The bases in the two strands are held together by Hydrogen bonds forming base pairs, Adenine pairs with thymine through two Hydrogen bonds and guanine with cytosine through three Hydrogen bonds. (iii) The two chains have an anti-parallel polarity, i.e. one chain has a 5’  3’ polarity, while the other has 3’  5’ polarity (iv) The two chains are coiled in a right-handed fashion and the pitch of the helix is 3.4 mm there are about 10 base pairs in each turn with 0.34 mm between two base pairs. (v) The plane of one base pair stacks over the other in the double helix. The length of DNA in E coli is approximately 1.36 mm, while that of humans 2.2 mm. 92 DELHI PUBLIC SCHOOL, HYDERABAD BIOLOGY c) Packaging of DNA.  In prokaryotes, with no well-defined nucleus, the DNA is organized in large loops held by certain positively charged proteins, in a region called Nucleoid.  In eukaryotes, the DNA is wrapped around  positively charged Histone Octamer into a structure called Nucleosome.  The Histone proteins are rich in basic amino acids Lysines and Arginines.  There are 5 types of Histones i.e. H1, H2A, H2B, H3 and H4.  Two copies of each of H2A, H2B, H3 and H4 form the Histone Octamer called nu- body.  The negatively charged DNA is wrapped around the positively charged Histone Octamer to form a structure called Nucleosome.  A typical Nucleosome consists of 200 bp of DNA helix.  DNA present between two adjacent Nucleosomes is called Linker DNA with about 80 bp.  The Nucleosomes are the repeating units that form chromatin fibres.  The chromatin fibres condense at metaphase stage of cell division to form chromosomes.  The packaging of chromatin at higher level requires additional set of proteins called non-Histone chromosomal (NHC) proteins.  In a nucleus, certain regions of the chromatin are loosely packed and they stain lighter than the other regions, these are called Euchromatin.  Euchromatin is transcriptionally very active, it replicates early and is more sensitive to mutations.  The other regions are tightly packed and they stain darker and are called Heterochromatin.  Heterochromatin is transcriptionally inactive, it replicates late and is less sensitive to mutations. RNA:  Ribonucleic acid (RNA) is the genetic material in some virus.  It is a long, unbranched, single stranded polymer of Ribonucleotides.  The 2’-OH group of Ribonucleotides is a reactive group, that makes RNA a catalyst.  It is evident that essential life processes such as metabolism, translation, splicing etc, have evolved around RNA, even before DNA has evolved as a genetic material.  Ribonuclease P for cleavage, SnRNPs (small nuclear Ribonucleo-proteins) for splicing, Peptidyl transferase (Ribozyme) for peptide bond formation etc are RNAs used as catalysts. 93 DELHI PUBLIC SCHOOL, HYDERABAD BIOLOGY  But RNA is being a catalyst is reactive and unstable. DNA has evolved from RNA with chemical modifications which make it stable and more suitable as a genetic material while is an Adapter, Structural molecule as well as a Catalyst in some cases. Structure of a Polynucleotide Chain (RNA).  In RNA also, each nucleotide has three components as in DNA.  The nitrogen bases are of two types: i) Purines (Adenine and Guanine) ii) Pyrimidines – (Cytosine and Uracil)  The sugar is Ribose which has an additional OH group on the 2’-position.  The nucleosides and nucleotides are called Ribonucleosides and Ribonucleotides respectively.  There are three types of RNA (a) Messenger RNA (mRNA)  It brings the genetic information of DNA transcribed on it for protein synthesis.  It is single-stranded. It constitutes 15% of the total RNA. (b) Transfer/Soluble RNA (tRNA/ sRNA).  It acts as an adapter molecule that reads the code on one hand and binds to the specific amino acid on the other hand. tRNA has a clover leaf like secondary structure but actually it is an inverted L-shaped compact molecule.  It has an ‘amino acid acceptor end’ (3’-end) which has base triplet CCA with –OH at the tip.  It has an ‘Anticodon-loop’, at the opposite end where the three bases are complimentary to the bases of the codon on mRNA for the particular amino acid.  