The DNA Questions in English

(A) $DNA$ stands for Deoxyribonucleic Acid. It is a long chain or polymer composed of repeating units called deoxyribonucleotides. Each deoxyribonucleotide consists of a deoxyribose sugar,a phosphate group,and a nitrogenous base. Therefore,$DNA$ is defined as a long polymer of deoxyribonucleotides.

(B) The correct matches based on the genetic material content are as follows:
$1$. Bacteriophage $\phi \times 174$ contains $5386$ nucleotides $(P-I)$.
$2$. Bacteriophage lambda contains $48502$ base pairs $(bp)$ $(Q-II)$.
$3$. Escherichia coli $(E. coli)$ contains $4.6 \times 10^6 \, bp$ $(R-IV)$.
$4$. Human haploid $DNA$ content is $3.3 \times 10^9 \, bp$ $(S-III)$.
Therefore,the correct matching sequence is $(P-I), (Q-II), (R-IV), (S-III)$.

(B) In a polynucleotide chain,nitrogenous bases are attached to the $1'$ carbon of the pentose sugar.
Purines are double-ring structures (Adenine and Guanine),while pyrimidines are single-ring structures (Cytosine,Thymine,and Uracil).
In the provided diagram,the shapes representing $P$ and $R$ are larger (double-ring structure),indicating they are purines.
The shapes representing $Q$ and $S$ are smaller (single-ring structure),indicating they are pyrimidines.
Therefore,$P$ and $R$ represent purine nitrogenous bases.

(B) polynucleotide chain consists of a sugar-phosphate backbone with nitrogenous bases attached to the sugar moiety.
Specifically,the backbone is formed by the alternating sequence of pentose sugar and phosphate groups linked by phosphodiester bonds.
The nitrogenous bases project inward from this backbone.

(C) The $DNA$ molecule is a polymer of nucleotides. Each nucleotide consists of a nitrogenous base,a pentose sugar (deoxyribose),and a phosphate group.
The phosphate group $(PO_4^{3-})$ is derived from phosphoric acid $(H_3PO_4)$.
In the $DNA$ backbone,the phosphate groups link the sugar molecules via phosphodiester bonds.
Because these phosphate groups retain acidic properties (they release protons,$H^+$),they confer an overall negative charge and acidic nature to the $DNA$ molecule.

(B) Erwin Chargaff observed that in $DNA$,the amount of Adenine $(A)$ is always equal to the amount of Thymine $(T)$,and the amount of Guanine $(G)$ is always equal to the amount of Cytosine $(C)$.
This is known as Chargaff's rule,which states that the ratio of $A+G$ to $T+C$ is $1$,or more specifically,the base pairing rules are $A=T$ and $G=C$.
These observations were crucial for Watson and Crick to propose the double-helical structure of $DNA$.

(A) The $DNA$ molecule consists of a sugar-phosphate backbone and nitrogenous bases.
While the sugar (deoxyribose) and the phosphate group are identical in all $DNA$ molecules,the sequence of nitrogenous bases ($Adenine$,$Thymine$,$Guanine$,and $Cytosine$) varies significantly among different organisms.
This specific sequence of nitrogenous bases acts as the genetic code,which determines the biological characteristics of an organism.
Therefore,the variation in $DNA$ is primarily due to the sequence of nitrogenous bases.

(D) In a $DNA$ polynucleotide chain,the sugar-phosphate backbone is formed by phosphodiester bonds.
At the $5^{\prime}$ end of the polynucleotide chain,the $5^{\prime}$ carbon of the deoxyribose sugar is attached to a free phosphate group.
At the $3^{\prime}$ end of the polynucleotide chain,the $3^{\prime}$ carbon of the deoxyribose sugar has a free hydroxyl $(-OH)$ group.
Therefore,the end with the free phosphate group is the $5^{\prime}$ end $(P = 5^{\prime})$ and the end with the free $-OH$ group is the $3^{\prime}$ end $(Q = 3^{\prime})$.

