Molecular Basis of Inheritance NEET Notes

Complete Molecular Basis of Inheritance Notes for Class 12 Biology | NEET Revision and NCERT Solutions

Nucleic acids (DNA & RNA) are the building blocks of genetic material.
DNA is the genetic material in most of the organisms.
RNA is the genetic material in some viruses. RNA mostly functions as messengers.

STRUCTURE OF POLYNUCLEOTIDE CHAIN

Polynucleotides are the polymer of nucleotides. DNA & RNA are polynucleotides. A nucleotide has 3 components:

1. A nitrogenous base

2. A pentose sugar (ribose in RNA & deoxyribose in DNA)

3. A phosphate group

Nitrogen bases are 2 types

Purines: It includes Adenine (A) and Guanine (G).
Pyrimidines: It includes Cytosine (C), Thymine (T) &Uracil (U). Thymine (5-methyl Uracil) present only in DNA) and Uracil only in RNA.
A nitrogenous base is linked to the pentose sugar through an N-glycosidic linkage to form nucleoside.
Nitrogen base + sugar + phosphate group = Nucleotide (deoxyribonucleotide). In RNA, every nucleotide residue has an additional –OH group present at 2'-position in the ribose.
2 nucleotides are linked through 3’-5’ phosphodiester bond to form dinucleotide.
When more nucleotides are linked it forms polynucleotide.

STRUCTURE OF DNA

Friedrich Meischer (1869): Identified DNA and named it as ‘Nuclein’.
James Watson &Francis Crick proposed double helix model of DNA. It was based on the X-ray diffraction data produced by Maurice Wilkins &Rosalind Franklin
DNA is made of 2 polynucleotide chains coiled in a right handed fashion. Its backbone is formed of sugar & phosphates. The bases project inside.
The 2 chains have anti-parallel polarity, i.e. one chain has the polarity 5’→3’ and the other has 3’→5’.
The bases in 2 stands are paired through H-bonds forming base pairs (bp).
A=T (2 hydrogen bonds) C≡G (3 hydrogen bonds)
Purine comes opposite to a pyrimidine. This generates uniform distance between the 2 strands.
Erwin Chargaff’s rule: In DNA, the proportion of A is equal to T and the proportion of G is equal to C.

i.e, [A] + [G] = [T] + [C] or [A] + [G] / [T] + [C] =1

Length of DNA = number of base pairs X distance between two adjacent base pairs.

Ф 174 (a bacteriophage) has 5386 nucleotides.
Bacteriophage lambda has 48502 base pairs (bp).
E. coli has 4.6x106 bp.
Haploid content of human DNA is 3.3x109 bp.
Number of base pairs in human = 6.6 x 109 Hence, the length of DNA = 6.6 x109 x 0.34x 10-9 = 2.2 m In E. coli, length of DNA =1.36 mm (1.36 x 10-3 m) ∴ The number of base pairs = 1.36x10−30.34 x 10−9 = 4 x 106 bp

PACKAGING OF DNA HELIX

In prokaryotes (E.g. E. coli), the DNA is not scattered throughout the cell. DNA, being negatively charged, is held with some positively charged proteins and form ‘nucleoid’.
In eukaryotes, there is a set of positively charged, basic proteins called histones
Histones are rich in positively charged basic amino acid residues lysinesand arginines.
8 histones form histone octamer.
Negatively charged DNA is wrapped around histone octamer to give nucleosome
A typical nucleosome contains 200 bp.
Therefore, the total number of nucleosomes in human = 6.6x109 bp200= 3.3x107
Nucleosomes constitute the repeating unit to form chromatin. Chromatin is the thread-like stained bodies.
Nucleosomes in chromatin = „beads-on-string‟.
Chromatin is packaged → chromatin fibres → coiled and condensed at metaphase stage → chromosomes.
Higher level packaging of chromatin requires non-histone chromosomal (NHC) proteins.
Chromatins include

Euchromatin: Loosely packed and transcriptionally active chromatin and stains light.
Heterochromatin: Densely packed and inactive region of chromatin and stains dark.

THE SEARCH FOR GENETIC MATERIAL

1. Griffith’s experiment (Transforming principle) Griffith used mice &Streptococcus pneumoniae. Streptococcus pneumoniae has 2 strains-

Smooth (S) strain (Virulent): Has polysaccharide mucus coat. Cause pneumonia.
Rough (R) strain (Non-virulent): No mucous coat. Does not cause Pneumonia.

