Answer

(b) (In a DNA strand the nucleotides are linked together by $3^{\prime}-5$ ’ phosphodiester linkage (bonds) to form a dinucleotide. More nucleotides can be joined in such a manner to form a polynucleotide chain.

• Glycosidic bonds: These bonds are typically found in carbohydrates, linking sugar molecules together. They are not involved in linking nucleotides in a DNA strand.

• Peptide bonds: These bonds are found in proteins, linking amino acids together. They do not play a role in the structure of DNA.

• Hydrogen bonds: These bonds occur between complementary nitrogenous bases (adenine-thymine and guanine-cytosine) in the DNA double helix, but they do not link nucleotides together within a single strand.

Answer

(c) A nitrogenous base is attached to the pentose sugar by an $\mathrm{N}$-glycosidic linkage to form a nucleoside, i.e., Nucleoside=Nitrogen base+Pentose sugar.

When a phosphate group is attached to the $5^{\prime}-\mathrm{OH}$ of a nucleoside through phosphodiester linkage, a nucleotide is formed, i.e., Nucleotide $=$ Nitrogen base + Pentose sugar + Phosphate $\left(\mathrm{PO}_{4}\right)$.

So, a nucleoside differs from a nucleotide as it lacks the phosphate group.

• (a) A nucleoside contains a nitrogenous base, so it does not lack the base.
• (b) A nucleoside contains a pentose sugar, so it does not lack the sugar.
• (d) A nucleoside does not specifically lack a hydroxyl group; the presence of hydroxyl groups depends on the type of pentose sugar (ribose or deoxyribose) in the nucleoside.

Answer

(c) Both deoxyribose and ribose belong to the class pentoses as it contains ’ $\mathrm{S}$ ’ carbon atoms.

Structure of (a) deoxyribose (b) ribose sugar

• (a) trioses: Trioses are sugars that contain three carbon atoms. Both deoxyribose and ribose contain five carbon atoms, not three, so they do not belong to the class of trioses.

• (b) hexoses: Hexoses are sugars that contain six carbon atoms. Both deoxyribose and ribose contain five carbon atoms, not six, so they do not belong to the class of hexoses.

• (d) polysaccharides: Polysaccharides are long chains of monosaccharide units linked together. Deoxyribose and ribose are individual monosaccharides, not long chains, so they do not belong to the class of polysaccharides.

Answer

(c) The diameter of the strand is always constant due to a pairing of purine (adenine and guanine) and pyrimidine (cytosine and thymine). This specific bonding gives uniform width to the DNA.

• (a) The antiparallel nature of DNA refers to the orientation of the two strands running in opposite directions (5’ to 3’ and 3’ to 5’). This is not directly related to the pairing of purines with pyrimidines.

• (b) The semiconservative nature of DNA replication means that each new DNA molecule consists of one old strand and one new strand. This concept is related to the mechanism of DNA replication, not the specific pairing of purines and pyrimidines.

• (d) Uniform length in all DNA is not a characteristic of DNA structure. DNA molecules can vary greatly in length depending on the organism and the specific DNA sequence. The pairing of purines with pyrimidines does not influence the overall length of the DNA molecule.

Answer

(c) DNA consists of a nitrogenous base, pentose sugar and a phosphate group. DNA has negative charge due to the presence of phosphate group $\left(\mathrm{PO}_{4}{ }^{3-}\right)$.

Histones are rich in the basic amino acid residues lysines and arginines, which carry positive charges in their side chains. Therefore, histones are positively charged.

• (a) both positive: This option is incorrect because DNA has a negative charge due to the presence of phosphate groups ($\mathrm{PO}_{4}{ }^{3-}$). While histones are positively charged due to the presence of basic amino acid residues like lysines and arginines, DNA cannot be positively charged.

• (b) both negative: This option is incorrect because histones are not negatively charged. Histones are rich in basic amino acids such as lysines and arginines, which carry positive charges. Therefore, histones are positively charged, not negatively charged.

• (d) zero: This option is incorrect because the net electric charge on DNA and histones is not zero. DNA has a negative charge due to its phosphate groups, and histones have a positive charge due to their basic amino acid residues. The charges do not cancel each other out to result in a net zero charge.

Answer

(c) The promoter is the binding site for RNA polymerase for initiation of transcription.

The promoter is located towards $5^{\prime}$-end (upstream) of the structural gene of coding strands and provides the binding site for RNA polymerase.

• (a) The promoter site is not located at the $3^{\prime}$ (downstream) end of the transcription unit; it is located at the $5^{\prime}$ (upstream) end. The terminator site is at the $3^{\prime}$ (downstream) end, not the $5^{\prime}$ (upstream) end.

• (b) This option is actually correct, as the promoter is at the $5^{\prime}$ (upstream) end and the terminator is at the $3^{\prime}$ (downstream) end of the transcription unit.

• (d) The promoter site is not located at the $3^{\prime}$ (downstream) end; it is located at the $5^{\prime}$ (upstream) end. The $3^{\prime}$ (downstream) end is where the terminator site is located.

Answer

(d) Sickle-cell anaemia is an autosome linked recessive trait. In this genetic disorder point mutation in $\beta$-globin chain results in change of glutamate (glutamic acid) to valine at sixth position. Only the homozygous individuals for $\mathrm{Hb}^{\mathrm{s}}$ 2, i.e., $\mathrm{Hb}^{\mathrm{s}} \mathrm{Hb}^{\mathrm{s}}$ show the diseased phenotype. The heterozygous individuals $\left(\mathrm{Hb}^{S} / \mathrm{Hb}^{A}\right)$ are carriers.

It is also known that heterozygotes, having both types of haemoglobin, show resistance to malaria infection because the body targets the $P$. falciparum (protozoan) infected cells for destruction of RBC.

• (a) It cannot be treated with iron supplements: This statement is actually correct. Sickle-cell anaemia is not caused by iron deficiency, so iron supplements are not an effective treatment. Therefore, this option is not incorrect.

• (b) It is a molecular disease: This statement is also correct. Sickle-cell anaemia is indeed a molecular disease caused by a point mutation in the $\beta$-globin chain of hemoglobin. Therefore, this option is not incorrect.

• (c) It confers resistance to acquiring malaria: This statement is correct as well. Heterozygous individuals (carriers) of the sickle-cell trait have a selective advantage in malaria-endemic regions because they show resistance to malaria infection. Therefore, this option is not incorrect.

Thinking Process

Three adjacent nitrogenous bases constitute a codon which specifies the placement of one amino acid in a polypeptide.

Answer

(d) Polypeptide synthesis is signalled by two initiation codons commonly AUG or methionine codon and rarely GUG or valine codon. Since there are 64 triplet codons and only 20 amino acids, the incorporation of some amino acids must be influenced by more than one codon.

Only tryptophan (UGG) and methionine (AUG) are specified by single codons. AUG codes for methionine in both prokaryotes and eukaryotes.

• (a) The statement “it codes for methionine only” is not incorrect in itself, but it is incomplete because it does not mention that AUG is also an initiation codon.

• (b) The statement “it is also an initiation codon” is not incorrect in itself, but it is incomplete because it does not mention that AUG codes for methionine in both prokaryotes and eukaryotes.