It also has Enzyme recognition site and Ribosome site for attachment of Ribosome. (c) Ribosomal RNA (rRNA) It forms the structure of ribosomes. It also plays a catalytic role during translation. Besides the above mentioned RNAs there are Sc RNA (small cellular RNA), SnRNA (small nuclear RNA). Replication of DNA.  Watson and Crick had proposed a scheme for replication of DNA, when they proposed the double helical structure for DNA. Semi-conservative Replication of DNA  The scheme suggested that the two strands would separate and each acts as a template for the synthesis of a new complementary strand. 94 DELHI PUBLIC SCHOOL, HYDERABAD BIOLOGY  After complete replication, each DNA molecule would have one parental and one newly synthesized strand.  The scheme is termed as Semi-conservative DNA replication. (a) Proof for Semi-conservative Replication of DNA.  Mathew Meselson and Franklin Stahl have performed an experiment using Escherichia coli to prove that DNA replication is semi-conservative.  They grew E coli in a medium containing 15NH4C1 until 15N was incorporated in the two strands of newly synthesized DNA, this heavy DNA can be separated from the normal (14N) DNA by centrifugation in Cesium Chloride (CsC1) density gradient.  Then they transferred the cells into a medium with normal 14NH4Cl and took out samples at various time intervals and extracted DNAs and centrifuged them to measure their densities.  The DNA extracted from the cells after one generation of transfer from the 15N medium to 14N medium (i.e. after 20 minutes) had an intermediate hybrid density.  The DNA extracted after two generations (i.e. after 40 minutes) consisted of equal amounts of ‘light’ DNA and ‘hybrid’ DNA.  Similar experiments were conducted by Taylor-et al in 1958, by using radioactive Thymidine, they proved that DNA on the chromosomes replicates in a semi- conservative manner. The Process of Replication of DNA:  The process involves a number of enzymes/catalysts, of which the main enzyme is DNA- dependent DNA- polymerase, that catalyses the polymerization of the Deoxy-ribonucleotides at a rate of approximately 2000 bp per second.  The process is also an energy-expensive process, Deoxy-ribounucleoside triphosphates serve the dual purpose of (i) Acting as substrate and (ii) Providing energy (from the two terminal phosphates). i) Origin of Replication:  The intertwined strands of DNA separate from a particular point called origin of replication.  Prokaryotes have single origin of replication, on the other hand, Eukaryotes have thousands of origins of replication. 95 DELHI PUBLIC SCHOOL, HYDERABAD BIOLOGY  ii) Activation of Deoxy-ribonucleotides:  Four types of deoxy-ribonucleotides- dAMP, dGMP, dTMP, dCMP are activated by enzyme phosphate, energy and enzyme phosphorylase into triphosphates. phosphorylase dAMP + 2H3PO4 ---------------------- dATP. energy Similarly dGMP, dTMP, and dCMP are also activated. iii) Unwinding of DNA helix:  Unwinding of DNA is brought about by enzyme Helicase.  The unwinding of DNA results in the formation of ‘Y’ shaped structure called Replication fork.  The exposed single strands are stabilized with the help of Single Strand Binding Proteins (SSBP).  Due to unwinding a super coiling develops at the end of DNA opposite to the replication fork.  The tension is released by enzyme Topoisomerase. In prokaryotes it is DNA Gyrase. iv) Formation of RNA primer: A new strand to be synthesized opposite to the parental strand. DNA Polymerase III is incapable of initiating the DNA synthesis, therefore an enzyme RNA Primase synthesizes a short RNA primer. The primer serves as the stepping stone to start error less replication. v) Elongation of New Strand:  The DNA-dependent DNA-polymerase catalyses polymerization of the nucleotides only in 5’  3’ direction.  Consequently, on one of the template strands (with 3’  5’ polarity) the synthesis of DNA is continuous, while on the other template strand (with polarity 5’  3’), the synthesis of DNA is discontinuous i.e. short stretches of DNA are synthesized.  These short fragments are called Okazaki fragments.  