Experiment:

S-strain → Inject into mice → Mice die
R-strain → Inject into mice → Mice live
S-strain (Heat killed) → Inject into mice → Mice live
S-strain (Hk) + R-strain (live) → Inject into mice → Mice die
He concluded that some ‘transforming principle’, transferred from heat-killed S-strain to R-strain. It enabled R-strain to synthesize smooth polysaccharide coat and become virulent. This must be due to the transfer of genetic material. 2. Biochemical characterization of transforming principle
- Oswald Avery, Colin MacLeod &Maclyn McCarty worked to determine the biochemical nature of ‘transforming principle’ in Griffith’s experiment.
- They purified biochemicals (proteins, DNA, RNA etc.) from heat killed S cells using suitable enzymes.
- They discovered that
Digestion of protein and RNA (using Proteases and RNases) did not affect transformation. So the transforming substance was not a protein or RNA.
Digestion of DNA with DNase inhibited transformation. It means that DNA caused transformation of R cells to S cells, i.e. DNA was the transforming substance.

3. Hershey-Chase Experiment (Blender Experiment)

- Hershey &Chase made 2 preparations of bacteriophage - In one, proteins were labeled with S35 by putting in medium containing radioactive sulphur (S-35). In the second, DNA was labeled with P32 by putting in a medium containing radioactive Phosphorous (P-32).
- These preparations were used separately to infect E. coli.
- After infection, the E. coli cells were gently agitated in a blender to separate the phage particles from the bacteria.
- Then the culture was centrifuged. Heavier bacterial cells are formed as a pellet at the bottom. Lighter viral components outside the bacterial cells remained in the supernatant.
- They found that:
- Supernatant contains viral protein labeled with S35, i.e. the viral protein had not entered the bacterial cells.
- The bacterial pellet contains radioactive P. This shows that viral DNA labeled with P32 had entered the bacterial cells. This proves that DNA is the genetic material.

PROPERTIES OF GENETIC MATERIAL

A genetic material must

Be able to generate its replica (Replication).
Chemically and structurally be stable.
Provide the mutations that are required for evolution.
Be able to express itself as ‘Mendelian Characters’.

DNA is a better genetic material

- Due to unstable nature of RNA, RNA viruses (E.g. Q.B bacteriophage, Tobacco Mosaic Virus etc.) mutate and evolve faster.
- For the storage of genetic information DNA is better due to its stability. But for the transmission of genetic information, RNA is better.
- RNA can directly code for the protein synthesis, hence can easily express the characters. DNA is dependent on RNA for protein synthesis

The 2 DNA strands are complementary. On heating, they separate. When appropriate conditions are provided they come together. (In Griffith’s experiment, when the bacteria were heat killed, some properties of DNA did not destroy).

RNA WORLD

RNA was the first genetic material.
It acts as genetic material and catalyst.
Essential life processes (metabolism, translation, splicing etc) evolved around RNA.
DNA evolved from RNA for stability.

CENTRAL DOGMA OF MOLECULAR BIOLOGY

It is proposed by Francis Crick. It states that the genetic information flows from DNA → RNA → Protein.

DNA REPLICATION

Replication is the copying of DNA from parental DNA.
Watson &Crick proposed Semi-conservative model of replication. It suggests that the parental DNA strands act as template for the synthesis of new complementary strands. After the completion of replication, each DNA molecule would have one parental and one new strand.
Matthew Messelson& Franklin Stahl (1958) experimentally proved Semi-conservative model.

Messelson& Stahl’s Experiment

They cultured E. coli in a medium containing 15NH4Cl (15N: heavy isotope of N). 15N was incorporated into both strands of bacterial DNA and the DNA became heavier.
Another preparation containing N salts labeled with 14N is also made. 14N was also incorporated in both strands of DNA and became lighter.
These 2 types of DNA can be separated by centrifugation in a CsCl density gradient.
They took E. coli cells from 15N medium and transferred to 14N medium. After one generation (i.e. after 20 minutes), they isolated and centrifuged the DNA. Its density was intermediate (hybrid) between 15N DNA and 14N DNA. This shows that the newly formed DNA one strand is old (15N type) and one strand is new (14N type). This confirms semi-conservative replication.
After II generation (i.e. after 40 minutes), there was equal amounts of hybrid DNA and light DNA. Taylor & colleagues (1958) performed similar experiments on Viciafaba(faba beans) using radioactive thymidine to detect distribution of newly synthesized DNA in the chromosomes. It proved that the DNA in chromosomes also replicate semiconservatively. The Machinery and Enzymes for Replication
DNA replication starts at a point called origin (ori).
A unit of replication with one origin is called a replicon.
During replication, the 2 strands unwind and separate by breaking H-bonds in presence of an enzyme, Helicase.
Unwinding of the DNA molecule at a point forms a ‘Y’-shaped structure called replication fork.
The separated strands act as templates for the synthesis of new strands.
DNA replicates in the 5’→3’ direction.
Deoxyribonucleoside triphosphates (dATP, dGTP, dCTP& TTP) act as substrate and also provide energy for polymerization.
Firstly, a small RNA primer is synthesized in presence of an enzyme, primase.
In the presence of an enzyme, DNA dependent DNA polymerase, many nucleotides join with one another to primer strand and form a polynucleotide chain (new strand).
The DNA polymerase forms one new strand (leading strand) in a continuous stretch in the 5’→3’ direction (Continuous synthesis).
The other new strand is formed in small stretches (Okazaki fragments) in 5’→3’ direction (Discontinuous synthesis).
The Okazaki fragments are then joined together to form a new strand by an enzyme, DNA ligase. This new strand is called lagging strand.
If a wrong base is introduced in the new strand, DNA polymerase can do proof reading.
E. coli completes replication within 38 minutes. i.e. 2000 bp per second.
In eukaryotes, the replication of DNA takes place at S-phase of the cell cycle. Failure in cell division after DNA replication results in polyploidy.

