The DNA Questions in English

(B) The $DNA$ (Deoxyribonucleic acid) molecule acts as the primary carrier of genetic information in almost all living organisms.
It stores the hereditary instructions required for the development,functioning,growth,and reproduction of organisms.
While $RNA$ molecules like $m-RNA$ and $t-RNA$ are involved in the process of protein synthesis,$DNA$ remains the stable repository of the genetic code.

(D) Histones are a group of positively charged,basic proteins that are associated with $DNA$ in the chromatin of eukaryotic cells.
They are rich in the basic amino acids $Lysine$ and $Arginine$.
These amino acids carry positive charges in their side chains at physiological $pH$,which allows the histones to bind tightly to the negatively charged phosphate backbone of the $DNA$ molecule.

(D) Both $DNA$ $(Deoxyribonucleic \text{ acid})$ and $RNA$ $(Ribonucleic \text{ acid})$ are nucleic acids composed of nucleotides.
Each nucleotide consists of three components: a nitrogenous base, a pentose sugar, and a phosphate group.
$DNA$ and $RNA$ share the phosphate group as a common structural component.
$DNA$ contains deoxyribose sugar and $RNA$ contains ribose sugar.
$DNA$ contains adenine, guanine, cytosine, and thymine, while $RNA$ contains adenine, guanine, cytosine, and uracil.

(C) According to Chargaff's rule,the amount of adenine $(A)$ is equal to the amount of thymine $(T)$,and the amount of guanine $(G)$ is equal to the amount of cytosine $(C)$.
Given that the percentage of adenine $(A)$ = $30\%$.
Therefore,the percentage of thymine $(T)$ = $30\%$.
The total percentage of $A + T = 30\% + 30\% = 60\%$.
The remaining percentage for $G + C$ is $100\% - 60\% = 40\%$.
Since $G = C$,the percentage of guanine $(G)$ = $40\% / 2 = 20\%$.
Thus,the correct option is $C$.

(B) The $X$-ray diffraction data produced by Maurice Wilkins and Rosalind Franklin provided crucial information about the structure of $DNA$.
According to the Watson-Crick model of $DNA$,which is based on this diffraction data,the $DNA$ molecule exists as a double helix.
The distance between the two strands of the $DNA$ helix is constant and is measured as $20\ \mathring{A}$ (or $2\ nm$).
Therefore,the diameter of the $DNA$ helix is $20\ \mathring{A}$.

(C) In $RNA$ molecules,the nitrogenous base Adenine $(A)$ forms a complementary base pair with Uracil $(U)$ via two hydrogen bonds. Unlike $DNA$,where Adenine pairs with Thymine $(T)$,$RNA$ contains Uracil instead of Thymine.

(A) In $1953$,James Watson and Francis Crick proposed the double-helical model of $DNA$ based on $X$-ray diffraction data produced by Maurice Wilkins and Rosalind Franklin. This model describes $DNA$ as two polynucleotide chains coiled around a common axis in a right-handed fashion,with the sugar-phosphate backbone on the outside and nitrogenous bases on the inside.

(C) $DNA$ (Deoxyribonucleic acid) consists of four nitrogenous bases: Adenine $(A)$,Guanine $(G)$,Cytosine $(C)$,and Thymine $(T)$.
Uracil $(U)$ is a nitrogenous base found in $RNA$ (Ribonucleic acid) instead of Thymine.
Therefore,Uracil is not found in $DNA$.

(C) In $DNA$,the nitrogenous bases follow Chargaff's rule of base pairing,where Adenine $(A)$ pairs with Thymine $(T)$ via two hydrogen bonds,and Guanine $(G)$ pairs with Cytosine $(C)$ via three hydrogen bonds.
In $RNA$,Uracil $(U)$ replaces Thymine $(T)$,so Adenine $(A)$ pairs with Uracil $(U)$.
The pair $A - C$ is incorrect because Adenine $(A)$ does not form a stable base pair with Cytosine $(C)$ under normal physiological conditions.

