The DNA Questions in English

(B) Exposure to $UV$ radiation,particularly $UV-B$,causes the formation of pyrimidine dimers,most commonly thymine dimers,in $DNA$.
This occurs because $UV$ light provides enough energy to form covalent bonds between adjacent thymine bases on the same strand of the $DNA$ molecule.
These dimers distort the $DNA$ structure,which can lead to mutations during $DNA$ replication if not repaired by the cell's repair mechanisms.

(B) The mechanism described refers to the movement of genetic elements known as transposons or 'jumping genes'.
This process,where a short segment of $DNA$ moves from one location to another within a chromosome or between chromosomes,is known as $DNA$ transposition.
$DNA$ replication is the process of copying $DNA$,$DNA$ hybridization involves the pairing of complementary strands,and $DNA$ recombination involves the exchange of genetic material between different molecules.

(A) James Watson and Francis Crick proposed the double-helical model of $DNA$ in the year $1953$. This discovery was based on $X$-ray diffraction data produced by Maurice Wilkins and Rosalind Franklin. The model provided a clear understanding of how genetic information is stored and replicated in living organisms.

(B) The structure of $DNA$ is characterized by the following features:
$(i)$ $DNA$ consists of two polynucleotide chains,making it a double helix.
(ii) The backbone of each strand is formed by the sugar-phosphate chain,not sugar and nitrogenous bases.
(iii) The two chains run in opposite directions,meaning they have anti-parallel polarity.
(iv) The nitrogenous bases of the two strands are paired via hydrogen bonds ($H$-bonds).
$(v)$ The helix is coiled in a right-handed fashion.
Therefore,statements $(i)$,(iii),(iv),and $(v)$ are correct. However,based on standard textbook options provided,the most accurate selection reflecting the core structural components is $(i)$ and (iv).

(C) $1$. Total number of bases in the $DNA$ fragment = $3.2 \ kb = 3.2 \times 1000 = 3200$ bases.
$2$. According to Chargaff's rule,the total number of purines equals the total number of pyrimidines $(A + G = T + C)$.
$3$. In a double-stranded $DNA$,$A = T$ and $G = C$. Therefore,$A + G = 3200 / 2 = 1600$.
$4$. Given that the number of adenine $(A)$ molecules = $820$.
$5$. Since $A = T$,the number of thymine $(T)$ molecules = $820$.
$6$. Using the equation $A + G = 1600$,we get $820 + G = 1600$.
$7$. $G = 1600 - 820 = 780$.
$8$. Since $G = C$,the number of cytosine $(C)$ molecules = $780$.

(C) According to Chargaff's rules for double-stranded $DNA$,the amount of adenine $(A)$ equals the amount of thymine $(T)$,and the amount of guanine $(G)$ equals the amount of cytosine $(C)$.
Therefore,$(A + G) = (T + C)$ and $(A + C) = (T + G)$.
Consequently,the ratios $(A + G) / (C + T) = 1$ and $(A + C) / (T + G) = 1$ are constant for all species.
However,the ratio $(G + C) / (A + T)$ varies significantly between different species,as it depends on the specific base composition of the organism's genome.

(D) The link between generations is provided by $DNA$ (Deoxyribonucleic acid), which is a type of $Nucleic \text{ } acid$.
$DNA$ acts as the genetic material that carries hereditary information from parents to offspring.
While chromosomes contain $DNA$, the fundamental chemical molecule responsible for the continuity of life and inheritance is $Nucleic \text{ } acid$.

(B) The given base ratio is $\frac{A+T}{G+C} = 0.3$.
We are given that the total amount of $A+T$ nitrogenous bases is $30,000$.
Substituting the value into the equation: $\frac{30,000}{G+C} = 0.3$.
Rearranging the equation to solve for $G+C$: $G+C = \frac{30,000}{0.3}$.
$G+C = \frac{300,000}{3} = 100,000$.
Therefore,the amount of $G+C$ nitrogenous bases is $100,000$.

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

(B) $DNA$ contains four nitrogenous bases: Adenine $(A)$,Guanine $(G)$,Cytosine $(C)$,and Thymine $(T)$.
$RNA$ also contains four nitrogenous bases: Adenine $(A)$,Guanine $(G)$,Cytosine $(C)$,and Uracil $(U)$.
In $RNA$,Uracil $(U)$ replaces Thymine $(T)$ which is found in $DNA$.