TRANSCRIPTION

It is the process of copying genetic information from one strand of the DNA into RNA.
Here, adenine pairs with uracil instead of thymine.
Both strands are not copied during transcription, because
The code for proteins is different in both strands. This complicates the translation.
If 2 RNA molecules are produced simultaneously this would be complimentary to each other, hence form a double stranded RNA. This prevents translation.
Transcription Unit

It is the segment of DNA between the sites of initiation and termination of transcription. It consists of 3 regions:

A promoter (Transcription start site): Binding site for RNA polymerase.
Structural gene: The region between promoter and terminator where transcription takes place.
A terminator: The site where transcription stops.

The DNA- dependent RNA polymerase catalyzes the polymerization only in 5’→3’direction.

3’→5’ acts as template strand. 5’→3’ acts as coding strand.
3’-ATGCATGCATGCATGCATGCATGC-5’ template strand. 5’-TACGTACGTACGTACGTACGTACG-3’ coding strand. Transcription unit and gene

Gene: Functional unit of inheritance. It is the DNA sequence coding for RNA molecule.

Cistron: A segment of DNA coding for a polypeptide.

Structural gene in a transcription unit is 2 types:

Monocistronic structural genes (split genes): It is seen in eukaryotes. Here, the coding sequences (expressed sequences or exons) are interrupted by introns (intervening sequences).
Polycistronic structural genes: It is seen in prokaryotes. Here, there are no split genes.

Steps of transcription in prokaryotes

Initiation: Here, the enzyme RNA polymerase binds at the promoter site of DNA. This causes the local unwinding of the DNA double helix. An initiation factor (σ factor) present in RNA polymerase initiates the RNA synthesis.
Elongation: The RNA chain is synthesized in the 5’-3’ direction. In this process, activated ribonucleosidetriphosphates (ATP, GTP, UTP & CTP) are added. This is complementary to the base sequence in the DNA template.
Termination: A termination factor (ρ factor) binds to the RNA polymerase and terminates the transcription.

In bacteria (Prokaryotes) transcription and translation can be coupled (Translation can begin before mRNA is fully transcribed) because
mRNA requires no processing to become active.
Transcription and translation take place in the same compartment (no separation of cytosol and nucleus).

In eukaryotes, there are 2 additional complexities:

1. There are 3 RNA polymerases:

RNA polymerase I: Transcribes rRNAs (28S, 18S & 5.8S).
RNA polymerase II: Transcribes the heterogeneous nuclear RNA (hnRNA). It is the precursor of mRNA.
RNA polymerase III: Transcribes tRNA, 5S rRNA and snRNAs (small nuclear RNAs).

2. The primary transcripts (hnRNA) contain both the exons and introns and are non-functional. Hence introns have to be removed. For this, it undergoes the following processes:

Splicing: From hnRNA introns are removed (by the spliceosome) and exons are spliced (joined) together.
Capping: Here, a nucleotide methyl guanosine triphosphate (cap) is added to the 5’ end of hnRNA.
Tailing (Polyadenylation): Here, adenylate residues (200-300) are added at 3’-end. It is the fully processed hnRNA, now called mRNA.

GENETIC CODE

It is the sequence of nucleotides (nitrogen bases) in mRNA that contains information for protein synthesis (translation).

20 AMINO ACIDS INVOLVED IN TRANSLATION

1. Alanine (Ala) 11. Leucine (Leu)

2. Arginine (Arg) 12. Lysine (Lys)

3. Asparagine (Asn) 13. Methionine (Met)

4. Aspartic acid (Asp) 14. Phenyl alanine (Phe)

5. Cystein (Cys) 15. Proline (Pro)

6. Glutamine (Gln) 16. Serine (Ser)

7. Glutamic acid (Glu) 17. Threonine (Thr)

8. Glycine (Gly) 18. Tryptophan (Trp)

9. Histidine (His) 19. Tyrosine (Tyr)

10. Isoleucine (Ile) 20. Valine (Val)

The codons for the various amino acids

George Gamow: Suggested that for coding 20 amino acids, the code should be made up of 3 nucleotides.
HarGobind Khorana: Developed the chemical method in synthesizing RNA molecules with defined combinations of bases (homopolymers& copolymers).
Marshall Nirenberg: Developed cell-free system for protein synthesis.
Severo Ochoa (polynucleotide phosphorylase) enzyme is used to polymerize RNA with defined sequences in a template independent manner.

