CBSE Class 12 Biology Molecular Basis Of Inheritance Notes

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Revision Notes for Class 12 Biology Chapter 6 Molecular Basis of Inheritance

Class 12 Biology students should refer to the following concepts and notes for Chapter 6 Molecular Basis of Inheritance in Class 12. These exam notes for Class 12 Biology will be very useful for upcoming class tests and examinations and help you to score good marks

Chapter 6 Molecular Basis of Inheritance Notes Class 12 Biology

Chapter 6. MOLECULAR BASIS OF INHERITANCE

DNA largest macromolecule made of helically twisted, two, antiparallel polydeoxyribonucleotide chains held together by hydrogen bonds.

→ X-ray diffraction pattern of DNA by Rosalind Franklin showed DNA a helix.

→ Components of DNA are (i) deoxyribose sugar, (ii) a phosphate, and (iii) nitrogen containing organic bases.

 DNA contains four different bases called adenine (A), guanine (G) cytosine (C), and thymine (T).

 These are grouped into two classes on the basis of their chemical structure: (i) Purines (with a double ring structure) and (ii) Pyrimidines (with a single ring structure)

→ 1953.James Watson and Francis Crick proposed three dimensional structure of DNA and won the Nobel prize.

 DNA double helix with sugar phosphate back bone on outside and paired bases inside.

→ Planes of the bases perpendicular to helix axis.

 Each turn has ten base pairs.( 34 A0)

 Diameter of helix 20 A0.

 Two strands of DNA antiparallel.

 DNA found both in nucleus and cytoplasm.

 Extranuclear DNA found in mitochondria and chloroplasts.

→ Two chains complementary

 Two chains held together by hydrogen bond.

 Adenine-Thymine pair has two hydrogen bonds.

 Guanine-Cytosine pair has three hydrogen bonds.

 Upon heating at temperature above 80-90 degree two strands uncoil and separate (Denaturation)

 On cooling two strands join together (renaturation /annealing)

→ DNA is mostly right handed and B form.

 Bacterial nucleoid consists of a single circular DNA molecule .

class_12-biology_concept_235

 DNA eukaryotes is wrapped around positively charged histone proteins to form nucleosome.

# Nucleosome contains 200 base pairs of DNA helix.

# Histone octamer =2(H2a+H2b+H3+H4)

# Linker DNA bears H1 protein

# Chromatin fibres formed by repeated units of nucleosomes.

# Non histone proteins required for packaging.

# Regions of chromatin, loosely packed and stains lightly called euchromatin.

# Regions of chromatin, densely packed and stains darkly is called heterochromatin.

DNA AS THE GENETIC MATERIAL

_Transformation experiment or Griffith effect.

• Griffith performed his experiments on Mice using Diplococcus pneumoniae.

• Two strains of bacteria are S-type and R-type cells.

• Experiments

_ Living S-strain Injected into mice →Mice killed

_ Living R-strain Injected into mice → Mice lived

_ Heat Killed S-strain Injected into mice →Mice lived

_ Living R-strain + Heat Killed S-strain Injected into mice→Mice killed

# Griffith concluded that R type bacteria is transformed into virulent form.

# Transformation - change in the genetic constitution of an organism by picking up genes present in the remains of its relatives.

BIOCHEMICAL CHARACTERISATION OF TRANSFORMING PRINCIPLE

# Proved by Oswarld Avery, Colin Macleod, Maclyn Mc Carty

class_12-biology_concept_236

From this we conclude that DNA is the genetic material.
Semi conservative nature of DNA Mathew Messelson and Franklin start.

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Replication of DNA In Eukaryotes:
Definition: "Process by which DNA produces daughter DNA molecules which are exact copies of the original DNA." In eukaryotes, DNA is double stranded. The two strands are complementary to each other because of their base sequences.