The discontinuously synthesized strands are later joined together by the enzyme DNA-ligase. Transcription:  Transcription is also governed by the complementarity of bases as in DNA (except that Uracil in place of Thymine is complementary to adenine)  Only one of the strands of the DNA acts as the template for RNA synthesis for the following reasons: 96 DELHI PUBLIC SCHOOL, HYDERABAD BIOLOGY  (i) If both the strands code for RNA, two different (complementary) RNA molecules and two different proteins would be formed, hence the genetic information-transfer machinery would become complicated. (ii)Since the two RNA molecules produced would be complementary to each other, they would wind together to form a double-stranded RNA without carrying out translation, that means the the process of transcription would become futile. A transcription unit in DNA has three regions: (i) A promoter (ii) Structural gene(s) (iii) A terminator.  The process is catalyzed by DNA-dependent RNA-polymerase, which catalyses the polymerization of nucleotides only in 5’  3’ direction.  The DNA strand with 3’  5’ polarity is called ‘Template strand’, while the other strand with 5’  3’ polarity is called ‘Coding strand’.  The coding strand is displaced and does not code for RNA, but reference points regarding transcription are made in relation to it.  The promoter refers to a particular sequence of DNA located towards the 5’ end (upstream) of coding strand, where the RNA polymerase becomes bound for transcription.  The terminator is a sequence of DNA located towards the 3’ end (downstream) of the coding strand, where the process of transcription would stop.  There are additional regulatory sequences that may be present upstream or downstream to the Promoter. (a) Transcription in prokaryotes.  In prokaryotes, the structural genes are Polycistronic and continuous.  In prokaryotes, there is a single DNA-dependent RNA polymerase, that catalyses the transcription of all the three types of RNA (mRNA, tRNA, rRNA).  RNA polymerase has 5 components( α2β,β’ω) and the σ factor binds to the promoter and initiates the process.  It uses Ribonucleoside triphosphates also called Ribonucleotides for polymerization on a DNA template following complementarity of bases.  The enzyme facilitates the opening of the DNA-helix and elongation continues.  Once the RNA polymerase reaches the terminator, the nascent RNA falls off and the RNA polymerase also separates, it is called termination of transcription and is facilitated by certain Termination factors (p). 97 DELHI PUBLIC SCHOOL, HYDERABAD BIOLOGY  In prokaryotes, the mRNA synthesized does not require any processing to become active and both transcription and translation occur in the same Cytosol, translation can start much before the mRNA is fully transcribed, i.e. transcription and translation can be coupled. Transcription in Eukaryotes:  In eukaryotes, the structural genes are Monocistronic and ‘Split’.  They have coding sequences called Exons that form part of mRNA and non- coding sequences, called Introns, that do not form part of the mRNA and are removed during splicing.  In eukaryotes, there are at least three different RNA polymerases in the nucleus apart from the RNA polymerase in the organelles , which function as follows: 98 DELHI PUBLIC SCHOOL, HYDERABAD BIOLOGY  RNA polymerase I transcribes rRNAs (26S, 18S and 5.8S)  RNA polymerase II transcribes the precursor of mRNA (called as heterogenous nuclear RNA (hnRNA) and RNA polymerase III catalyses transcription of tRNA.  The primary transcript contains both Exons and Introns and it is subjected to process, called Splicing, where the Introns are removed and the Exons are joined in a definite order form mRNA  The hnRNA undergoes ,two additional processes called ‘Capping’ and ‘Tailing’  In Capping, Methyl guanosine triphosphate is added to the 5’ end of hnRNA.  In tailing, Adenylate residues (about 200-300) are added at the 3’ –end of hnRNA.  The fully processed hnRNA is called functional mRNA and is released from the nucleus into the cytoplasm. Genetic Code:  Genetic code refers to the relationship between the sequence of nucleotides on mRNA and the sequence of amino acids in the polypeptide.  