MOLECULAR BASIS OF INHERITANCE NOTES CLASS 12

 

MOLECULAR BASIS OF INHERITANCE


MOLECULAR BASIS OF INHERITANCE NOTES CLASS   12



The DNA and RNA World 

1. Over the times after Mendel, the nature of the inheritable material was delved, performing in the realisation that DNA is the inheritable material in maturity of organisms. 

2. Deoxyribonucleic acid (DNA) and Ribonucleic acid (RNA) are the two types of nucleic acid plant in living systems. Nucleic acids are polymers of nucleotides. 

3. DNA acts as a inheritable material in utmost organisms, whereas RNA acts as a inheritable material in some contagions. 

4. RNA substantially functions as runner. RNA has other functions as appendage, structural or as a catalytic patch. 

Molecular Basis of Inheritance Class 12 Handwritten Notes

Structure of Polynucleotide Chain 

(i) A nucleotide has three corridor, i.e. a nitrogenous base, a pentose sugar (deoxyribose in DNA and ribose in RNA) and a phosphate group. 




(ii) Nitrogenous bases are purines, i.e. adenine, guanine and pyrimidines, i.e. cytosine, uracil and thymine. 



(iii) Cytosine is common for both DNA and RNA and thymine is present in DNA. Uracil is present in RNA at the place of thymine. 




(iv) A nitrogenous base is linked to the pentose sugar through a N-glycosidic relation to form a nucleoside, i.e. adenosine and guanosine,etc. 




(v) When a phosphate group is linked to 5 ′ — OH of a nucleoside through phosphodiester relation, a corresponding nucleotide is formed. 

(vi) Two nucleotides are linked through 3 ′-> 5 ′ phosphodiester relation to form a dinucleotide. 

(vii) Several nucleotides can be joined to form a polynucleotide chain. 

(viii) The backbone in a polynucleotide chain is formed due to sugar and phosphates. 



(ix) The nitrogenous bases linked to sugar half design from the backbone. 

(x) The base dyads are reciprocal to each other. 

6. In case of RNA, every nucleotide residue has an fresh — OH group present at 2- position in the ribose. Also, the uracil is plant at the place of thymine (5-methyl uracil). 

 Discoveries Related to Structure of DNA 

 (i) Friedrich Meischer in 1869, first linked DNA as an acidic substance present in the nexus and named it as‘nuclein’. 

 (ii) James Watson and Francis Crick, proposed a veritably simple double helix model for the structure of DNA in 1953 grounded onX-ray diffraction data. 

 (iii) Erwin Chargaff proposed that for a double- stranded DNA, the rates between adenine and thymine and guanine and cytosine are constant and equals to one. 


 Molecular Basis of Inheritance Class 12 PDF notes


Salient Features of Double-helix Structure of DNA 

 (i) DNA is a long polymer of deoxyribonucleotides. It's made up of two polynucleotide chains, where the backbone is constituted by sugar-phosphate and the bases design outside. 

 (ii) The two chains haveanti-parallel opposition, i.e. 5 ′> 3 ′ for one, 3 ′> 5 ′ for another. 

 (iii) The bases in two beaches are paired through hydrogen bond (H — bonds) forming base dyads (bp). Adenine forms two hydrogen bonds with thymine from contrary beachfront andvice-versa. Guanine bonds with cytosine by three H — bonds. Due to this, purine always comes opposite to a pyrimidine. This forms a invariant distance between the two beaches. 

 (iv) The two chains are curled in a right-handed fashion. The pitch of the helix is3.4 nm and there are roughly 10 bp in each turn. Due to this, the distance between a base brace in a helix is about0.34 nm. 

 (v) The aeroplane of one base brace heaps over the other in double helix. This confers stability to the spiral structure in addition to H — bonds. 


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The length of a DNA double helix is about2.2 measures (6.6 x 109 bp x0.34 x 10-9 m/ bp) Thus, it needs special packaging in a cell. 