(A) Thymine is a pyrimidine base found in $DNA$.
Chemically,it is known as $5-$methyl uracil.
Uracil is found in $RNA$,while in $DNA$,the $5^{th}$ carbon of the uracil ring is methylated to form thymine.

(D) The $X$-ray diffraction data for $DNA$ was produced by Maurice Wilkins and Rosalind Franklin. This critical data provided the structural insights that allowed James Watson and Francis Crick to propose the double helix model of $DNA$ in $1953$.

(A) In molecular biology,a template is a strand of nucleic acid ($DNA$ or $RNA$) that serves as a pattern or guide for the synthesis of a complementary strand. During processes like replication or transcription,the sequence of nucleotides on the template strand determines the sequence of nucleotides in the newly synthesized polymer,ensuring genetic information is accurately copied.

(D) In the $B-DNA$ double helix model proposed by Watson and Crick, the distance between two consecutive base pairs is $0.34 \, nm$ (or $3.4 \, \mathring{A}$).
This distance is calculated by dividing the total pitch of one full turn of the helix $(3.4 \, nm)$ by the number of base pairs per turn $(10)$.
Therefore, the distance between two adjacent base pairs is $0.34 \, nm$.

(C) The structure of $DNA$ ($B$-$DNA$) consists of a double helix.
One full turn of the $DNA$ helix corresponds to a distance of $3.4\,nm$ (or $34\,\mathring{A}$).
Each turn contains $10$ base pairs $(bp)$.
Therefore,the distance between two consecutive base pairs is $0.34\,nm$ $(3.4\,nm / 10)$.
Thus,one turn has $10\,bp$ and a length of $3.4\,nm$.

(C) In a $DNA$ molecule,the two polynucleotide 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.
The label $P$ in the figure points to the three dotted lines between Guanine $(G)$ and Cytosine $(C)$,which represent the three hydrogen bonds connecting these complementary bases.

(D) The provided figure shows the double-helical structure of $DNA$. In this structure,the horizontal rungs connecting the two sugar-phosphate backbones represent the nitrogenous base pairs. The label $P$ points to these horizontal rungs,which are formed by hydrogen bonding between complementary nitrogenous bases (e.g.,Adenine with Thymine,Guanine with Cytosine). Therefore,$P$ represents a base pair.

(B) The distance between two consecutive base pairs in $DNA$ is $0.34 \,nm$ or $0.34 \times 10^{-6} \,mm$.
Total length of $DNA = 2.72 \,mm$.
Number of base pairs = $\frac{\text{Total length of DNA}}{\text{Distance between two consecutive base pairs}}$.
Number of base pairs = $\frac{2.72 \,mm}{0.34 \times 10^{-6} \,mm}$.
Number of base pairs = $\frac{2.72}{0.34} \times 10^6$.
Number of base pairs = $8 \times 10^6$ base pairs.

(B) The packaging of $DNA$ in eukaryotes involves several levels of organization:
$1$. $DNA$ helix wraps around histone octamers to form nucleosomes (beads-on-a-string structure).
$2$. These nucleosomes further coil and condense to form chromatin fibers.
$3$. These fibers are further organized into chromatin,which eventually condenses during cell division to form the highly compact structure known as the chromosome.
Therefore,the correct sequence is: $DNA$ $\rightarrow$ Nucleosome $\rightarrow$ Chromatin $\rightarrow$ Chromatin fibers $\rightarrow$ Chromosome.