(B) gene is a segment of $DNA$ (Deoxyribonucleic acid).
$DNA$ is a polymer of nucleotides,which are linked together by phosphodiester bonds to form long chains known as polynucleotides.
Therefore,a gene is composed of polynucleotides.

(B) The $DNA$ molecule consists of two polynucleotide chains that run in opposite directions,known as antiparallel orientation.
This antiparallel arrangement is primarily maintained by the specific hydrogen bonding between the nitrogenous bases of the two strands.
Specifically,Adenine $(A)$ pairs with Thymine $(T)$ via two hydrogen bonds,and Guanine $(G)$ pairs with Cytosine $(C)$ via three hydrogen bonds.
While phosphodiester bonds form the backbone of each individual strand,the antiparallel nature of the double helix is defined by the hydrogen bonding pattern between the complementary bases.

(D) $DNA$ (Deoxyribonucleic acid) contains 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 found in $DNA$.

(B) Nucleic acids are polymers of nucleotides. Each nucleotide consists of a nitrogenous base,a pentose sugar,and a phosphate group.
Nitrogenous bases 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 $(U)$.
In $DNA$,the purine bases are Adenine and Guanine,while the pyrimidine bases are Cytosine and Thymine.

(B) $DNA$ (Deoxyribonucleic acid) is a fundamental molecule that carries genetic information in living organisms.
Within the eukaryotic cell,$DNA$ is tightly packaged with histone proteins to form a complex structure known as chromatin.
Chromatin is the primary form in which $DNA$ exists within the nucleus,making it the essential component of the genetic material.
While $DNA$ is found inside the nucleus,the nucleus itself is an organelle,not a component of $DNA$.
Ribosomes are involved in protein synthesis,and chloroplasts are organelles for photosynthesis,neither of which are components of $DNA$.

(B) The $DNA$ double helix consists of two polynucleotide chains that run in opposite directions,which is referred to as antiparallel polarity.
This orientation is primarily determined by the $3'-5'$ phosphodiester linkages that connect the nucleotides in the sugar-phosphate backbone.
While $H$-bonds are responsible for holding the two strands together via complementary base pairing,the antiparallel nature itself is a consequence of the chemical orientation of the phosphodiester bonds in the backbone.

(B) The genetic material in most living organisms is $DNA$ (Deoxyribonucleic acid).
Chromosomes are composed of $DNA$ and proteins (primarily histones).
$DNA$ contains the instructions (genes) necessary for the development,functioning,growth,and reproduction of organisms.
During cell division,$DNA$ is replicated and passed on to the next generation,ensuring the inheritance of genetic traits.

(C) Nitrogenous bases are classified into two types: purines and pyrimidines.
Purines include Adenine $(A)$ and Guanine $(G)$,which are double-ring structures.
Pyrimidines include Cytosine $(C)$,Thymine $(T)$,and Uracil $(U)$,which are single-ring structures.
Therefore,Guanine is a purine,not a pyrimidine.

(C) The $DNA$ double helix consists of two polynucleotide chains that run in opposite directions,which is known as antiparallel polarity.
One strand runs in the $5' \rightarrow 3'$ direction,while the other runs in the $3' \rightarrow 5'$ direction.
This means that at one end of the double helix,one strand has a free $5'$-phosphate group while the other has a free $3'$-hydroxyl group,and vice versa at the other end.
Therefore,the phosphate groups at the start of the two $DNA$ strands are in opposite positions.

(D) The $DNA$ (Deoxyribonucleic acid) molecule consists of a sugar-phosphate backbone.
This backbone is formed by the alternating arrangement of $2$-deoxyribose sugar molecules and phosphate groups.
The nitrogenous bases (purines and pyrimidines) are attached to the $1'$ carbon of the deoxyribose sugar and project inward,but they do not form the structural backbone itself.
Therefore,the correct answer is sugar and phosphate.