Semi-conservative method of DNA replication Important points:
(i) Most common method of DNA replication.
(ii) Takes place in the nucleus where the DNA is present in the chromosomes.
(iii) Replication takes place in the S-phase (synthesis phase) of the interphase nucleus.
(iv) Deoxyribose nucleotides needed for formation of new DNA strands are present in nucleoplasm. At the time of replication, the two strands of DNA first separate. Each strand then acts as a template for the formation of a new strand. A new strand is constructed on each old strand, and two exactly identical double stranded DNA molecules are formed. In each new DNA molecule, one strand is old (original) while the other is newly formed. Hence, Watson and Crick described this method as semi-conservative replication. (A) An overall process of DNA replication showing replication fork and formation of new strands template and lagging template.

The various steps involved in this process are summarized as follows:
i. Mechanism of replication starts at a specific point of the DNA molecule, called origin.
ii. At origin, DNA strand breaks because of an incision (nick). This is made by an enzyme called incision enzyme (endonuclease).
iii. The hydrogen bonds joining the two strands are broken by the enzyme.
iv. The two strands start unwinding. This takes place with the help of a DNA unwinding enzyme Helicases.
Two polynucleotide strands are thus separated.
v. The point where the two strands separate appears like a fork or a Y-shape. This is described as a replicating fork.
vi. A new strand is constructed on each old strand. This takes place with the help of a small RNA primer molecule which is complimentary to the DNA at that point.
vii. Each old DNA strand acts as a template (site) for the construction of new strand. The RNA primer attaches itself to the old strand and attracts the enzymes (DNA polymerase III) which add new nucleotides through base complementation. The deoxyribose nucleotides are present in the surrounding nucleoplasm. New DNA strand is thus constructed opposite to each old strand.
viii. Formation of new complementary strand always begins at the 3' end of the template strand (original strand)
and progresses towards the 5' end (ie in 3' - 5' direction). Since the new strand is antiparallel to the template strand, it is obvious that the new strand itself is always developed in the, 5'-3' direction. For this reason when the two original strands separate (then with respect to the origin of separation), one acts as 3'-5' template while the other acts as 5'- 3' template.
ix. Of the two, the replication of 3'-5' template begins first. Hence the new strand formed on it is called the leading strand. The other template (5'-3') must begin replication at the fork and progress back toward the previously transcribed fragment. The new strand formed on it is called the lagging strand.
x. Replication of the lagging strand takes place in small fragments called Okazaki fragments. These are then connected together by the enzyme ligase.
xi. Replication may take place in only one direction on the DNA helix (unidirectional) or in two directions (bidirectional).
xii. At the end of the process, two double stranded DNA molecules are formed from the original DNA molecule.

Three major types of RNA:

1. Messenger RNA or mRNA- has the information to make a protein. It is very unstable and comprises
~5% of total RNA polymer. Its length is highly variable, of the range 7503000 nucleotides.

2. Transfer RNA or tRNA- small molecule, about 90 nucleotides long. It is highly folded into an elaborate 3-d structure and comprises about 15% of total RNA.

3. Ribosomal RNA or rRNA- 80% of the total RNA, is associated with subcellular structures called ribosomes in which the polymer length varies from 120-3000 nucleotides and is folded into an elaborate structure which give ribosomes their shape.

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Non ambiguous — Particular codon will always code for same amino acid.
Degenerate — Number of codons can code for one amino acid.
Universal — Specific codon codes for same amino acid in all organisms

Translation:-
♦ Process of joining of amino acids by peptide bond to form a polypeptide

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* AA—Amino acid
*ATP—Adenosine Triphosphate
*E—Pyrophosphate
AA—AMP-E-Amino acid adenylate enzyme complex
AA—t RNA—Amino acyl-t RNA complex

LAC OPERON
*Discovered by Jacob and Manod.
*Experimented on E.coli.
Refer to figure number 6.14 of page 117 of text Book

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VApplication of Human genome project
-: Identification of defective genes.
-: Opportunity to offer early treatment.
-: Identification of genes that confer susceptibility to certain disease.
-: Prediction of protein that the genes produce.
-: Drug designing to enhance or inhibit the activities of the proteins.

