The DNA Questions in English

(C) Nucleosomes are the fundamental repeating units of eukaryotic chromatin,consisting of a segment of $DNA$ wound around a histone protein core.
Each nucleosome core particle contains $146$ base pairs of $DNA$ wrapped around an octamer of histone proteins.
The $DNA$ segment that connects two adjacent nucleosomes is known as linker $DNA$.
This linker $DNA$ is associated with histone $H1$,which helps in the stabilization of the chromatin structure.

(C) nucleosome is the basic structural unit of $DNA$ packaging in eukaryotes. It consists of a segment of $DNA$ wound around a core of histone proteins. The histone octamer core is composed of two copies each of four types of histone proteins: $H_{2A}$,$H_{2B}$,$H_3$,and $H_4$. Therefore,the octamer is represented as $2(H_{2A} + H_{2B} + H_3 + H_4)$. The $H_1$ histone protein is not part of the octamer core; instead,it binds to the $DNA$ where it enters and leaves the nucleosome,helping to stabilize the structure.

(C) In prokaryotes,the $DNA$ is organized in large loops held by proteins.
The $DNA$ is reduced to $30 \ \mu m$ in diameter due to folding and looping of the $DNA$ strands.
The statement claiming this reduction is due to 'coiling and supercoiling' is incorrect,as coiling and supercoiling are responsible for further compaction to $2 \ \mu m$.

(A) The structure of chromatin,where $DNA$ is wrapped around histone octamers to form nucleosomes,was proposed by Roger Kornberg in $1974$. This model explains how long $DNA$ molecules are packaged into the compact structure of eukaryotic chromosomes using histone and non-histone proteins.

(C) One nucleosome approximately contains $200$ base pair long $DNA$ helix,of which about $146$ bp long segment is wound around each histone octamer,and the remaining base pairs contribute as linker $DNA$,which connects successive nucleosomes in a solenoid fibre.

(C) Statement $(a)$ is correct: $DNA$ is a double-stranded,helically coiled molecule.
Statement $(b)$ is incorrect: The two strands of $DNA$ are complementary and anti-parallel,not parallel.
Statement $(c)$ is correct: $DNA$ contains deoxyribose sugar and is acidic due to the presence of phosphate groups.
Statement $(d)$ is incorrect: $DNA$ contains four nitrogenous bases (Adenine,Guanine,Cytosine,and Thymine),not three.
Therefore,statements $(a)$ and $(c)$ are correct.

(B) Chromatin is classified into two types based on its staining properties and transcriptional activity: $Euchromatin$ and $Heterochromatin$.
$Euchromatin$ is loosely packed, transcriptionally active, and stains lightly.
$Heterochromatin$ is densely packed, transcriptionally inactive, and stains darkly.
Studies on the composition of these regions indicate that $Heterochromatin$ is significantly more condensed and contains a higher density of $DNA$ compared to $Euchromatin$.
Specifically, $Heterochromatin$ is approximately $9-12$ times more rich in $DNA$ content per unit volume compared to $Euchromatin$.

(B) The clover leaf model is the secondary structure of $tRNA$ (transfer $RNA$).
It consists of three loops: the $D-loop$, the $anticodon-loop$, and the $T\psi C-loop$.
This structure is essential for its function in protein synthesis, where it carries specific amino acids to the ribosome based on the codon sequence of $mRNA$.

(C) The structure of $DNA$ consists of two polynucleotide chains that form a double helix.
These two strands are complementary to each other,meaning the sequence of bases in one strand determines the sequence in the other due to specific base pairing rules ($A$ pairs with $T$,and $G$ pairs with $C$).
Furthermore,the two strands run in opposite directions,meaning one strand runs in the $5' \rightarrow 3'$ direction,while the other runs in the $3' \rightarrow 5'$ direction. This orientation is referred to as antiparallel.

(D) nucleosome consists of a core particle and a linker $DNA$. The core particle is an octamer of histone proteins,which contains two molecules each of $H2A$,$H2B$,$H3$,and $H4$.
These eight histone molecules form the core around which the $DNA$ is wrapped.
The $H1$ histone protein is known as the linker histone,which binds to the $DNA$ where it enters and leaves the nucleosome core particle.
Therefore,$H1$ is present as a single molecule per nucleosome,not as a double molecule.