 (i) In prokaryotic cells (which don't have a defined nexus), similar asE.coli, DNA ( being negatively charged) is held with some proteins (that have positive charges) in a region called as nucleoid. The DNA in nucleoid is organised in large circles held by proteins. 

 (ii) In eukaryotes, there's a set of appreciatively charged proteins called histones that are rich in introductory amino acid remainders, lysines and arginines (both positive). 

Histones are organised to form a unit of eight motes called histone octamer. The negatively charged DNA is wrapped around the appreciatively charged histone octamer to form a structure called nucleosome. 

(iii) A typical nucleosome contains 200 bp of DNA helix. 

Nucleosomes constitute the repeating unit of a structure in nexus called chromatin (thread-suchlike stained structure).

Under electron microscope, the nucleosomes in chromatin can be seen as globules-on- string.

This structure in chromatin is packaged to form chromatin fibres that further coils and condense to form chromosomes at metaphase stage. 

(iv) The packaging of chromatin at advanced position requires fresh set of proteins which are inclusively called Non-Histone Chromosomal (NHC) proteins. 

(v) In a nexus, some regions of chromatin are approximately packed ( stains light) called euchromatin (transcriptionally active chromatin). In some regions, chromatin is densely packed ( stains dark) called heterochromatin ( inactive chromatin). 


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Transforming Principle 

 (i) Frederick Griffith (1928) carried out a series of trials with Streptococcus pneumoniae (bacterium causing pneumonia). 

 (ii) According to him, when the bacteria are grown on a culture plate, some produce smooth candescent colonies (S), while others produce rough (R) colonies. 

 (iii) This is because the S-strain bacteria have a mucous (polysaccharide) fleece, while R-strain does not. 

 (iv) Mice infected with S-strain ( malign) die from pneumonia but mice infected with R-strain don't develop pneumonia. 

 (v) Griffith killed bacteria by heating and observed that heat- killed S-strain bacteria fitted into mice didn't kill them. On fitting admixture of heat- killed S and live R bacteria, the mice failed. He recovered living S-bacteria from dead mice. 

 (vi) From this trial, he concluded that the‘R-strain bacteria’ had been converted by the heat- killed S-strain bacteria. Some transubstantiating principle transferred from heat- killed S-strain, had enabled the R-strain to synthesise a smooth polysaccharide fleece and come malign. This must be due to transfer of the inheritable material. 

 Biochemical Nature of Transforming Principle 

 (i) Oswald Avery, Colin MacLeod and Maclyn McCarty, worked to determine the biochemical nature of transubstantiating principle in Griffith’s trial. 

 (ii) They purified biochemicals (proteins, RNA and DNA, etc) from heat- killed S- cells and discovered that DNA alone from S-bacteria caused R-bacteria to be converted. 

 (iii) They also discovered that protease (protein digesting enzyme) and RNAases (RNA- digesting enzymes) didn't affect metamorphosis. 

 (iv) Digestion with DNAse did inhibit metamorphosis, indicating that DNA caused metamorphosis. 

 (v) They concluded that DNA is the heritable material. But, still all the biologists weren't induced. 

 MOLECULAR BASIS OF INHERITANCE NOTES CLASS 12

DNA is the Inheritable Material 

 (i) Alfred Hershey and Martha Chase (1952) gave unambiguous evidence that DNA is the inheritable material. 

 (ii) In their trials, bacteriophages ( contagions that infect bacteria) were used. 

 (iii) They grew some contagions on a medium that contained radioactive phosphorus and some others on sulphur containing radioactive medium. 

 (iv) Contagions grown in the presence of radioactive phosphorus contained radioactive DNA but not radioactive protein because DNA contains phosphorus but protein does not. In the same way, contagions grown on radioactive sulphur contained radioactive protein, but not radioactive DNA because DNA doesn't contain sulphur. 