(C) $DNA$ (Deoxyribonucleic acid) and $RNA$ (Ribonucleic acid) differ in two primary components:
$1$. Sugar: $DNA$ contains $2'-deoxyribose$ sugar,whereas $RNA$ contains $ribose$ sugar.
$2$. Nitrogenous base: $DNA$ contains $thymine$ $(T)$ as a pyrimidine,while $RNA$ contains $uracil$ $(U)$ instead of $thymine$.
Therefore,both the sugar and the nitrogenous base are different between $DNA$ and $RNA$.

(A) Nitrogenous bases in nucleic acids are classified into two types: purines and pyrimidines.
Purines include Adenine $(A)$ and Guanine $(G)$,which are double-ring structures.
Pyrimidines include Cytosine $(C)$,Thymine $(T)$,and Uracil $(U)$,which are single-ring structures.
Therefore,$T$ (Thymine) and $C$ (Cytosine) are pyrimidines.

(A) The distance between two consecutive base pairs in $DNA$ is $0.34 \text{ nm}$ or $0.34 \times 10^{-9} \text{ m}$.
Given,the number of base pairs = $10,000$ or $10^4$.
The total length of $DNA$ = (Number of base pairs) $\times$ (Distance between two consecutive base pairs).
Length = $10^4 \times 0.34 \times 10^{-9} \text{ m}$.
Length = $0.34 \times 10^{-5} \text{ m}$ or $3.4 \times 10^{-6} \text{ m}$.
Since $3.4 \times 10^{-6} \text{ m}$ is the calculated value,and option $A$ is $3.4 \times 10^{-6} \text{ m}$,the correct answer is $A$.

(B) James Watson and Francis Crick proposed the double-helical model of $DNA$ in $1953$.
This specific form of $DNA$ is characterized by a right-handed helix with a diameter of $20 \text{ Å}$ and a pitch of $34 \text{ Å}$.
This structure is widely known as the $B-DNA$ form, which is the most common and stable conformation of $DNA$ found in living cells under physiological conditions.

(C) In the structure of $DNA$,nitrogenous bases pair with each other through hydrogen bonds.
According to Watson and Crick's model,Guanine $(G)$ always pairs with Cytosine $(C)$ via three hydrogen bonds.
Conversely,Adenine $(A)$ pairs with Thymine $(T)$ via two hydrogen bonds.
Therefore,the number of $H$-bonds between Guanine and Cytosine is $3$.

(D) $tRNA$ (transfer $RNA$) is known as soluble $RNA$ $(sRNA)$ because it is the smallest type of $RNA$ and remains in the soluble fraction of the cell during centrifugation. It plays a crucial role in protein synthesis by carrying specific amino acids to the ribosome.

(A) In the $DNA$ double helix structure,the two strands are held together by hydrogen bonds ($H$-bonds) between the nitrogenous bases.
According to Chargaff's rule and the base-pairing principle,Adenine $(A)$ always pairs with Thymine $(T)$ via two $H$-bonds,and Guanine $(G)$ always pairs with Cytosine $(C)$ via three $H$-bonds.
These hydrogen bonds provide stability to the $DNA$ double helix structure.

(C) In a $DNA$ double helix, the base pairing follows Chargaff's rule: $Adenine$ $(A)$ pairs with $Thymine$ $(T)$ via $2$ hydrogen bonds, and $Guanine$ $(G)$ pairs with $Cytosine$ $(C)$ via $3$ hydrogen bonds.
Given the sequence of one strand is $A-T-C-G-A$.
The complementary strand will be $T-A-G-C-T$.
The hydrogen bonds between the strands are:
$A$ pairs with $T$ ($2$ bonds)
$T$ pairs with $A$ ($2$ bonds)
$C$ pairs with $G$ ($3$ bonds)
$G$ pairs with $C$ ($3$ bonds)
$A$ pairs with $T$ ($2$ bonds)
Total number of $H$-bonds = $2 + 2 + 3 + 3 + 2 = 12$.

(D) According to Chargaff's rule for $DNA$ structure,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:
Number of thymine $(T)$ = $80$
Number of adenine $(A)$ = $T = 80$
Number of guanine $(G)$ = $90$
Number of cytosine $(C)$ = $G = 90$
Total number of nucleotides = $A + T + G + C$
Total number of nucleotides = $80 + 80 + 90 + 90 = 340$.

