The DNA Questions in English

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

(B) $Deoxyribonucleic acid$ $(DNA)$ acts as the connecting link between two generations.
This is because $DNA$ is the genetic material that stores hereditary information.
It replicates and is passed from parents to offspring during reproduction,ensuring the transfer of genetic traits across generations.

(D) Heterocyclic nitrogenous bases (purines and pyrimidines) present in $DNA$ and $RNA$ exhibit a characteristic absorption of ultraviolet $(UV)$ light.
This absorption is primarily due to the conjugated double bond systems within the ring structures of these bases.
The peak absorption for these nitrogenous bases occurs at approximately $260 \ nm$.
This property is widely used in molecular biology for the quantification of nucleic acids using spectrophotometry.

(C) The correct answer is $C$.
$1$. $tRNA$ binds to an amino acid at its $3'$ end (acceptor arm),not the $5'$ end.
$2$. The cloverleaf model represents the two-dimensional structure of $tRNA$,while the $L$-shaped model represents the three-dimensional structure.
$3$. $tRNA$ contains an anticodon that recognizes the codon on $mRNA$,not the other way around.
Therefore,statement $C$ is false because it incorrectly states that $tRNA$ has a codon that recognizes an anticodon on $mRNA$.

(B) According to Chargaff's rules for double-stranded $DNA$:
$1$. The ratio of adenine to thymine is constant and equals $1$ $(A/T = 1)$.
$2$. The ratio of guanine to cytosine is constant and equals $1$ $(G/C = 1)$.
$3$. The sum of purines $(A + G)$ equals the sum of pyrimidines $(C + T)$.
$4$. The ratio $(A + T) / (G + C)$ is constant for a given species but is not necessarily $1$. Therefore,the statement $A + T = G + C$ is not universally correct.

(B) $tRNA$ (transfer $RNA$) has a secondary structure that resembles a clover leaf. This model was proposed by $R.W. Holley$ in $1965$. The structure is formed due to extensive intramolecular base pairing,which results in the formation of double-stranded stems and single-stranded loops.

(B) The distance between two consecutive base pairs in a $DNA$ helix is $0.34 \ nm$ $(0.34 \times 10^{-9} \ m)$.
Given that the total number of base pairs is $1,00,000$ ($10^5$ base pairs).
The total length of the $DNA$ molecule is calculated as: $\text{Total length} = \text{Number of base pairs} \times \text{Distance between two base pairs}$.
$\text{Total length} = 1,00,000 \times 0.34 \ nm = 34,000 \ nm$.
This can be expressed in scientific notation as $3.4 \times 10^4 \ nm$.

(D) 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 guanine $(G)$ is equal to cytosine $(C)$.
Given: $G = 75$,therefore $C = 75$.
Given: $T = 75$,therefore $A = 75$.
The total number of nucleotides in the $DNA$ segment is the sum of all four nitrogenous bases: $A + T + G + C = 75 + 75 + 75 + 75 = 300$.

(A) The hairpin structure in $RNA$ is formed by intramolecular base pairing,where a single strand folds back on itself. The loop at the end of the hairpin is a single-stranded region that does not participate in base pairing. While the sequence can vary,$Uracil$ $(U)$ is often found in these loops,but among the given options,$Adenine$ $(A)$ is frequently involved in structural motifs or specific binding sites within these loops. However,it is important to note that $Thymine$ $(T)$ is not present in $RNA$ as it is replaced by $Uracil$ $(U)$. Therefore,$Thymine$ is incorrect. Given the context of standard biological questions regarding $RNA$ structure,$Adenine$ is the most plausible answer among the choices provided.

(C) $DNA$ (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 found in $DNA$.

(D) $tRNA$ (transfer $RNA$) molecules are historically and commonly referred to as soluble $RNA$ $(sRNA)$ because they remain in the supernatant fraction after high-speed centrifugation of cell homogenates. They are also known as adapter $RNA$ because they bridge the gap between the genetic code in $mRNA$ and the amino acids in a protein.

(A) Supercoiling is a mechanism used to compact $DNA$ within the cell.
In prokaryotes,the circular $DNA$ molecule is supercoiled to fit within the nucleoid region.
In eukaryotes,$DNA$ is wrapped around histone proteins to form nucleosomes,which are further coiled and supercoiled to form chromatin fibers.
Therefore,supercoiled $DNA$ is found in both prokaryotes and eukaryotes.

