The DNA Questions in English

(C) The length of $DNA$ is calculated by multiplying the total number of base pairs by the distance between two consecutive base pairs.
Given:
Total number of base pairs = $6.6 \times 10^9 \ bp$
Distance between two base pairs = $0.34 \ nm = 0.34 \times 10^{-9} \ m$
Length of $DNA$ = $(\text{Total number of base pairs}) \times (\text{Distance between two base pairs})$
Length of $DNA$ = $(6.6 \times 10^9) \times (0.34 \times 10^{-9} \ m)$
Length of $DNA$ = $6.6 \times 0.34 \ m$
Length of $DNA$ = $2.244 \ m \approx 2.2 \ m$.
Therefore,the correct option is $C$.

$(4 \times 10^6 \; BP)$ The length of $DNA$ is calculated by multiplying the number of base pairs by the distance between two consecutive base pairs.
Given:
Length of $E. \; coli$ $DNA$ $= 1.36 \; mm = 1.36 \times 10^{-3} \; m$.
Distance between two consecutive base pairs $= 0.34 \; nm = 0.34 \times 10^{-9} \; m$.
Number of base pairs $= \frac{\text{Total length of DNA}}{\text{Distance between two consecutive base pairs}}$.
Number of base pairs $= \frac{1.36 \times 10^{-3} \; m}{0.34 \times 10^{-9} \; m} = 4 \times 10^{6} \; bp$.

(N/A) $1$. Bacteriophages: These are viruses that infect bacteria. They are commonly used in experiments to study genetic material.
$2$. $DNA$ (Deoxyribonucleic acid): It acts as the genetic material in most organisms. It is a long polymer of deoxyribonucleotides.

(N/A) $1$. Nucleoside: $A$ nucleoside is formed when a nitrogenous base is linked to a pentose sugar through an $N$-glycosidic linkage.
$2$. Histones: Histones are a set of positively charged basic proteins. They are rich in the basic amino acid residues,specifically lysines and arginines.

(N/A) $1$. $Histone$ $octamer$: Histones are positively charged basic proteins that are organized to form a unit of eight molecules,known as a histone octamer. The negatively charged $DNA$ is wrapped around this positively charged histone octamer to form a structure called a nucleosome.
$2$. $Euchromatin$: In a typical nucleus,some regions of chromatin are loosely packed. These regions stain light and are referred to as euchromatin. Euchromatin is considered transcriptionally active chromatin because the $DNA$ is accessible for the transcription process.

(N/A) Gene: $A$ gene is defined as the functional unit of inheritance. It is a segment of $DNA$ that codes for a polypeptide,$rRNA$,or $tRNA$ molecule.
Split gene: In eukaryotes,the coding sequences or expressed sequences are defined as exons. Exons are interrupted by introns or intervening sequences that do not appear in mature or processed $RNA$. This arrangement of genes containing both exons and introns is known as a split gene,which is a characteristic feature of eukaryotic $DNA$.

(A) The full form of $NHC$ is $Non-histone \text{ } chromosomal \text{ } proteins$.
These are a diverse group of proteins associated with $DNA$ in the chromatin, which play crucial roles in the regulation of gene expression, $DNA$ replication, and chromosome structure.

(A) The full form of $mRNA$ is messenger $RNA$. It carries the genetic information from $DNA$ to the ribosome for protein synthesis.
The full form of $tRNA$ is transfer $RNA$. It acts as an adapter molecule that brings specific amino acids to the ribosome during the process of translation.

(A) $DNA$ stands for Deoxyribonucleic acid.
$RNA$ stands for Ribonucleic acid.
These are the two primary types of nucleic acids found in living organisms, responsible for storing and transmitting genetic information.

(B) In $1953$,James Watson and Francis Crick proposed the double helix model of $DNA$ based on $X$-ray diffraction data produced by Rosalind Franklin and Maurice Wilkins. This model explained the structure of $DNA$ and how it could store and replicate genetic information.

(A) $DNA$ (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 present in $DNA$.