TECHNIQUE FOR DNA FINGER PRINTING
• Technique developed by Dr.Alec Jeffreys.
• Process is also known as DNA typing/DNA profiling

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QUESTIONS

ONE MARK QUESTION

1. Name the genetic material in TMV.

2. Write the scientific name of the plant on which Taylor et al performed their experiment.

3. What would be the proportion of light and hybrid density DNA molecules after 80 minutes of a single cell of E. coli growth?

4. When does DNA replicate in the cell cycle ?

5. Name the amino acids having only one codon.

TWO MARK QUESTION

1. What is meant by semiconservative nature of DNA replication?

2. What are the functions of DNA polymerase?

3. What is frame shift mutation ? Name the type of mutation that does not affect protein synthesis .

4. What are the untranslated regions (UTRs) ?

5. Briefly describe polymorphism.

6. What do you mean by phosphodiester bond?
Ans: The bond which is formed between the 3’-OH of one deoxyribonucleotide and 5’-phosphate residue of an adjacent deoxyribonucleotide.

7. What type of transcription is found in retrovirus? Name the enzyme.
Ans: in retrovirus the genetic information flows from RNA to DNA and is called reverse transcription while the enzyme involved is called reverse transcriptase.

8. What would happen if histones were to be mutated and made rich in amino acids aspartic acid and glutamic acid in place of basic amino acids such as lysine and arginine?
Ans: If histone proteins were rich in acidic amino acids instead of basic amino acids then they would not have any role in DNA packaging in eukaryotes as DNA is also negatively charged molecule. The packaging of DNA around the nucleosome would not happen. Consequently, the chromatin fibre would not be formed.

9. Explain what happens in frameshift mutation. Name one disease caused by the disorder.
Ans: Mutation in which addition/insertion or deletion of one or two bases changes the reading frame from the site of mutation is called frameshift mutation. It may result in polypeptide with different sequences of amino acids. Disease caused by frameshift mutation - sickle-cell anemia.

10. Comment on the utility of variability in number of tandem repeats during DNA fingerprinting.
Ans: Tandemness in repeats provides many copies of the sequence for fingerprinting and variability in nitrogen base sequence in them. Being individual-specific, this proves to be useful in the process of DNA fingerprinting.

11. Why is lactose considered an inducer in lac operon?
Ans: Lactose binds to repressor molecule and prevents it from binding with the operator, as a result RNA polymerase binds to promoter-operator region to transcribe the structural genes. Thus the lac operon is switched on.

12. If a double-stranded DNA has 20 % of cytosine, calculate the % of adenine in the DNA.
Ans: cytosine = 20%, therefore guanine = 20%
According to Chargaff’s rule,
A+T = 100 – (G+C)
A+T = 100 – 40. Since both adenine and thymine are in equal amount.
Therefore, Thymine= Adenine = 60%/2 = 30%

13. What is cistron?
Ans: Region of the DNA template (gene) coding for a single protein is called cistron.

THREE MARK QUESTIONS

1.Describe the discontinuous synthesis of DNA.

2. How is Lac operon “switched on” in an E.coli cell ?

3.Name the three RNA Polymerases found in eukaryotes and mention their functions.

4.Explain the two major approaches involved in the sequencing of genomes.

FIVE MARKS QUESTIONS

1.Describe the salient features of the double helical model of DNA.

2. Bring out the salient features of genetic code .

3.Describe in detail the steps in the technique of DNA finger printing.

4.Describe the process of replication of DNA.