It was George Gamow, who suggested that the code must be made up of three bases, in order to code for the twenty different amino acids, with only four bases (A, T, G, C), ;there would be (4’) or (4 x 4 x 4) = 64 triplet codons.  Har Gobind Khorana could synthesize RNA molecules with definite combinations of bases (Homo-polymers and Copolymers).  Marshal Nirenberg made a cell-free system for protein synthesis, that helped in deciphering the genetic code.  Severo Ochoa discovered enzyme polynucleotide phosphorylase, that could Polymerase RNA with a definite sequence in template-independent manner.  The checkerboard pattern of genetic code was prepared as given below: 99 DELHI PUBLIC SCHOOL, HYDERABAD BIOLOGY  The salient features of genetic code are given below: (i) The codons are triplets and there are Sixty-four codons, sixty-one codons code for twenty amino acids and three codons (UAA, UAG, UGA) do not code for any amino acid, but function as stop Termination codons. (ii)Each codon codes for only one/particular amino acid and so the genetic code is ‘Unambiguous’ and ‘Specific’. (iii) Since some amino acids are coded buy more than one codon, the genetic code is said to be ‘Degenerate’. (iv)The codons are read in a Continuous manner in the 5’  3’ direction and have no punctuations. (comma less). (v)The genetic code is universal, i.e. the codons code for the same amino acid in any organism, be it a bacterium or a human being. (vi) AUG has a dual functions of coding for Methionine and acting as Initiation codon. Mutations: (a) Point Mutation.  This type of mutation involves a change in single base pair.  An example of point mutation is a change of single base pair in the gene for Beta globin chain (of Haemoglobin) that results in substitution of Glutamate by Valine, it causes a disease called Sickle-cell anaemia. (b) Frame Shift mutation.  It is the type of mutation where addition/insertion or deletion of one or two bases changes the reading frame from the site of mutation, resulting in a protein with a different set of amino acids.  This also forms the basis of proof that codon is a triplet and codons are read in a continuous manner.  Insertion or deletion of three or its multiples of bases does not alter the reading frame, but one/more amino acids are added or deleted in the protein translated. (c) Silent mutation.  If a base change in a codon does not alter the amino acid coded, the mutation is said to be Silent Mutation. Translation:  In this process, amino acids become joined together by peptide bonds, to form polypeptides.  The formation of peptide bonds requires energy and hence in the first phase, the amino acids are activated. (a) Activation of amino acids.  In this process, a particular amino acid becomes activated and attached to the 3’ end of a specific tRNA molecule.  The reaction is catalyzed by the enzyme Amino-acyl-tRNA synthetase. Amino acid  AA  ATP E  AA AMP ENZ  PPi nzyme AA AMP ENZ  tRNA AA tRNA  AMP  PPi 100 DELHI PUBLIC SCHOOL, HYDERABAD BIOLOGY (b) Initiation of polypeptide synthesis.  The ribosome in its inactive state exists as two subunits – a large subunit and a small subunit.  When the small subunit encounters the mRNA, translation begins.  Initiation is brought about by Initiation factors eIFs (9 of them) in Eukaryotes and IFs in (3 of them) in Prokaryotes.  The mRNA binds o the small subunit of ribosome, following base pair rule between the bases of mRNA and those on rRNA, it is catalyzed by ‘initiation factors’.  There are two sites on the larger subunit, the P-site and the A-site.  The small subunit (with the tRNA) attaches to the large subunit in such a way that the initiation codon (AUG) comes on the P-site. (c) Elongation of polypeptide chain.  A second tRNA charged with an appropriate amino acid binds to the A-site of the ribosome.  A peptide (CO-NH) bond is formed between the carboxyl group of Methionine and the amino group of the second Amino acid, this reaction is catalyzed by the enzyme Peptidyl transferase.  The ribosome moves from codon to codon along the mRNA in the 5’  3’ direction.  