• (c) The statement “it codes for methionine in both prokaryotes and eukaryotes” is not incorrect in itself, but it is incomplete because it does not mention that AUG is also an initiation codon.

Answer

(d) RNA was the first genetic material. There is now enough evidence to suggest that essential life processes (such as metabolism, translation, splicing, etc.), evolved around RNA.

RNA used to act as a genetic material as well as a catalyst (there are some important biochemical reactions in living systems that are catalysed by RNA catalysts and not by protein enzymes). But, RNA being a catalyst was reactive and hence unstable.

Therefore, DNA has evolved from RNA with chemical modifications that make it more stable. DNA being double-stranded and having complementary strand further resists changes by evolving a process of repair.

• Protein: Proteins are essential for various cellular functions, including catalysis of biochemical reactions, structural support, and signaling. However, they do not have the capability to store and transmit genetic information, which is a fundamental requirement for the first genetic material.

• Carbohydrates: Carbohydrates primarily serve as energy sources and structural components in cells. They lack the ability to store genetic information and do not have the necessary structural complexity to perform the role of genetic material.

• DNA: DNA is the primary genetic material in most current life forms due to its stability and ability to store genetic information. However, it is believed to have evolved from RNA. DNA requires complex enzymes for replication and is not capable of catalyzing reactions on its own, which suggests it was not the first genetic material.

Answer

(b) In eukaryotes, the monocistronic structural genes have interrupted coding sequences i.e., the genes in eukaryotes are split. The coding sequences or expressed sequences are defined as exons.

These sequences (exons) appear in mature or processed RNA. The exons are interrupted by introns or intervening sequences which do not appear in mature or processed RNA.

• (a) Exons and introns do not appear in the mature RNA: This is incorrect because exons, which are the coding sequences, do appear in the mature mRNA after the introns are removed during RNA processing.

• (c) Introns appear but exons do not appear in the mature RNA: This is incorrect because introns are non-coding sequences that are spliced out during RNA processing, and exons, which are the coding sequences, remain in the mature mRNA.

• (d) Both exons and introns appear in the mature RNA: This is incorrect because introns are removed during the RNA splicing process, leaving only exons in the mature mRNA.

Answer

(c) In humans, chromosome 1 has highest genes (2968 approx.) and the $Y$ has the fewest (231 approx.) genes.

• Option (a): Chromosome 21 does not have the highest number of genes. It has one of the lowest gene counts among human chromosomes. The Y chromosome does have the fewest genes, but chromosome 21 is not the one with the highest number of genes.

• Option (b): Chromosome 1 does indeed have the highest number of genes, but the X chromosome does not have the fewest. The Y chromosome has fewer genes than the X chromosome.

• Option (d): The X chromosome does not have the highest number of genes. While it has more genes than the Y chromosome, it does not surpass chromosome 1 in gene count. The Y chromosome does have the fewest genes, but the X chromosome is not the one with the highest number of genes.

Answer

(d) It was only in 1953 that James Watson and Francis Crick, based on the X-ray diffraction data produced by Maurice Wilkins and Rasalind Franklin, proposed a very simple but famous double helix model for the structure of DNA.

Erwin Chargaff observed that for a double-stranded DNA, the ratios between adenine and thymine and guanine and cytosine are constant and equals one.

On the other bond Matthew Meselson and Franklin Stahl in 1958 performed experiments on E.coli to prove that DNA replication is semi-conservative. But had no contribution it the development of double helix model.

• Rosalind Franklin: She produced critical X-ray diffraction images of DNA, which provided key insights into the helical structure of DNA. Her work was fundamental to the development of the double helix model.

• Maurice Wilkins: He also worked on X-ray diffraction studies of DNA and shared his findings with Watson and Crick, which helped them in formulating the double helix model.

• Erwin Chargaff: He discovered the base pairing rules, known as Chargaff’s rules, which state that in double-stranded DNA, the amount of adenine equals thymine and the amount of guanine equals cytosine. This was crucial for understanding the structure of DNA.

Answer

(b) The enzyme called DNA polymerase progressively adds deoxyribonucleotides to the free $3^{\prime}$-end of the growing polynucleotide chain so, that replication of the $3^{\prime}-5^{\prime}$ strand of the DNA molecule is continuous (growth of the new strand in $5^{\prime} \rightarrow 3^{\prime}$ direction).

So, to prevent polymerisation of nucleotides $3^{\prime} \mathrm{OH}$ group in deoxyribose should be replaced/removed.

• (a) Replace purine with pyrimidines: This modification would not prevent polymerization because the polymerization process is dependent on the formation of phosphodiester bonds between the sugar and phosphate groups of nucleotides, not on the specific type of nitrogenous base (purine or pyrimidine) present.

• (c) Remove/Replace 2’ $\mathrm{OH}$ group with some other group in deoxy ribose: The 2’ $\mathrm{OH}$ group is already absent in deoxyribose (which distinguishes it from ribose in RNA). Modifying the 2’ position would not affect the polymerization process, which relies on the 3’ $\mathrm{OH}$ group for the formation of the phosphodiester bond.

• (d) Both (b) and (c): While removing/replacing the 3’ $\mathrm{OH}$ group would prevent polymerization, modifying the 2’ $\mathrm{OH}$ group is irrelevant in the context of DNA polymerization. Therefore, this option is partially correct but includes an unnecessary modification.

Thinking Process

The replication of $3^{\prime} \rightarrow 5^{\prime}$ strand is continuous and it is called leading strand, while the replication of second strand ( $5^{\prime} \rightarrow 3^{\prime}$ strand) of the DNA molecules is discontinuous and it is known as the lagging strand.

Answer

(b) DNA polymerase adds deoxyribonucleotides to the free 3 ‘-end of the growing polynucleotide chain so, that replication of the $3^{\prime} \rightarrow 5^{\prime}$ strand of the DNA molecule is continuous (growth of the new strand in $5^{\prime} \rightarrow 3^{\prime}$ direction)

Since, DNA dependent DNA polymerase catalyses polymerisation only in one direction $\left(5^{\prime} \rightarrow 3^{\prime}\right)$, discontinuous synthesis of DNA occurs in the other strand.

• (a) DNA molecule being synthesised is very long: The length of the DNA molecule does not determine whether synthesis is continuous or discontinuous. The key factor is the directionality of DNA polymerase activity, not the length of the DNA.

• (c) it is a more efficient process: Discontinuous synthesis is not necessarily more efficient. It is a consequence of the unidirectional activity of DNA polymerase and the antiparallel nature of DNA strands, not an efficiency consideration.

• (d) DNA ligase has to have a role: While DNA ligase is involved in joining Okazaki fragments on the lagging strand, its role is a consequence of discontinuous synthesis, not a cause of it. The primary reason for discontinuous synthesis is the directionality of DNA polymerase.

Answer

(b) The DNA dependent RNA polymerase helps in DNA replication by catalysing the polymerisation in only one direction, i.e., $5^{\prime} \rightarrow 3^{\prime}$.