(C) $RNA$ stands for Ribonucleic Acid.
It is a polymer of nucleotides.
Each nucleotide in $RNA$ consists of a nitrogenous base,a phosphate group,and a pentose sugar.
The specific pentose sugar present in $RNA$ is Ribose sugar,which contains an $-OH$ group at the $2'$ carbon position,distinguishing it from the Deoxyribose sugar found in $DNA$.

(B) Thymine ($5$-methyluracil) is a pyrimidine nitrogenous base specifically found in $DNA$.
In $RNA$,uracil is present instead of thymine.
Therefore,thymine is a characteristic component of $DNA$.

(C) According to $Chargaff's$ rule and the structure of $DNA$ proposed by $Watson$ and $Crick$,nitrogenous bases pair specifically through hydrogen bonds.
$Adenine$ $(A)$ always pairs with $Thymine$ $(T)$ via two hydrogen bonds.
$Guanine$ $(G)$ always pairs with $Cytosine$ $(C)$ via three hydrogen bonds.
Therefore,the correct answer is $Thymine$.

(B) $DNA$ $(Deoxyribonucleic \text{ acid})$ molecule consists of two polynucleotide chains that are coiled around a common axis to form a double helix structure.
These two strands run in an anti-parallel direction, meaning one strand runs in the $5' \to 3'$ direction and the other in the $3' \to 5'$ direction.
The nitrogenous bases of the two strands are linked together by hydrogen bonds, forming the rungs of the ladder-like structure.

(A) polynucleotide is a long chain polymer composed of many nucleotide monomers covalently bonded together in a chain. Each nucleotide consists of three components: a nitrogenous base,a pentose sugar (ribose or deoxyribose),and a phosphate group. Therefore,polynucleotides are polymers of nucleotides.

(A) Nitrogenous bases in $DNA$ and $RNA$ 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 $(C)$,Thymine $(T)$,and Uracil $(U)$.
Therefore,among the given options,Adenine is a purine.

(C) According to the base-pairing rules established by Watson and Crick,nitrogenous bases in $DNA$ pair specifically through hydrogen bonds.
Cytosine $(C)$ always pairs with Guanine $(G)$ via three hydrogen bonds.
Adenine $(A)$ pairs with Thymine $(T)$ via two hydrogen bonds in $DNA$.
Therefore,the correct option is $C$.

(B) In the structure of $DNA$,nitrogenous bases pair specifically according to Chargaff's rules. Adenine $(A)$ always pairs with thymine $(T)$ via two hydrogen bonds,while guanine $(G)$ pairs with cytosine $(C)$ via three hydrogen bonds. Therefore,adenine is connected to thymine by two hydrogen bonds.

(B) Nitrogenous bases are classified into two types: Purines and Pyrimidines.
Purines include Adenine $(A)$ and Guanine $(G)$,which are present in both $DNA$ and $RNA$.
Pyrimidines include Cytosine $(C)$,Thymine $(T)$,and Uracil $(U)$.
Cytosine $(C)$ is common to both $DNA$ and $RNA$.
Thymine $(T)$ is present only in $DNA$,while Uracil $(U)$ is present only in $RNA$.
Therefore,Cytosine is the pyrimidine base found in both $DNA$ and $RNA$.

(C) In the structure of $DNA$, nitrogenous bases pair specifically through hydrogen bonds.
$Adenine$ $(A)$ pairs with $Thymine$ $(T)$ via $2$ hydrogen bonds.
$Guanine$ $(G)$ pairs with $Cytosine$ $(C)$ via $3$ hydrogen bonds.
Therefore, there are $3$ hydrogen bonds between $Guanine$ and $Cytosine$.

(A) nucleoside is formed by the attachment of a nitrogenous base to the $1'$ position of a pentose sugar (ribose or deoxyribose) via an $N$-glycosidic linkage.
Therefore,the nitrogenous base is attached to the first carbon $(C-1')$ of the pentose sugar.