Amino acids are added one by one in the sequence of the codons and become joined together to form polypeptide. d) Termination of Polypeptide synthesis:  When one of the Termination codons comes at the A-site, it does not code for any Amino acid and there is no tRNA molecule for it.  As a result, the polypeptide synthesis (or elongation of polypeptide) stops.  The Polypeptide synthesized is released from the ribosome, catalyzed by a ‘release factor’. 101 DELHI PUBLIC SCHOOL, HYDERABAD BIOLOGY Regulation of Gene Expression:  In prokaryotes, the agene expression is controlled by the initiation of transcription.  In eukaryotes, the regulation can be exerted at four levels. (i) Transcriptional level (formation of primary transcript) (ii) Processing level (splicing) (iii) Transport of mRNA from nucleus to cytoplasm and (iv) Translational level.  The regulation at transcriptional level was elucidated by Jacob and Monod.  All the genes controlling one metabolic pathway constitute an Operon.  A few examples are lac-operon, trp-operon, val-operon, his-operon, etc. in E.coli.  An Operon consists of the following components: (a) Structural gene(s):  They transcribe the mRNA for the amino acid sequence of proteins (enzymes) (b) Promoter gene:  The promoter gene is a sequence of DNA, where the RNA polymerase binds and initiates transcription. (c) Operator:  The operator is a sequence of bases on DNA adjacent to the promoter.  The accessibility of promoter gene for RNA polymerase is regulated by proteins called Repressor.  In most cases a specific Repressor protein binds to the operator. (d) Regulator gene:  This codes for the Repressor protein that binds to the Operator and ‘switches off’ the transcription unit.  So the regulatory gene is also represented as ‘I’ gene (inhibitory gene). (e) Inducer:  The substance/substrate that prevents the repressor from binding to the operator, is called an Inducer, it keeps the ‘switch on’ and transcription continues. Lac-operon in Escherichia coli.  It is an inducible Operon, where Lactose is the inducer, it is the substrate for the enzyme β-galactosidase.  The components of lac-operon and their functions are as follows: (a) Structural genes.  There are three Structural genes (z, y, a) which transcribe a Polycistronic mRNA.  Gene ‘z’ codes for β-galactosidase (b-gal), that catalyses the hydrolysis of Lactose into Galactose and Glucose.  Gene’ y ‘codes for Permease, which increases the permeability of cell to β-galactosides (Lactose).  Gene ‘a’ codes for Transacetylase, that catalyses the transacetylation of Lactose into its active form. 102 DELHI PUBLIC SCHOOL, HYDERABAD BIOLOGY (b) Promoter.  It is a sequence of bases near to the structural genes, it is the site where RNA- polymerase binds for transcription. (c) Operator.  It is a sequence of bases of DNA near the promoter, where a Repressor always binds. It functions as a switch for the operon. (d) Repressor.  It is a protein coded by ‘I’ gene, synthesized all the time constitutively.  It binds to the operator gene and prevents the RNA polymerase from transcribing. (e) Inducer.  Lactose is the inducer that inactivates the repressor and prevents it from binding to the operator.  This allows an access for the RNA polymerase to the promoter and transcription continues.  Thus the substrate, lactose regulates the lac-operon. Human Genome Project (HGP).  Human Genome Project was a 13-year International collaborated research programme, that was launched in the year 1990 and completed in 2006.  This project was coordinated by the U.S. Department of Energy and the National Institute of Health.  During the early years of the project, the Welcome Trust (U.K) became a major partner, other countries like Japan, Germany, China and France contributed significantly.  Its aim was to find out the complete DNA sequence of the human genome.  The two factors that made this possible are: 103 DELHI PUBLIC SCHOOL, HYDERABAD BIOLOGY  (i) Genetic engineering techniques, with which it was possible to isolate and close any segment of DNA. (ii) Availability of simple and fast techniques, for determining the DNA sequences.  