(A) In $B-DNA$, the structure is a double helix with a constant diameter of $2.0\ nm$ $(20\ \text{\AA})$.
Each full turn of the helix consists of $10$ base pairs.
The distance between two adjacent base pairs is $0.34\ nm$ $(3.4\ \text{\AA})$.
Therefore, the length of one full turn of the helix is $10 \times 0.34\ nm = 3.4\ nm$.

(C) The molecular weight of nucleic acids depends on their size and complexity.
$t-RNA$ (transfer $RNA$) is the smallest molecule among the three,typically consisting of $70-90$ nucleotides.
$m-RNA$ (messenger $RNA$) varies in length depending on the gene it encodes,but generally has a higher molecular weight than $t-RNA$.
$r-RNA$ (ribosomal $RNA$) and $DNA$ (deoxyribonucleic acid) are much larger. However,in the context of typical cellular components,$DNA$ molecules (chromosomes) have a significantly higher molecular weight than individual $RNA$ molecules.
Therefore,the correct order of increasing molecular weight is $t-RNA < m-RNA < DNA$.

(B) The $DNA$ double helix model proposed by Watson and Crick states that the two strands of $DNA$ run in opposite directions, meaning they are antiparallel ($5' \to 3'$ and $3' \to 5'$).
Furthermore, the nitrogenous bases on the two strands are complementary to each other, where Adenine $(A)$ pairs with Thymine $(T)$ via two hydrogen bonds, and Guanine $(G)$ pairs with Cytosine $(C)$ via three hydrogen bonds.
Therefore, the two strands are antiparallel and complementary.

(A) Nucleic acids are polymers of nucleotides.
$DNA$ (Deoxyribonucleic acid) is a polymer of deoxyribonucleotides,where the sugar component is deoxyribose.
$RNA$ (Ribonucleic acid) is a polymer of ribonucleotides,where the sugar component is ribose.
Therefore,the polymer of deoxyribonucleotides is $DNA$.

(A) Both $DNA$ $(Deoxyribonucleic \text{ acid})$ and $RNA$ $(Ribonucleic \text{ acid})$ are polynucleotides.
They are formed by the polymerization of nucleotides, which consist of a nitrogenous base, a pentose sugar, and a phosphate group.
Option $A$ is correct because both are polymers of nucleotides.
Option $B$ is incorrect because $DNA$ contains thymine while $RNA$ contains uracil.
Option $C$ is incorrect because $DNA$ contains deoxyribose sugar while $RNA$ contains ribose sugar.
Option $D$ is incorrect because while $DNA$ is the primary genetic material in most organisms, $RNA$ acts as genetic material only in some viruses.

(D) In a nucleic acid,individual nucleotides are linked together to form a polynucleotide chain.
This linkage occurs between the $3'$-carbon atom of the sugar moiety of one nucleotide and the $5'$-phosphate group of the adjacent nucleotide.
This specific covalent linkage is known as a $3'-5'$ phosphodiester bond.
Therefore,the correct answer is $D$.

(B) The structure of $DNA$ is a double helix where the two strands are held together by hydrogen bonds between specific nitrogenous bases.
According to the principle of base pairing (Chargaff's rules),Adenine $(A)$ always pairs with Thymine $(T)$ via two hydrogen bonds,and Guanine $(G)$ always pairs with Cytosine $(C)$ via three hydrogen bonds.
Because of this specific pairing,the two strands are said to be complementary to each other.
Therefore,if the sequence of one strand is known,the sequence of the other strand can be predicted based on this complementary relationship.

(B) The $DNA$ double helix consists of two polynucleotide chains that run in opposite directions,known as antiparallel orientation.
This antiparallel nature is primarily determined by the specific chemical structure of the sugar-phosphate backbone and the base pairing rules.
Specifically,the $5' \rightarrow 3'$ direction of one strand is opposite to the $3' \rightarrow 5'$ direction of the other strand.
While $H$-bonds (hydrogen bonds) are responsible for holding the two strands together by pairing nitrogenous bases,the antiparallel orientation itself is a structural consequence of the phosphodiester bonds linking the nucleotides in a specific $5'-3'$ polarity.