(B) The $DNA$ molecule consists of two polynucleotide chains that form a double helix.
These two strands are antiparallel,meaning they run in opposite directions ($5' \rightarrow 3'$ and $3' \rightarrow 5'$).
Furthermore,they are complementary,which means the sequence of nitrogenous bases in one strand determines the sequence of nitrogenous bases in the other strand due to specific base pairing ($A$ pairs with $T$,and $G$ pairs with $C$).

(B) When $DNA$ is heated,the hydrogen bonds between the complementary base pairs break,causing the double helix to denature or melt.
$G-C$ base pairs are held together by three hydrogen bonds,whereas $A-T$ base pairs are held together by only two hydrogen bonds.
Therefore,$DNA$ molecules with a higher $G : C$ content require more energy (higher temperature) to denature.
By comparing the $G : C$ ratio,one can determine the thermal stability and the extent of denaturation of $DNA$ samples.

(D) The given base sequence is $CAG, AGG, CGA, CCA$.
In nucleic acids,$DNA$ contains the nitrogenous base Thymine $(T)$,while $RNA$ ($mRNA$,$tRNA$,$rRNA$) contains Uracil $(U)$ instead of Thymine.
The provided sequence contains only Cytosine $(C)$,Adenine $(A)$,and Guanine $(G)$.
Since neither Thymine $(T)$ nor Uracil $(U)$ is present in the sequence,it is impossible to distinguish whether this segment belongs to $DNA$ or $RNA$.
Therefore,the data is not sufficient to conclude the nature of the nucleic acid strand.

(B) The double helix model proposed by Watson and Crick describes the $B-DNA$ structure.
$B-DNA$ is the most common form of $DNA$ found in living cells under physiological conditions.
It is a right-handed helix with a diameter of approximately $20$ $\mathring{A}$ and a pitch of $34$ $\mathring{A}$ per turn, containing $10$ base pairs per turn.

(C) Barbara McClintock discovered jumping genes (transposons) in $1940$ in maize $(Zea \text{ } mays)$.
She referred to these as controlling elements or mobile genetic elements.
For her pioneering work on these genetic elements, she was awarded the Nobel Prize in $1983$.

(A) According to the base-pairing rules in $DNA$ (Chargaff's rules),$Adenine$ $(A)$ always pairs with $Thymine$ $(T)$ and $Guanine$ $(G)$ always pairs with $Cytosine$ $(C)$.
Given the sequence of one strand: $A-G-C-T-A-A-G-C-C$.
The complementary strand will be formed by replacing $A$ with $T$,$T$ with $A$,$G$ with $C$,and $C$ with $G$.
Therefore,the complementary sequence is $T-C-G-A-T-T-C-G-G$,which is $TCGATTCGG$.

(C) The synthesis of proteins in a cell is governed by the central dogma of molecular biology.
$DNA$ acts as the master blueprint that contains the genetic information in the form of a specific sequence of nucleotides.
This sequence is transcribed into $mRNA$,which is then translated into a specific sequence of amino acids to form a protein.
Therefore,the primary structure and function of any protein are determined by the specific sequence of nucleotides in the $DNA$ molecule.

(B) The correct answer is $(b)$. Heterochromatin is a tightly packed form of $DNA$ that is genetically inactive or inert. Because it is so densely coiled,the transcriptional machinery cannot access the $DNA$ sequences,meaning it does not undergo transcription to produce messenger $RNA$ $(mRNA)$ for protein synthesis. In contrast,euchromatin is loosely packed and is transcriptionally active.

(D) gene is traditionally defined as the functional unit of inheritance.
According to the classical definition,a gene is the unit of function (cistron),the unit of mutation (muton),and the unit of recombination (recon).
Among the given options,'Unit of function' is the most comprehensive definition as it refers to the cistron,which codes for a specific polypeptide or functional $RNA$ molecule.

(D) gene is defined as the functional unit of inheritance. It is a specific segment or portion of $DNA$ that contains the instructions for synthesizing a functional product,such as a protein or an $RNA$ molecule. Therefore,the term 'gene' refers to a portion of $DNA$.