• (a) Initiation: While RNA polymerase is involved in the initiation of transcription, it is not the only factor required. Initiation also involves other proteins and factors such as sigma factors in prokaryotes or transcription factors in eukaryotes that help the RNA polymerase to recognize the promoter region and start transcription.

• (c) Termination: Termination of transcription involves specific sequences in the DNA and sometimes additional proteins (like rho factor in prokaryotes) that signal the RNA polymerase to stop transcription. RNA polymerase alone does not catalyze the termination process.

• (d) All of these: This option is incorrect because, as mentioned, RNA polymerase is not solely responsible for all steps of transcription. While it plays a crucial role in elongation, initiation and termination require additional factors and sequences.

Thinking Process

Regulation of gene expression refers to a very broad term that may occur at various levels.

Answer

(b) Considering that gene expression results in the formation of a polypeptide, it can be regulated at several levels. In eukaryotes, the regulation could be exerted at

(i) transcriptional level (formation of primary transcript)

(ii) processing level (regulation of splicing)

(iii) transport of $m \mathrm{RNA}$ from nucleus to the cytoplasm

(iv) translational level

While, in prokaryotes, control of the rate of transcriptional initiation is the predominant site for control of gene expression.

• (a) DNA-replication: Control of gene expression does not primarily occur at the level of DNA replication. DNA replication is the process of copying the entire genome before cell division, and while it is crucial for cell proliferation, it does not directly regulate which genes are expressed or how much they are expressed.

• (c) translation: While translation is a step in gene expression where mRNA is decoded to produce a polypeptide, it is not the primary level at which gene expression is controlled. In both prokaryotes and eukaryotes, the regulation of gene expression predominantly occurs at the transcriptional level, which determines whether an mRNA is produced in the first place.

• (d) None of these: This option is incorrect because gene expression is indeed regulated, and the primary level of regulation is at the transcriptional level, as stated in the provided answer.

Thinking Process

Regulatory protein is a term used in genetics to describe a protein involved in regulating gene expressions. There are often needed to switch a gene on (activator) or to switch off a gene (represssor).

Answer

(d) Regulatory sequences (proteins) control the functions of structural genes and are at called regulatory genes. The important regulatory genes are promoters, terminators, operators and repressor.

To regulate the process of transcription, transcription factors (a sequence of specific DNA-binding factor) alone or with other proteins, promoter (as on activator) or stop as a repress or the binding site of RNA polymerase to DNA.

• (a) They only increase expression: This statement is incorrect because regulatory proteins can also decrease expression by acting as repressors, not just increase it.

• (b) They only decrease expression: This statement is incorrect because regulatory proteins can also increase expression by acting as activators, not just decrease it.

• (c) They interact with RNA polymerase but do not affect the expression: This statement is incorrect because the primary role of regulatory proteins is to affect gene expression by either increasing or decreasing the transcription process.

Answer

(a) Chromosome 1 was the last completed chromosome, sequenced two decades after the beginning of the human Genome Project (hGP). It is the designation for the largest human chromosome.

• Chromosome 11: This chromosome was sequenced earlier than Chromosome 1. It is not the largest human chromosome and was not the last to be completed in the Human Genome Project.

• Chromosome 21: This chromosome is one of the smallest human chromosomes and was sequenced before Chromosome 1. It was not the last to be completed.

• Chromosome-X: Although significant due to its role in sex determination and various genetic conditions, Chromosome-X was sequenced before Chromosome 1 and was not the last human chromosome to be completely sequenced.

Thinking Process

RNA is a single chain polyribonucleotide which functions as carrier of coded genetic or hereditary information from DNA to cytoplasm for taking part in protein and enzyme synthesis.

Answer

(d) $r \mathrm{RNA}, m \mathrm{RNA}$ and $t \mathrm{RNA}$ are major classes of RNAs that are involved in gene expression. $r$ RNAs bind protein molecules and give rise to ribosomes. mRNA carries coded information for translation into polypeptide formation. tRNA is called soluble or adaptor RNA and carries amino acids to $m$ RNA during protein synthesis.

• Option (a) is incorrect because it only mentions the role of mRNA in carrying genetic information from DNA to ribosomes, but does not account for the roles of rRNA and tRNA.
• Option (b) is incorrect because it only describes the function of tRNA in carrying amino acids to ribosomes, but does not include the functions of mRNA and rRNA.
• Option (c) is incorrect because it only highlights the role of rRNA as a constituent component of ribosomes, but does not consider the functions of mRNA and tRNA.

Answer

(b) According to Chargaff’s rules of base pairing,

(i) The amount of adenine is always equal to the amount of thymine and the amount of guanine is always equal to the amount of cytosine.

(ii) Adenine is joined to thymine with two hydrogen bonds and guanine is joined to cytosine by three hydrogen bonds.

(iii) The ratio of adenine to thymine and that of guanine to cytosine is always equal to one,

i.e.,

$$ \frac{A}{T}=\frac{G}{C}=1 $$

In the given organism, the DNA is not following the Chargaff’s rule, hence it can be concluded that it is a single-stranded DNA, not double-stranded.

• (a) it is a double-stranded circular DNA: This option is incorrect because, according to Chargaff’s rules, in double-stranded DNA, the amount of adenine (A) should be equal to the amount of thymine (T), and the amount of guanine (G) should be equal to the amount of cytosine (C). In the given data, the proportions of A and T are not equal (A = 29%, T = 17%), and the proportions of G and C are also not equal (G = 17%, C = 32%). Therefore, it cannot be double-stranded DNA, whether circular or linear.

• (c) it is a double-stranded linear DNA: This option is incorrect for the same reason as option (a). The given proportions of nucleotides do not follow Chargaff’s rules, which are applicable to all double-stranded DNA, whether circular or linear. The unequal proportions of A and T, and G and C, indicate that the DNA is not double-stranded.

• (d) No conclusion can be drawn: This option is incorrect because a conclusion can indeed be drawn based on the given data. The deviation from Chargaff’s rules clearly indicates that the DNA is single-stranded. Therefore, it is possible to conclude that the DNA is single-stranded, making this option incorrect.

Answer

(c) In some viruses, like retroviruses (e.g., HIV), an enzyme called reverse transcriptase is used to generate complementary DNA (cDNA) from an RNA template. This process is termed reverse transcription.

• (a) A-DNA: A-DNA is a right-handed double helix structure of DNA, which is one of the many possible double helical structures of DNA. It is not related to the process of synthesizing DNA from an RNA template.

• (b) B-DNA: B-DNA is the most common form of DNA in cells, characterized by its right-handed helix. It is the standard form of DNA but does not specifically refer to DNA synthesized from an RNA template.

• (d) $r$ DNA: Recombinant DNA (rDNA) refers to DNA molecules formed by laboratory methods of genetic recombination to bring together genetic material from multiple sources. It is not related to the natural process of synthesizing DNA from an RNA template in viruses.

Answer

(d) Meselson and Stahl found that DNA of the first generation was hybrid or intermediate $\left({ }^{15} \mathrm{~N}\right.$ and $\left.{ }^{14} \mathrm{~N}\right)$. It settled in caesium chloride at a level higher than the fully labelled DNA of parent bacteria $\left({ }^{15} \mathrm{~N}^{15} \mathrm{~N}\right)$. The second generation of bacteria after 40 minutes, contained two types of DNA, $50 \%$ light $\left(\mathrm{N}^{14} \mathrm{~N}^{14}\right)$ and $50 \%$ intermediate $\left(\mathrm{N}^{15} \mathrm{~N}^{14}\right)$.