(A) In a nucleotide,the nitrogenous base is linked to the $1'$ carbon of the pentose sugar (ribose or deoxyribose) through an $N$-glycosidic linkage to form a nucleoside.
Subsequently,the phosphate group is linked to the $5'$-$OH$ of the nucleoside through a phosphoester linkage to form a nucleotide.
Therefore,the bond connecting the nitrogenous base to the pentose sugar is the $N$-glycosidic bond.

(D) In a nucleotide,the pentose sugar is a $5$-carbon sugar. The nitrogenous base is attached to the $1'$ carbon position of the sugar. The phosphate group is attached to the $5'$ carbon position of the pentose sugar via a phosphoester bond. Therefore,the correct position for the phosphate attachment is the $5^{th}$ carbon.

(B) In $DNA$,the four nitrogenous bases are Adenine $(A)$,Guanine $(G)$,Cytosine $(C)$,and Thymine $(T)$.
$dAMP$ (deoxyadenosine monophosphate),$dGMP$ (deoxyguanosine monophosphate),and $dTMP$ (deoxythymidine monophosphate) are the deoxyribonucleotides found in $DNA$.
$dUMP$ (deoxyuridine monophosphate) is not a standard component of $DNA$; instead,Uracil $(U)$ is found in $RNA$ as a ribonucleotide $(UMP)$.
Therefore,$dUMP$ is the odd one out.

(A) In $RNA$,the four nitrogenous bases are Adenine $(A)$,Guanine $(G)$,Cytosine $(C)$,and Uracil $(U)$.
$5-$methyl uracil is another name for Thymine $(T)$,which is found in $DNA$ but not in $RNA$.
Therefore,$5-$methyl uracil is the nitrogenous base that is not present in $RNA$.

(D) $RNA$ (Ribonucleic acid) and $DNA$ (Deoxyribonucleic acid) differ in two primary structural components:
$1$. Pentose Sugar: $RNA$ contains ribose sugar,whereas $DNA$ contains deoxyribose sugar.
$2$. Nitrogenous Bases: $RNA$ contains Uracil $(U)$ instead of Thymine $(T)$,which is found in $DNA$.
Since the question asks for the difference,both the pentose sugar and the pyrimidine base (Uracil vs Thymine) are key distinguishing factors. Therefore,option $D$ is the most comprehensive answer.

(D) The backbone of a polynucleotide chain (like $DNA$ or $RNA$) is formed by the sugar-phosphate chain.
$1$. Each nucleotide consists of a nitrogenous base,a pentose sugar,and a phosphate group.
$2$. The sugar of one nucleotide is linked to the phosphate group of the adjacent nucleotide through a phosphodiester bond.
$3$. This alternating sequence of sugar and phosphate molecules creates the structural backbone of the polynucleotide strand,while the nitrogenous bases project inward from this backbone.

(B) The $X-ray$ diffraction data for $DNA$ was produced by Maurice Wilkins and Rosalind Franklin.
This data was crucial for James Watson and Francis Crick to propose the double helix model of $DNA$ in $1953$.

(C) James Watson and Francis Crick proposed the famous double helix model of $DNA$ in $1953$.
This model was based on $X$-ray diffraction data produced by Maurice Wilkins and Rosalind Franklin.
For this discovery,Watson,Crick,and Wilkins were awarded the Nobel Prize in Physiology or Medicine in $1962$.

(C) According to Chargaff's rules for double-stranded $DNA$,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 $(C)$.
Therefore,the ratios $A/T = 1$ and $G/C = 1$.
However,the ratio of Adenine to Guanine $(A/G)$ or Adenine to Cytosine $(A/C)$ is not constant and varies between different species.
Thus,the ratio of Adenine and Guanine or Adenine and Cytosine is not constant and not equal to each other.

(C) The double helix model of $DNA$ proposed by James Watson and Francis Crick was primarily based on the $X$-ray diffraction data produced by Rosalind Franklin and Maurice Wilkins.
Erwin Chargaff's observations regarding the base pairing rules ($A=T$ and $G=C$) also provided crucial evidence for the structure.
However,the structural model itself was fundamentally derived from the diffraction patterns obtained by Rosalind Franklin.