5. What is satellite DNA ? Name their types. Mention their basis for classification of satellite DNA.

6. What are the differences between DNA and RNA?
Ans: 

DNARNA
1. Polymer of deoxyribonucleotides
consisting of two antiparallel strands
1. Polymer of ribonucleotides consisting of
only a single strand
2. Purine nucleotides are- adenine and guanine. Pyrimidine nucleotides are cytosine and thymine.2. Purine nucleotides are- adenine and guanine. Pyrimidine nucleotides are
cytosine and uracil.
3. Main function is to carry all the hereditary characteristics.3. Main function is to perform protein synthesis
4. Mainly present in nuclear material of
chromatin fibre, mitochondria and chloroplast.
4. Mainly present in cytoplasm, nucleolus
and chromosome

7. What are B-DNA, A-DNA and Z-DNA?
Ans:

B-DNAA-DNAZ-DNA
1. Most predominant form of DNA, the conformation described by Watson and Crick, present under physiological conditions in the body, right handed double helix.1. In lower concentration of salts or in a partially dehydrated state, this form is present, found in some Gram positive bacteria, right handed double helix1. It has been discovered
in synthetically made
oligodeoxynucleotides, left handed double helix.
2. Base pair per turn 10.2. Base pair per turn 112. Base pair per turn 12.
3. Diameter 2 nm.3. Diameter 2.6 nm.3. Diameter 1.8 nm.

8. What do you mean by grooves of DNA?
Ans: DNA backbone is somewhat tilted from its vertical axis, it has two uneven grooves or furrowings i.e., one major groove (about 12 A°) and one minor groove (about 6 A°). They are the protein binding sites of DNA.

9. Recall the experiment done by Frederick Griffith. If the RNA, instead of DNA was the genetic material, would the heat killed strain of strain of streptococcus have transformed the r-strain into virulent strain? Explain your answer.
Ans: RNA is more labile and prone to degradation (owing to the presence of 2’-OH group in its ribose). Hence heat-killed S-strain may not have retained its ability to transform the R-strain.

10. What do you mean by selfish DNA?
Ans: DNA whose role appears to be to mediate its own replicationand survival within the genome, e.g. some satellite DNA, and transposable elements.

11. What are the differences between euchromatin and heterochromatin?
Ans:

euchromatinheterochromatin
1. During interphase certain areas in
chromatin are loosely coiled and stain less intensely.
1. During interphase certain areas in chromatin remain tightly coiled or condensed and hence stain darkly
2. These contain the genes or the coding DNA.2. These contain non-coding DNA like the repetitive DNA

 

 MOLECULAR BASIS OF INHERITANCE

•The cell is the fundamental unit of the living organisms has a complex network of many structures being coordinated by the large molecule-DNA

DNA is a long polymer of nucleotides.

 A nucleotide has three parts – deoxyribose sugar, nitrogenous base and phosphate group.

 There are two types of nitrogenous bases namely two Purines(double ring structure)- Adenine (A) and Guanine(G) and two pyrimidines (single ring structure)-Thymine (T) and Cytosine (C)

 class_12-biology_concept_262

 The molecule consists of large number of nucleotides joined together between the sugars and the phosphates by 3’-5’ phosphodiester bonds.

 The sugar-phosphate linked structure forms the backbone of the molecule

 We can see the presence of OH- group at the 3’C and PO4- group at the 5’C.

 The molecule is double helical right handed structure (B-DNA) consisting of two antiparallel polynucleotide strands and held by hydrogen bonds.

 Two strands are complementary to each other ie., adenine(A) in strand with pair with thymine (T) in the opposite strand and vice versa. Like wise, Cytosine (C) will pair with Guanine (G) . We can understand that the base sequence in of the strand is known, the base sequence of the complementary strand can be deduced.

One strand shows the polarity 5’ → 3’ and the other will show 3’ → 5’

 Adenine pairs with Thymine by two hydrogen bonds and guanine pairs with Cytosine by three hydrogen bonds. 

A — T OR T — A

C≡ G OR G  C

 The double strand molecule has alternate major groove and minor groove held by proteins called histones (in the case of eukaryotes).

 The biochemical work of Erwin Chargaff reveals that A+T/C+G= 1 and so A+G = T+C. Therefore, purines are always equal to pyrimidines (1:1). We must keep In mind that A+T= C+G.

The purine and pyrimidine bases are spaced by 0.34nm(3.4Ao) apart which gives 10 base pairs in one complete turn of the backbone. It results in 3.4nm (34Ao) per each complete turn of the helix. 