The third generation of bacteria after 60 minutes contained two types of DNA, $25 \%$ intermediate $\left(\mathrm{N}^{15} \mathrm{~N}^{14}\right)$ and $75 \%$ light $\left(\mathrm{N}^{14} \mathrm{~N}^{14}\right)$ in 1:3 ratio. The fourth generation after 80 minutes contained $12.5 \% \mathrm{~N}^{15} \mathrm{~N}^{14}$ and $87.5 \% \mathrm{~N}^{14} \mathrm{~N}^{14}$ DNA in $1: 7$ ratio.

• Option (a) $1: 1: 0$: This option is incorrect because it suggests that there is an equal amount of $15_{N} / 15_{N}$ and $15_{N} / 14_{N}$ DNA, and no $14_{N} / 14_{N}$ DNA. However, after four generations, the amount of $14_{N} / 14_{N}$ DNA should be significantly higher than the hybrid $15_{N} / 14_{N}$ DNA, and there should be no $15_{N} / 15_{N}$ DNA left.

• Option (b) $1: 4: 0$: This option is incorrect because it suggests that there is a significant amount of $15_{N} / 15_{N}$ DNA and a higher amount of $15_{N} / 14_{N}$ DNA, with no $14_{N} / 14_{N}$ DNA. After four generations, the $15_{N} / 15_{N}$ DNA should be completely absent, and the majority of the DNA should be $14_{N} / 14_{N}$.

• Option (c) $0: 1: 3$: This option is incorrect because it suggests that there is no $15_{N} / 15_{N}$ DNA, a small amount of $15_{N} / 14_{N}$ DNA, and a larger amount of $14_{N} / 14_{N}$ DNA. While it correctly indicates the absence of $15_{N} / 15_{N}$ DNA, the ratio of $15_{N} / 14_{N}$ to $14_{N} / 14_{N}$ DNA should be $1: 7$ after four generations, not $1: 3$.

Answer

(a) 5’ - A T G A A T G - 3’ (coding strand)

$\downarrow$

5’- T A C T T A C - 3’ (complementary strand)

$\downarrow$

5’ -A U GA A $\cup G-3^{\prime}(R N A)$

• Option (b) is incorrect because the sequence $5^{\prime}-\cup A C \cup \cup A C-3^{\prime}$ does not match the expected RNA transcript sequence derived from the given DNA coding strand. The correct RNA sequence should be complementary to the DNA template strand and should use uracil (U) instead of thymine (T).

• Option (c) is incorrect because the sequence $5^{\prime}$ - $\mathrm{C}$ A $\cup \cup \mathrm{CA} \cup-3^{\prime}$ does not match the expected RNA transcript sequence. The RNA sequence should be complementary to the DNA template strand and should use uracil (U) instead of thymine (T).

• Option (d) is incorrect because the sequence $5^{\prime}-\mathrm{G} \cup \mathrm{A} A \mathrm{G} \cup \mathrm{A}-3^{\prime}$ does not match the expected RNA transcript sequence. The RNA sequence should be complementary to the DNA template strand and should use uracil (U) instead of thymine (T).

Thinking Process

In prokaryotes, the structural gene is polycistronic and continuous. In bacteria (prokaryotes), the transcription of all the three types of RNA (mRNA, tRNA and rRNA) is catalysed by single DNA dependent enzyme, called the RNA polymerase.

Answer

(c) In $E$. coli bacterium, the RNA polymerase has co-factors $\beta, \beta^{\prime}, \alpha, \alpha^{\prime}$ and $\omega$ along with $\sigma$ (sigma) factor, to catalyse the process. The transcription is completed in three steps. Initiation $\sigma$ (sigma) factor recognises the start signal and promotor region on DNA which then along with RNA polymerase binds to the promoter to initiate transcription.

Transcription start site

Schematic structure of a transcription unit

Elongation The RNA polymerase after initiation of RNA transcription loses the $\sigma$ factor but continues the polymerisation of ribonucleotides to form RNA.

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 separates. This process is called termination which is facilitated by a termination factor $\rho$ (rho)

In prokaryotes, mRNA does not require any processing, so both transcription and translation occur in the cytosol. It can be said that transcription and translation are coupled together

Representation of initiation, elongation and termination are as given

Termination

Process of transcription in bacteria

• Option (a) the promoter, structural gene and the terminator region:

  • The RNA polymerase holoenzyme does not transcribe the promoter region. The promoter is a specific DNA sequence where RNA polymerase binds to initiate transcription, but it is not transcribed into RNA. The transcription process starts at the transcription start site, which is downstream of the promoter.

• Option (b) the promoter and the terminator region:

  • Similar to option (a), the promoter region is not transcribed. The RNA polymerase holoenzyme binds to the promoter to initiate transcription but does not transcribe it. Additionally, the terminator region is transcribed, but the promoter is not.

• Option (d) the structural gene only:

  • This option is incorrect because the RNA polymerase holoenzyme transcribes both the structural gene and the terminator region. The terminator region is essential for signaling the end of transcription, and it is transcribed into RNA before the transcription process is terminated.

Answer

(a) $5^{\prime}-\mathrm{A} \cup \mathrm{G}-3^{\prime}$ (codon in $m \mathrm{RNA}$ )

$5^{\prime}-\cup|| \mid$

• Option (b) 5’ - CAU - 3’: This option is incorrect because the sequence is written in the 5’ to 3’ direction, but the tRNA anticodon should be complementary and antiparallel to the mRNA codon. The correct anticodon sequence should be written in the 3’ to 5’ direction to pair with the mRNA codon 5’ - AUG - 3'.

• Option (c) $5^{\prime}-\mathrm{A} \cup \mathrm{G}-3^{\prime}$: This option is incorrect because it does not represent a valid nucleotide sequence. The notation “A ∪ G” is not a standard representation of a nucleotide sequence and does not correspond to any known codon or anticodon.

• Option (d) 5’ - GUA - 3’: This option is incorrect because the sequence is not complementary to the mRNA codon 5’ - AUG - 3’. The correct anticodon sequence should be complementary to AUG, which would be UAC, but written in the 3’ to 5’ direction.

Answer

(b) AA-binding site (amino acid binding site) lies at the $3^{\prime}$ end opposite the anticodon and has CCA-OH group. It is the site where amino acid attaches to the tRNA.

• (a) The $5^{\prime}$-end of the tRNA is not the site where the amino acid attaches. The $5^{\prime}$-end typically has a phosphate group and is involved in the overall structure of the tRNA molecule, but it does not play a role in amino acid attachment.

• (c) The anticodon site is a region of the tRNA that contains a specific sequence of three nucleotides complementary to the mRNA codon. This site is crucial for the recognition of the mRNA codon during translation, but it is not involved in the attachment of the amino acid.

• (d) The DHU loop (D-arm) is a region of the tRNA that contains dihydrouridine residues. It is involved in the proper folding and stability of the tRNA molecule, but it does not serve as the attachment site for the amino acid.