 (v) Radioactive phages were allowed to attach toE. coli bacteria. As the infection progressed, viral fleeces were removed from the bacteria by agitating them in a blender. The contagion patches were separated from the bacteria by spinning them in a centrifuge. 

(vi) Bacteria which were infected with contagions that had radioactive DNA were radioactive, indicating that DNA was the material that passed from the contagion to the bacteria. 

(vii) Bacteria that were infected with contagions that had radioactive proteins weren't radioactive. This indicated that the proteins didn't enter the bacteria from contagions. It proved that DNA is a inheritable material that's passed from contagion to bacteria. 

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 Properties of Genetic Material 

 (i) It came establised that DNA is the inheritable material from the Hershey-Chase trial. 

 (ii) In some contagions, RNA was also reported as inheritable material,e.g. Tobacco mosaic contagions, QB bacteriophage,etc. 

 (iii) Characteristics of a Inheritable Material 

 (a) It should be suitable to replicate. 

 (b) It should be chemically and structurally stable. 

 (c) It should give compass for slow changes (mutation) that are needed for elaboration. 

( d) It should be suitable to express itself in the form of‘Mendelian characters’. 

 (iv) According to the below mentioned rules, both the nucleic acids (DNA and RNA) have the capability to direct duplications. 

 Stability can be explained in DNA as the two beaches being reciprocal if separated by hotting come together in applicable conditions. 

 (v) The 2 ′ — OH group present at every nucleotide in RNA is a reactive group and makes RNA labile and fluently degradable, hence it's reactive. 

 (vi) DNA is chemically less reactive and structurally more stable when compared to RNA. Thymine also confers fresh stability to DNA. So, among the two nucleic acids, the DNA is a predominant inheritable material. 

 (vii) Both RNA and DNA are suitable to change. Contagions having RNA genome and having shorter life span mutate and evolve briskly. 

 (viii) DNA is dependent on RNA for protein conflation, while RNA can directly decode for it. The protein synthesising ministry has evolved around RNA. This concluded that the DNA being more stable is suitable for storehouse of inheritable information, while for the transmission of inheritable information, RNA is suitable. 

Francis Crick proposed the central dogma in molecular biology, which states that the inheritable information flows from 

15. Replication Scheme for replication of DNA nominated as semi conservative DNA replication was proposed by Watson and Crick (1953). According to it, 

(i) The two beaches would separate and act as a template for the conflation of new reciprocal beaches.. 

(ii) After replication, each DNA patch would have one maternal and one recently synthesised beachfront. 

16. Experimental evidence that DNA replicates semiconservatively, comes first fromE. coli and latterly from advanced organisms, similar as plants and human cells.

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Matthew Meselson and Franklin Stahl performed the following trials to prove this in 1958. 

 (i)E. coli was grown in a medium containing 15NH4C1 as the only nitrogen source for numerous generations. 15N got incorporated into recently synthesised DNA (and other nitrogen containing composites). This heavy DNA patch could be distinguished from the normal DNA by centrifugation in a cesium chloride (CsCl) viscosity grade. 

 (ii) They also transferred the cells into a medium with normal 14NH4Cl and took samples at colorful definite intervals as the cells multiplied and uprooted the DNA that remained as double stranded helices. DNA samples were separated singly on CsCl slants to measure DNA consistence. 

 (iii) The DNA that was uprooted from the culture, one generation (after 20 min) after the transfer from 15 N to 14N medium had a mongrel or intermediate viscosity. DNA uprooted from the culture after another generation (after 40 min) was composed of equal quantities of this mongrel DNA and of light DNA. 

 (iv) Veritably analogous trials were carried out by Taylor and Associates on Vicia faba (faba sap) using radioactive thymidine and the same results, i.e. DNA replicates semiconservatively, were attained as in earlier trials. 

