Given, Cytosine = 20%
Percentage of Guanine = 20%
Now according to Chargafl's rule,
A + T = 100 - (G +C) 
 A + T = 100 - 40
Percentage of Thymine = Percenlage of Adenine

(i) The sequence of amino acids is Met-Phe-Phe-Phe- Phe-Phe-Phe.
(ii) The sequence of nucleotides can be:
(i) AUG UUU UUU UUU UUU UUU UUU
           (or)
(ii) AUG UUC UUC UUC UUC UUC UUC
            (or)
(iii) AUG UUU UUU UUU UUC UUC UUC
(iii) Yes, there is difficulty in predicting the sequence of nucleotides, as many more combinations are possible; this difficulty is because amino acid, phenylalanine is coded by two codons, UUU and UUC.
(iv) The two properties of genetic code that can be applied to this are:
(i) Genetic code is unambiguous and specific, i.e. one codon codes for only one particular amino acid, e.g. AUG codes for methionine, UUU codes for phenylalanine, UUC codes for phenylalanine.
(ii) Genetic code is degenerate, i.e. one amino acid is coded by more than one codon, e.g. Phenylalanine is coded by two codons, UUU and UUC.

5'-UACGUACGUACGUACGUACGUACG-3'.

Given, length of E.coli DNA =1.36 mm = 1.36x 10^-3 m.
Distance between two consecutive base pairs =0.34 nm = 0.34 x 10^-9m
Hence, for E.coli total number of base pairs = \(\frac{1.36 \times 10^{-3}}{0.34 \times 10^{-9}}\) =  4 x 10⁶ bp

The differences between unambiguous and universal genetic codes are as follows

(ii) The differences between degenerate and initiator codes are as follows

a) Transcription is the process in which RNA is synthesized from DNA.
Transcription is a 3 step process-
1) Initiation- RNA polymerase binds to DNA sequence called the promoter and providing the single-stranded template for transcription process.
2)  Elongation- RNA polymerase reads the template strand and adds bases leading to elongation of the chain.
3) Termination- In this step terminator sequences mark the end of transcription.

The replication fork of DNA formed during DNA replication.

Structure of a nucleosome

A nucleosome is found in the nucleus of the cell.

Amino acids are activated in the presence of ATP and linked to (cognate) t-RNA.
Carries amino acid to the site synthesis/reaches amino acids to the respective codon.

Codingstrand-5'
TACGTACGTACGTACGTACGTACG 3'
mRNAstrand-5'
UACGUACGUACGUACGUACGUACG 3'

Nucleases remove unwanted nucleotides while ligases rejoin useful of nucleic acid (RNA)

RNA polymerase I and III catalyse the transcription of r-RNA and t-RNA respectively

1. AUGUCG - at the 5' end before the initiation codon.
2. GAGGAA-at the 3' end after the termination codon.
3. They are normally located in front of the initiation codon at the 5' end and behind the termination codon at the 3' end of the coding strand of DNA.

1.It is a DNA sequence that provides the binding site for RNA polymerase for transcription.
2.It is located towards the 5' end (upstream) of the structural gene (of the coding strand).

Both the strands of DNA are not copied during transcription for the following reasons:
(i) If both the strands of DNA are copied, two different RNAs (complementary to each other) and hence two different polypeptides would be formed; if a segment of DNA produces two polypeptides, the genetic information machinery becomes complicated.
(ii) The two complementary RNA molecules (produced simultaneously) would form a double-stranded RNA rather than getting translated into polypetides.
(iii) RNA polymerase carries out polymerisation in the 5' ~ 3' direction and hence the DNA strand with 3' ~ 5' polarity acts as the template strand.

1. The polynucleotide chain has at the 5th position of the pentose sugar, a free phosphate moiety;This end is called 5' end.
2. The 3^rd position of the pentose of the sugar has a OH-group attached; this is called the 3' end.

Single nucleotide polymorphism helps in
(i) finding the chromosomal locations of disease-associated genes or sequences of DNA and
(ii) tracing human history

Dual function of AUG codon
1. It codes for amino acid,methionine
2. It functions as the initiation codon
3. It is transcribed by TAC on DNA
4. Its anticodon is UAC.

1. When one of the termination codons (UAA, UAG, UGA) comes at the A-site, it does not code for any amino acid and there is no tRNA molecule for it.
2. As a result, the polypeptide synthesis (or elongation of polypeptide) stops.
3. The polypeptide synthesised is released from the ribosome, catalysed by a 'release factor'.

Role of Ribosomes
(i) Ribosomes are the main cellular factory of protein synthesis the large subunit has two sites (P and A) for binding of amino acids,so that they are close to each other for formation of peptide bond.
(ii)They also act as catalyst (23s rRNA)in prokaryotes for formation of peptide bonds.ATP provides energy for the activation of amino acids.

(i) The deoxyribonucleoside triphosphates are the building blocks for the DNA-strand(polynucleotide chain), as substrate.
(ii) These also serve as energy source in the form of ATP and GTP from there two terminal phosphates.

