The DNA Questions in English

(A) $RNA$ (Ribonucleic acid) is a polynucleotide chain. Each nucleotide unit in $RNA$ consists of a ribose sugar,a phosphate group,and a nitrogenous base (Adenine,Uracil,Guanine,or Cytosine). Therefore,$RNA$ is a polymer of ribonucleotides.

(A) According to Chargaff's rules of base pairing in $DNA$:
$A$ (Adenine) pairs with $T$ (Thymine) via $2$ hydrogen bonds.
$G$ (Guanine) pairs with $C$ (Cytosine) via $3$ hydrogen bonds.
Therefore,cytosine always pairs with guanine.

(D) $DNA$ transposition is the process which involves the movement of $DNA$ elements from one site in the genome to another.
It is mediated by transposase enzymes.
These short segments of $DNA$ with a remarkable capacity to move from one location in a chromosome to another are called transposons, jumping genes, transposable elements, or mobile genetic elements.
These were first discovered by Barbara McClintock in maize $(Zea \, mays)$, for which she was awarded the Nobel Prize in Physiology or Medicine.

(C) Histones are basic proteins found in eukaryotic chromosomes.
These proteins are rich in the positively charged basic amino acids,specifically lysine and arginine.
There are five main types of histones,namely $H_{1}, H_{2}A, H_{2}B, H_{3},$ and $H_{4}$,which are found in almost all eukaryotic cells.

(B) Jumping genes or movable genetic elements, also known as transposons, were discovered by Barbara McClintock $(1902-1992)$ in maize $(Zea \, mays)$.
These genetic elements are capable of moving from one location to another within the genome, which is why they are referred to as 'jumping genes'.
Barbara McClintock was awarded the Nobel Prize in Physiology or Medicine in $1983$ for this discovery.

(C) In $DNA$,the two strands are held together by hydrogen bonds between the nitrogenous bases. Adenine $(A)$ pairs with Thymine $(T)$ via two hydrogen bonds,and Guanine $(G)$ pairs with Cytosine $(C)$ via three hydrogen bonds. These bonds provide stability to the double-helical structure of $DNA$.

(A) $X$-ray crystallography is the study of molecular structure by examining diffraction patterns produced by $X$-rays beamed through a crystalline form of the molecules. It is widely used in biochemistry to determine the molecular structure of complex molecules such as proteins and $DNA$.

(D) Chargaff's rules are applicable to double stranded $DNA$ because,according to these rules,the amount of adenine is equal to the amount of thymine $(A = T)$ and the amount of guanine is equal to the amount of cytosine $(G = C)$.
This base pairing equality is a fundamental characteristic of the double-helical structure of $DNA$.

(D) The structure of $DNA$ is a double helix.
Each full turn of the $DNA$ helix measures approximately $3.4 \ nm$ in length.
There are $10$ base pairs present in each full turn of the $DNA$ double helix.

(A) According to the Watson and Crick model of $DNA$,it is a double helical molecule.
Each turn of the helix has a length of $34 \; \mathring{A}$ $(3.4 \; nm)$.
There are $10$ base pairs present in each full turn of the helix.
The distance between two adjacent base pairs is $3.4 \; \mathring{A}$ $(0.34 \; nm)$.
Therefore,the correct statement is that it has $10$ base pairs and $34 \; \mathring{A}$ distance for every turn.

(C) nucleosome is the basic structural unit of eukaryotic chromatin.
It consists of a segment of negatively charged $DNA$ molecule wrapped around a positively charged histone protein core.
The histone core is an octamer,composed of two copies each of four histone proteins: $H2A, H2B, H3,$ and $H4$.
Therefore,option $C$ is the correct description.

(C) nucleotide is an organic molecule consisting of a nucleoside (a nitrogenous base and a pentose sugar) connected to a phosphate group.
These nucleotides polymerize to form nucleic acids,such as $DNA$ or $RNA$.
Nucleic acids serve as the primary molecules that store and transmit genetic information in all living organisms.

(B) Histones are positively charged,basic proteins that are rich in the basic amino acids lysine and arginine.
Because they are positively charged,they bind tightly to the negatively charged phosphate backbone of $DNA$,which is acidic in nature.