Salient features of genetic code

Triplet code (three-letter code).
61 codons code for amino acids. 3 codons (UAA, UAG &UGA) do not code for any amino acids. They function as stop codons (Termination codons or non-sense codons).
Genetic code is universal. E.g. From bacteria to human UUU codes for Phenylalanine. Some exceptions are found in mitochondrial codons, and in some protozoans.
No punctuations b/w adjacent codons (comma less code). The codon is read in mRNA in a contiguous fashion.
Genetic code is Non-overlapping.
A single amino acid is represented by many codons (except AUG for methionine & UGG for tryptophan). Such codons are called degenerate codons.
Genetic code is unambiguous and specific. i.e. one codon specifies only one amino acid.
AUG has dual functions. It codes for Methionine (met), and also acts as initiator codon. In eukaryotes, methionine is the first amino acid and formyl methionine in prokaryotes.

TYPES OF RNA

- mRNA (messenger RNA): Provide template for translation (protein synthesis).
- rRNA (ribosomal RNA): Structural & catalytic role during translation. E.g. 23S rRNA in bacteria acts as ribozyme.
- tRNA (transfer RNA or sRNA or soluble RNA): Brings amino acids for protein synthesis and reads the genetic code.
tRNA- the adapter molecule tRNA has
An Anticodon (NODOC) loop that has bases complementary to the code.
An amino acid acceptor end to which amino acid binds.
- - For initiation, there is another tRNA called initiator tRNA.
- - There are no tRNAs for stop codons.
- - Secondary (2-D) structure of tRNA looks like a clover-leaf. 3-D structure looks like inverted ‘L’.

TRANSLATION (PROTEIN SYNTHESIS)

It takes place in ribosomes. Includes 4 steps

1. Charging of tRNA (aminoacylation of tRNA)

Formation of peptide bond requires energy obtained from ATP.

For this, amino acids are activated (amino acid + ATP) and linked to their cognate tRNA in the presence of aminoacyl tRNA synthetase. So the tRNA becomes charged.

2. Initiation

It begins at the 5’-end of mRNA in the presence of an initiation factor.
The mRNA binds to the small subunit of ribosome. Now the large subunit binds to the small subunit to complete the initiation complex.
Large subunit has 2 binding sites for tRNA- aminoacyl tRNA binding site (A site) and peptidyl site (P site).
Initiation codon for methionine is AUG. So methionyl tRNAcomplex would have UAC at the Anticodon site.

3. Elongation

At the P site the first codon of mRNA binds with anticodon of methionyl tRNA complex.
Another aminoacyl tRNA complex with an appropriate amino acid enters the ribosome and attaches to A site. Its anticodon binds to the second codon on the mRNA and a peptide bond is formed between first and second amino acids in presence of an enzyme, peptidyl transferase.
First amino acid and its tRNA are broken. This tRNA is removed from P site and second tRNA at the A site is pulled to P site along with mRNA. This is called translocation.
Then 3rd codon comes into A site and a suitable tRNA with 3rd amino acid binds at the A site. This process is repeated.
A group of ribosomes associated with a single mRNA for translation is called a polyribosome (polysomes).

4. Termination

When aminoacyl tRNA reaches the termination codon like UAA, UAG &UGA, the termination of translation occurs. The polypeptide and tRNA are released from the ribosomes.
The ribosome dissociates into large and small subunits at the end of protein synthesis.
An mRNA has additional sequences that are not translated (untranslated regions or UTR). UTRs are present at both 5’-end (before start codon) and 3’-end (after stop codon). They are required for efficient translation process

REGULATION OF GENE EXPRESSION

Gene expression results in the formation of a polypeptide. In eukaryotes, the regulation includes the following levels:

1. Transcriptional level (formation of primary transcript)

2. Processing level (regulation of splicing)

3. Transport of mRNA from nucleus to the cytoplasm

4. Translational level.

The metabolic, physiological and environmental conditions regulate expression of genes. E.g.
In E. coli the enzyme, beta-galactosidase hydrolyses lactose into galactose and glucose. In the absence of lactose, the synthesis of beta-galactosidase stops.
The development and differentiation of embryo into adult are a result of the expression of several set of genes.

OPERON CONCEPT

“Each metabolic reaction is controlled by a set of genes”
All the genes regulating a metabolic reaction constitute an Operon. E.g. lac operon, trp operon, ara operon, his operon, val operon etc.
When a substrate is added to growth medium of bacteria, a set of genes is switched on to metabolize it. This is called induction.
When a metabolite (product) is added, the genes to produce it are turned off. This is called repression.