 

MOLECULAR BASIS OF INHERITANCE NOTES CLASS 12

 

DNA replication ministry and enzymes process of replication requires a set of catalysts (enzymes). 

 (i) The main enzyme is DNA-dependent DNA polymerase, since it uses a DNA template to catalyse the polymerisation of deoxynucleotides. The average rate of polymerisation by these enzymes is roughly 2000 bp/ second. 

 (ii) These polymerases has to catalyse the response with high degree of delicacy because any mistake during replication would affect into mutations. is the process of copying inheritable information from one beachfront of the i DNJS. into RNA. 

The principle of complementarity It also facilitates opening of the helix and continues extension. 

(v) Once the polymerase reaches the terminator region, the incipient RNA falls off, so also the RNA polymerase. This results in termination of recap. 
 
(vi) RNA polymerase is only able of catalysing the process of extension. 
 It associates transiently with inauguration- factor (a) and terminator factor (b), to initiate and terminate the recap, independently. Therefore, catalysing all the three way. 

(vii) Since, the mRNA doesn't bear any processing to come active and also since recap and restatement take place in the same cube, numerous times the restatement can begin much before the mRNA is completely transcribed. As a result, recap and restatement can be coupled in bacteria. 

 23. Recap in eukaryotes have fresh complications than prokaryotes. 

(i) There are at least three RNA polymerases in the nexus other than the RNA polymerase in organelles. The RNA polymerase I transcribes rRNAs (28S, 18S and5.8 S). RNA polymerase III is responsible for recap of fRNA, 5srRNA and SnRNAs ( small nuclear RNAs). RNA polymerase II transcribes precursor of mRNA, the heterogenous nuclear RNA (/ mRNA). 
 
(ii) Another complexity is that, the primary reiterations contain both the exons and the introns and arenon-functional. Hence, subject to a process called splicing. In this process, introns are removed and exons are joined in a definite order. 

( iii)/ mRNA undergoes fresh processing called as circumscribing and trailing. In circumscribing, an unusual nucleotide is added to the 5 ′- end of/ mRNA. In trailing, adenylate remainders (200-300) are added at 3 ′- end in a template. It's the completely reused/ mRNA, now called mRNA, that's transported out of the nexus for restatement process. 

Significance of these complications are 

(i) The split gene arrangements represent an ancient point of genome. 
(ii) The presence of introns is reminescent of age. 
(iii) The process of splicing represents the dominance of RNA world. 
Machinery & Enzymes for Replication 
 
Machinery & Enzymes for Replication

Enzymes play an important part acting as catalysts during the process of DNA replication. Some of the important enzymes are DNA polymerase 
  • Helicase 
  • Primase 
  • DNA ligase 
Energy source is demanded to give energy during the replication process. Deoxyribonucleoside triphosphates act as substrates & give energy for polymerization response. &,Enzymes_For_Replication 

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DNA polymerase 
 
DNA polymerase creates DNA from nucleotides. It reads the being DNA beaches to produce two new beaches that match the living bones. 

This enzyme is demanded every time a cell divides so that one dupe of DNA can be passed to each son cell. 

DNA polymerase is a largely effective enzyme, as it can replicate a large number of base dyads in a veritably short time. 
Rate of replication or Rate of polymerization is approx 2000 bp per second. A aggregate of4.6 * 106 base dyads are replicated within 18 twinkles. 

DNA polymerase also beget with high degree of delicacy. A mistake is made formerly in every 1 billion base dyads copied. DNA polymerase evidence reads to check for crimes. Still, these crimes if remain can beget mutations. 
 
MOLECULAR BASIS OF INHERITANCE NOTES CLASS 12

Helicase 

Enzyme Helicase unwinds DNA from tightly double stranded structure. Only after the beaches are separated, DNA polymerase can do its job of creating the new beaches. This enzyme separates the beaches by breaking the hydrogen bonds between the bases of the two beaches. 
 
