The DNA Questions in English

(D) Coliphage $T_2$ is a type of bacteriophage that infects the bacterium $Escherichia \text{ } coli$.
It contains a linear, double-stranded $DNA$ $(dsDNA)$ molecule as its genetic material, which is enclosed within a protein capsid.

(C) provirus is a viral genome that is integrated into the $DNA$ of a host cell. In this state,the viral $DNA$ remains latent and is replicated along with the host cell's genome during cell division. This is a characteristic feature of the lysogenic cycle in bacteriophages and retroviruses.

(A) The bacteriophage $\phi \times 174$ is a well-known virus that contains a single-stranded $DNA$ molecule as its genetic material.
Therefore,the correct option is $A$ $(ss\, DNA)$.

(B) The correct answer is $(b)$.
Single-stranded $DNA$ is found in certain bacteriophages such as $\phi \times 174$,$S-13$,$F-1$,$M-13$,and in some viruses like Parvoviruses.
In contrast,$E. coli$,$\lambda$ phage,and $T_4$ phage contain double-stranded $DNA$ as their genetic material.

(C) The genome of $E. coli$ consists of approximately $4.6$ million base pairs. Based on the average size of a bacterial gene,this amount of genetic material is estimated to code for approximately $4000$ to $4300$ different proteins. Therefore,the most accurate range among the given options is $3000$ to $4000$ proteins,though it is often cited as being close to $4000$.

(B) Histone proteins are a group of proteins that are rich in basic amino acids,specifically $Lysine$ and $Arginine$.
These amino acids carry positive charges in their side chains at physiological $pH$.
Because $DNA$ is negatively charged due to the phosphate groups in its backbone,the positively charged histone proteins facilitate the packaging of $DNA$ into structural units called nucleosomes.
Therefore,histone proteins are basic in nature.

(C) Banding is a cytogenetic technique used to visualize the structure of chromosomes. In $1969$,scientists developed banding techniques where chromosomes are treated with specific stains to produce a pattern of light and dark bands. Each band corresponds to a specific region of the chromosome,which is essential for the identification and mapping of chromosomes.

(C) The supercoiled structure of eukaryotic chromosomes is explained by the $Nucleosome$ model.
In eukaryotes,$DNA$ is wrapped around a core of histone proteins to form a structure called a $Nucleosome$.
These nucleosomes further coil and fold to form higher-order chromatin structures,ultimately resulting in the highly condensed supercoiled structure of the chromosome during cell division.

(B) Eukaryotic telomeric $DNA$ consists of short,tandemly repeated sequences.
These sequences are characterized by clusters of $G$-residues (Guanine) in one strand and $C$-residues (Cytosine) in the complementary strand.
Specifically,the $3'$ end of the $G$-rich strand forms a single-stranded overhang,which is essential for telomere stability and protection.
Therefore,telomeres are primarily known for being Guanine-rich repeats.

(C) $DNA$ differs from $RNA$ primarily in two aspects:
$1$. Sugar: $DNA$ contains $2'$-deoxyribose sugar,whereas $RNA$ contains ribose sugar.
$2$. Pyrimidines: $DNA$ contains thymine $(T)$ as a nitrogenous base,while $RNA$ contains uracil $(U)$ instead of thymine.
Therefore,the correct answer is $C$.

(A) The two strands of the $DNA$ double helix are held together by hydrogen bonds between the nitrogenous bases.
Specifically, there is a double hydrogen bond between adenine and thymine $(A = T)$ and a triple hydrogen bond between cytosine and guanine $(C \equiv G)$.

(A) According to Chargaff's rules and the structure of the $DNA$ double helix proposed by Watson and Crick,nitrogenous bases pair specifically through hydrogen bonds.
In $DNA$,guanine $(G)$ always forms three hydrogen bonds with cytosine $(C)$.
Adenine $(A)$ always forms two hydrogen bonds with thymine $(T)$.
Therefore,guanine pairs with cytosine.

(A) The two strands of $DNA$ molecules run in opposite or antiparallel directions.
This structural orientation is maintained because the nitrogenous bases of the two strands are joined together by hydrogen bonds ($H$-bonds) following the principle of complementary base pairing.
While phosphodiester bonds hold the nucleotides together within a single strand,the antiparallel nature is specifically defined by the orientation required for the hydrogen bonding between base pairs.