Lac operon in E. coli:

The operon controlling lactose metabolism. It consists of

A regulatory or inhibitor (i) gene: Codes for the repressor.
3 structural genes:
1. z gene: Codes for galactosidase (hydrolyze lactose to galactose and glucose).
2. y gene: Codes for permease (increase permeability of the cell to lactose).
3. a gene: Codes for a transacetylase.

- The genes present in the operon function together in the same or related metabolic pathway. There is an operator region for each operon.
- If there is no lactose (inducer), Lac operon remains switched off. The regulator gene synthesizes mRNA to produce the repressor protein; this protein binds to the operator genes and blocks RNA polymerase movement. So the structural genes are not expressed.
- If lactose is provided in the growth medium, the lactose is transported into the E. coli cells by the action of permease. Lactose (inducer) binds with repressor protein.
So repressor protein cannot bind to operator gene. The operator gene becomes free and induces the RNA polymerase to bind with promoter gene. Then transcription starts. Regulation of lac operon by repressor is called negative regulation.

In the absence of inducer:

So repressor protein cannot bind to operator gene. The operator gene becomes free and induces the RNA polymerase to bind with promoter gene. Then transcription starts. Regulation of lac operon by repressor is called negative regulation.

In the presence of inducer:

HUMAN GENOME PROJECT (HGP)

The entire DNA in the haploid set of chromosome of an organism is called a Genome.
In Human genome, DNA is packed in 23 chromosomes.
Human Genome Project (1990-2003) is the first effort in identifying the sequence of nucleotides and mapping of all the genes in human genome.
Human genome contains about 3x109 bp.

Goals of HGP

a. Identify all the estimated genes in human DNA

b. Determine the sequences of the 3 billion chemical base pairs that make up human DNA.

c. Store this information in databases.

d. Improve tools for data analysis.

e. Transfer related technologies to other sectors.

f. Address the ethical, legal and social issues (ELSI) that may arise from the project.

HGP was closely associated with Bioinformatics.
Bioinformatics: Application of computer science and information technology to the field of biology & medicine. Usually applies in analyzing DNA sequence data. Methodologies of HGP: 2 major approaches.
Expressed Sequence Tags (ESTs): Focused on identifying all the genes that are expressed as RNA.
Sequence annotation: Sequencing whole set of genome containing all the coding & non-coding sequence and later assigning different regions in the sequence with functions.

Procedure:

Isolate total DNA from a cell → Convert into random fragments → Clone in suitable host (e.g. BAC & YAC) foramplification → Fragments are sequenced using Automated DNA sequencers (using Frederick Sanger method) → Sequences are arranged based on overlapping regions → Alignment of sequences using computer programs Genetic and physical maps on the genome were generated using information on polymorphism of restriction endonuclease recognition sites and some repetitive DNA sequences (microsatellites).

Salient features of Human Genome

a. Human genome contains 3164.7 million nucleotide bases.

b. Total number of genes= about 30,000.

c. Average gene consists of 3000 bases, but sizes vary. Largest known human gene (dystrophin on X-chromosome) contains 2.4 million bases.

d. 99.9% nucleotide bases are identical in all people. 0.1% is what makes each of us unique.

e. Functions of over 50% of discovered genes are unknown.

f. Chromosome I has most genes (2968) and Y has the fewest (231).

g. Less than 2% of the genome codes for proteins.

h. Repeated sequences make up very large portion of human genome. Repetitive sequences are stretches of DNA sequences that are repeated many times. They have no direct coding functions, but they shed light on chromosome structure, dynamics and evolution.

i. About 1.4 million locations where single-base DNA differences (SNPs- Single nucleotide polymorphism or ‘snips’) occur in humans.

DNA FINGERPRINTING (DNA PROFILING)

It is the technique to identify the similarities of the DNA fragments of 2 individuals.
Developed by Alec Jeffreys (1985).
Basis of DNA fingerprinting
- DNA carries some non-coding sequences called repetitive sequence [variable number tandem repeats (VNTR)].
- Number of repeats is specific from person to person.
- The size of VNTR varies from 0.1 to 20 kb.
- Repetitive DNA are separated from bulk genomic DNA as different peaks during density gradient centrifugation.
- The bulk DNA forms a major peak and the other small peaks are called as satellite DNA.
Satellite DNA is classified into many categories, (micro-satellites, mini-satellites etc) based on base composition (A:T rich or G:C rich), length of segment and number of repetitive units.
An inheritable mutation observed in a population at high frequency is called DNA polymorphism (variation at genetic level).
Polymorphism is higher in non-coding DNA sequence. Because mutations in these sequences may not have any immediate effect in an individual’s reproductive ability.
These mutations accumulate generation after generation and cause polymorphism. For evolution & speciation, polymorphisms play important role.