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Primase 
 
This enzyme creates a short scrap of RNA ( manual) paired with the template DNA beachfront. This enzyme initiates the process of creation of new beaches. DNA polymerase can not initiate the process on its own. Thus, primase initiates the same.
 
MOLECULAR BASIS OF INHERITANCE NOTES CLASS 12

Process of DNA replication 
 
Replication can not be initiated in any arbitrary part of DNA. Region in a DNA where replication initiates is nominated as‘ Origin of Replication’. 
 
Step 1. 
 
Enzyme Helicase breaks hydrogen bonds, therefore separating the two beaches of DNA. Replication chopstick structure is formed. 

Step 2. 
 
Nonstop conflation takes place in the Leading beachfront.
In this beachfront, DNA is synthesized in the same direction as the growing replication chopstick. Observe the direction of movement of Helicase & DNA polymerase. 

Disontinuous_Synthesis 

  • Spastic conflation takes place in the Lagging beachfront. Conflation in this beachfront is more complicated than the Leading beachfront. 
  • DNA polymerase can add new free nucleotides to the 3’ end of the new beachfront. In the dragging beachfront, no free 3’-OH end is available. Thus, DNA polymerase is unfit to initiate the process.
  •  Enzyme Primase initiate the process by creating a small RNA scrap called Manual. DNA polymerase also extends the primed parts adding free nucleotides. RNA manuals are replaced with DNA. 
  • Therefore, we've DNA fractions. Observe the direction of conflation in Lagging beachfront (It's contrary to the direction of growing replication chopstick). 
  • DNA Ligase now joins the DNA fractions and forms a complete DNA. 
  • These DNA fractions are nominated as‘Okazaki fractions’after the name of the scientist who first described the process of Spastic conflation on Lagging beachfront. 
  • This entire process of DNA replication occurs during S- phase of cell cycle in eukaryotes. Research is still going on for further detail on the replication process.
class 12 biology chapter 6 notes

Transcription
 
  • The process of copying inheritable information from one beachfront of the DNA into RNA is nominated as recap. 
  • In transcription only a member of DNA and only one of the beaches is copied into RNA because Still, they would decode for RNA patch with different sequences and the sequences of amino acids in the enciphered protein would be different, if both beaches act a template. 
  • the two RNA motes would be reciprocal to each other and would form a double stranded RNA which would help restatement. 
MOLECULAR BASIS OF INHERITANCE NOTES CLASS 12

Transcription unit 
 
  •  A  transcription unit consists of 
  1. A Protagonist 
  2. The Structural gene 
  3. A Terminator 

  • The two beaches of the DNA in the structural gene of a recap unit is nominated as template beachfront and rendering beachfront. 
  • The beachfront that has the opposition 3'→ 5' acts as a template, and is appertained as template beachfront. 
  • The other beachfront which has the opposition (5'→ 3') is appertained as rendering beachfront. 
  • The protagonist and terminator adjoin the structural gene in a recap unit. 
  • The protagonist is located towards 5'- end (upstream) of the structural gene which provides binding point for RNA polymerase. 
  • The terminator is located towards 3'- end (downstream) of the rendering beachfront which defines the end of the process of recap. 

Structure of a gene 

  • A gene is defined as the functional unit of heritage.  
  • A gene also appertained as a cistron can be defined as a member of DNA rendering for a polypeptide. 
  • The structural gene in a recap unit could be said as monocistronic substantially in eukaryotes or polycistronic substantially in bacteria or prokaryotes. 
  • Exons are the coding sequences or expressed sequences that appear in mature or reused RNA. 
  • Introns are the intermediating sequences which intrude exons and don't appear in mature or reused RNA. 
Types of RNA 
 
 There are three major types of RNAs 

  • mRNA ( runner RNA), 
  • tRNA (transfer RNA), 
  • rRNA (ribosomal RNA). 
 