(D) According to Chargaff's rules for double-stranded $DNA$:
$1$. The amount of Adenine $(A)$ is equal to the amount of Thymine $(T)$,so $A = T$.
$2$. The amount of Cytosine $(C)$ is equal to the amount of Guanine $(G)$,so $C = G$.
$3$. Consequently,the sum of purines $(A + G)$ is equal to the sum of pyrimidines $(T + C)$,so $A + G = T + C$.
$4$. The ratio $(A + T) / (G + C)$ is constant for a given species but varies between different species. Therefore,the statement $A + T = G + C$ is not a universal rule and is generally incorrect.

(C) According to $Chargaff's$ rule,in a double-stranded $DNA$ molecule,the amount of Adenine $(A)$ is equal to Thymine $(T)$,and the amount of Cytosine $(C)$ is equal to Guanine $(G)$.
Given: $A = 166$ and $C = 144$.
Therefore,$T = A = 166$ and $G = C = 144$.
The total number of nitrogenous bases is the sum of all four bases: $A + T + G + C = 166 + 166 + 144 + 144 = 620$.
The number of base pairs is half the total number of bases: $620 / 2 = 310$.

(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 amount of guanine $(G)$ is $20 \%$,then the amount of cytosine $(C)$ must also be $20 \%$.
The total percentage of $G + C$ is $20 \% + 20 \% = 40 \%$.
Since the total percentage of all four bases $(A + T + G + C)$ is $100 \%$,the sum of $A + T$ is $100 \% - 40 \% = 60 \%$.
Since $A = T$,the amount of thymine $(T)$ is $60 \% / 2 = 30 \%$.

(D) In the $DNA$ double helix structure,the two polynucleotide chains are antiparallel to each other.
These two chains are held together by hydrogen bonds between the nitrogenous bases.
Specifically,adenine $(A)$ pairs with thymine $(T)$ via two hydrogen bonds,and guanine $(G)$ pairs with cytosine $(C)$ via three hydrogen bonds.
Therefore,the correct option is $D$.

(B) According to Chargaff's rule,the total amount of purines equals the total amount of pyrimidines,and the sum of all four bases is $100\%$.
Given that the sum of cytosine $(C)$ and guanine $(G)$ is $40\%$,we have $C + G = 40\%$.
Since the total percentage of all bases $(A + T + C + G)$ is $100\%$,we can write $A + T + 40\% = 100\%$.
Therefore,$A + T = 100\% - 40\% = 60\%$.
According to the base pairing rules,the amount of adenine $(A)$ is equal to the amount of thymine $(T)$.
Thus,$2A = 60\%$,which means $A = 30\%$.
Therefore,the content of adenine is $30\%$.

(A) In the $B-DNA$ structure,the distance between two consecutive base pairs is $0.34 \, nm$ $(3.4 \, \mathring{A})$.
One full turn of the $DNA$ helix contains $10$ base pairs.
Therefore,the pitch of one turn is calculated as $10 \times 0.34 \, nm = 3.4 \, nm$ (or $34 \, \mathring{A}$).
Since $3.4 \, nm$ is equivalent to $34 \, \mathring{A}$,and the options provided list $3.4 \, nm$ as the correct value for the pitch,option $A$ is the correct answer.

(B) In the $B-DNA$ model proposed by Watson and Crick, the structure is a double helix.
Each full turn of the helix measures approximately $3.4 \, nm$ $(34 \, \text{\AA})$.
The distance between two consecutive base pairs is $0.34 \, nm$ $(3.4 \, \text{\AA})$.
Therefore, the number of base pairs per turn is calculated as $\frac{3.4 \, nm}{0.34 \, nm} = 10$ base pairs.

(A) In a polynucleotide chain,the sugar-phosphate backbone is formed by the linkage of nucleotides.
Each nucleotide consists of a pentose sugar,a nitrogenous base,and a phosphate group.
The $5'$ carbon of the pentose sugar is attached to the phosphate group,while the $3'$ carbon is attached to the $OH$ group.
Therefore,at the $5'$ end of a polynucleotide chain,a free phosphate group is present.