(A) The molecular weight of $DNA$ varies significantly across different organisms.
In yeast $(Saccharomyces \ cerevisiae)$,the molecular weight of the $DNA$ molecule is approximately $2.56 \times 10^9$ Daltons.
This value is a standard biological reference point for the genome size of this unicellular fungus.

(A) The $DNA$ molecule is negatively charged due to the phosphate backbone,while histones are positively charged proteins rich in basic amino acids like lysine and arginine. The interaction between $DNA$ and histones is primarily electrostatic. By increasing the ionic strength of the solution using a high concentration of salt (like $6\ M\ NaCl$),the electrostatic attraction between the negatively charged $DNA$ and positively charged histones is disrupted,causing the histones to dissociate from the $DNA$.

(C) $t-RNA$ (transfer $RNA$) contains both common nitrogenous bases (Adenine,Guanine,Cytosine,and Uracil) and unusual or modified nitrogenous bases (such as dihydrouridine,pseudouridine,and ribothymidine).
These modified bases are primarily located in the loops of the $t-RNA$ cloverleaf structure.
They play a crucial role in the stability and recognition of $t-RNA$ during the process of protein synthesis.

(A) The $t-RNA$ molecule is often represented as a clover-leaf structure in its two-dimensional form. However,in its actual three-dimensional configuration,it appears as an $L$-shaped molecule. This compact $L$-shaped structure is essential for its function in protein synthesis,allowing it to fit into the ribosome.

(B) The $DNA$ molecule wraps around the histone octamer to form a nucleosome.
Each nucleosome consists of approximately $146$ base pairs of $DNA$ wrapped around the histone core.
The $DNA$ makes approximately $1.65$ turns around the histone octamer.
Based on structural studies,the $DNA$ wraps around the histone octamer at an angle of approximately $90^\circ$ relative to the axis of the nucleosome.

(A) The nucleosome core particle consists of an octamer of histone proteins $(H_2A, H_2B, H_3, H_4)$ wrapped by $DNA$.
$H_1$ histone is known as the linker histone.
It binds to the $DNA$ at the entry and exit sites of the nucleosome,helping to stabilize the structure and link adjacent nucleosomes together to form higher-order chromatin structures.

(D) nucleosome is the basic structural unit of chromatin in eukaryotes.
It consists of a segment of $DNA$ wound around a core of histone proteins.
The core particle is composed of an octamer of histone proteins ($H2A$,$H2B$,$H3$,and $H4$,two of each).
Therefore,the correct composition is histone proteins and $DNA$.

(B) Foetal Haemoglobin $(Hb F)$ is the primary form of haemoglobin present in the developing foetus.
It consists of two alpha $(\alpha)$ polypeptide chains and two gamma $(\gamma)$ polypeptide chains.
Therefore, its composition is represented as $\alpha_2 \gamma_2$.
In contrast, adult haemoglobin $(Hb A)$ consists of two alpha $(\alpha)$ and two beta $(\beta)$ chains $(\alpha_2 \beta_2)$.

(C) In the context of prokaryotic organisms,a $Gene$ is defined as a segment of $DNA$ that codes for a polypeptide.
This functional unit of $DNA$ is referred to as a $Cistron$.
Since a single $Gene$ corresponds to a single $Cistron$ in prokaryotes (monocistronic/polycistronic context notwithstanding,the fundamental definition equates the two),they are often used interchangeably in basic genetic terminology.

(D) The biological unit of heredity is the $Gene$.
Genes are segments of $DNA$ that contain the instructions for building proteins,which determine the traits of an organism.
They are passed from parents to offspring and are responsible for the inheritance of characteristics.
Therefore,the correct option is $D$.

(B) mutation is defined as a permanent alteration in the $DNA$ sequence that makes up a gene. The smallest possible change that can result in a mutation is a point mutation,which involves a change in a single nucleotide base pair within the $DNA$ sequence. This can lead to a change in the corresponding amino acid or have no effect,depending on the nature of the change.

(B) The term 'nucleoprotein' was first coined by Richard Altmann in $1889$. He described these substances as complexes of nucleic acids and proteins found in the cell nucleus.