 A protein acquires a charge depending on the abundance of amino acids residues that have positively charged groups in their side chains. Histones are rich in basic amino acids residues-lysines and arginine both of these amino acids carry positive charges in their side chains,.Hense, histones are positively charged molecules.

(a) 5' ATGGGGCTC 3'    sense
3' TACCCCGAG 5'         antisense
RNA 5' AUGGGGCUC 3' sense
3' UACCCCGAG 5'        antisense
(b) The two strands of RNA (i.e. sense and antisense) being complementary will bind with each other and form double stranded RNA. As a result its translation and protein expression would be inhibited.

(i) i-Regulatory gene, p - Promotor gene.
(ii) Inducer is lactose.
Functions
(a) Enters the cell and binds to the repressor and inactivates it.
(b) As a result, repressor cannot bind to the operator .This allows  RNA polymerase to have access to the promotor and transactions proceeds.

The template strand as 3'-TACGTAGTACGTTAGTCC-5' and the coding strand reads as 5'-ATGCATGCATGCAATCAGG-3' Therefore,the total number of Thymine are=10.
The Carboxyl group(-COOH) of one amino acids reacts with an amino group(-NH₂)of other amino acid to form a peptide bond(-CO-NH).Make a peptide bond between the two amino acids by removing water molecule in the given figure

  

(a)
\(\\ \overline { ^{ 3' }TAC\quad CAT\quad TAG\quad GAT\\ _{ 5' }\underline { AUG\quad GUA\quad AUC\quad CUA } _{ mRNA } } ^{ 5' }\)

In eukaryotes, there are three RNApolymerases.
(i) RNA-polymerase I catalyses transcription of rRNAs (28 S, 18 Sand 5.8 S).
(ii) RNA- polymerase II catalyses transcription of precursor of mRNA; it is called hnRNA.
(iii) RNA- polymerase III catalyses tRNA, 5srRNA and snRNAs.

1. Central Dogma of molecular biology
2. It is the present day representation of the central dogma of molecular biology.
3. RNA is made from DNA by the process of transcription.
4. With the discovery of retroviruses, there is reverse transcription, i.e. the synthesis of DNA on RNA template, catalysed by the enzyme reverse transcriptase
5. RNA is translated into a polypeptide, making use of amino acids, ribosomes and enzymes.

1. The viruses/bacteriophages grown on radioactive sulphur \(\left( ^{ 35 }{ S } \right) \)    contained radioactive protein but not radioactive DNA, because DNA does not contain sulphur.
2.When these viruses were allowed to infect bacteria, the bacteria did not contain radioactivity, because proteins did not enter the bacteria; hence protein is not the genetic material.
3. The viruses grown on radioactive phosphorus  \(\left( ^{ 32 }{ P } \right) \)   contained radioactive DNA, becauseDNA contains phosphorus and not proteins. 
4. When these viruses were allowed to infect the bacteria, the bacteria were radioactive, indicating that DNA is the genetic material that has passed from the virus into bacteria.

1. In the mammalian cells (or eukaryotes) there is a set of positively-charged basic proteins, called histones.
2. Histones are organised to form a unit of eight molecules, called histone octamer.
3. The negatively charged DNA is wrapped around the positively charged histone octamer to form a structure, called nucleosome.
4. A typical nucleosome contains 200 bp of DNA helix.
5. The nucleosomes constitute the repeating units of chromatin, which appear as beads-on-string structure under an electron microscope.
6. These are further packaged to form the chromatin fibres, which condense to form chromosomes.
7. The packaging of chromatin at higher levels requires additional set of proteins called non-histone chromosomal (NHC) proteins.

(i) 'BAC' - Bacterial Artificial Chromosome.
'YAC' - Yeast Artificial Chromosome.
(ii) They are the commonly used vectors for cloning the DNA fragments in the hosts like bacteria and yeast.
(iii) The cloning results into amplification of each fragment of DNA, so that they could be sequenced with ease.

(i) ATCGTACTA.
(ii) Base pairs are held together by weak hydroyen bodsin a DNA molecule. Adenine pairs with thymine by two H-bonds and guanine pairs with cytosine by three H-bonds.
(iii) According to basc complementarity rule proposed by Erwin Chargalf for a double-stranded DNA, the ratios between adenine-thymine and guanine-cytosine are constant and equal to one.

In Eukaryotes:
1. The structural genes are split. They have coding sequences (exons) interspersed with non-coding sequences (introns).
2. The primary transcript of RNA undergoes a process called splicing, by which the introns are removed and the exons are joined together.
3. The hnRNA (precursor of mRNA) undergoes capping and tailing to become mRNA.
4. In capping, methyl guanosine triphosphate is added to the 5' end of hnRNA.
5. In tailing, adenylate residues (about 200-300) are added at the 3' end.
6. The fully processed mRNA is released from the nucleus into the cytoplasm.