(A) $DNA$ was first identified by Friedrich Miescher in $1869$. He isolated it from the nuclei of pus cells and called it 'nuclein'.

(B) $DNA$ (Deoxyribonucleic acid) is a polymer of nucleotides. Each nucleotide consists of a nitrogenous base,a pentose sugar (deoxyribose),and a phosphate group. The phosphate group is derived from phosphoric acid $(H_3PO_4)$. Due to the presence of these phosphate groups,which release hydrogen ions $(H^+)$ in an aqueous solution,$DNA$ exhibits acidic properties. Therefore,the acidic nature of $DNA$ is primarily attributed to the phosphoric acid component.

(A) The genetic material of $E. coli$ (Escherichia coli) is a single,double-stranded $(ds)$ circular $DNA$ molecule. This is a characteristic feature of most prokaryotic organisms,where the $DNA$ is not enclosed within a nucleus and exists as a circular chromosome.

(C) In the structure of $B-DNA$,the helix makes a complete turn every $3.4 \ nm$ (or $34 \ \mathring{A}$).
Each turn consists of $10$ base pairs.
The distance between two adjacent base pairs is $0.34 \ nm$ (or $3.4 \ \mathring{A}$).
Therefore,the length of one full turn is $10 \times 0.34 \ nm = 3.4 \ nm$.

(B) $DNA$ (Deoxyribonucleic acid) is a double-stranded molecule that contains the genetic blueprint for living organisms.
$1$. It contains $Deoxyribose$ sugar,whereas $RNA$ contains $Ribose$ sugar.
$2$. $DNA$ contains the nitrogenous base $Thymine$ instead of $Uracil$ (which is found in $RNA$).
$3$. $DNA$ is typically double-stranded,while $RNA$ is single-stranded.
$4$. Therefore,the presence of $Deoxyribose$ sugar is a characteristic property of $DNA$.

(A) Thymine is a pyrimidine base found in $DNA$.
Its chemical structure is $5$-methyl uracil.
In uracil,the carbon at the $5$th position is attached to a hydrogen atom,whereas in thymine,this hydrogen is replaced by a methyl group $(-CH_3)$.
Therefore,thymine is chemically identified as $5$-methyl uracil.

(B) The double-helical model of $DNA$ proposed by James Watson and Francis Crick in $1953$ is specifically referred to as the $B-DNA$ form.
This form is the most common conformation of $DNA$ found in living cells under physiological conditions.
It has a right-handed helix with a diameter of approximately $20 \ \mathring{A}$ and a pitch of $3.4 \ \text{nm}$ per turn,containing $10$ base pairs per turn.

(D) In $1953$,James Watson and Francis Crick proposed the double-helical model of $DNA$ based on $X$-ray diffraction data produced by Maurice Wilkins and Rosalind Franklin. This model provided a structural basis for understanding how genetic information is stored and replicated in living organisms.

(B) Nucleic acids are composed of nitrogenous bases,pentose sugars,and phosphate groups.
Nitrogenous bases 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 $(U)$.
In $DNA$,the purine bases are Adenine $(A)$ and Guanine $(G)$.

(C) According to the base-pairing rules of $DNA$ (Chargaff's rules),Adenine $(A)$ pairs with Thymine $(T)$ and Cytosine $(C)$ pairs with Guanine $(G)$.
Also,the two strands of $DNA$ are antiparallel,meaning if one strand runs in the $3' \rightarrow 5'$ direction,the complementary strand will run in the $5' \rightarrow 3'$ direction.
Given template strand: $3'-ATTCGTAC-5'$.
Matching bases: $A$ pairs with $T$,$T$ pairs with $A$,$C$ pairs with $G$,$G$ pairs with $C$.
Complementary sequence: $5'-TAAGCATG-3'$.

(C) The $DNA$ molecule maintains a uniform diameter of approximately $20 \text{ Å}$ throughout its length.
This uniformity is primarily due to the specific base pairing rules proposed by Watson and Crick.
In $DNA$, a purine (a double-ring structure, such as Adenine or Guanine) always pairs with a pyrimidine (a single-ring structure, such as Thymine or Cytosine).
Because a purine is always paired with a pyrimidine, the total width of the base pair remains constant (one double ring + one single ring = three rings total), which ensures the uniform diameter of the $DNA$ double helix.