PACKAGING OF DNA

 The nucleoplasm of the nucleus (Eukaryotes) has the mesh of chromatin. The chromatin fibre appears to a long string like structure having beads. The beads represents nucleosomes.

 Interestingly, it appears like a rugby ball when viewed the three dimensional shape.

 Nucleosome cores consist of an octamer of two molecules each of 4 histones namely H2A. H2B, H3, and H4.A fifth kind histone(H1) is located on the linker DNA between the nucleosome particles. The core is wrapped by superhelical strand of DNA having 165bp in two turns.

 Histones has rich of basic amino acids lysines and arginines , it is positively charged and you are aware that DNA is negatively charged particle.

 Further packaging is done by Non-histone chromosomal proteins(NHC). 

 DNA is revealed as purple colour by the feulgen stain. It shows light regions (Euchromatin) and dark regions (heterochromatin). Remember, euchromatin is the regions of active DNA.

IN SEARCH OF GENETIC MATERIAL
TRANSFORMING PRINCIPLE
 Fredrick Griffith proposed this principle of Transformation by conducting the experiment on mice by injecting two strains namely R (rough) and S(smooth having capsule made of polysaccharide) . R strain is non-virulent and S is virulent.

 Live R strain Mice Alive

 Live S strain Mice Die

 Heat-killed S strain Mice Alive

 Heat killed S strain + live R strain Mice Mice die

Griffith concluded that live R strain became transformed as virulent due to heat-killed S strain.

IDENTIFICATION OF THE TRANSFORMING PRINCIPLE
CBSE Class 12 Biology - Molecular Basis Of Inheritance notes

 It was tested by Oswald Avery, Colin Mcleod and Maclyn McCarty to determine the biochemical nature.

 It is concluded that the transforming principle is DNA 

Further proof by Hershey and Chase experiment

 They did the experiment by using phage having radioactive labelled S35 and other set of the phage having P32.

The flow chart is shown WITH STEPS INVOLVED

CBSE Class 12 Biology - Molecular Basis Of Inheritance notes

E.coli has P32 and not S35 indicating the DNA .

DNA REPLICATION
 DNA replication takes place during the `S’ phase of the interphase in the cell-cycle(glance class XI-NCERT text Book)

 It is semi-conservative in nature ie., in each DNA molecule, one strand is old and the other newly formed strand.

EXPERIMENTAL PROOF
 The semi-conservative type of replication was confirmed by Meselson and Stahl in 1958.

 E.coli was grown in N15 medium (heavy density non-radioactive isotope) for many generations. And centrifuged with CsCl2 density gradient and found as-

CBSE Class 12 Biology - Molecular Basis Of Inheritance notes

 Later, DNA with N15 were transferred to a N14 medium and allowed to replicate.(Replication duration is 20 min).

 After 20 minutes, it is shown as-

CBSE Class 12 Biology - Molecular Basis Of Inheritance notes

 After 40 minutes, it reveals-

CBSE Class 12 Biology - Molecular Basis Of Inheritance notes

 The various steps involved are as follows:

- The process starts at a specific point called origin of replication(ori). The strand breaksfollowed by cleaving of hydrogen bonds.

- The two strands starts un winding by the enzyme helicase.

- It appears like `Y; shaped and it refers as replication fork.

The DNA polymerases which is responsible for synthesizing new DNA to the original strand (template strand) cannot function without the 3’OH group. Therefore, it is provided by short segment of RNA (primer) catalysed by the RNA polymerase. - The synthesis of the new DNA takes place in the 5’ 3’. The two template strands show 3’ 5’ and the other one has 5’ 3’ polarity. The replication begins at .3’ 5’ template first and the new DNA strand is called leading strand. In the opposite strand, it progresses from back towards the previously transcribed segment. This is the lagging strand.

- The lagging strand consists of small fragments called Okazaki fragments. These are connected by the enzyme ligase.

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RNA WORLD

• RNA serves as non-genetic material in eukaryotes /prokaryotes but it can behave as genetic material in few viruses referred as plant viruses and retroviruses.