(C) nucleosome consists of $DNA$ wrapped around a histone octamer.
In a typical nucleosome,the $DNA$ segment consists of approximately $146$ base pairs.
Since the distance between two adjacent base pairs in a $B-DNA$ helix is $0.34 \, nm$ $(3.4 \, \mathring{A})$,the total length of the $DNA$ can be calculated as:
Length $= 146 \times 0.34 \, nm \approx 49.64 \, nm$.
However,considering the linker $DNA$ and the complete structure,the standard value often cited in textbooks for the core $DNA$ length is approximately $146$ base pairs,which corresponds to roughly $50 \, nm$. Among the given options,$17 \, nm$ is incorrect,$3.4 \, nm$ is the length of one turn ($10$ base pairs),and $34 \, nm$ is a common approximation used in some contexts for the total length of the $DNA$ associated with the nucleosome core and linker. Given the standard curriculum,$34 \, nm$ is the most appropriate choice among the provided options.

(B) Euchromatin is the loosely packed region of chromatin that stains lightly and is transcriptionally active. Thus,it corresponds to $I, IV, V$.
Heterochromatin is the densely packed region of chromatin that stains darkly and is transcriptionally inactive. Thus,it corresponds to $II, III, VI$.
Therefore,the correct matching is Euchromatin: $I, IV, V$ and Heterochromatin: $II, III, VI$.

(B) The image shows a nucleosome,which is the basic unit of $DNA$ packaging in eukaryotes.
It consists of a segment of $DNA$ wrapped around a histone octamer.
The length of $DNA$ wrapped around the histone octamer is approximately $146$ base pairs,which is often rounded to $147$ or $150$ base pairs in many textbook contexts for simplicity.
Among the given options,$150$ base pairs is the standard accepted value for the $DNA$ wrapped around the histone core in a nucleosome.

(A) histone octamer is the fundamental unit of chromatin structure,known as a nucleosome.
It consists of a complex of $8$ histone proteins.
These proteins are $2$ molecules each of $H_2A, H_2B, H_3,$ and $H_4$.
$H_1$ histone is not part of the octamer core but binds to the $DNA$ where it enters and leaves the nucleosome.

(A) Histones are basic proteins that are rich in the basic amino acids $Lysine$ and $Arginine$.
These amino acids carry positive charges in their side chains.
Because $DNA$ is negatively charged due to its phosphate backbone,these positively charged histone proteins help in the packaging of $DNA$ into structures called nucleosomes.

(C) $DNAs$ is simply the plural form of the $DNA$ molecule,referring to multiple strands or copies of Deoxyribonucleic Acid.
$DNAase$ (or $DNase$) is an enzyme that catalyzes the hydrolytic cleavage of phosphodiester linkages in the $DNA$ backbone,effectively breaking down or degrading $DNA$ molecules.

(D) The statement "It has not escaped our notice that the specific pairing we have postulated immediately suggests a possible copying mechanism for the genetic material" was made by $James \ Watson$ and $Francis \ Crick$ in their $1953$ paper published in the journal $Nature$. This famous sentence refers to the complementary base pairing ($A$ with $T$ and $G$ with $C$) in the $DNA$ double helix structure, which provides a logical basis for how genetic information is replicated during cell division.

(B) The $tRNA$ molecule exhibits a clover-leaf shape in its two-dimensional representation,which is known as its secondary structure. In its actual three-dimensional form,it appears as an inverted $L$-shaped molecule,which represents its tertiary structure. Therefore,the clover-leaf model specifically refers to the secondary structure of $tRNA$.

(D) $tRNA$ (transfer $RNA$) is also known as $sRNA$ (soluble $RNA$) because it was historically isolated in the soluble fraction of the cell during centrifugation. It acts as an adapter molecule that carries specific amino acids to the ribosome during protein synthesis.

(D) $DNA$ is a hydrophilic molecule because of the presence of the phosphate backbone,which makes it polar.
Due to its hydrophilic nature,$DNA$ cannot pass through cell membranes,which are composed of a hydrophobic lipid bilayer.
$DNA$ is negatively charged due to the phosphate groups $(PO_4^{3-})$ present in its backbone.
Therefore,all the given statements are correct.