(D) Histone proteins are basic proteins rich in the amino acids lysine and arginine.
They carry a positive charge in their side chains.
$DNA$ is a negatively charged molecule due to the presence of phosphate groups in its backbone.
Histones help in neutralizing the negative charge of $DNA$ and facilitate the packaging of $DNA$ into compact structures called nucleosomes,which are essential for stabilizing the chromosome structure.

(B) The correct answer is $B$.
In a $DNA$ molecule,nitrogenous bases follow Chargaff's rule of base pairing.
According to this rule,Adenine $(A)$ always pairs with Thymine $(T)$ via two hydrogen bonds.
Similarly,Cytosine $(C)$ always pairs with Guanine $(G)$ via three hydrogen bonds.
Therefore,Adenine and Thymine is the correct base pair present in $DNA$.

(C) In $DNA$ molecules,nitrogenous bases form specific hydrogen bonds with each other. Adenine $(A)$ pairs with Thymine $(T)$ via two hydrogen bonds $(A = T)$. Cytosine $(C)$ pairs with Guanine $(G)$ via three hydrogen bonds $(C \equiv G)$. Therefore,the correct representation is $A = T$ and $C \equiv G$.

(B) $DNA$ (Deoxyribonucleic acid) is primarily found as a double-stranded molecule in most living organisms,forming a double helix structure.
However,in certain viruses,such as the $\phi X174$ bacteriophage,$DNA$ exists in a single-stranded form.
Therefore,while $DNA$ is typically double-stranded,it is rarely single-stranded in specific biological contexts.

(D) Nucleic acids,specifically $DNA$ and $RNA$,possess the unique ability to undergo self-replication.
$DNA$ acts as the genetic material in most organisms and can replicate itself to pass genetic information to the next generation.
Enzymes are proteins that catalyze reactions,carbohydrates are energy sources or structural components,and water is an inorganic solvent; none of these possess the inherent chemical structure required for self-replication.

(C) The distance between two consecutive base pairs in a $DNA$ double helix is $0.34 \, nm$ or $3.4 \, \mathring A$.
For $23$ base pairs,there are $22$ gaps between them.
However,in standard calculations for the length of a $DNA$ segment,the formula used is $L = n \times d$,where $n$ is the number of base pairs and $d$ is the distance between adjacent base pairs $(3.4 \, \mathring A)$.
Length $= 23 \times 3.4 \, \mathring A = 78.2 \, \mathring A$.

(D) In the structure of nucleic acid bases,the number of hydrogen bond donor and acceptor sites determines the binding capability.
$Adenine$ $(A)$ typically forms $2$ hydrogen bonds with $Thymine$ $(T)$ in $DNA$.
$Guanine$ $(G)$ forms $3$ hydrogen bonds with $Cytosine$ $(C)$.
Among the options,$Adenine$ is characterized by having two specific hydrogen bonding sites (one donor and one acceptor) that allow it to pair specifically with $Thymine$ via two hydrogen bonds.

(B) In a $DNA$ double helix structure,the distance between two adjacent base pairs is approximately $3.4 \, \mathring{A}$ $(0.34 \, nm)$.
This distance is a fundamental characteristic of the $B-DNA$ form,which is the most common form of $DNA$ found in living organisms.

(C) The $DNA$ double helix structure, as proposed by Watson and Crick, consists of a right-handed helix.
Each full turn of the $DNA$ helix measures approximately $3.4 \, nm$ $(34 \, \text{\AA})$.
There are $10$ base pairs (nucleotide pairs) present in each complete turn of the $DNA$ helix.
Therefore, the distance between two adjacent base pairs is approximately $0.34 \, nm$ $(3.4 \, \text{\AA})$.

(C) $DNA$ (Deoxyribonucleic acid) and $RNA$ (Ribonucleic acid) are both nucleic acids.
Both are long-chain polymers composed of repeating units called nucleotides.
$A$ nucleotide consists of a nitrogenous base,a pentose sugar,and a phosphate group.
Therefore,the fundamental structural similarity is that both are polymers of nucleotides.

(A) $DNA$ (Deoxyribonucleic acid) is a long chain molecule known as a polymer.
Its monomeric units are called nucleotides.
$A$ nucleotide consists of three components: a nitrogenous base,a pentose sugar (deoxyribose),and a phosphate group.
Therefore,$DNA$ is a polymer of nucleotides.