(C) According to $Chargaff's$ rule,in a double-stranded $DNA$,the amount of adenine $(A)$ is equal to thymine $(T)$,and the amount of cytosine $(C)$ is equal to guanine $(G)$.
Given that cytosine $(C)$ = $20\; \%$,then guanine $(G)$ must also be $20\; \%$.
The total percentage of $C + G = 20\; \% + 20\; \% = 40\; \%$.
Since the total percentage of all four bases is $100\; \%$,the remaining percentage for adenine and thymine is $100\; \% - 40\; \% = 60\; \%$.
Since $A = T$,the percentage of adenine is $60\; \% / 2 = 30\; \%$.

(B) histone octamer is a complex of eight positively charged histone proteins (two of each $H2A, H2B, H3$ and $H4$) that aid in the packaging of $DNA$.
Negatively charged $DNA$ wraps around these histone octamers to form the nucleosome.
The $DNA$ is held there by ionic bonds.
Linker histone $H1$ binds to each nucleosome where the $DNA$ enters and exits,and this draws a string of nucleosomes closer together to form the $10 \ nm$ fibre.
The nucleosomes in chromatin are seen as a beads-on-string structure when viewed under an electron microscope.

(B) Double-stranded $DNA$ is significantly more stable than single-stranded $RNA$,which helps protect the genetic code.
Having a second complementary strand provides a template for repair in the event of mutations or chemical damage,ensuring the integrity of the genetic information over long periods.

(C) $DNA$ (deoxyribonucleic acid) consists of $3$ different components: a phosphate group,a $5$-carbon deoxyribose sugar,and a nitrogenous base.
Nitrogenous bases are classified into two types based on their structure:
$1$. Purines: These are $9$-membered,double-ringed structures. Adenine $(A)$ and Guanine $(G)$ are purines.
$2$. Pyrimidines: These are $6$-membered,single-ringed structures. Cytosine $(C)$,Thymine $(T)$,and Uracil $(U)$ are pyrimidines.
Therefore,the pair consisting of Adenine and Guanine is a purine pair.

(D) The two strands of a double helix model of $DNA$ are held together by hydrogen bonds between nitrogenous bases,which help to stabilize the structure.
Specifically,$Adenine$ $(A)$ pairs with $Thymine$ $(T)$ via two hydrogen bonds,and $Guanine$ $(G)$ pairs with $Cytosine$ $(C)$ via three hydrogen bonds.

(B) Nucleotides consist of three components: a nitrogenous base,a pentose sugar,and a phosphate group.
In a polynucleotide chain,the phosphate group of one nucleotide is linked to the $3'$-carbon of the pentose sugar of the adjacent nucleotide by a phosphodiester bond.
This bond formation involves the $5'$-phosphate group of one nucleotide and the $3'$-hydroxyl group of the next,creating the sugar-phosphate backbone of the $DNA$ or $RNA$ molecule.

(A) Nitrogenous bases are classified into Purines and Pyrimidines.
Purines include Adenine $(A)$ and Guanine $(G)$,which are common to both $DNA$ and $RNA$.
Pyrimidines include Cytosine $(C)$,Thymine $(T)$,and Uracil $(U)$.
$DNA$ contains Cytosine $(C)$ and Thymine $(T)$.
$RNA$ contains Cytosine $(C)$ and Uracil $(U)$.
Therefore,the nitrogenous bases common to both $DNA$ and $RNA$ are Adenine $(A)$,Guanine $(G)$,and Cytosine $(C)$.

(B) In a polynucleotide chain,such as $DNA$ or $RNA$,the nucleotides are linked together to form a backbone.
This linkage occurs between the $3'-$carbon atom of the sugar molecule of one nucleotide and the $5'-$carbon atom of the sugar molecule of the adjacent nucleotide through a phosphate group.
This specific covalent linkage is known as a phosphodiester bond.
As shown in the diagram,the phosphate group $(P)$ connects two sugar molecules $(S)$,forming the phosphodiester bond.

(B) In a nucleoside,the nitrogenous base (purine or pyrimidine) is linked to the pentose sugar through an $N$-glycosidic bond.
Specifically,the $N-1$ nitrogen of the pyrimidine or the $N-9$ nitrogen of the purine forms a covalent bond with the $C-1'$ carbon of the sugar molecule.
Therefore,the correct bond formed is an $N$-glycosidic bond.