(D) nucleosome core particle consists of an octamer of histone proteins ($H2A, H2B, H3, H4$ - two of each) around which $146$ base pairs of $DNA$ are wrapped.
When the linker $DNA$ (which connects adjacent nucleosomes) is included,the total length of $DNA$ associated with one nucleosome unit is approximately $200$ base pairs.
The $H1$ histone protein binds to the linker $DNA$,making the total number of histone molecules involved in the nucleosome complex equal to $9$ ($8$ in the core + $1$ linker histone).

(B) According to Chargaff's rule 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 $(C)$.
Given: $A = 30 \%$.
Since $A = T$,then $T = 30 \%$.
The sum of all bases is $A + T + G + C = 100 \%$.
Substituting the values: $30 \% + 30 \% + G + C = 100 \%$.
$60 \% + G + C = 100 \%$.
$G + C = 100 \% - 60 \% = 40 \%$.
Since $G = C$,we can write $2C = 40 \%$.
Therefore,$C = 20 \%$.
Thus,the percentage of cytosine is $20 \%$.

(A) Statement $I$ is incorrect because Bacteriophage lambda has $48502$ base pairs $(bp)$.
Statement $\phi \times 174$ (phi $X$ $174$) bacteriophage has $5386$ nucleotides.
Statement $II$ is correct as $E. \text{ coli}$ contains $4.6 \times 10^6$ $bp$.
Statement $III$ is correct as the haploid content of human $DNA$ is $3.3 \times 10^9$ $bp$.
Therefore, only statement $I$ is incorrect.

(D) Histones are positively charged,basic proteins that are associated with $DNA$ in eukaryotic cells.
They are rich in the basic amino acids $Lysine$ and $Arginine$.
These amino acids carry positive charges in their side chains,which allow them to interact strongly with the negatively charged phosphate backbone of the $DNA$ molecule.
Therefore,the correct option is $D$.

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

(C) The total number of base pairs in a human diploid cell is $6.6 \times 10^9$ bp.
Since a sperm cell is haploid,it contains half the amount of $DNA$,which is $3.3 \times 10^9$ bp.
The distance between two consecutive base pairs is $0.34 \ \text{nm}$ or $0.34 \times 10^{-9} \ \text{m}$.
The total length of the $DNA$ double helix is calculated as: $\text{Length} = \text{Number of base pairs} \times \text{Distance between base pairs}$.
$\text{Length} = (3.3 \times 10^9) \times (0.34 \times 10^{-9} \ \text{m}) = 1.122 \ \text{m}$.
Therefore,the length is approximately $1.1 \ \text{m}$.

(C) According to Chargaff's rule,in double-stranded $DNA$,the amount of Adenine $(A)$ is equal to Thymine $(T)$,and the amount of Guanine $(G)$ is equal to Cytosine $(C)$.
Given: $G = 35 \%$.
Since $G = C$,therefore $C = 35 \%$.
The sum of all four bases is $100 \%$,so $A + T + G + C = 100 \%$.
Substituting the values: $A + T + 35 \% + 35 \% = 100 \%$.
$A + T + 70 \% = 100 \%$.
$A + T = 30 \%$.
Since $A = T$,we can write $2T = 30 \%$.
Therefore,$T = 15 \%$.

(A) The total length of $DNA$ in $E. coli$ is calculated by multiplying the total number of base pairs by the distance between two consecutive base pairs.
$E. coli$ has $4.6 \times 10^6$ base pairs.
The distance between two consecutive base pairs is $0.34 \times 10^{-9} \ m$.
Length = $(4.6 \times 10^6) \times (0.34 \times 10^{-9} \ m) \approx 1.36 \times 10^{-3} \ m = 1.36 \ mm$.
Therefore,the correct option is $A$.

(D) In eukaryotic genes,the coding sequences that appear in mature $mRNA$ are called $Exons$.
$Introns$ are the intervening,non-coding sequences that are removed during the process of $RNA$ splicing.
$Promoters$ and $Regulators$ are sequences involved in the control of gene expression,not the coding sequences themselves.
Therefore,the correct answer is $D$.

(A) The chemical name for $5$-Methyl uracil is Thymine.
Thymine is a pyrimidine nitrogenous base found in $DNA$.
It is structurally similar to Uracil,which is found in $RNA$,but with a methyl group $(-CH_3)$ attached at the $5^{th}$ position of the pyrimidine ring.
Therefore,the correct option is $A$.