(C) In genetics,the sites within a gene or genome where mutations occur at a rate significantly higher than the average are known as $Hot$ $spots$. These regions are highly susceptible to mutation due to specific $DNA$ sequences or structural features.

(A) Genetic mutations are defined as permanent alterations in the nucleotide sequence of the genome of an organism.
Since $DNA$ serves as the primary genetic material in almost all living organisms,mutations primarily occur in the $DNA$ sequence.
While $RNA$ viruses can undergo mutations in their $RNA$ genome,in the context of general biological principles and the central dogma,genetic mutations are fundamentally changes in $DNA$.

(D) In $1979$,Har Gobind Khorana and his colleagues successfully synthesized the first biologically functional gene in the laboratory. This gene coded for the tyrosine suppressor $tRNA$ of $E. coli$. The synthesized gene consisted of $207$ nucleotide pairs. This was a landmark achievement in the field of synthetic biology and genetic engineering.

(C) Ultraviolet $(UV)$ radiation,particularly $UV-B$,has a high energy level that can be absorbed by the nitrogenous bases of $DNA$.
When $DNA$ is exposed to $UV$ light,it causes the formation of covalent bonds between adjacent pyrimidine bases (specifically thymine or cytosine) on the same strand.
This process is known as the formation of pyrimidine dimers.
These dimers distort the $DNA$ helix,which can lead to errors during $DNA$ replication and transcription,potentially causing mutations or cell death.

(B) Molecular biology is the branch of biology that deals with the structure and function of the macromolecules (e.g.,proteins and nucleic acids) essential to life. It focuses on the physicochemical study of these biological molecules and their interactions within the cell to understand the molecular basis of biological activity. Therefore,it is defined as the physicochemical study of biological molecules.

$(A)$ The '$nucleosome$' core particle consists of an octamer of histone proteins, which includes two molecules each of '$H_2A$', '$H_2B$', '$H_3$', and '$H_4$'.
These eight histone proteins form the core around which the $DNA$ is wrapped.
The '$H_1$' histone protein is known as the linker histone, which binds to the $DNA$ at the entry and exit points of the '$nucleosome$' but is not part of the core particle itself.
Therefore, '$H_1$' is the histone absent in the '$nucleosome$' core particle.

(D) nucleosome is the basic repeating unit of eukaryotic chromatin.
$1$. It consists of a core of eight histone proteins ($H_2A, H_2B, H_3, H_4$ in two copies each) around which $DNA$ is wrapped.
$2$. The $DNA$ wraps around the histone octamer approximately $1.65$ to $2$ times.
$3$. Each nucleosome typically contains about $200$ base pairs of $DNA$ (including the linker $DNA$).
Therefore,all the given statements are correct.

(A) Genes are the functional units of heredity and are composed of segments of $DNA$. These genes are linearly arranged on structures called chromosomes,which are found within the nucleus of eukaryotic cells. Therefore,the correct option is $A$.

(B) In $DNA$,the four nitrogenous bases are $Adenine$ $(A)$,$Guanine$ $(G)$,$Cytosine$ $(C)$,and $Thymine$ $(T)$.
In $RNA$,$Thymine$ is replaced by $Uracil$ $(U)$.
Therefore,$RNA$ contains $Adenine$,$Guanine$,$Cytosine$,and $Uracil$ instead of $Thymine$.

(A) The Feulgen reaction is a staining technique used to identify chromosomal material or $DNA$ in cell nuclei.
It involves mild acid hydrolysis of $DNA$ using $1M$ $HCl$ at $60^{\circ}C$.
This process removes purine bases from the $DNA$ molecule,which exposes the aldehyde groups of the deoxyribose sugar.
These free aldehyde groups then react with the Schiff's reagent to produce a magenta or purple color,confirming the presence of $DNA$.

(A) In the context of chromatin structure,the inactive region of $DNA$ is known as $Heterochromatin$.
$Heterochromatin$ is highly condensed and tightly packed,which makes it transcriptionally inactive.
Because it is so densely packed,the amount of $DNA$ per unit volume in these regions is considered to be high compared to the $Euchromatin$ (active region),which is loosely packed.