(A) nucleosome consists of a core particle and linker $DNA$.
The core particle of a nucleosome is composed of an octamer of histone proteins ($H2A, H2B, H3, H4$ - two of each) around which approximately $146$ bp of $DNA$ is wrapped.
Option $A$ is correct because the histone octamer forms the central core of the nucleosome.
Option $B$ is incorrect because a complete nucleosome (including linker $DNA$) contains about $200$ bp of $DNA$,not just the core particle.
Option $C$ is incorrect as non-histone proteins are not part of the nucleosome core.
Option $D$ is incorrect because linker $DNA$ connects adjacent nucleosomes and is not part of the core particle itself.

(D) The packaging of $DNA$ is significantly more complex in eukaryotes compared to prokaryotes. In prokaryotes,$DNA$ is organized in large loops held by proteins in a region called the nucleoid,involving positively charged proteins. In eukaryotes,the organization is much more elaborate,involving a set of positively charged basic proteins called histones. $DNA$ is wrapped around the histone octamer to form a structure called a nucleosome. Further packaging requires additional sets of proteins that are collectively referred to as Non-Histone Chromosomal $(NHC)$ proteins.

(B) The $E. coli$ genome consists of $4.6 \times 10^6$ base pairs.
The distance between two consecutive base pairs is $0.34 \times 10^{-9} \; m$ or $0.34 \times 10^{-6} \; mm$.
Therefore,the total length of $DNA$ is calculated as: $4.6 \times 10^6 \times 0.34 \times 10^{-6} \; mm = 1.36 \; mm$.

(A) The two strands of $DNA$ are antiparallel and complementary to each other.
Given strand: $5^{\prime}-ATGCATCG-3^{\prime}$.
The complementary strand will run in the $3^{\prime} \rightarrow 5^{\prime}$ direction as $3^{\prime}-TACGTAGC-5^{\prime}$.
To write this in the $5^{\prime} \rightarrow 3^{\prime}$ direction,we reverse the sequence: $5^{\prime}-CGATGCAT-3^{\prime}$.
Therefore,the correct sequence is $5^{\prime}-CGATGCAT-3^{\prime}$.

(D) In the context of chromatin organization,proteins associated with $DNA$ are classified into two main types: Histones and Non-Histone Chromosomal $(NHC)$ proteins.
Histones are basic proteins rich in lysine and arginine,which are responsible for the basic packaging of $DNA$ into nucleosomes.
$NHC$ proteins are a diverse group of proteins that are required for the higher-level packaging of chromatin and also function in gene regulation and $DNA$ replication.
Therefore,the correct statement regarding $NHC$ proteins is that they are required for the packaging of chromatin at higher levels.

(D) In $DNA$,the nitrogenous base is attached to the pentose sugar via an $N$-glycosidic linkage.
For purines (Adenine and Guanine),the linkage is between the $C-1^{\prime}$ of the sugar and the $N-9$ position of the base,known as a $\beta-1^{\prime}-9-N$-glycosidic bond.
For pyrimidines (Cytosine and Thymine),the linkage is between the $C-1^{\prime}$ of the sugar and the $N-1$ position of the base,known as a $\beta-1^{\prime}-1-N$-glycosidic bond.
The nucleotides are linked together by $3^{\prime}-5^{\prime}$ phosphodiester bonds.
Therefore,a $\beta-1^{\prime}-2-N$-glycosidic bond is not present in $DNA$.

(C) Reverse transcriptase is an enzyme found in retroviruses that catalyzes the synthesis of complementary $DNA$ $(cDNA)$ from an $RNA$ template. This process is known as reverse transcription. Unlike standard transcription,which goes from $DNA$ to $RNA$,this enzyme allows the viral $RNA$ genome to be converted into $DNA$,which can then be integrated into the host cell's genome.

(A) mobile genetic element is any genetic component capable of moving itself,with or without duplication,from one site in a genome to another.
Mobile genetic elements include plasmids,viruses,transposable genetic elements (transposons),short interspersed elements,and pathogenicity islands.
The term 'transposon' was introduced by $R. W. Hedges$ and $A. E. Jacob$ in $1974$ to describe 'controlling elements' or 'jumping genes',which were originally discovered by $Barbara$ $McClintock$ in $1950$ while studying maize.

(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 Adenine $(A)$ = $30\%$,then Thymine $(T)$ must also be $30\%$.
The total percentage of $A + T = 30\% + 30\% = 60\%$.
The remaining percentage for $G + C$ is $100\% - 60\% = 40\%$.
Since $G = C$,each must be $40\% / 2 = 20\%$.
Therefore,$T = 30\%, G = 20\%, C = 20\%$.