Steps of DNA fingerprinting (Southern Blotting Technique)

a. Isolate DNA (from any cells like blood stains, semen stains or hair roots).

b. Make copies (amplification) of DNA by polymerase chain reaction (PCR).

c. Digest DNA by restriction endonucleases.

d. Separate DNA fragments by gel electrophoresis.

e. Treat with alkali solution (NaOH) to denature DNA bonds in the gel into single strands.

f. Transfer (blotting) single stranded DNA fragments to synthetic membranes such as nitrocellulose or nylon, and then baked in a vacuum oven at 80oC for 3-5 hours (to fix the DNA fragment on the membrane).

g. Nitrocellulose filter paper is placed in a solution containing radioactive labeled single stranded DNA probe. The DNA probe binds with the complimentary sequences of the DNA fragment on the membrane to form a hybridized DNA.

h. The filter paper is washed to remove unbound probe.

i. The hybridized DNA is photographed on to an X-ray film by autoradiography. The image (in the form of dark & light bands) obtained is called DNA fingerprint.

Application of DNA fingerprinting

Forensic tool to solve paternity, rape, murder etc.
For the diagnosis of genetic diseases.
To determine phylogenetic status of animals.

NCERT Solutions

Students learned about inheritance patterns and the genetic basis for such patterns in the previous chapter. The nature of those 'factors' regulating the pattern of inheritance was not clear at the time of Mendel. The nature of the putative genetic material was investigated over the next hundred years, culminating in the discovery that DNA - deoxyribonucleic acid - is the genetic material, at least for the majority of organisms. They learned in Class XI that nucleic acids are polymers of nucleotides. DNA, the Search for Genetic Material, RNA World, Replication, Transcription, Genetic Code, Translation, Regulation of Gene Expression, Human Genome Project, and DNA Fingerprinting are also covered. Search for genetic material and DNA are two topics covered in Chapter Molecular Basis of Inheritance.

1. Group the following as nitrogenous bases and nucleosides:

Adenine, Cytidine, Thymine, Guanosine, Uracil and Cytosine.

Solution:

Following is the grouping:

Nitrogenous base

Nucleosides

Adenine

Cytidine

Thymine

Guanosine

Uracil

Cytosine

2. If a double stranded DNA has 20 per cent of cytosine, calculate the percent of adenine in the DNA.

Solution:

As per Chargaff’s rule, DNA molecules are required to have an equal ratio of purine(adenine and guanine) and pyridine(cytosine and thymine). This is to say that the number of adenine molecules is equivalent to the cytosine molecule.

Percentage of adenosine = percentage of thymine,

Percentage of guanine = percentage of cytosine

Hence, according to the law, if the double stranded DNA has 20% of cytosine, it should have 20% of guanine. Therefore, the percentage of G + percentage of C = 40%

The other 60% indicates both A + T percentage molecule. As adenine and thymine is always found in equal numbers, the adenine content is 30%.

3. If the sequence of one strand of DNA is written as follows:

5′ -ATGCATGCATGCATGCATGCATGCATGC-3′

Write down the sequence of complementary strand in 5’→3′ direction.

Solution:

With regards to base sequence, DNA strands are complementary to each other. Therefore, if the sequence of one strand of DNA is written as:

5′ -ATGCATGCATGCATGCATGCATGCATGC-3′

The sequence of the complementary strand in 3′ -> 5′ is as follows:

3′ – TACGTACGTACGTACGTACGTACGTACG – 5′

Subsequently, sequence of complementary strand in 5’→3′ direction is written as:

5′ – GCATGCATGCATGCATGCATGCATGCAT – 3′

4. If the sequence of the coding strand in a transcription unit is written as follows:

5′ -ATGCATGCATGCATGCATGCATGCATGC-3′

Write down the sequence of mRNA.

Solution:

In a transcription unit, if the coding strand is as follows:

5′ -ATGCATGCATGCATGCATGCATGCATGC-3′

Then the sequence which codes for nothing is the same as mRNA except that in mRNA, uracil replaces thymine. Hence, the mRNA sequence is as written below:

5′ -AUGCAUGCAUGCAUGCAUGCAUGCAUGC-3′

5. Which property of DNA double helix led Watson and Crick to hypothesise semi-conservative mode of DNA replication? Explain.

Solution:

Watson and Crick noted that the two DNA strands are anti-parallel, complementary to each other with regards to their base sequences which facilitates each strand to serve as a template to synthesize a new strand. This organization of the DNA molecule furthered the hypothesis that replication of DNA is semi-conservative. In other words, the double stranded DNA molecule splits and inturn each of the separated strand serves as a template to synthesize a new complementary strand. Subsequently, each DNA molecule will have one parental strand and a newly synthesized daughter strand. As only one parental strand is conserved in each of the daughter molecule, the mode of replication is termed as semi-conservative.