Thinking Process

The cellular factory responsible for synthesising proteins is the ribosome.

Answer

(a) The ribosome consists of structural RNAs and about 80 different proteins. In its inactive state, it exist as two subunits, a large subunit and a small subunit. When the smaller subunit encounters the mRNA, the process of translation of the mRNA to protein begins .

• (b) The larger ribosomal sub-unit: The larger ribosomal subunit does not initiate the translation process. It joins the smaller subunit only after the mRNA has bound to the smaller subunit and the initiator tRNA has been positioned correctly.

• (c) The whole ribosome: The whole ribosome does not bind to the mRNA initially. Translation initiation begins with the binding of the mRNA to the smaller ribosomal subunit, followed by the assembly of the larger subunit to form the complete ribosome.

• (d) No such specificity exists: This statement is incorrect because there is a specific sequence of events in translation initiation. The mRNA specifically binds to the smaller ribosomal subunit first, which is a crucial step in the initiation process.

Answer

(a)

Jacob and Monod model of an inducible operon

In case of lactose presence

(i) Lactose acts as an inducer which binds to the repressor and forms an inactive repressor.

(ii) The repressor fails to bind to the operator region.

(iii) The RNA polymerase binds to the operator and transcript lac $m R N A$.

(iv) lac mRNA is polycistronic, i.e., produces all three enzymes, $\beta$-galactosidase, permease and transacetylase.

(v) The lac operon is switched on.

In case of lactose absence

(i) When lactose is absent, $i$ gene regulates and produces repressor mRNA which translate repression.

(ii) The repressor protein binds to the operator region of the operon and as a result prevents RNA polymerase to bind to the operon.

(iii) The operon is switched off.

• (b) Repressor binds to operator: This option is incorrect because when the repressor binds to the operator, it prevents RNA polymerase from binding to the promoter and transcribing the lac operon genes. Therefore, the lac operon is switched off, not on.

• (c) RNA polymerase binds to the operator: This option is incorrect because RNA polymerase does not bind to the operator; it binds to the promoter region. The operator is the site where the repressor binds. When the repressor is bound to the operator, RNA polymerase cannot proceed with transcription.

• (d) Lactose is present and it binds to RNA polymerase: This option is incorrect because lactose does not bind to RNA polymerase. Instead, lactose binds to the repressor, causing a conformational change that prevents the repressor from binding to the operator, thereby allowing RNA polymerase to transcribe the lac operon genes.

Answer

Functions of histones in DNA packaging are

(i) Histones as units of octamer participate in primary packaging of DNA.

(ii) Basic histone proteins neutralise the acidic DNA molecule.

Answer

Densely packed and dark stained chromatin regions are called hetorochromation, while, loosely packed light stained regions are called euchromatin.

Euchromation is transcriptionally active and is transcribed into mRNA. Due to very tight coiling heterochromatin can not be transcribed and is inert/inactive form.

Answer

In bacteria, three types of DNA polymerases are there. All of them can add nucleotides in $5^{\prime} \rightarrow 3^{\prime}$ direction. They process exonuclease activity as well. DNA polymerase III can proofread the newly synthesised strand and senses the wrong base insertions.

It deletes wrong bases and helps correct the mistake by putting in the right one, DNA polymerase. The only mistake it cannot corrects substitution of uracil in place of thymin.

It can repair any damages done to DNA by UV exposure, etc., or the left over proofreading mistakes. It detects mutation caused by UV, removes mismatched pairs and puts back the right ones.

Answer

Synthesis of DNA always takes place in $5^{\prime} \rightarrow 3^{\prime}$ direction. In a double stranded DNA both strands are anti parallel and complementary. During DNA synthesis as both strands act as templates, only one strand, i.e., $3^{\prime} \rightarrow 5^{\prime}$ can synthesis complementary strand in $5^{\prime} \rightarrow 3^{\prime}$ direction.

The other strand, i.e., $5^{\prime} \rightarrow 3^{\prime}$ has to be synthesised in small stretches in opposite direction as replication fork moves to right. That is why DNA synthesis is discontinuous on one of the parental strands of DNA. These small stretches called Okazaki fragments are joined together by DNA ligase enzyme that closes the nicks.

Answer

According to base complementary rules,

(a) 5’TTACGTCGATAATCC-3'

(b) 5’CGAUUAUCGACGUAA-3'

RNA uses the base uracil $(U)$ rather than thymine $(T)$. So, in RNA the base pairs are

Adenine $(A)$ pairs with uracil $(U)$

Guanine (G) pairs with cytosine (C).

Answer

DNA polymorphism refers to the variation in DNA arising through mutation at non-coding sequences.

A special type of polymorphism, called VNTR (Variable Number of Tandem Repeats), is composed of repeated copies of a DNA sequence that lie adjacent to one another on the chromosome. Since, polymorphism is the basis of genetic mapping of human genome, therefore, it forms the basis of DNA fingerprinting too.

The single nucleotide polymorphisms are used in locating diseases and tracing of human history as well as in case of paternity testing.

Thinking Process

It is the case of sickle-cell anaemia.

Answer

Due to point mutation in $\beta$-globin chain of haemoglobin molecule, glutamic acid (Glu) is replaced by valine ( $\mathrm{Val}$ ) at the sixth position.

Under stress condition erythrocytes lose their circular shape and become sickle-shaped. As a result, the cells cannot pass through narrow capillaries. Blood capillaries are clogged and thus affect blood supply to different organs.

Answer

Sometimes cattle or even human beings give birth to their young ones that are having extremely different of organs like limbs/position of eye etc. It happens due to the disturbance in coordinated regulation of expression in sets of genes, which are associated with organ development.

Answer

DNA polymerase enzyme is highly specific to recognise only deoxy ribonucleoside triphosphates. Therefore, it cannot hold RNA $\beta$-nucleotides.

Answer

The enzymes involved in DNA replication other than DNA polymerase and ligase are listed below with their functions.

(i) Helicase - Opens the helix

(ii) Topoisomerases - Removes the super coiling of DNA

(iii) Primase — Synthesises RNA primer

(iv) Telomerase - To synthesis the DNA of telomeric end of chromosomes.

Answer

In some viruses, RNA is the genetic material.

e.g., Tobacco mosaic viruscs, QB bacteriophage, HIV, influenza virus, etc.

Short Answer Type Questions

Answer

In Griffith’s experiment, transformation can be defined as a change in the genetic constitution of an organism by picking out up DNA from the environment (from dead organisms).

Transformation helps in identification of DNA as a genetic material. When heat was used to kill the virulent bacteria, they died but not their genetic material (DNA). This DNA when picked up by non-virulent bacteria made them capable of causing infection.

Since, ability to cause infection could be passed on by these organisms to their progeny, it was concluded that DNA was the material that was inherited.

Answer

Oswald, Avery, Colin MacLeod and Maclyn McCarty revealed biochemical nature of the transforming principle.

They reported Griffith’s experiment in an in vitro system in order to determine biochemical nature of transforming principle.