(A) 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,the nitrogenous bases present in $RNA$ are Adenine,Guanine,Cytosine,and Uracil.

(C) Telomerase is a specialized enzyme that adds specific nucleotide sequences to the ends of eukaryotic chromosomes,known as telomeres.
It is a ribonucleoprotein complex,meaning it consists of both a protein component ($TERT$ - Telomerase Reverse Transcriptase) and an $RNA$ component ($TERC$ - Telomerase $RNA$ Component) that serves as a template for the synthesis of telomeric $DNA$ repeats.

(C) The term $Genome$ refers to the complete set of genetic material $(DNA)$ present in a cell or organism. In diploid organisms,a $Genome$ represents the haploid set of chromosomes inherited from one parent. Therefore,a complete set of chromosomes inherited as a unit from one parent is defined as a $Genome$.

(B) The functional unit of inheritance is the $Gene$.
$Gene$s are specific segments of $DNA$ that contain the instructions for synthesizing proteins or functional $RNA$ molecules.
They are responsible for the transmission of hereditary traits from parents to offspring.

(C) The unit of crossing over (recombination) is known as $Recon$.
$Cistron$ is defined as the unit of function (gene).
$Muton$ is defined as the unit of mutation.
$Recon$ is the smallest unit of $DNA$ capable of undergoing recombination (crossing over).

(B) The functional unit of mutation is known as a $Muton$.
$Muton$ is defined as the smallest unit of $DNA$ that can undergo a mutation.
$Recon$ is the unit of recombination,and $Cistron$ is the unit of function (coding for a polypeptide chain).

(B) $tRNA$ (transfer $RNA$) is known as the smallest type of $RNA$ molecule. It typically consists of a single strand of nucleotides folded into a specific cloverleaf structure. The length of a standard $tRNA$ molecule ranges from approximately $75$ to $90$ nucleotides.

(B) 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 Guanine is $19\%$,the percentage of Cytosine must also be $19\%$ because $G = C$.
This rule applies to the $DNA$ of all cells within an organism,whether they are germ cells (like sperms) or somatic cells.

(B) According to Chargaff's rule,in a double-stranded $DNA$ molecule,the amount of guanine $(G)$ is equal to cytosine $(C)$,and the amount of adenine $(A)$ is equal to thymine $(T)$.
Given that $G = 10\%$,then $C = 10\%$.
The total percentage of all four bases is $100\%$.
Therefore,$A + T = 100\% - (G + C) = 100\% - (10\% + 10\%) = 80\%$.
Since $A = T$,we have $2T = 80\%$,which means $T = 40\%$.
Thus,the percentage of thymine is $40\%$.

(C) According to the solenoid model of chromatin organization,the $11 \ nm$ nucleosome chain coils into a higher-order structure known as the $30 \ nm$ chromatin fibre.
This structure is formed by the solenoid coiling of the nucleosome chain.
In this helical arrangement,there are approximately $6$ nucleosomes per turn of the coil.

(D) The isolation and purification of a specific $DNA$ segment (the $lac$ operon) from a living organism was first achieved by Jonathan Beckwith and his colleagues in $1969$. They used specialized transducing phages to isolate the $lac$ genes from $Escherichia \text{ } coli$.

(C) Micrococcal nuclease is an endo-exonuclease that preferentially digests the linker $DNA$ present between nucleosomes.
It is widely used in molecular biology to study chromatin structure by cleaving the $DNA$ at the exposed linker regions,thereby releasing individual nucleosomes or nucleosome arrays.

(A) According to the Watson-Crick model of $DNA$ structure,the $DNA$ molecule exists as a double helix.
The distance between the two strands of the $DNA$ helix is constant,which is approximately $20 \ \mathring{A}$ $(2 \ nm)$.
Therefore,the mean diameter of the $DNA$ double helix is $20 \ \mathring{A}$.

(D) When a $DNA$ molecule is heated to high temperatures (typically $82-90^{\circ}C$),the hydrogen bonds between the complementary nitrogenous bases break. This process is known as $DNA$ denaturation or melting,which causes the double-stranded $DNA$ to separate into two single-stranded chains. Therefore,the double helix uncoils and separates.

(D) In a $B-DNA$ double helix,the distance between two adjacent base pairs is approximately $3.4\mathring A$.
Each full turn of the helix contains $10$ base pairs.
Therefore,the length of one full turn (pitch) of the $DNA$ helix is $10 \times 3.4\mathring A = 34\mathring A$.

