Molecular Basis of Inheritance

Q1. Describe the structure of double helix model of DNA.

Ans:
1. DNA is made up of two polynucleotide chains, where the backbone is made up of sugar and phosphate groups
2. The nitrogenous bases project towards the centre.
3. The two Chains have anti-parallel polarity. It means, if one chain has the polarity 5'→3', the other has 3'→5'.
4. Both Strands of DNA are helically coiled in a right-handed fashion.
5. The nitrogen basses (adenine, guanine, cytosine, thymine) are paired through hydrogen bond forming
base pairs (bp).
6. Adenine pairs with thymine by two hydrogen bonds and guanine pairs with cytosine with three hydrogen bonds. As a result, always a purine comes opposite to a pyrimidine.
7. The pitch of the Helix is 3.4 nm and there are roughly 10 bp in each turn. Consequently, the distance between a bp in a helix is approximately 0.34 nm.

Q2. Mention and describe the salient features of genetic code.

Ans:
1. The codons are triplet: Out of 64 codons, 61 code for 20 amino acids and 3 codons (UAA, UGA, UAG) do not code for any amino acid, hence function as stop or terminating codons.
2. Genetic codes are unambiguous and specific: One codon code for only one particular amino acid.
3. Genetic codes are degenerate: Some amino acids are code by more than one codon.
4. Genetic codes are without punctuations and comma less: The codon is read on mRNA in a continuous fashion.
5. Genetic code is nearly universal: i.e., a particular codon codes for the same amino acid in all organisms except in mitochondria and few protozoa.
6. AUG is a dual function codon, its codes for methionine (met) and it is also acts as initiator codon.

Q3. List out the goals of human genome project.

Ans:
1. Identify all the approximately 20,000 to 25,000 genes in human DNA.
2. Determine the sequences of the 3 billion chemical base pairs that make up human DNA.
3. Store this information in database.
4. Improve tools for data analysis.
5. Transfer related Technologies and information to other sectors, such as industries.
6. Address the ethical, legal and social issues [ELSI] that may arise from the project.

Q4. Mention any five salient features of human genome project.

Ans:
1. The Human Genome contains 3164.7 million bases pairs.
2. The average gene consists of 3000 bases; the largest known human gene being dystrophin at 2.4 million bases.
3. The total number of genes is estimated to be 30,000 and 99.9% nucleotide bases are exactly the same in all people.
4. The functions are unknown for over 50% of the discovered genes.
5. Less than 2% of the genome codes for proteins.
6. The Human genome contains large repeated sequences.
7. Chromosome 1 has most genes (2968) and the Y has the fewest genes (231).
8. Scientists have identified about 1.4 million locations where single base DNA sequence differences called single nucleotide polymorphism or SNPs occur in humans.

Q5. Explain the process of translation in eukaryotes.

Ans:
➢ Charging of tRNA (aminoacylation of tRNA) – activation of amino acids in the presence of ATP and their linking to specific tRNA.
➢ Initiation - Binding of the ribosome (small sub-unit) to mRNA at the initiator codon (AUG).
Binding of the initiator tRNA carrying the amino acid (methionine) to the initiator codon to initiate protein synthesis.
➢ Elongation of polypeptide chain - Movement of ribosome from codon to codon along the mRNA and adding of amino acids (linked to tRNA) one by one.
➢ Termination – binding of the release factor to stop codon located at the 3’-end of mRNA, terminating translation and releasing the polynucleotide from ribosome.

Q6. Explain the process of transcription in prokaryotes.

Ans:
The process of copying genetic information from one strand of the DNA into RNA is termed as transcription.

In bacteria, the transcription of all the three types of RNA [mRNA, tRNA and rRNA] is catalysed by single DNA-dependent RNA-polymerase enzyme.

The RNA polymerases able to catalyse all the three steps of transcription; they are, initiation, elongation and termination.

1. Initiation: Sigma (σ) factor recognises the start signal and promoter region on DNA which then along with RNA polymerase binds to the promoter to initiate transcription.
2. Elongation: The RNA polymerase after initiation of RNA transcription loses the σ factor but continues the polymerization of ribonucleotides to form RNA.
3. Termination: Once the RNA polymerase reaches the termination region of DNA, the RNA polymerase is separated from DNA-RNA hybrid, as a result nascent RNA separate.
This process is called termination which is the facilitated by a termination factor Rho (ρ).

Q7. Describe the process of DNA replication with the help of a diagram.

Ans:
➢ DNA replication begins at a unique and fixed point called origin of replication or ‘ori’.
➢ Various enzymes are required as catalysts during DNA replication in living cell.
➢ The complementary strands of DNA double helix are separated by two enzymes, DNA helicase and DNA gyrase. This is called unwinding of double-stranded DNA.
➢ Unwinding of double-stranded DNA forms a Y-shaped configuration in the DNA duplex, which is called replicating fork.
➢ An enzyme called DNA-dependent DNA polymerase catalyse polymerisation only in one direction, i.e., 5` 3`.
➢ On one strand (3` 5`), the replication is continuous and it is called leading strand, while the replication of another strand (5` 3`) is discontinuous and it is known as the lagging strand.
➢ The replication of lagging strand generates small polynucleotide fragments called ‘Okazaki fragments’.
➢ These Okazaki fragments are joined together by an enzyme called DNA ligase.

Q8. Describe the structure of an adaptor molecule (t-RNA)

Ans:
tRNA has 5 arms or loops:
1. Anticodon loop: That has bases complementary to the code.
2. Amino acid acceptor end: To which amino acids bind.
3. T loop: Which helps in binding to ribosome.
4. D loop:Which helps in binding aminoacyl synthetase.
5. Variable loop
Each tRNA is specific for a particular amino acid.

Q9. Draw the schematic structure and explain the different regions of transcription unit.

Ans:
The transcription unit of DNA contains three regions in the DNA:
a) The promoter: It is the binding site for RNA polymerase for initiation of transcription.
b) The structural gene: It codes for enzymes or protein for structural functions.
c) The terminator: It is the region where transcription ends.

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Q10. Explain the regulation of Lac Operon in the absence and presence of lactose as an inducer.

Ans:
In the absence of inducer (Lactose):
The regulator gene i regulates and produce repressor protein. The repressor protein binds to the operator region of the operon and prevents RNA polymerase from transcribing the operon. Hence, no enzymes are formed.
This type of operon is said to be “Switched off” .

In the presence of inducer:
Lactose as an inducer, binds to the repressor and repressor becomes inactive.
Now the repressor fails to bind to the operator region.
This allows RNA polymerase binds to the operator and transcribes lac mRNA.
Lac mRNA is polycistronic, produces all three enzymes called β-galactosidase, permease and transacetylase.
Now the operon is termed as “Switched on” .

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Q11. Draw a neat labeled diagram of nucleosome.

Ans:
Structure of nucleosome

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