They reported that DNA from the heat-killed S-type bacteria caused the transformation of non-virulent R-type bacteria into virulent S-type bacteria. They also discovered that proteases and RNase did not affect transformation while DNase inhibited the process. They concluded that DNA is the hereditary material.

Thinking Process

Meselson and Stahl used heavy isotope of ${ }^{15} \mathrm{~N}$ in the nutrient medium to grow Escherichia coli (E. coli), for several generations.

Answer

They performed experiments on $E$. coli to prove that DNA replication is semi-conservative. They first grew the bacteria in a medium containing ${ }^{15} \mathrm{NH}_{4} \mathrm{Cl}$ (in which ${ }^{15} \mathrm{~N}$ is the heavy istope of nitrogen) for many generations.

Then they transferred the cells into a medium with normal ${ }^{14} \mathrm{NH}_{4} \mathrm{Cl}$ (in which ${ }^{14} \mathrm{~N}$ is the lighter isotope) and took the samples at various definite time intervals as the cells multiplied. The extracted DNAs were centrifuged and measured to get their densities.

The DNA extracted from the culture after one generation of transfer from then ${ }^{15} \mathrm{~N}$ mediun to ${ }^{14} \mathrm{~N}$ mediun, (i.e., after 20 minutes E.coli divides every 20 minutes) showed an intermediate hybrid density, i.e., both heavy and light nitrogen, which proved the semi-conservative nature of DNA.

Answer

A cistron is stretch of base sequences that codes for one polypeptide chain including adjacent control regions. It may also code for a tRNA, $r$ RNA molecule or may perform other specific functions including regulating functions of other cistrons.

This term has replaced the definition of a gene. Monocistronic transcription unit will have all the regulatory and coding sequences for a single polypeptide, whereas polycistronic may have coding sequences for more than one polypeptide.

In eukaryotic cells almost all the messenger RNAs are monocistronic. In prokaryotes, lac operon coding sequence would be an example of polycistronic DNA region.

Answer

Salient features of human genome

(i) The human genome contains 3164.7 million nucleotide bases.

(ii) The average gene consists of 30000 the largest know human gene being dystrophin at 2.4 Million bases.

(iii) The total number of genes is estimated to be 30000 and $99.9 \%$ nucleotide bases are exactly the same in all people.

(iv) The functions are unknown for over $50 \%$ of the discovered genes.

(v) Less than $2 \%$ of the genome codes for proteins.

(vi) The human genome contains large repeated sequences.

(vii) The repeated sequence is thought to have no direct coding functions but they throw light on chromosome structures, dynamics and evolution.

(viii) Chromosome I has most genes (2968) and the $\mathrm{Y}$ has the fewest genes (231).

(ix) Scientists have identified about 1.4 million locations where single base DNA sequence differences called SNPs or Single Nucleotide Polymorphisms occur in humans.

Answer

While replicating, the entire DNA molecule to keep the whole molecule stabilised does not open in one go because it would be highly expens energetically. Actually unwiding creates tension in the molecule as uncoiled parts

Actually, unwinding creates tension in the molecule as uncoiled parts start forming super coils due to the interaction of exposed nucleotides.

Instead, helicase enzyme acts on the double strand at ori site (origin of replication) and a small stretch is unzipped. Immediately, it is held and stabilised by single strand binding proteins

Slowly with the help of enzymes, exposed strands are copied as a point of unwinding moves and ahead in both directions.

It gives an appearance of Y-shaped structure which is called replication fork.

The two functions that the monomer units of NTPS play are

(i) They pair up with exposed nucleotides of the template strand and make phosphodiester linkages and release a pyrophosphate.

(ii) Hydrolysis of this pyrophosphate by enzyme pyrophosphatase releases energy that will facilitate making hydrogen bonds between free nucleotides and bases of the template strand.

Answer

Retroviruses do not follow central dogma of biology (DNA $\rightarrow$ RNA $\rightarrow$ Protein) because their genetic material is not DNA. Instead they have RNA that is converted to DNA by the enzyme reverse transcriptase.

Answer

The length of DNA double helix $=2 \times 10^{9} \times 0.44 \times 10^{-9} / \mathrm{bp}$.

Answer

If histones were mutated and made rich in acidic amino acids. They will not be able to serve the purpose of keeping the DNA coiled around them. This is because DNA is negatively charged molecule and histones are positively charged because of basic amino acids.

So, they are attracted to each other. If histones become negatively charged, instead of binding, they will rather repel DNA. The packaging of DNA in eukaryotes would not happen. Consequently, the chromatin fibre would not be formed.

Answer

RNA is more liable and prone to degradation (owing to the presence of 2’OH group in its ribose). Hence, heat-killed S-stain may not have retained its ability to transform the R-strain into virulent form if RNA was its genetic material.

Answer

Use of ${ }^{15} \mathrm{~N}$ will be inappropriate because method of detection of ${ }^{32} \mathrm{P}$ and ${ }^{15} \mathrm{~N}$ different ${ }^{32} \mathrm{P}$ being a radioactive isotope while ${ }^{15} \mathrm{~N}$ is non-radioactive but is the heavier isotope of nitrogen).

Even if ${ }^{15} \mathrm{~N}$ was radioactive then its presence would have been detected, both inside the cell $\left({ }^{15} \mathrm{~N}\right).$ incorporated as introgenous base in DNA) as well as in the supernatant, because

${ }^{15} \mathrm{~N}$ would also get incorporated in amino group of amino acids in proteins. Hence, the use of ${ }^{15} \mathrm{~N}$ would not give any conclusive results.

Answer

Some amino acids are coded by more then one codon (known as degeneracy of codons), hence, on deducing a nucleotide sequence from an amino acid sequence, multiple nucleotide sequence will be obtained, e.g., lle (Isoleucine) has three codons AUU, AUC, AUA. Hence, a dipeptide Met-lle can have the following nucleotide sequence.

(i) AUG-AUU

(ii) AUG-AUC

(iii) AUG-AUA

And if, we deduce amion acid sequence from the above nucleotide sequences, all the three will code for Met-lle.

Answer

The statement is correct. Because of degeneracy of codons, mutations at third base of codon, usually does not result into any change is phenotype. This is called silent mutations.

On other hand, if codon is changed in away that now it specifies another amino acid, it may other the protein function as it happens in cse of $\beta$-globulin of haemoglobin protein. Where a substitution of valine instead of glutamic acid causes change in its structure and function, and resulting into sickle-cell trait.

Answer

In the complete absence of expression of lac operon, permease will not be synthesised which is essential for transport of lactose from medium into the cells. And if lactose cannot be transported into the cell, then it cannot act as inducers. Hence, cannot relieve the lac operon from its repressed state.

Thinking Process

In 1990, US department of energy and National Institute of Health Embarked and Coordinated on the project of sequencing human genome called HGP or Human Genome Project.

Answer

The sequencing of human genome helped in enhancing the basic understanding of genetics and immunity to various disorders. Various genes that cause genetic disorders were identified with the help of this project.

It was found that more than 1200 genes are responsible for common human cardiovascular diseases, endocrine diseases (like diabetes), neurological, disorders (like Alzeimer’s disease, cancers and many more. These diseases can be treated easily by knowing the particular gene responsible for the particular disease.