(A) probe is a single-stranded $DNA$ or $RNA$ molecule used to detect the presence of a complementary sequence in a sample.
Hybridization occurs through base pairing between complementary nucleotides,where $A$ pairs with $T$ (or $U$ in $RNA$) and $C$ pairs with $G$.
Crucially,$DNA$ strands are antiparallel,meaning the $3'-5'$ end of one strand aligns with the $5'-3'$ end of the complementary strand.
Given the probe sequence: $3'-A-T-C-A-G-C-5'$.
The complementary strand must be antiparallel $(5'-3')$ and follow base-pairing rules:
$3' \rightarrow 5'$ (Probe)
$5' \rightarrow 3'$ (Target)
$A$ pairs with $T$
$T$ pairs with $A$
$C$ pairs with $G$
$A$ pairs with $T$
$G$ pairs with $C$
$C$ pairs with $G$
Therefore,the complementary sequence is $5'-TAGTCG-3'$.
Thus,the correct option is $A$.

(C) The correct matches are based on the standard values provided in the $NCERT$ textbook for the Molecular Basis of Inheritance:
$1$. Bacteriophage $\phi \times 174$ has $5386$ nucleotides $(a-ii)$.
$2$. Bacteriophage lambda has $48502$ base pairs $(b-i)$.
$3$. Escherichia coli $(E. coli)$ has $4.6 \times 10^6$ base pairs $(c-iv)$.
$4$. The haploid content of human $DNA$ is $3.3 \times 10^9$ base pairs $(d-iii)$.
Therefore,the correct matching is $(a) - (ii), (b) - (i), (c) - (iv), (d) - (iii)$.

(B) According to Chargaff's rule,the amount of adenine $(A)$ is equal to the amount of thymine $(T)$,and the amount of cytosine $(C)$ is equal to the amount of guanine $(G)$.
Given that $A = 30\%$,therefore $T$ must also be $30\%$.
Given that $C = 20\%$,therefore $G$ must also be $20\%$.
Thus,the percentage composition of $T$ is $30\%$ and $G$ is $20\%$.

(A) Statement $I$ is incorrect because $DNA$ is negatively charged due to the presence of phosphate groups,and it is held by positively charged proteins in prokaryotes.
Statement $II$ is correct because in eukaryotes,the negatively charged $DNA$ wraps around the positively charged histone octamer to form a nucleosome.
Therefore,Statement $I$ is false and Statement $II$ is true.

(A) Histones are basic proteins that are associated with $DNA$ to form nucleosomes.
These proteins are rich in positively charged basic amino acids,specifically $Lysine$ and $Arginine$.
The positive charge of these amino acids allows histones to bind tightly to the negatively charged phosphate backbone of the $DNA$ molecule.

(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 $(A)$ = $20 \%$,then thymine $(T)$ must also be $20 \%$.
The total percentage of purines $(A + G)$ or pyrimidines $(T + C)$ in a $DNA$ molecule is always $50 \%$.
Since $T + C = 50 \%$ and we know $T = 20 \%$,we can calculate $C = 50 \% - 20 \% = 30 \%$.
Therefore,the total percentage of pyrimidines $(T + C)$ is $20 \% + 30 \% = 50 \%$.

(D) The correct values for genetic material length as per $NCERT$ are:
$1$. Bacteriophage $\phi \times 174$ has $5386$ nucleotides.
$2$. Bacteriophage lambda has $48502$ base pairs $(bp)$.
$3$. $\text{E. coli}$ has $4.6 \times 10^6 \ bp$.
$4$. The haploid content of human $\text{DNA}$ is $3.3 \times 10^9 \ bp$.
Option $D$ states $3.3 \times 10^6 \ bp$,which is incorrect; the correct value is $3.3 \times 10^9 \ bp$.

(C) Mobile genetic elements, also known as transposons, are $DNA$ sequences that can change their position within a genome. Retrotransposons, a specific class of these elements, replicate through an $RNA$ intermediate. They are first transcribed into $RNA$ and then reverse-transcribed back into $DNA$ to be inserted at a new genomic location.