(B) Histones are basic proteins that are positively charged.
They are rich in basic amino acids,specifically Lysine and Arginine.
These proteins are organized to form a unit of $8$ molecules known as a histone octamer.
Since they are basic proteins,their $pH$ is alkaline (basic),not acidic.
Therefore,the statement that the $pH$ of histones is slightly acidic is incorrect.

(A) Statement $(a)$ is correct: Euchromatin is loosely packed and stains lightly.
Statement $(b)$ is incorrect: Heterochromatin is densely packed and transcriptionally inactive,whereas euchromatin is transcriptionally active.
Statement $(c)$ is correct: The negatively charged $DNA$ is wrapped around the positively charged histone octamer to form a nucleosome.
Statement $(d)$ is correct: Histone proteins are basic proteins rich in the basic amino acids lysine and arginine.
Statement $(e)$ is incorrect: $A$ typical nucleosome contains $200 \ bp$ of $DNA$ helix,not $400 \ bp$.
Therefore,the correct statements are $(a), (c),$ and $(d)$.

(D) The distance between two consecutive base pairs in a $DNA$ molecule is $0.34 \text{ nm}$ or $0.34 \times 10^{-9} \text{ m}$.
Given the total length of the $DNA$ molecule is $1.1 \text{ m}$.
The number of base pairs is calculated by dividing the total length by the distance between two consecutive base pairs:
$\text{Number of base pairs} = \frac{\text{Total length}}{\text{Distance between two base pairs}}$
$\text{Number of base pairs} = \frac{1.1 \text{ m}}{0.34 \times 10^{-9} \text{ m}}$
$\text{Number of base pairs} \approx 3.235 \times 10^{9} \text{ bp}$.
Rounding this to the nearest option,we get $3.3 \times 10^{9} \text{ bp}$.

(A) $DNA$ (Deoxyribonucleic acid) contains the nitrogenous bases Adenine $(A)$,Guanine $(G)$,Cytosine $(C)$,and Thymine $(T)$.
$RNA$ (Ribonucleic acid) contains the nitrogenous bases Adenine $(A)$,Guanine $(G)$,Cytosine $(C)$,and Uracil $(U)$.
Therefore,the correct sequence for $DNA$ and $RNA$ is $A, G, C, T$ and $A, G, C, U$ respectively.

(C) nucleoside is formed by the combination of a nitrogenous base and a pentose sugar.
In $DNA$,the pentose sugar is deoxyribose.
When deoxyribose sugar combines with nitrogenous bases (Adenine,Guanine,Cytosine,and Thymine),they form deoxyadenosine,deoxyguanosine,deoxycytidine,and deoxythymidine,respectively.
These are collectively known as the nucleosides of $DNA$.

(D) The compounds mentioned (deoxyadenylic acid,deoxyguanylic acid,deoxycytidylic acid,and deoxythymidylic acid) are the monomeric units of $DNA$.
These compounds are formed by the combination of a nitrogenous base,a deoxyribose sugar,and a phosphate group.
$A$ molecule consisting of a nitrogenous base,a sugar,and a phosphate group is called a nucleotide.
Since these specific molecules contain deoxyribose sugar,they are the nucleotides of $DNA$.

(A) The structure labeled $P$ is a double-ring structure,which is characteristic of a purine. Specifically,it is Adenine,which has an amino group $(-NH_2)$ attached to the purine ring.
The structure labeled $Q$ is a single-ring structure,which is characteristic of a pyrimidine. Specifically,it is Uracil,which contains two carbonyl groups $(C=O)$ and lacks the methyl group found in Thymine.
Therefore,$P$ is Adenine and $Q$ is Uracil.

(D) Nitrogenous bases in nucleic acids are classified into two categories based on their chemical structure:
$1$. Purines: These are double-ring structures (bicyclic). Adenine $(A)$ and Guanine $(G)$ are purines.
$2$. Pyrimidines: These are single-ring structures (monocyclic). Cytosine $(C)$,Thymine $(T)$,and Uracil $(U)$ are pyrimidines.
Therefore,the correct classification is Purines $= A, G$ and Pyrimidines $= C, T, U$.

(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 $A = 20 \%$,then $T = 20 \%$.
The sum of all four bases is $A + T + G + C = 100 \%$.
Substituting the known values: $20 \% + 20 \% + G + C = 100 \%$.
$40 \% + G + C = 100 \%$.
$G + C = 60 \%$.
Since $G = C$,we have $2G = 60 \%$.
Therefore,$G = 30 \%$.