(B) The term $genome$ refers to the complete set of genetic material present in an organism.
In sexually reproducing organisms,the $genome$ is defined as the $haploid$ set of chromosomes present in a gamete.
Therefore,option $(b)$ is the correct answer.

(A) Jumping genes,also known as transposable elements,are $DNA$ sequences that can change their position within a genome.
These elements were first discovered by Barbara McClintock in maize.
In modern genetics,these are formally referred to as $Transposons$.

(A) $DNA$ is the primary hereditary material in most organisms. It is organized into structures called chromosomes,which act as the carriers of this genetic information from one generation to the next. Therefore,chromosomes are considered the carriers of hereditary material.

(D) Telomerase is a specialized enzyme that adds specific $DNA$ sequence repeats to the $3'$ end of $DNA$ strands in the telomere regions.
It is a ribonucleoprotein complex,meaning it consists of both a protein component (which acts as a reverse transcriptase) and an $RNA$ component (which serves as a template for the synthesis of telomeric $DNA$ repeats).

(C) . Woodcock $(1973)$ observed the structure of chromatin under an electron microscope.
He termed each beaded structure on the chromosome as a "nu body" or nucleosome.

(B) . $A$ $Gene$ is the smallest unit of heredity and is capable of duplicating its genetic material through the process of faithful replication. This ensures the precise distribution of genetic information among new cells during cell division.

(D) The nucleosome core particle consists of an octamer of histone proteins.
This octamer is composed of two molecules each of four types of histones: $H_2A, H_2B, H_3,$ and $H_4$.
$H_1$ histone is not part of the core; it binds to the $DNA$ where it enters and leaves the nucleosome,helping to stabilize the structure.

(B) nucleosome is the fundamental unit of $DNA$ packaging in eukaryotes,consisting of a segment of $DNA$ wound around a histone protein core.
The histone octamer core is composed of two copies each of histones $H2A$,$H2B$,$H3$,and $H4$.
The $DNA$ wraps around this core approximately $1.65$ times.
The diameter of the nucleosome particle is approximately $110 \ \mathring{A}$ $(11 \ nm)$.
Therefore,the correct option is $B$.

(B) The $DNA$ connecting two adjacent nucleosomes is known as linker $DNA$.
In eukaryotes,the nucleosome core particle consists of approximately $146$ base pairs of $DNA$ wrapped around a histone octamer.
The linker $DNA$,which connects these core particles,typically varies in length depending on the species and cell type.
However,it generally ranges from $15$ to $100$ base pairs (nucleotide pairs) in length.
Therefore,the correct range is $15$ to $100$.

(D) Nucleosomes are the fundamental repeating structural units of eukaryotic chromatin,which package $DNA$ into chromosomes.
$1$. $DNA$ wraps around a core of eight histone proteins to form a nucleosome.
$2$. These nucleosomes appear like 'beads-on-a-string' under an electron microscope.
$3$. Therefore,nucleosomes are considered the basic structural units of chromosomes.

(B) The correct answer is $B$. $A$ gene is defined as a functional unit of $DNA$ that contains the information for the synthesis of a specific protein or polypeptide chain,encoded by the sequence of nitrogenous bases or nucleotides.

(A) Genes are segments of $DNA$ that code for specific proteins or functional $RNA$ molecules.
Chemically,$DNA$ is a polymer of nucleotides,which are linked together by phosphodiester bonds to form long chains.
Therefore,genes are chemically described as polynucleotides.

(D) The terms $cistron$,$recon$,and $muton$ were proposed by $Seymour \ Benzer$ in $1955$.
These terms were introduced to define the functional,recombinational,and mutational units of a gene,respectively,based on his studies on the $rII$ region of the $T4$ bacteriophage.
$Cistron$ refers to the functional unit of a gene.
$Recon$ refers to the smallest unit of recombination.
$Muton$ refers to the smallest unit of mutation.

(C) gene is a segment of genetic material that codes for a functional product. In most living organisms,genes are composed of $DNA$. However,in many viruses,the genetic material is $RNA$. Therefore,a gene can be made up of either $DNA$ or $RNA$ depending on the organism.

(C) The concept of 'jumping genes' or 'transposons' was discovered by Barbara McClintock in maize $(Zea mays)$. She observed that these genetic elements could move from one location to another within the genome. For this groundbreaking discovery, she was awarded the Nobel Prize in Physiology or Medicine in $1983$.