Answer

The total number of genes is estimated at 25000 much lower than previous estimates of 140000 that had been based on extrapolations from gene-rich areas as opposed to a composite of gene-rich and gene-poor areas.

Almost all (99.9 %) nucleotide bases are exactly the same in all people. Functions for over $50 \%$ discovered genes are not known yet. Scientist have identified about 1.4 million locations where single-base DNA difference (SNPs or Single Nucleotide Polymorphisms) occur in humans.

This information promises to revolutionise the processes of finding chromosomal locations for disease-associated sequence and tracing human history.

Answer

Human genome helps to find out the complete genome sequence of the human. It has many advantages and disadvantages.

Some important advantages

It provides the knowledge of the effects of variations of DNA among individuals can revolutionise the ways to diagnose, treat and prevent many diseases that affect humans. It also provides clues to the understanding of human biology. It helps to find out the human evolution. Identification through DNA forensics is also possible.

Some important disadvantages

People might discover and untreatable genetic disease. People may abuse the knowledge obtained from the HGP. Problem can occur for the ownership of the genetic test result and the patenting of human genes and DNA. People believe that they are special and unique in their own ways and may wish to remain like that.

Answer

Bacteriophage does not have repetitive sequences such as VNTRs in its genome, as its genome is very small and have all the coding sequence. DNA finger printing is not done for phages.

Answer

Further polymerisation would not occur, as the $3^{\prime} \mathrm{OH}$ on sugar is not there to add a new nucleotide for forming ester bond.

Answer

Watson and Crick had the following informations which helped them to develop a model of DNA.

(i) Chargaff’s Law suggesting $A=T$ and $C=G$

(ii) Wilkins and Rosalind Franklin’s work on DNA crystal’s X-ray diffraction studies about DNAs physical structure.

Watson and Crick proposed

(a) Pattern of complementary bases pair

(b) Semi-conservative replication

(c) Mutation through tautomerism

Answer

(i) Methylated guanine cap helps in binding of mRNA to smaller ribosomal sub-unit during initiation of translation.

(ii) Poly-A tail provides longevity to $m R N A$ ’s life. Tail length and longevity of $m R N A$ are positively correlated.

Answer

Functional $m R N A$ of structural genes need not always include all of its exons. This alternate splicing of exons is sex-specific, tissue-specific and even developmental stage-specific. By such alternate splicing of exons, a single gene may encode for several isoproteins and/ or proteins of similar class

In absence of such a kind of splicing, there should have been new genes for every protein/isoprotein. Such an extravagancy has been avoided in natural phenomena by way of alternate splicing.

Answer

Tandemness in repeats provides many copies of the sequence for finger-printing and variability in nitrogen base sequences present in them. Being individual-specific, this proves to be useful in the process of DNA fingerprinting.

Answer

Hershey and Chase conducted experiments on bacteriophage to prove that DNA is the genetic material.

Hershey and Chase experiment

(i) Some bacteriophage virus were grown on a medium that contained radioactive phosphorus $\left({ }^{32} \mathrm{P}\right)$ and some in another medium with radioactive sulphur $\left({ }^{35} \mathrm{~S}\right)$.

(ii) Viruses grown in the presence of radioactive phosphorus $\left({ }^{32} \mathrm{P}\right)$ contained radioactive DNA.

(iii) Similar viruses grown in presence of radioactive sulphur $\left({ }^{35} \mathrm{~S}\right)$ contained radioactive protein.

(iv) Both the radioactive virus types were allowed to infect $E$. coli separately.

(v) Soon after infection, the bacterial cells were gently agitated in blender to remove viral coats from the bacteria.

(vi) The culture was also centrifuged to separate the viral particle from the bacterial cell.

Observations and Conclusions

(i) Only radioactive ${ }^{32} \mathrm{P}$ was found to be associated with the bacterial cell, whereas radioactive ${ }^{35} \mathrm{~S}$ was only found in surrounding medium and not in the bacterial cell.

(ii) This indicates that only DNA and not protein coat entered the bacterial cell.

(iii) This proves that DNA is the genetic material which is passed from virus to bacteria and not protein.

If both DNA and proteins contained phosphorus and sulphur, the result might change.

In case (i)

Radioactive ${ }^{35} \mathrm{~S}$ and + Bacteriophage ${ }^{32} \mathrm{P}$ labelled protein capsule $\rightarrow$ No radioactive

${ }^{35} \mathrm{~S}$ and ${ }^{32} \mathrm{P}$ Detected in cells + Radioactivity $\left({ }^{35} \mathrm{~S}\right.$ and $\left.{ }^{32} \mathrm{P}\right)$ detected in supernatant

In case (ii)

Radioactive ${ }^{35} \mathrm{~S}$ and ${ }^{32} \mathrm{P}$ lebelled DNA + Bacteriophage $\rightarrow$ Radioactive ${ }^{32} \mathrm{P}$ and ${ }^{35} \mathrm{~S}$

Detected in cells + No radioactivity detected in supernatant

Answer

A molecule that can act as a genetic material must fulfil the following

(i) It should be able to generate its replica (replication).

(ii) It should chemically and structurally be stable.

(iii) It should provide the scope for slow changes (mutation) that are required for evolution.

(iv) It should be able to express itself in the form of Mendelian.

Biochemical differences between DNA and RNA

(i) Both nucleic acid (DNA and RNA) are able to direct their duplication proteins fails for the first criteria.

(ii) RNA is reactive, it also acts are catalyst, hence DNA is less reactive and structurally more stable than RNA.

(iii) Presence of thymine at the place of uracil also confers additional stability to DNA.

Thinking Process

Post-transcriptional modifications include the modification of the mRNA transcript synthesised by RNA polymerase II (in eukaryotes).

Answer

Post-transcriptional Modifications

The primary transcripts are non-functional, containing both the coding region, exon and non-coding region, intron in RNA and are called heterogenous RNA or hnRNA.

In eukaryotes, three types of RNA polymerases are found in the nucleus

(i) RNA polymerase I transcribes $r$ RNAs (28 S and $5.8 \mathrm{~S}$ ).

(ii) RNA polymerase II transcribes the precursor of $m R N A$ (called heterogeneous nuclear RNA or hnRNA).

(iii) RNA polymerase III transcribes $t$ RNA, $5 \mathrm{~S} r \mathrm{RNA}$ and snRNAs (small nuclear RNAs).

The hnRNA undergoes two additional processes called capping and tailing

In capping, an unusual nucleotide, methyl guanosine triphosphate is added to the $5^{\prime}$-end of hnRNA.

In tailing, adenylate residues (about 200-300) are added at 3’-end in a template independent manner.

Now the hnRNA undergoes a process where the introns are removed and exons are joined to form $m R N A$ by the process called splicing.

Diagram representation of a post transcriptional modification in eukaryotes

Note In prokaryotes, $m R N A$ does not require any processing.

Thinking Process

Translation is the process of synthesis of protein from $m R N A$ with the help of ribosome. A translational unit in mRNA from $5^{\prime} \rightarrow 3^{\prime}$ comprises of a start codon, region coding for a polypeptide, a stop codon and Untranslated Regions (UTRs) at both $5^{\prime}$-end and 3 -end for efficient process.