All three RNAs are demanded to synthesize a protein in a cell. The mRNA provides the template, tRNA brings amino acids and reads the inheritable law, and rRNAs play structural and catalytic part during restatement. 







Structure of tRNA 
 
The tRNA, also called as sRNA ( answerable RNA) has a part as an appendage patch. 

tRNA has an anticodon circle that has bases reciprocal to the law. 
 
It has an amino acid accepter end to which it binds to amino acids. 

The secondary structure of tRNA looks like a clover- splint. 
In factual structure, the tRNA is compact patch which looks like reversedL. 
 
Process of Transcription

In prokaryotes transcription takes place in three way 
 
Inauguration 
 
RNA polymerase binds to protagonist and initiates recap. Inauguration factor or sigma (σ) recognizes the protagonist of the DNA. 
 
Extension 

RNA polymerase facilitates opening of the helix and continues extension. 
RNA polymerase uses nucleoside triphosphates as substrate and polymerizes in a template depended fashion following the rule of complementarity. 
 
Only a short stretch of RNA remains bound to the enzyme. 
 
Termination 

Once the polymerases reaches the terminator region RNA polymerase binds with the termination- factor (ρ) to terminate recap. 
 
The incipient RNA falls off with the RNA polymerase which results in termination of recap. 

The recap and restatement can be coupled in bacteria as the mRNA doesn't bear any processing to come active, and also recap and restatement take place in the same cube.

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In eukaryotes 
 
There are two fresh complications in eukaryotes. 

The first complexity is that there are at least three RNA polymerases in the nexus. 

The RNA polymerase I transcribes rRNAs (28S, 18S, and5.8 S) 
 
The RNA polymerase III is responsible for recap of tRNA, 5srRNA, and snRNAs ( small nuclear RNAs). 
 
The RNA polymerase II transcribes precursor of mRNA, the miscellaneous nuclear RNA (hnRNA). 
 
The alternate complexity is that the primary reiterations contain both the exons and the introns and arenon-functional. 
 
Primary reiterations are subordinated to a process called splicing where the introns are removed and exons are joined in a defined order. 
 
hnRNA suffer two fresh processing called as circumscribing and trailing. 

In circumscribing an unusual nucleotide (methyl guanosine triphosphate) is added to the 5'- end of hnRNA. 

In trailing, adenylate remainders (200-300) are added at 3'- end in a template independent manner and the completely reused hnRNA is called mRNA 

mRNA is transported out of the nexus for restatement. 
 Significance of complications 
 
The split-gene arrangements represent presumably an ancient point of the genome. 

The presence of introns is evocative of age, and the process of splicing represents the dominance of RNA- world. 

MOLECULAR BASIS OF INHERITANCE NOTES CLASS12
 
Inheritable law 
 
The sequence of nucleotides on DNA which determines the sequence of amino acids in a polypeptide chain is nominated as Inheritable law. 

The process of restatement requires transfer of inheritable information from a polymer of nucleotides to a polymer of amino acids but there's no complementarity between nucleotides and amino acids which led to the proposition of a inheritable law that could direct the sequence of amino acids during conflation of proteins. 
 
The salient features of inheritable law are as follows The codon is triplet, 61 codons law for amino acids and 3 codons don't decode for any amino acids, hence they serve as stop codons. 

One codon canons for only one amino acid therefore it's unequivocal and specific. 

Some amino acids are enciphered by further than one codon, hence the law is degenerate. 

The codon is read in mRNA in a conterminous fashion, there are no punctuations. 
The law is nearly universal. For illustration, from bacteria to mortal UUU would decode for Phenylalanine (phe). 

AUG has binary functions, it codes for Methionine ( met), and it also act as generator codon. 

Mutation and inheritable law 
 
The connections between genes and DNA are best understood by mutation studies. 

Point mutation is the insertion or omission of a single gene in the structural gene. 