(A) Erwin Chargaff proposed the base-pairing rules for $\text{DNA}$ in $1950.$
He observed that in any double-stranded $\text{DNA}$ molecule,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 $(C)$.
This leads to the ratios $\frac{A}{T} = 1$ and $\frac{G}{C} = 1,$ or $\frac{A+G}{T+C} = 1.$
These observations are known as Chargaff's rules.

(A) In the $\text{DNA}$ double helix model proposed by Watson and Crick,the two polynucleotide chains are held together by hydrogen bonds between nitrogenous bases.
According to Chargaff's rule and the structural constraints of the double helix,a purine (adenine or guanine) always pairs with a pyrimidine (thymine or cytosine).
$A$ purine is a double-ring structure,while a pyrimidine is a single-ring structure.
By pairing a purine with a pyrimidine,the total width of the base pair remains constant (approximately $20 \text{ Å}$),which maintains the uniform diameter of the $\text{dsDNA}$ molecule.
Therefore,the constant distance between the two chains is a direct consequence of the purine-pyrimidine pairing.
Thus,both $A$ and $R$ are true,and $R$ is the correct explanation of $A$.

(D) The stability of the $\text{DNA}$ double helix is primarily maintained by two factors:
$1$. The hydrogen bonds between the nitrogenous bases of the two strands provide specific base pairing ($A=T$ and $G≡C$),which holds the strands together.
$2$. The base stacking interactions,where the planar surfaces of the base pairs stack over one another,contribute significantly to the thermodynamic stability of the helix.
Since both these features are correct,the correct option is $(D)$.

(A) In a double-stranded $\text{DNA}$ molecule,the two strands are complementary to each other.
According to base-pairing rules,Adenine $(A)$ pairs with Thymine $(T)$,and Guanine $(G)$ pairs with Cytosine $(C)$.
If the first strand has $15 \% A, 15 \% T, 40 \% G$,and $30 \% C$,the complementary strand must have bases that pair with these.
Specifically,$A$ on the first strand pairs with $T$ on the second,$T$ on the first pairs with $A$ on the second,$G$ on the first pairs with $C$ on the second,and $C$ on the first pairs with $G$ on the second.
Therefore,the complementary strand will have:
$15 \% T$ (pairing with $15 \% A$),
$15 \% A$ (pairing with $15 \% T$),
$40 \% C$ (pairing with $40 \% G$),
$30 \% G$ (pairing with $30 \% C$).
Rearranging these values,we get $15 \% A, 15 \% T, 30 \% G$,and $40 \% C$.

(B) Statement $(a)$ is correct: Euchromatin is loosely packed and stains light.
Statement $(b)$ is correct: Heterochromatin is densely packed and is transcriptionally inactive.
Statement $(c)$ is correct: $A$ histone octamer consists of two molecules each of $H2A, H2B, H3,$ and $H4$.
Statement $(d)$ is incorrect: Packaging at higher levels requires non-histone chromosomal $(NHC)$ proteins,not additional histones.
Statement $(e)$ is incorrect: $A$ human diploid cell contains $6.6 \times 10^9$ base pairs,not $6.6 \times 10$ nucleosomes.
Therefore,the correct statements are $(a), (b),$ and $(c)$.

(A) In one complete turn of $B-DNA,$ there are $10$ base pairs.
Each base pair consists of one purine and one pyrimidine.
Therefore,there are $10$ purine bases and $10$ pyrimidine bases in one complete turn.
Since the question asks for the number of purine bases,the answer is $10$.

(B) The given diagram represents a nucleosome structure.
In this structure:
$A$ represents the $DNA$ molecule that wraps around the histone core.
$B$ represents the histone octamer,which is a complex of eight histone proteins ($H_2A, H_2B, H_3, H_4$ in pairs).
$C$ represents the linker histone,specifically Histone-$H_1$,which binds to the $DNA$ where it enters and leaves the nucleosome core particle.
Therefore,the correct identification is $A \rightarrow$ $DNA$,$B \rightarrow$ Histone octamer,$C \rightarrow$ Histone-$H_1$.