The sequence of bases in the template strand because of complementary base pairing governs the sequence of bases in the daughter strand. Subsequently, this property of DNA led Watson and Crick to hypothesize the semi-conservative mode of replication.

6. Depending upon the chemical nature of the template (DNA or RNA) and the nature of nucleic acids synthesized from it (DNA or RNA), list the types of nucleic acid polymerases.

Solution:

The list includes two different types of nucleic acid polymerases:

DNA-dependent DNA polymerases
DNA-dependent RNA polymerases

To synthesize a new strand, the DNA-dependent DNA polymerases uses a DNA template. But, DNA-dependent RNA polymerases utilizes a DNA template strand to synthesize RNA.

7. How did Hershey and Chase differentiate between DNA and protein in their experiment while proving that DNA is the genetic material?

Solution:

To prove DNA is the genetic material, Hershey and Chase studied and worked on bacteriophage and E.Coli. To label protein coat and DNA of the bacteriophage, they made use of different radioactive isotopes. In a medium containing radioactive phosphorous (32P), they cultivated some bacteriophages to detect DNA and few more on a medium comprising radioactive sulphur (35S) to detect protein. The labelled radioactive phages were then made to infect the bacteria – E.coli. Once infected, the protein coat of the bacteriophage was segregated from the bacterial cell by mixing and then subjecting to the centrifugation process.

It was observed that in the supernatant, as the protein coat was lighter, the infected bacteria got settled at the bottom of the centrifuge tube. In case I – supernatant was observed to be radioactive indicating that protein did not enter in the bacterial cell when infected. But in case II – the bacterial cells were radioactive as they possess radioactive DNA. Thus it was proved that DNA is a genetic material as it was transferred from virus to bacteria.

8. Differentiate between the following:

(a) Repetitive DNA and Satellite DNA

(b) mRNA and tRNA

Solution:

(a) Repetitive DNA and Satellite DNA

Repetitive DNA

Satellite DNA

These are DNA sequences containing small segments repeated several times

These are repetitive DNA sequences containing highly repetitive DNA

Vary in length from several base pairs to hundreds and thousands

Shorter in length and are close to hundred base pairs long

Can be segregated from bulk DNA by density gradient centrifugation because of which they appear as light bands

Can be segregated from bulk DNA through density gradient centrifugation because of which they appear as dark bands and small peaks.

(b) mRNA and tRNA

mRNA

tRNA

The messenger RNA serves as a template for the transcription process

The transfer RNA serves as an adaptor molecule carrying a particular amino acid to the mRNA to synthesize polypeptide

mRNA is a linear molecule

Resembles a clover shaped leaf

Gets attached to the ribosome only

Gets attached at one end to the ribosome and at the other end to an amino acid

Template strand

Coding strand

Serves as a template for the mRNA synthesis during transcription

Complementary strand of template strand

Consists of a sequence that is complementary to the mRNA

Consists of a sequence that is identical to the mRNA except that thymine in DNA is replaced by uracil IN mRNA

Template strand runs from 3′ to 5′

Coding strand runs from 5′ to 3′

9. List two essential roles of ribosome during translation.

Solution:

Two essential roles of ribosome during translation are:

Ribosomes are sites where synthesis of proteins occurs from individual amino acids. It consists of two subunits – larger subunit serves as an amino acid binding site whereas smaller subunit attaches to the mRNA forming a protein synthesizing complex
Since large subunit of ribosome has two different sites to attach to tRNA, it facilitates amino acids to come closer for peptide bond formation. Also, ribosome behaves as a catalyst for the formation of peptide bond. Example – 23s r-RNA acts as a ribozyme in bacteria

10. In the medium where E. coli was growing, lactose was added, which induced the lac operon. Then, why does lac operon shut down some time after addition of lactose in the medium?

Solution:

A segment of DNA which comprises of three adjacent structural genes – a promoter gene, an operator gene, and a regulator gene is known as Lac operon. It is functional in a coordinated manner in order to metabolize lactose into galactose and glucose. Lactose serves as an inducer in lac operon which binds to the repressor, inactivating it. Once the lactose is bound to the repressor, RNA polymerase binds to the area of promoter. Therefore, the structural genes express their product, corresponding enzymes are yielded. Lactose is metabolized due to the action of enzymes into galactose and glucose. Later on, the level of inducer decreases as enzymes completely metabolize them which results in the synthesis of the repressor from the regulator gene. Thus, the repressor is bound to the operator gene, restricting RNA polymerase from transcribing the operon, subsequently transcription is ceased. This kind of regulation is termed as negative regulation.