(A) $DNA$ (Deoxyribonucleic acid) is the primary genetic material in almost all living organisms. It stores the complete biological information required for the development,functioning,growth,and reproduction of an organism. This information is encoded in the sequence of nucleotide bases,which is transcribed into $RNA$ and translated into proteins to perform various life processes.

(D) The Feulgen reaction is a staining technique used in histology to identify chromosomal material or $DNA$ in cell specimens.
It relies on the acid hydrolysis of $DNA$,which exposes aldehyde groups on the deoxyribose sugar.
These aldehyde groups then react with the Schiff reagent to produce a distinct magenta or purple color.
Since $RNA$ does not contain deoxyribose,it does not undergo this specific reaction,making the Feulgen test highly specific for $DNA$.

(A) Statement $A$ is correct because every somatic cell in the human body contains the complete genome,including the genes for insulin production. These genes are present in all cells but are only expressed in the beta cells of the pancreas.
Statement $B$ is incorrect because nucleosomes are composed of $DNA$ wrapped around a histone octamer,not nucleotides.
Statement $C$ is incorrect because the histone octamer forms the core,and $DNA$ wraps around it.
Statement $D$ is incorrect because the structure that produces asters during cell division is the centrosome,not the centromere.

(D) Ionizing radiations such as $X$-rays and Gamma rays,as well as non-ionizing radiations like $UV$-rays,are known to cause damage to $DNA$.
$X$-rays and Gamma rays cause double-strand breaks in the $DNA$ molecule.
$UV$-rays cause the formation of pyrimidine dimers (specifically thymine dimers),which disrupt the replication and transcription processes of $DNA$.
Therefore,all the listed radiations are capable of damaging $DNA$.

(B) The term 'living chemical' is often used to describe $DNA$ because it carries the genetic information necessary for life,replication,and protein synthesis,yet it is a chemical molecule. It acts as the blueprint for all living organisms.

(D) The sugar present in $DNA$ $(Deoxyribonucleic \text{ acid})$ is $2-deoxyribose$.
Ribose sugar has the molecular formula $C_5H_{10}O_5$.
In $2-deoxyribose$, one oxygen atom is removed from the second carbon position of the ribose sugar.
Therefore, the molecular formula of $2-deoxyribose$ is $C_5H_{10}O_4$.

(A) The information required for protein synthesis is stored in the $DNA$ (Deoxyribonucleic acid),which is a type of nucleic acid located within the nucleus of the cell. This genetic information is transcribed into $mRNA$ and then translated into proteins by ribosomes. Therefore,the primary storage of this information is the nucleic acids found in the nucleus.

(D) According to Chargaff's rules and the Watson-Crick model of $DNA$,adenine $(A)$ always pairs with thymine $(T)$ via two hydrogen bonds.
Cytosine $(C)$ always pairs with guanine $(G)$ via three hydrogen bonds.
Uracil $(U)$ is found in $RNA$ instead of thymine and pairs with adenine in $RNA$ molecules.
Therefore,in $DNA$,adenine $(A)$ pairs with thymine $(T)$.

(A) In the structure of $DNA$,nitrogenous bases pair specifically according to Chargaff's rules.
$G$ (Guanine) always pairs with $C$ (Cytosine) through $3$ hydrogen bonds.
$A$ (Adenine) always pairs with $T$ (Thymine) through $2$ hydrogen bonds.
Therefore,there are $3$ hydrogen bonds between $G$ and $C$.

(B) $1$. $RNA$ is typically single-stranded,not double-stranded.
$2$. According to the $Watson-Crick$ model,the two strands of $DNA$ are coiled together in a specific right-handed helical manner.
$3$. In $DNA$,purines ($Adenine$,$Guanine$) always pair with pyrimidines ($Thymine$,$Cytosine$) via hydrogen bonds,not purine-purine or pyrimidine-pyrimidine.
$4$. The distance between the two strands of $DNA$ (the diameter of the helix) is approximately $20 \ \mathring A$,while the distance per turn is $34 \ \mathring A$. Therefore,statement $B$ is the only correct statement.