EXAMPLE: point mutation is a change of single base brace in the gene for beta globin chain that results in the change of amino acid residue glutamate to valine, which results into a diseased condition called as sickle cell anaemia. 

Frame shift mutation is the insertion and omission of three or its multiple bases which fit or cancel one or multiple codon hence one or multiple amino acids, and reading frame remains unaltered from that point onwards. 

Illustration-cystic fibrosis. 
 
Restatement 
 
Restatement reers to the process of polymerization of amino acids to form a polypeptide. 

The order and sequence of amino acids are defined by the sequence of bases in the mRNA and the amino acids are joined by a bond which is known as a peptide bond. 

Conformation of a peptide bond requires energy and therefore amino acids are actuated in the presence of ATP and linked to their connate tRNA by the process of charging of tRNA or aminoacylation of tRNA. 

Still, a peptide bond forms which is enhanced by the presence of a catalyst similar as ribosome, If charged tRNAs are brought near enough. 

Ribosome in its inactive state exists as two subunits; a large subunit and a small subunit. 

There are two spots in the large subunit, for posterior amino acids to bind to and therefore come close enough to each other for the conformation of a peptide bond. 
 
 A translational unit in mRNA is the sequence of RNA that's adjoined by the launch codon (AUG) and the stop codon and canons for a polypeptide. 

 An mRNA also has some fresh sequences that aren't restated and are appertained as untranslated regions (UTR). 

The UTRs are present at both 5'- end (before launch codon) and at 3'- end (after stop codon) which are needed for effective restatement process. 
  
The ribosome moves from codon to codon along the mRNA. 
Amino acids are added one by one, restated into Polypeptide sequences mandated by DNA and represented by mRNA. 
 

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Regulation of gene expression 
 
Gene expression results in the conformation of a polypeptide and it can be regulated at several situations similar as transcriptional position ( conformation of primary paraphrase), 

  • processing position ( regulation of splicing), 
  • transport of mRNA from nexus to the cytoplasm, 
  • translational position. 

In prokaryotes, control of the rate of transcriptional inauguration is the predominant point for control of gene expression. 
 
In a recap unit, the exertion of RNA polymerase at a given protagonist is in turn regulated by commerce with appurtenant proteins which can act both appreciatively (activators) and negatively (repressors).

Regulation of gene expression can be studied with the help of Lac operon. 
 
 MOLECULAR BASIS OF INHERITANCE NOTES CLASS 12

Lac operon 
 
  • Lac refers to lactose in lac operon. 
  • The lac operon consists of one nonsupervisory gene the i gene which codes for the repressor of the lac operon and three structural genes (z, y, and a). 
  • The z gene canons for beta-galactosidase (β- girl), which hydrolyses disaccharide, lactose into galactose and glucose. 
  • The y gene canons for permease, which increases permeability of the cell to β-galactosides. 
  • The a gene encodes a transacetylase. 
  • Lactose is nominated as debaser as lactose is the substrate for the enzyme beta-galactosidase and it regulates switching on and off of the operon. 

In the absence of debaser 

  • The repressor of the operon is synthesized ( each-the- time – constitutively) from the i gene. 
  • The repressor protein binds to the driver region of the operon and prevents RNA polymerase from transcribing the operon. 
 
In the presence of debaser 
 
  • The repressor is inactivated by commerce with the debaser which allows RNA polymerase access to the protagonist and recap proceeds. 
  • Regulation of lac operon by repressor is appertained to as negative regulation. 
Human genome design 

  • The scientific design which deal with the study of base sequences of DNA motes of complete set of chromosomes is called mortal genome design. 
  • HGP was nearly associated with the rapid-fire development of a new area in biology called as 
  • Pretensions of Human Genome Project 
  • Identify all the roughly- genes in mortal DNA. 
  • Determine the sequences of the 3 billion chemical base dyads that make up mortal DNA. 
  • Store this information in databases; 
  • Ameliorate tools for data analysis; 
  • Transfer related technologies to other sectors, similar as diligence; 
  • Address the ethical, legal, and social issues (ELSI) that may arise from the design. 
 