(D) According to Chargaff's rule,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 $(C)$.
Given: Total base pairs = $50$,so total nitrogen bases = $100$.
Number of Thymine $(T)$ = $20$.
Therefore,number of Adenine $(A)$ = $20$.
Number of $A-T$ pairs = $20$.
Number of $G-C$ pairs = Total base pairs - $A-T$ pairs = $50 - 20 = 30$.
Hydrogen bonds between $A$ and $T$ = $2 \times 20 = 40$.
Hydrogen bonds between $G$ and $C$ = $3 \times 30 = 90$.
Total hydrogen bonds = $40 + 90 = 130$.

(C) The correct answer is $C$.
Histones are positively charged proteins because they contain a very high proportion of basic amino acids,specifically $Lysine$ and $Arginine$.
These amino acids possess positively charged side chains at physiological $pH$.
Because of this positive charge,histones can effectively bind to the negatively charged phosphate backbone of $DNA$ to form nucleosomes.

(A) The correct answer is $(A)$.
The diagram shows a nucleosome.
$A$ nucleosome is the basic structural unit of $DNA$ packaging in eukaryotes.
It consists of a segment of negatively charged $DNA$ wrapped around a positively charged histone octamer (a complex of eight histone proteins).
The $H_1$ histone protein helps in stabilizing the structure by binding to the $DNA$ where it enters and leaves the nucleosome.

(B) The correct answer is $B$.
In a typical nucleosome,the $DNA$ helix is wrapped around a histone octamer.
This structure consists of approximately $200$ base pairs $(bp)$ of $DNA$.

(C) The correct answer is $C$.
Nitrogenous bases found in $DNA$ are Adenine $(A)$,Guanine $(G)$,Cytosine $(C)$,and Thymine $(T)$.
In $RNA$,Thymine is replaced by Uracil $(U)$.
Therefore,Thymine is the nitrogenous base found exclusively in $DNA$ and not in $RNA$.

(C) The genome sizes of various organisms are as follows:
$1$. $\phi \times 174$ bacteriophage has $5386$ nucleotides.
$2$. Bacteriophage lambda has $48502$ base pairs (bp).
$3$. $E. coli$ has $4.6 \times 10^6$ base pairs (bp).
$4$. Human (haploid content) has $3.3 \times 10^9$ base pairs (bp).
Matching these with the given columns:
$(A) \rightarrow (s)$
$(B) \rightarrow (t)$
$(C) \rightarrow (p)$
$(D) \rightarrow (q)$
Therefore, the correct matching is $A-s, B-t, C-p, D-q$.

(B) The four nitrogenous bases present in $DNA$ are adenine,guanine,cytosine,and thymine.
Thymine is chemically known as $5$-methyl uracil.
Unlike $RNA$,which contains uracil,$DNA$ contains thymine as its specific nitrogenous base.

(A) Total length of $DNA$ = $3.2 \text{ Kbp} = 3200 \text{ bp}$.
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 $(C)$.
Given,$A = 820$,therefore $T = 820$.
Total $A + T$ content = $820 + 820 = 1640$.
Total $G + C$ content = Total base pairs - Total $A + T$ content = $3200 - 1640 = 1560$.
Since $G = C$,the number of cytosine bases = $\frac{1560}{2} = 780$.

(B) Molecular biology is a branch of biology that deals with the molecular basis of biological activity in and between cells,including molecular synthesis,modification,mechanisms,and interactions. It primarily focuses on the structure and functions of the macromolecules (polymers of life) such as $DNA$,$RNA$,and proteins,which are essential for life processes.

(A) The correct answer is $A$.
In humans,the primary transcript,known as $hnRNA$ (heterogeneous nuclear $RNA$),undergoes processing to form mature $mRNA$.
The average length of $hnRNA$ in humans is approximately $3000$ bases.

(C) The correct option is $(C)$.
In a $DNA$ double helix,the two strands are antiparallel and complementary to each other.
The template strand runs in the $3' \rightarrow 5'$ direction,while the coding strand runs in the $5' \rightarrow 3'$ direction.
According to the base-pairing rules,$A$ pairs with $T$ and $G$ pairs with $C$.
Given template strand: $3'-ATGCATGCATGCATGC-5'$
Complementary coding strand: $5'-TACGTACGTACGTACG-3'$

(A, C, D, E) In eukaryotic cells,$DNA$ is wrapped around a histone octamer (composed of two molecules each of $H2A$,$H2B$,$H3$,and $H4$) to form a nucleosome.
Histones are positively charged basic proteins,as they are rich in the basic amino acids lysine and arginine.
Therefore,statements $A$,$C$,$D$,and $E$ are correct.
Statement $B$ is incorrect because histones are positively charged,not negatively charged.