(C) In a $B-DNA$ double helix, the distance between two consecutive base pairs is $0.34 \, \text{nm}$ (nanometers).
This is equivalent to $3.4 \, \mathring{A}$.
Since $1 \, \text{nm} = 10^{-9} \, \text{m}$ and $1 \, \mathring{A} = 10^{-10} \, \text{m}$, the value $0.34 \, \text{nm}$ is the standard measurement used in molecular biology for the rise per base pair.

(B) Watson and Crick proposed the double-helical structure of $DNA$.
$1$. The two strands of $DNA$ are antiparallel,meaning one runs in the $5' \rightarrow 3'$ direction and the other in the $3' \rightarrow 5'$ direction.
$2$. Both strands form a right-handed helix. Therefore,the statement that they are not right-handed is incorrect.
$3$. According to Chargaff's rule,purines (Adenine and Guanine) always pair with pyrimidines (Thymine and Cytosine) via hydrogen bonds.
$4$. The pitch of the helix is $3.4 \ nm$ or $34 \ \mathring{A}$,and there are $10$ base pairs per turn.

(D) Deoxyribonucleic acid $(DNA)$ is a large molecule composed of long chains of repeating units called nucleotides.
Since it consists of many repeating monomeric units (nucleotides) linked together by phosphodiester bonds,it is classified as a polymer.
Specifically,it is a polynucleotide.

(A) The bacteriophage $\phi \times 174$ is a well-known virus that infects bacteria.
According to the $NCERT$ textbook,the genome of $\phi \times 174$ consists of $5386$ nucleotides.
This is a single-stranded $DNA$ virus.
Therefore,the correct option is $A$.

(B) The genome size of $E. coli$ (Escherichia coli) is $4.6 \times 10^6 \, bp$.
- Human $DNA$ consists of $3.3 \times 10^9 \, bp$ (haploid content).
- $\phi \times 174$ bacteriophage has $5386$ nucleotides.
- Bacteriophage lambda has $48502 \, bp$.

(D) The $DNA$ molecule consists of two polynucleotide chains that form a double helix.
These two strands are held together by hydrogen bonds between the nitrogenous bases.
Specifically,adenine $(A)$ pairs with thymine $(T)$ via two hydrogen bonds,and guanine $(G)$ pairs with cytosine $(C)$ via three hydrogen bonds.
These hydrogen bonds are crucial for maintaining the structural integrity and stability of the helical shape of the $DNA$ molecule.

(D) The length of $DNA$ in different organisms is as follows:
$1$. $\phi \times 174$ bacteriophage: $5386$ nucleotides.
$2$. $E. coli$: $4.6 \times 10^6$ base pairs.
$3$. Human ($haploid$ content): $3.3 \times 10^9$ base pairs.
Comparing these values, the ascending order is: $\phi \times 174$ ($5386$ bp) < $E. coli$ ($4.6 \times 10^6$ bp) < Human ($3.3 \times 10^9$ bp).
Therefore, the correct sequence is $\phi \times 174 \rightarrow E. coli \rightarrow$ Human.

(C) The Central Dogma of molecular biology was proposed by Francis Crick in $1958$.
It states that the flow of genetic information in a biological system follows the path: $DNA \rightarrow RNA \rightarrow Protein$.
This process involves replication of $DNA$,transcription of $DNA$ into $RNA$,and translation of $RNA$ into $Protein$.

(B) In a $DNA$ double helix,the distance between two consecutive base pairs is approximately $0.34 \ nm$ (nanometers).
Since $1 \ nm = 10^{-9} \ m$,the distance can be expressed as $0.34 \times 10^{-9} \ m$.
Therefore,the correct option is $B$.

(D) The total length of $DNA$ in a double-helical structure is determined by multiplying the total number of nitrogen base pairs by the distance between two consecutive base pairs.
In $B-DNA$,the distance between two consecutive base pairs is approximately $0.34 \ nm$ $(0.34 \times 10^{-9} \ m)$.
Therefore,the formula is: $\text{Total length of } DNA = \text{Total number of base pairs} \times \text{Distance between two consecutive base pairs}$.