Methodologies 

  • To relating all the genes that expressed as RNA appertained to as Expressed Sequence Markers (ESTs).
  • Simply sequencing the whole set of genome that contained all the coding andnon-coding sequence, and latterly assigning different regions in the sequence with functions is called as Sequence Reflection.
  • The total DNA from a cell is insulated and converted into arbitrary fractions of fairly lower sizes and reproduced in suitable host using specialised vectors. 
  • The cloning redounded into modification of each piece of DNA scrap. 
  • The generally used vectors are BAC (bacterial artificial chromosomes), and YAC ( incentive artificial chromosomes). 
  • The fractions were sequenced using automated DNA sequencers. 
  • Specialized computer grounded programmes were developed for the alignment of the sequences. 
  • The sequences were latterly annotated and were assigned to each chromosome. 
  • The sequence of chromosome 1 was completed only in May 2006. 

Automated DNA sequencer 

Salient Features of Human Genome 

  • The mortal genome contains3164.7 million nucleotide bases. 
  • The average gene consists of 3000 bases with the largest known mortal gene being dystrophin at2.4 million bases. 
  • The largest known mortal gene-dystrophin 
  • The total number of genes is estimated at9 per cent nucleotide bases are exactly the same in all people. 
  • The functions are unknown for over 50 per cent of discovered genes. 
  • Lower than 2 per cent of the genome canons for proteins. 
  • Repeated sequences make up veritably large portion of the mortal genome. 
  • Repetitious sequences are stretches of DNA sequences that are repeated numerous times. 
  • Chromosome 1 has most genes (2968), and the Y has the smallest (231). 
  • Scientists have linked about1.4 million locales where single base DNA differences (SNPs – single nucleotide polymorphism) do in humans. 
MOLECULAR BASIS OF INHERITANCE NOTES CLASS 12

Operations of HGP 
 
All the genes in a genome can be studied together. Helps to understand how knockouts of thousands of genes and proteins work together in connected networks. helps to diagnose and treat inheritable conditions. 
 
DNA characteristic 
 
  • The process of comparison of DNA from different sources to establish the identity is called DNA characteristic. 

  • DNA characteristic involves relating differences in some specific regions in DNA sequence called as repetitious DNA. 

  • RepetitiousDNA are separated from bulk genomic DNA as different peaks during viscosity grade centrifugation. 

  • The bulk DNA forms a major peak and the other small peaks are appertained to as satellite DNA. 

  • Satellite DNA is of two types grounded on base composition, length of member, and number of repetitious units 
  1. micro-satellites 
  2. mini-satellites 
 chapter 6 biology class 12 ncert notes

Satellite DNA 

  • Satellite DNA sequences typically don't decode for any proteins, but they form a large portion of mortal genome. 
  • Satellite DNA sequence show high degree of polymorphism and form the base of 
  • DNA characteristic. 
  • An heritable mutation being in a population at high frequence, is appertained to as DNA polymorphism. 
  • Repeated nucleotide sequences in thenon-coding DNA of an existent is called Variable Number of Tandem Repeats (VNTR). 
  • The size of VNTR varies in size from 0.1 to 20 kb. 
  • DNA characteristic includes the following way 
  1. insulation of DNA 
  2. digestion of DNA by restriction endonucleases 
  3. separation of DNA fractions by electrophoresis 

  • transferring( blotting) of separated DNA fractions to synthetic membranes, similar as nitrocellulose or nylon.
  • hybridization using labelled VNTR inquiry discovery of hybridized DNA fractions by autoradiography
MOLECULAR BASIS OF INHERITANCE NOTES CLASS 12

 Fragments of DNA Operations 

  •  In identification of culprits. 
  •  In determining population and inheritable diversities. 
  •  In working maternal controversies. 



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