Human Genome Project was called a mega project for the following facts: (i) The human genome has approximately 3.3 x 109 bp, if the cost of sequencing is US $3 per bp, the approximate cost is about US $9 billions. (ii) If the sequences obtained were to be stored in typed form books and if each page contains 1000 letters and each book contained 1000 pages, then 3300 such books would be needed to store the complete information. (iii) The enormous quantity of data expected to be generated also necessitates the use of high speed computational device for data storage, retrieval and analysis.  The project was closely associated with a new branch of Biology, called Bioinformatics. Goals of HGP.  Some major important goals of HGP are to : (i) Identify all the genes (approximately 20000-25000) in human DNA (ii)Determine the sequences of the three billion base pairs present in human DNA (iii) Store this information in databases. (iv) Improve the tools for data analysis (v) Transfer the technologies to other sectors (like industries). (vi) Address the ethical, legal and social issues (ELSI), that may arise from this project. Advantages/Uses of HGP. (i) Knowledge of the effects of variations of DNA among individuals can revolutionize the ways to diagnose, treat and even prevent a number of diseases/disorders that affect human beings. (ii) It provides clues to the understanding of human biology. Methodologies of HGP.  The methods involved have two major approaches: (i) One approach, called Expressed Sequence Tags (ESTs), focused on identifying all the genes that expressed as RNA. (ii)Second approach, called Sequence Annotation, was to simply sequence the whole set of genome, that included all the Coding and Non-coding sequences and later assigning functions to different regions in the sequence.  The total DNA from the cell is isolated and converted into random fragments of relatively smaller sizes.  These fragments are then cloned in suitable hosts using specialized vectors, the commonly used hosts are bacterial and yeast and the vectors are Bacterial Artificial Chromosomes (BAC) and Yeast Artificial Chromosomes (YAC). 104 DELHI PUBLIC SCHOOL, HYDERABAD BIOLOGY  The fragments are then sequenced using Automated DNA Sequences, which work on the principle developed by Frederick Sanger.  The sequences were then arranged on the basis of certain overlapping regions present in them, this required the generation of overlapping fragments for sequencing.  Specialized computer based programmes were developed for alignment of the sequences.  These sequences were annotated and assigned to the respective chromosomes.  The next task was to assign the genetic and physical maps on the genome, this was generated using the information on polymorphism of Restriction Endonuclease recognition sites and certain repetitive DNA sequences, called Microsatellites. Salient features of Human Genome. Following are some of the salient observations derived from HGP. (i) The human genome contains 3164.7 million nucleotides (base pairs). (ii)The size of the genes varies, an average gene consists of 3000 bases, while the largest gene, Dystrophin consists of 2.4 million bases. TDF is the smallest gene with 14 bases. (iii) The total number of genes is estimated as 30,000 and 99.9% of the nucleotides are the same in all humans. (iv) The functions of over 50% of the discovered genes are not known. (v) Only less than 2% of the genome codes for proteins. (vi) Repetitive sequences make up a large portion of the human genome. (vii) Chromosome 1 has 2968 genes (the maximum) and the Y-chromosome has 231 genes (the least). (ix) Scientists have identified about 1.4 million locations, where DNA differs in single base in human beings, these are called Single Nucleotide Polymorphisms (SNPs). Application/Future challenge of HGP. (i) Having the complete sequence of human genome, will enable a radically new approach to biological research i.e. a systematic approach on a much broader scale. (ii) All the genes in a genome or all the transcripts in a particular tissue/organ/tumor can be studied. (iii) It will be possible to understand how the enormous number of genes and proteins work together in interconnected networks in the chemistry of life. DNA-Fingerprinting:  The technique of DNA-fingerprinting was developed by Dr. Alec Jeffrey, in an attempt to find out markers for inherited diseases, the process is also known as DNA-typing or DNA-profiling.  