(A) The distance between two consecutive base pairs in a $DNA$ double helix is $0.34 \ nm$ (or $3.4 \ \mathring{A}$).
Given that the total number of base pairs is $10,000$.
The total length of the $DNA$ molecule is calculated as:
$\text{Length} = \text{Number of base pairs} \times \text{Distance between two base pairs}$
$\text{Length} = 10,000 \times 0.34 \ nm = 3400 \ nm$.
Therefore,the correct option is $A$.

(C) In the $B-DNA$ structure,one full turn of the helix measures approximately $3.4 \ nm$ $(34 \ \mathring{A})$.
Each base pair is spaced at a distance of $0.34 \ nm$ $(3.4 \ \mathring{A})$.
Therefore,the number of base pairs per turn is calculated as $3.4 \ nm / 0.34 \ nm = 10$ base pairs.
Thus,there are $10$ base pairs in one full turn of the $DNA$ helix.

(D) Chargaff's rule states that in a double-stranded $DNA$ 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 $(C)$.
Mathematically,this implies that the ratio of $(A + T)$ to $(G + C)$ is constant for a given species.
Therefore,the correct representation of Chargaff's rule is $\frac{A + T}{G + C} = \text{constant}$.

(B) In $DNA$ molecules, the nitrogenous bases follow specific base-pairing rules known as Chargaff's rules.
Adenine $(A)$, which is a purine, always pairs with Thymine $(T)$, which is a pyrimidine, via two hydrogen bonds $(A=T)$.
Guanine $(G)$, which is a purine, always pairs with Cytosine $(C)$, which is a pyrimidine, via three hydrogen bonds $(G \equiv C)$.
Therefore, the correct base pairs are $A-T$ and $G-C$.

(D) Nitrogenous bases are organic molecules containing nitrogen,carbon,hydrogen,and oxygen.
They are classified into purines and pyrimidines.
Phosphorus is a component of the phosphate group in nucleotides,but it is not a part of the nitrogenous base structure itself.
Therefore,phosphorus is the correct answer.

(B) Genetic information in living organisms is stored and transmitted through nucleic acids,specifically $DNA$ (deoxyribonucleic acid) and $RNA$ (ribonucleic acid).
These molecules are polymers composed of long chains of repeating units called nucleotides.
$A$ single nucleotide consists of three components: a nitrogenous base,a pentose sugar (deoxyribose or ribose),and a phosphate group.
Therefore,the correct answer is $B$ (Nucleotides).

(C) $DNA$ molecule is a double-stranded structure consisting of base pairs.
Each base pair consists of two nucleotides, one on each strand.
Therefore, the total number of nucleotides is calculated as: $\text{Number of nucleotides} = \text{Number of base pairs} \times 2$.
Given that there are $10,000$ base pairs, the calculation is: $10,000 \times 2 = 20,000$ nucleotides.
Thus, the correct answer is $20,000$.

(B) In $RNA$ (and $DNA$),the individual nucleotides are linked together to form a polynucleotide chain through $3'-5'$ phosphodiester bonds.
These bonds are formed between the $3'$-hydroxyl group of the sugar of one nucleotide and the $5'$-phosphate group of the next nucleotide.
Therefore,the correct answer is phosphodiester bonds.

(A) According to the base pairing rules of $DNA$ (Chargaff's rules),$Adenine$ $(A)$ pairs with $Thymine$ $(T)$ and $Guanine$ $(G)$ pairs with $Cytosine$ $(C)$.
Given the sequence: $GAT, TAG, CAT, GAC$.
$1$. $G$ pairs with $C$,$A$ pairs with $T$,$T$ pairs with $A$. So,$GAT$ becomes $CTA$.
$2$. $T$ pairs with $A$,$A$ pairs with $T$,$G$ pairs with $C$. So,$TAG$ becomes $ATC$.
$3$. $C$ pairs with $G$,$A$ pairs with $T$,$T$ pairs with $A$. So,$CAT$ becomes $GTA$.
$4$. $G$ pairs with $C$,$A$ pairs with $T$,$C$ pairs with $G$. So,$GAC$ becomes $CTG$.
Therefore,the complementary sequence is $CTA, ATC, GTA, CTG$.