(A) Genes are segments of $DNA$ that contain the instructions for building proteins.
These instructions are encoded in the specific linear sequence of nucleotides along the $DNA$ molecule.
Therefore,the correct representation of a gene is a sequence of nucleotides.

(A) The genetically active region of chromatin is known as $Euchromatin$.
$Euchromatin$ is loosely packed and remains in an expanded state,which allows the transcriptional machinery to access the $DNA$ easily.
Because of this open structure,it is actively involved in the process of transcription,making it genetically active.

(A) The nucleosome core particle consists of an octamer of histone proteins,which includes two molecules each of $H_2A$,$H_2B$,$H_3$,and $H_4$.
$H_1$ histone is known as the linker histone,which binds to the $DNA$ where it enters and leaves the nucleosome core particle,helping to stabilize the higher-order chromatin structure.
Therefore,$H_1$ is not a part of the nucleosome core particle.

(A) The $DNA$ molecule is negatively charged and is wrapped around a positively charged histone octamer to form a structure called a nucleosome.
An octamer consists of two molecules each of four types of histone proteins: $H2A$,$H2B$,$H3$,and $H4$.

(D) The nucleosome core particle consists of an octamer of histone proteins.
This octamer is formed by two molecules each of four types of histones: $H2A$,$H2B$,$H3$,and $H4$.
Therefore,the total number of histone proteins in the core particle is $2 \times 4 = 8$.

(D) Heredity refers to the transmission of genetic characters from parents to offspring.
$1$. $DNA$ (Deoxyribonucleic acid) is the chemical basis of heredity,acting as the primary genetic material.
$2$. Genes are specific segments of $DNA$ that code for functional products (proteins or $RNA$) and are the functional units of heredity.
$3$. Chromosomes are the structures composed of $DNA$ and proteins that carry these genes and ensure their accurate segregation during cell division.
Since all three are fundamentally involved in the storage,expression,and transmission of genetic information,they are collectively considered carriers of heredity.

(D) Salivary gland chromosomes,also known as polytene chromosomes,are found in the salivary glands of dipteran larvae (e.g.,$Drosophila$).
These chromosomes are formed through a process called endoreduplication,where the $DNA$ undergoes multiple rounds of replication without cell division,resulting in thousands of chromatids aligned side-by-side.
This process creates distinct banding patterns (chromomeres) that are visible under a light microscope.
Because these banding patterns are constant and specific to certain gene loci,they are extremely useful for cytogenetic gene mapping.

(D) There are five types of histone proteins involved in the packaging of $DNA$ into chromatin within a chromosome.
These are $H2A$,$H2B$,$H3$,$H4$,and $H1$.
The core histone octamer consists of two molecules each of $H2A$,$H2B$,$H3$,and $H4$,while $H1$ acts as a linker histone that binds to the $DNA$ where it enters and leaves the nucleosome.

(A) The term $genome$ refers to the complete set of genetic material present in an organism or a cell.
In biological terms,a genome corresponds to the $haploid$ set of chromosomes of a species,which contains all the genes necessary for the organism's development and function.
Therefore,the correct option is $A$.

(D) . Transposons are genetic elements that can change their position within the genome. They were originally discovered in maize plants by $B. McClintock$. These elements are often referred to as "jumping genes" and can influence gene expression by turning it on or off or by causing mutations when they insert into a gene.

(C) The famous sentence, "It has not escaped our notice that the specific pairing we have postulated immediately suggests a possible copying mechanism for the genetic material," appears in the $1953$ paper published in the journal Nature by James Watson and Francis Crick. This statement describes the semi-conservative nature of $DNA$ replication based on the base-pairing rules they discovered.

(B) In the lysogenic cycle,the viral $DNA$ enters the host cell and integrates into the host bacterial genome.
Once integrated,it is referred to as a prophage.
The viral $DNA$ replicates along with the host $DNA$ during each cell division without causing immediate lysis of the host cell.
Therefore,it remains persistent and attached to the bacterial $DNA$.

(C) The correct answer is $(c)$ $\phi \times 174$.
Bacteriophage $\phi \times 174$ is a well-known example of a virus that contains a single-stranded $DNA$ molecule as its genetic material.
In contrast,bacteriophages like $T_2$,$T_4$,and $\lambda$ contain double-stranded $DNA$ as their genetic material.