DNA-fingerprinting involves identifying differences in some specific short nucleotide repeats, called Variable Number Tandem Repeats (VNTRs), that vary in number from person to person and are inherited. 105 DELHI PUBLIC SCHOOL, HYDERABAD BIOLOGY  The VNTRs of two persons may be of same length and sequence at certain sites, but vary at others.  The procedure of DNA-fingerprinting includes the following major steps: (i) Extraction. DNA is extracted from the cells in a high-speed, refrigerated centrifuge (ii) Amplification. Many copies of the extracted DA are made by Polymerase Chain Reaction (PCR) (iii)Restriction Digestion. DNA is cut into fragments with restriction enzymes into precise reproducible sequences. (iv) Separation of DNA sequences/restriction fragments. The cut DNA fragments are introduced and passed through Electrophoresis set-up containing Agarose polymer gel, the separated fragments can be visualized by staining them with a dye that shows fluorescence under ultraviolet radiation. The DNA fragments are sorted out and arranged according to the size and electric charge. (v) Southern Blotting. The separated DNA sequences are transferred onto a Nitrocellulose or Nylon membrane or sheet placed over the ge (vi) Hybridization. The nylon membrane is immersed in a bath and Radio-active probes (DNA segments of known sequence) are added, these probes target a specific nucleotide sequence that is complimentary to them. (vii) Autoradiography. The nylon membrane is pressed on to an x-ray film and dark bands develop at the probe sites, which resemble the bar-codes. DNA fingerprinting technique is used for the following purposes: (i) To identify criminals in the forensic laboratories. (ii) To determine the real parents i.e. to identify the true biological father or mother in case of disputes. (iii) To verify whether an immigrant is really a close relative (as he/she claims) of the mentioned resident. (iv) To identify racial groups to rewrite the biological evolution. 106 DELHI PUBLIC SCHOOL, HYDERABAD BIOLOGY 107 DELHI PUBLIC SCHOOL, HYDERABAD BIOLOGY  Ch: 6 Molecular Basis of Inheritance Previous Years’ Questions 1. It is established that RNA is the first genetic material. Explain giving three reasons. 2. Why is DNA considered a better hereditary material than RNA? 3. Study the given portion of double stranded polynucleotide chain carefully. Identify a,b,c, and the 5’-end of the chain. 4. Draw a schematic representation of Dinucleotide. Label the following: i) The components of a nucleotide. ii) 5’ end iii) N-glycosidic linkage. iv) Phosphodiester linkage. 5. Describe Frederick Griffith’s experiment on Streptococcus pneumonia. Discuss the conclusion he arrived at. OR In a series of experiments with Streptococcus and mice, F. Griffith concluded that R-strain bacteria had been transformed. Explain. 6. Explain the role of 35S and 32P in the experiments conducted by Hershey and Chase. 7. Answer the following questions based on Meselson and Stahl’s experiment. (a) Write the name of the chemical substance used as a source of Nitrogen in the Experiment performed by them. (b) Why did the scientists synthesize the light and the heavy DNA molecules in the organism used in the experiment. 8. (a) What is this diagram representing? (b) Name the parts a, b and ac. (c) In the eukaryotes, the DNA molecules are organized within the nucleus. d) How is the DNA molecule organized in a bacterial cell in absence of a nucleus. 9. (a) Draw a neat labelled diagram of a Nucleosome. (b) Mention what enables Histones to acquire a positive charge. 10. (a) Equation.  A  DNA  B m RNA C Pr oteins 108 DELHI PUBLIC SCHOOL, HYDERABAD BIOLOGY  Look at the above sequence and mention the events (A), (B) and (C) (b) What does Central Dogma in Molecular Biology? How does it differ in some viruses? 