Answer

There are three-stages of protein synthesis

(i) Initiation

Assembly of Ribosomes on mRNA In prokaryotes, initiation requires the large and small ribosome subunits, the $m R N A$, initiation $t$ RNA and three Initiation Factors (IFs).

Activation of Amino Acid Amino acids become activated by binding with aminoacyl tRNA synthetase enzyme in the presence of ATP.

$ \text{Amino acid} (AA) + ATP \xrightarrow[ \text{synthetases}] {Aminoacyl - tRNA} \text{AA-AMP-Enzyme complex} + P _i$

Transfer of Amino Acid to tRNA The AA-AMP-enzyme complex formed reacts with specific $t$ RNA to form aminoacyl $t$ RNA complex.

AA-AMP-Enzyme complex + tRNA $\rightarrow$ AAtRNA + AMP + Enzyme

The cap region of $m R N A$ binds to the smaller subunit of ribosome.

The ribosome has two sites, A-site and P-site.

The smaller subunit first binds the initiator tRNA then and then binds to the larger subunit so, that initiation codon (AUG) lies on the P-site.

The initiation $t \mathrm{RNA}$, i.e., methionyl tRNA then binds to the P-site.

(ii) Elongation

Another charged aminoacyl tRNA complex binds to the A-site of the ribosome.

Peptide bond formation and movement along the mRNA called translocation. A peptide bond is formed between carboxyl group (- $\mathrm{COOH}$ ) of amino acid at P-site and amino group ( $-\mathrm{NH}$ ) of amino acid at A-site by the enzyme peptidyl transferase.

The ribosome slides over mRNA from codon to codon in the $5^{\prime} \rightarrow 3^{\prime}$ direction.

According to the sequence of codon, amino acids are attached to one another by peptide bonds and a polypeptide chain is formed.

(iii) Termination

When the A-site of ribosome reaches a termination codon which does not code for any amino acid, no charged $t$ RNA binds to the A-site.

Dissociation of polypeptide from ribosome takes place which is catalysed by a ‘release factor’.

There are three termination codons, i.e., UGA, UAG and UAA.

Process of translation

Answer

The concept of operon was first proposed in 1961, by Jacob and Monad. An operon is a unit of prokaryotic gene expression which includes coordinately regulated (structural) genes and control elements which are recognised by regulatory gene product.

Components of an Operon

(i) Structural gene The fragment of DNA which transcribe mRNA for polypeptide synthesis.

(ii) Promoter The sequence of DNA where RNA polymerase binds and initiates transcription of structural genes is called promoter.

(iii) Operator The sequence of DNA adjacent to promoter where specific repressor protein binds is called operator.

(iv) Regulator gene The gene that codes for the repressor protein that binds to the operator and suppresses its activity as a result of which transcription will be switched off.

(v) Inducer The substrate that prevents the repressor from binding to the operator, is called an inducer. As a result transcription is switched on. It is a chemical of diverse nature like metabolite, hormone substrate, etc.

Inducible Operon System

An inducible operon system is a regulated unit of genetic material which is switched on in response to the presence of a chemical. e.g., the lactose or lac-operon of E.coli.

The lactose operon The lac $z, y$, a genes are transcribed from a lac transcription unit under the control of a single promoter. They encode enzyme required for the use of lactose as a carbon source. The lac $i$ gene product, the lac repressor, is expressed from a separate transcription unit upstream from the operator.

lac operon consists of three structural genes ( $z, y$ and $a$), operator, promoter and a separate regulatory gene.

The three structural genes $(a, y$ and $a)$ transcribe a polycistronic mRNA.

Lac operon

Gene $z$ codes for $\beta$-galactosidase ( $\beta$-gal) enzyme which breaks lactose into galactose and glucose.

Gene $y$ codes for permease, which increases the permeability of the cell to lactose.

Gene a codes for enzyme transacetylase, which catalyses the transacetylation of lactose in its active form.

When Lactose is Absent

(i) When lactose is absent, $i$ gene regulates and produces repressor mRNA which translate repression.

(ii) The repressor protein binds to the operator region of the operon and as a result prevents RNA polymerase to bind to the operon.

(iii) The operon is switched off.

When Lactose is Present

(i) Lactose acts as an inducer which binds to the repressor and forms an inactive repressor.

(ii) The repressor fails to bind to the operator region.

(iii) The RNA polymerase binds to the operator and transcript lac mRNA.

(iv) lac mRNA is polycistronic, i.e., produces all three enzymes, $\beta$-galactosidase, permease and transacetylase.

(v) The lac operon is switched on.

Answer

DNA fingerprinting is the technique used in solving the paternity dispute for a child. DNA fingerprinting is a technique of determining nucleotide sequences of certain areas of DNA which are unique to each individual.

The basis of DNA fingerprinting is DNA polymorphism. Although the DNA from different individuals is more alike than different, there are many regions of the human chromosomes that exhibit a great deal of diversity. Such variable sequences are termed ‘polymorphic’ (meaning many forms).

A special type of polymorphism, called VNTR (Variable Number of Tandem Repeats), is composed of repeated copies of a DNA sequence that lie adjacent to one another on the chromosome. Since, polymorphism is the basis of genetic mapping of human.

Answer

Sequencing of human genome has made it possible to understand the link between various genes and their functions. If there are any gene defects that express as disorders or that increase the susceptibility of an individual to a disease then specific gene therapies can be worked out

Methodologies of human genome sequencing

The methods involve two major approaches

(i) Expressed Sequence Tags (ESTs) This method focusses on identifying all the genes that are expressed as RNA.

(ii) Sequence annotation It is an approach of simply sequencing the whole set of genome that contains all the coding and non-coding sequences, and later assigning different regions in the sequence with functions.

For sequencing, first the total DNA from cell is i.e., solated and broken down in relatively small sizes as fragments.

There DNA fragments are cloned in suitable host using suitable vectors. When bacteria is used as vector, they are called Bacterial Artificial Chromosomes (BAC) and when yeast is used as vector, they are called Yeast Artificial Chromosomes (YACs).

Frederick Sanger developed a principle according to which the fragments of DNA are sequenced by automated DNA sequences.

On the basis of overlapping regions on DNA fragments, these sequences are arranged accordingly. For alignment of these sequences, specialised computer-based programmes were developed.

Finally, the genetic and physical maps of the genome were constructed by collecting information about certain repetitive DNA sequences and DNA polymorphism, based on endonuclease recognition sites.

Answer

Dr. Alec Jeffreys developed the technique of DNA fingerprinting in an attempt to identify DNA marker for inherited diseases.

DNA fingerprinting uses short nucleotide repeats called Variable Number Tandem Repeats (VNTRs) as markers. VNTRs vary from person to person and are inherited from one generation to the next. Only closely individuals have similar VNTRs.

Answer

In above case, as E.coli is a mutant for DNA ligase, it will result in no further joining of Okazaki fragments on lagging strand.

This will ultimately result into the formation of both high molecular weight fragments (on leading strands) and low molecular weight fragments (on lagging strand). Hence, only the graph (a) could be the appropriate result after centrifugation.