(B) Rosalind Franklin and Maurice Wilkins used $X$-ray diffraction data to study the structure of $DNA$. Their findings provided critical evidence that $DNA$ has a helical structure,which was later used by Watson and Crick to propose the double helix model of $DNA$.

(B) According to Chargaff's rules,in a double-stranded $DNA$ molecule,the amount of Adenine $(A)$ equals the amount of Thymine $(T)$,and the amount of Guanine $(G)$ equals the amount of Cytosine $(C)$.
Therefore,$A = T$ and $G = C$.
This implies that the ratio $(A + G) / (T + C)$ is equal to $1$,because $(A + G) = (T + C)$.
This ratio is constant for all species,although the base composition $(A + T) / (G + C)$ varies significantly between different species.

(C) In the structure of nucleosomes,the histone octamer is wrapped by $DNA$. The $DNA$ makes approximately $1.65$ turns around the histone core. Studies on the interaction between histone proteins and $DNA$ show that the $DNA$ helix interacts with the histone surface,and the orientation of the $DNA$ relative to the histone core involves specific angular positioning. Specifically,the $DNA$ path around the histone octamer is oriented at an angle of $45^o$ to the helix axis.

(A) $DNA$ acts as the primary carrier of genetic information in almost all living organisms.
It stores biological information in the form of sequences of nucleotides,which constitute genes.
$A$ gene is considered the fundamental unit of inheritance,responsible for the transmission of traits from one generation to the next.

(B) According to Chargaff's rule,the amount of Adenine $(A)$ is equal to Thymine $(T)$,and the amount of Guanine $(G)$ is equal to Cytosine $(C)$.
Also,the sum of purines $(A + G)$ is equal to the sum of pyrimidines $(T + C)$,which constitutes $50\%$ of the total $DNA$ base composition.
Given that Adenine $(A)$ = $30\%$,then Thymine $(T)$ = $30\%$.
Since $A + G = 50\%$,we have $30\% + G = 50\%$.
Therefore,$G = 50\% - 30\% = 20\%$.
Thus,the percentage of Guanine is $20\%$.

(D) In biological systems,$DNA$ methylation is an epigenetic mechanism that involves the addition of a methyl group $(-CH_3)$ to the $DNA$ molecule. This process typically occurs at the $5'$-position of the cytosine ring,resulting in $5$-methylcytosine $(5mC)$. Therefore,$DNA$ is primarily methylated at the $C$-residue (cytosine).

(B) Nitrogenous bases in $DNA$ are classified into two types: Purines and Pyrimidines.
Purines are double-ring structures,which include Adenine $(A)$ and Guanine $(G)$.
Pyrimidines are single-ring structures,which include Cytosine $(C)$ and Thymine $(T)$ in $DNA$.
Therefore,the purines of $DNA$ are Adenine and Guanine.

(C) According to the Watson-Crick model,the $DNA$ molecule consists of two long,parallel polynucleotide chains that are spirally coiled around a common axis in a regular manner to form a double helix structure.

(C) According to the base-pairing rules in $DNA$ (Chargaff's rule),Adenine $(A)$ always pairs with Thymine $(T)$,and Guanine $(G)$ always pairs with Cytosine $(C)$.
Given the sequence of one strand: $A-G-C-T-T-C-G-A$.
The complementary strand will be formed by pairing:
$A \rightarrow T$
$G \rightarrow C$
$C \rightarrow G$
$T \rightarrow A$
$T \rightarrow A$
$C \rightarrow G$
$G \rightarrow C$
$A \rightarrow T$
Therefore,the sequence of the other chain is $TCGAAGCT$.

(D) $mRNA$ stands for messenger $RNA$ (Ribonucleic Acid).
$RNA$ is a polymer composed of repeating units called ribonucleotides.
Each ribonucleotide consists of a ribose sugar,a phosphate group,and a nitrogenous base (Adenine,Guanine,Cytosine,or Uracil).
Therefore,$mRNA$ is a polymer of ribonucleotides.

(A) The standard form of $DNA$ found in living organisms is $B-DNA$.
$B-DNA$ exhibits a right-handed double helical structure.
In this structure,the two polynucleotide chains are coiled around a common axis in a right-handed direction.