11. The base sequence in one of the strands of DNA is: TAGGATGAT (i) Give the base sequence of its complementary strand. (ii) How are these base pairs held together in a DNA molecule? (iii) Explain the base complimentary rule. Name the scientist who framed this rule. 12. Draw a labelled schematic sketch of replication fork of DNA. Explain the role of the enzymes involved in DNA replication. 13. Describe the discontinuous synthesis of DNA 14. (a) Draw a schematic representation of the structure of a Transcription unit and show the following in it: i) Direction in which the transcription occurs. ii) Polarity of the two strands involved iii) Template strand. iv) Terminator gene. (b) Mention the function of Promoter gene in transcription. 15. Describe the initiation process of transcription in bacteria. 16. Given below is a part of the template strand of a structural gene: TAC CAT TAG GAT (a) Write its transcribed mRNA strand with its polarity. (b) Explain the mechanism involved in initiation of transcription of this strand. 17. (i) Name the enzyme that catalyses the transcription of hnRNA. (ii) Why does the hnRNA need to undergo changes? List the changes that hnRNA undergoes and where in the cell such changes take place. 18. Explain the dual function of AUG codon. Give the sequence of bases it is transcribed from and its Anticodon. 19. Identify giving reasons, the salient features of genetic code by studying the following nucleotide sequence of mRNA strand and the polypeptide translated from it. AUG UUU UCU UUU UUU UCU UAG Met – Phe – Ser – Phe – Phe – Ser 20. ‘Unambiguous’, ‘Universal’ and ‘Degenerate’ are some of the terms used for the genetic code. Explain the salient features of each one of them. 21. Explain the process of charging of tRNA. Why is it essential for Translation? 22. Draw the structure of a tRNA charged with Methionine. 23. (a) Name the scientist who called tRNA an adaptor molecule. (b) Draw a clover leaf structure of tRNA showing the following: (i) Tyrosine attached to its amino acid site. (ii) Anticodon for this amino acid in its correct site (codon for tyrosine is UGA). 109 DELHI PUBLIC SCHOOL, HYDERABAD BIOLOGY 24.24. (a) Name the molecule ‘X’ synthesized by ‘I’ gene. How does this molecule get inactivated. (b) Which one of the structural genes codes for β-galactosidase? (c) When will the transcription of this gene stop? 25. How are the structural genes activated in the lac operon in E.coli? Explain. 26. Draw a schematic diagram of lac operon in its ‘switched off’ position. Label the following: (i) The structural genes. (ii) Repressor bound to its correct position. (iii) Promoter gene. (iv) Regulator gene. 27. Write the full form of VNTR. How is VNTR different from a probe? 28. Name the first two steps in DNA finger printing. Describe them briefly. 29. What is amplification with reference to DNA finger printing? 30. What is Satellite DNA in a genome. Explain their role in DNA fingerprinting. 31. Write the full form of SNPs, BAC and YAC. 32. What is meant by R-strain with which Griffith carried out his experiment on Diplococcus pneumonae? What did he prove from these experiments? 33. The base sequence on one of the strands of DNA is ATGTGTATA. (i) Give the base sequence of its complementary strand. (ii) If an RNA strand is transcribed by this strand, what would be the base sequence of RNA. (iii) In what other respect, an RNA molecule differs from a DNA molecule? 34. Retrovirus do not follow Central Dogma. Comment. 35. An mRNA strand has a serious of codons out of which three are given below: (i) AUG (ii) UUU (iii) UAG (a) What will their DNA codon be translated. (b) What are the DNA codons that would have transcribed these RNA codons? 36. Name indicating their functions, a few additional enzymes, other than DNA polymerase and Ligase, that are involved in the replication of DNA with high degree of processing and accuracy. 37. A tRNA is charged with amino acid Phenylalanine. (i) At what end of the tRNA is the amino acid attached. (ii) What is the mRNA codon that codes for Phenylalanine? (iii) Name the enzyme responsible for this attachment. 110 DELHI PUBLIC SCHOOL, HYDERABAD BIOLOGY