• Having the presence of the 2’OH group in the ribose sugar makes it highly unstable, reactive and so it shows catalytic role.

TYPES OF RNA

• mRNA- Messenger RNA is transcribed from DNA by the RNA polymerase II (Eukaryotes) and by single large holoenzyme (prokaryotes). mRNA carries the information of the DNA in the form of triplet bases called codons to the ribosome for protein synthesis. . It has a start codon as AUG (it is an amino acid –methionine) and ends with any stop codon (UGA, UAG, UAA). It is unstable and comprises less than 5% of the total RNA.

• tRNA- transfer RNA- It appears like a clover leaf or hair-pin pattern. It is the smallest among the RNA and comprises 10-20% of the total RNA. It carries amino acids to the mRNA during the protein synthesis. It may be less than 61 amino acids because tRNA shows wobble hypothesis. ( According to Crick, the third base of the codon wobbles with the first base of the anti-codon of the tRNA.). RNA polymerase III is needed for the synthesis of tRNA.(eukaryotes)

• rRNA – ribosomal RNA- it comprise 80 % of the total RNA. It is found in the ribosome and provides proper binding site for the mRNA of the ribosome. It is produced by the RNA polymerase I .(eukaryotes).

TRANSCRIPTION

• The process of synthesis of mRNA from the DNA by the help of DNA dependent RNA polymerase.Let’s see the process in prokaryotes and in eukaryotes

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TRANSLATION

The story of protein synthesis has come to the climax stage, till now it can be show as

DNA - Has the language of nucleotides
            ↓

mRNA - Has the language of nucleotides
            ↓
Polypeptide- Has the language of amino acids.
            ↓

STEPS INVOLVED

 

• CHARGING OF AMINO ACID

The activation of the amino acid is done by the enzyme aminoacylsynthetase with the help of ATP

Amino-acyl synthetase

AA+ ATP → AA ~AMP + PPi

• CHARGING OF tRNA (aminoacylation of tRNA)

It involves the transfer of activated amino acid to tRNA

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• TERMINATION- It stops at the site of stop codons as no tRNA arrives.
                      Complete polypeptide is released by the release factor.

REGULATION OF GENE EXPRESSION

The schematic representation of Lac operon – Three genes z y a , promoter and operator.

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HUMAN GENOME PROJECT (HGP)
• It is the entire sequencing of the Human Genome .

TECHNIQUES
- Expressed sequence tags
- Sequence annotation

TOOL
- Restriction enzymes
- Vectors- YAC, BAC
- Dideoxynucleotides (dATP , dTTP,dCTP, dGTP)
- Primers
- Polyacrylamide gel
- Nitrocellulose membrane
- DNA probes

METHOD
Sanger’s dideoxy gel electrophoresis
FINDINGS

 

DNA FINGERPRINTING
It was developed by Alec Jeffres.
It is based on the principle of DNA polymorphism consisting of repetitive DNA including satellite DNA. It is two types STR and VNTR
VNTR (mini satellites ) is significant in this method.

STEPS
1. EXTRACTION OF DNA .
2. AMPLICATION OF DNA USING PCR
3. DIGESTION OF DNA USING RESTRICTION ENZYMES
4. SEPARATION OF DNA FRAGMENTS BY GEL ELECTROPHORESIS.
5. HYBRIDISATION
6. AUTORADIOFRAPHY BY X-RAY DIFFRACTION

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Chapter 10 Microbes In Human Welfare
CBSE Class 12 Biology Microbes In Human Welfare Notes
Chapter 11 Biotechnology Principles and Processes
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Chapter 12 Biotechnology and Its Application
CBSE Class 12 Biology Biotechnology And Its Application Notes
Chapter 13 Organisms and Populations
CBSE Class 12 Biology Organisms And Populations Notes
Chapter 15 Biodiversity and Conservation
CBSE Class 12 Biology Biodiversity And Conservation Notes
Chapter 16 Environmental Issues
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