(A) The $DNA$ molecule consists of two polynucleotide chains. Each nucleotide is composed of a deoxyribose sugar,a phosphate group,and a nitrogenous base. The backbone of the $DNA$ strand is formed by the alternating sequence of deoxyribose sugar and phosphate groups linked by phosphodiester bonds. The nitrogenous bases project inward from this backbone.

(D) In the double-helical structure of $DNA$,the backbone is formed by the sugar-phosphate chain,which is located on the outside of the helix.
The nitrogenous bases (adenine,guanine,cytosine,and thymine) are attached to the sugar and are directed inwards towards the central axis.
These nitrogenous bases form hydrogen bonds with their complementary bases on the opposite strand,maintaining the stability of the $DNA$ structure.

(B) In the structure of $DNA$,nitrogenous bases pair with each other through hydrogen bonds.
According to Chargaff's rules and the Watson-Crick model of $DNA$,adenine $(A)$ always pairs with thymine $(T)$ via $2$ hydrogen bonds.
Conversely,guanine $(G)$ pairs with cytosine $(C)$ via $3$ hydrogen bonds.
Therefore,the correct number of hydrogen bonds between adenine and thymine is $2$.

(C) In the structure of $DNA$,nitrogenous bases pair specifically through hydrogen bonds.
According to Watson and Crick's base-pairing rules,guanine $(G)$ always pairs with cytosine $(C)$ via $3$ hydrogen bonds.
Conversely,adenine $(A)$ pairs with thymine $(T)$ via $2$ hydrogen bonds.
Therefore,the correct number of hydrogen bonds between guanine and cytosine is $3$.

(A) According to the Watson-Crick model of $DNA$ structure:
$1$. The $DNA$ molecule forms a double helix.
$2$. One full turn of the helix (pitch) measures $34 \mathring A$ $(3.4 \ nm)$.
$3$. There are $10$ base pairs present in each full turn of the $DNA$ helix.
$4$. Therefore,the distance between two adjacent base pairs is $3.4 \mathring A$ $(0.34 \ nm)$.
$5$. Thus,the correct values are $34 \mathring A$ and $10$ base pairs.

(B) According to the Watson-Crick model of $DNA$ structure,the $DNA$ molecule consists of two polynucleotide strands that form a double helix.
The diameter of this double helix is constant throughout its length.
The distance between the two polynucleotide strands (the diameter of the $DNA$ helix) is $20 \mathring A$ $(2 \text{ nm})$.
Therefore,the correct option is $B$.

(B) dinucleotide is formed when two nucleotides are linked together by a phosphodiester bond.
This bond is formed between the $3'-OH$ group of the sugar moiety of one nucleotide and the $5'$-phosphate group of the sugar moiety of the next nucleotide.
This linkage creates a sugar-phosphate backbone in the nucleic acid chain.

(D) gene is the fundamental physical and functional unit of heredity.
It is a specific sequence of nucleotides in $DNA$ that encodes the synthesis of a gene product,either $RNA$ or protein.
In classical genetics,it is referred to as a 'factor' that controls the expression of specific traits in an organism.
Therefore,all the given statements correctly describe a gene.

(A) The term $Genome$ refers to the complete set of genetic material $(DNA)$ present in an organism or cell. It includes all the genes as well as the non-coding sequences of the $DNA$. Therefore,it represents the entire genetic blueprint of an organism.

(D) $DNA$ molecule is a polymer of nucleotides. Each nucleotide consists of three components: a nitrogenous base $(NB)$,a pentose sugar $(S)$,and a phosphate group $(P)$.
In a nucleotide,the nitrogenous base is attached to the $1'$ carbon of the sugar,and the phosphate group is attached to the $5'$ carbon of the sugar.
These nucleotides are linked together by phosphodiester bonds to form a polynucleotide chain. The backbone of the $DNA$ strand is formed by the alternating sugar and phosphate groups $(S-P-S-P...)$,with the nitrogenous bases $(NB)$ attached to each sugar molecule.
Looking at the options,the structure where the phosphate group $(P)$ is attached to the sugar $(S)$,and the nitrogenous base $(NB)$ is also attached to the sugar $(S)$ in a repeating sequence represents the correct backbone structure of $DNA$.