11. Explain (in one or two lines) the function of the followings:

(a) Promoter

(b) tRNA

Solution:

(a) Promoter

The section of DNA which aids in initiating transcription process and serving as a binding site for RNA polymerase is the promoter

(b) tRNA

The transfer RNA or tRNA is a small clover shaped RNA which reads the genetic code that is located on mRNA. It conveys particular amino acids to the mRNA ribosome while proteins are translated. Specific tRNAs are present for specific amino acids.

In eukaryotes, these are coding sequences of DNA which transcribe proteins. In between, exons comprise of long non-coding sections of DNA known as introns.

12. Why is the Human Genome project called a mega project?

Solution:

The Human genome project was ought to be a mega project as the scale and the goals of the project were humongous. It had goals to sequence every base pair located in the human genome which took around 13 years for completion and accomplished in 2003. This large scale project intended to develop new technologies and produce new information in the stream of genomic studies. Subsequently, it unlocked scope for several new areas and possibilities such as in the stream of biotechnology, genetics, medical sciences etc which hints at comprehending different aspects of human biology.

13. What is DNA fingerprinting? Mention its application.

Solution:

The technique of DNA fingerprinting is helpful in identifying and analyzing variations in different individuals at the DNA level. It is based on the principle of variability and polymorphism in DNA sequences.

Its applications are as follows:

To identify potential crime suspects in forensic science
Helpful in establishing family and paternity relationships
Useful in identifying and preserving the commercial varieties of livestock and crops
Useful to discover and know more about the evolutionary history of an entity, thus tracing the linkages between different entities.

14. Briefly describe the following:

(a) Transcription

(b) Polymorphism

(d) Bioinformatics

Solution:

(a) Transcription

It is the process to synthesize RNA from DNA template. A section of DNA gets duplicated into mRNA in the process. The transcription process originates from the promoter region of the template DNA and ends at the terminator region. The section of DNA between these two regions is termed as transcription unit. The transcription process necessitates DNA template, RNA polymerase enzyme, four types of ribonucleotides and a few cofactors, namely Mg2+.

Three significant events occurring during the transcription process are:

Initiation
Elongation
Termination

The process of transcription is initiated when the DNA-dependent RNA polymerase and several initiation factors bind at the promoter region of the template strand at the double stranded DNA. The RNA polymerase passes along the DNA, causing the DNA duplex to unwind to two distinct strands. One of the strands is known as sense strand which serves as a template for mRNA synthesis. The RNA polymerase enzyme uses nucleoside triphosphates (dNTPs) as raw material to polymerize them for the formation of mRNA as per the complementary bases positioned in the template DNA. This phenomena of opening of helix and elongation of polynucleotide chain lasts until the enzyme arrives at the terminator region. Once the RNA polymerase arrives at the terminator section, the freshly produced mRNA transcripted in addition to the enzyme is liberated. Additionally, another factor known as terminator factor is essential for the transcription process to end.

(b) Polymorphism

It is a form of genetic variation wherein different nucleotide sequences can be present at different sites in a molecule of DNA. There is a high frequency of this heritable mutation to be observed in a population which emerges as a result of mutation either in the germ cells or somatic cells. The germ cell mutation can be passed from parents to the offsprings which leads to the accumulation of different mutations in a population causing variation and polymorphism in the population. This has a crucial role to play in the evolution and speciation process.

It is the process wherein amino acids are polymerized for the formation of a polypeptide chain, a ribosome by reading mRNA molecule. In mRNA, the triplet sequence of base pairs defines the order and sequence of amino acids in a polypeptide chain.

This phenomena includes three steps:

Initiation
Elongation
Termination

The tRNA gets charged during the initiation of the translation when the amino acids binds to it with the help of ATP. The initiation code – AUG that is located on the mRNA is identified only by the charged tRNA. For the process of translation, the ribosome acts as an actual site, containing two distinct sites in a large subunit to attach to the following amino acids. The smaller subunit of ribosome is bound to the mRNA at AUG(initiation code) subsequently by the large subunit. The process of translation is then said to be initiated.

In the elongation process, the ribosome passes one codon downstream with the mRNA in order to leave some space to bind another charged tRNA. Amino acid brings tRNA which gets linked with the former amino acid via peptide bond. The process continues leading to the formation of a polypeptide chain. When the ribosome arrives, one or more STOP codon (UAA, UGA, and UAG). The phenomena of translation ends, polypeptide chain releases and the ribosomes get detached from mRNA.

(d) Bioinformatics

It is the application of statistical and computational methodologies to the molecular biology stream. It explains practical issues that arise from the analysis and management of biological data. The stream of bioinformatics developed after accomplishing human genome project as huge amount of data was produced during the process which has to be stored and managed for easy access and analysis for later use. Thus, bioinformatics includes creating biological databases which stores huge data of biology. It comes up with few tools for efficient and easy access to data which can be used. Also, bioinformatics brought in new algorithms and statistical techniques to figure out the dynamics between data, to predict structure of protein and their functionalities and to group protein sequences into their associated families.