(C) The correct matches based on the genetic material content are as follows:
$1$. $\phi \times 174$ bacteriophage has $5386$ nucleotides (bases) - $(a-ii)$.
$2$. Human (diploid) $DNA$ content is $6.6 \times 10^9$ base pairs (b.p) - $(b-iii)$.
$3$. $E. coli$ has $4.6 \times 10^6$ base pairs (b.p) - $(c-iv)$.
$4$. $\lambda$-phage has $48502$ base pairs (b.p) - $(d-i)$.
Therefore,the correct sequence is $a-ii, b-iii, c-iv, d-i$.

(B) According to Chargaff's rule,the amount of purines is always equal to the amount of pyrimidines in a double-stranded $DNA$ molecule.
This is expressed as the ratio of purines to pyrimidines being $1:1$.
Since the number of purine bases is given as $40$,the number of pyrimidine bases must also be $40$.

(A) In a $B-DNA$ double helix,the distance between two consecutive base pairs is approximately $3.4 \ \mathring{A}$ $(0.34 \ \text{nm})$.
Given that the $dsDNA$ strand contains $100$ base pairs.
The total length of the $DNA$ molecule is calculated by multiplying the number of base pairs by the distance between each pair.
Length $= 100 \times 3.4 \ \mathring{A} = 340 \ \mathring{A}$.

(B) Erwin Chargaff observed that the amount of purines $(A + G)$ is always equal to the amount of pyrimidines $(T + C)$ in a double-stranded $DNA$ molecule.
This is known as Chargaff's rule, which states that the ratio of adenine to thymine is $1:1$ and the ratio of guanine to cytosine is $1:1$.

(C) $RNA$ contains four nitrogenous bases: Adenine $(A)$,Guanine $(G)$,Cytosine $(C)$,and Uracil $(U)$. Thymine $(T)$ is found in $DNA$ but is replaced by Uracil $(U)$ in $RNA$. Therefore,Thymine is not a component of $RNA$.

(B) The bacteriophage $\phi X 174$ is a well-known virus that infects $E. coli$.
Its genetic material consists of a single-stranded $DNA$ molecule,commonly referred to as $ssDNA$.
Unlike most organisms that use double-stranded $DNA$ $(dsDNA)$ as their genetic material,$\phi X 174$ is a classic example of a virus with a single-stranded $DNA$ genome.

(C) nucleotide of $DNA$ consists of three components: a nitrogenous base,a deoxyribose sugar,and a phosphate group.
$DNA$ contains the nitrogenous bases Adenine $(A)$,Guanine $(G)$,Cytosine $(C)$,and Thymine $(T)$.
$RNA$ contains the nitrogenous base Uracil $(U)$ instead of Thymine and uses ribose sugar instead of deoxyribose sugar.
Option $A$ is incorrect because it uses ribose sugar.
Option $B$ is incorrect because it uses Uracil and ribose sugar.
Option $C$ is correct because it contains Adenine (a $DNA$ base),deoxyribose sugar,and a phosphate group.
Option $D$ is incorrect because it uses Uracil.

(C) The total length of the $DNA$ segment is $340 \text{ Å}$.
Since the distance between two consecutive base pairs in a $DNA$ helix is $3.4 \text{ Å}$,the total number of base pairs is $340 \text{ Å} / 3.4 \text{ Å} = 100 \text{ base pairs}$.
This means there are $200$ total nitrogenous bases ($100$ on each strand).
According to Chargaff's rule,the number of adenine $(A)$ equals the number of thymine $(T)$,and the number of guanine $(G)$ equals the number of cytosine $(C)$.
Given that there are $20$ thymine bases,there must be $20$ adenine bases.
The total number of $A + T$ bases is $20 + 20 = 40$.
The remaining bases are $G + C = 200 - 40 = 160$.
Since $G = C$,the number of guanine bases is $160 / 2 = 80$.