Mix Example-Molecular Basis of Inheritance Questions in English

(B) The genetic mapping and complete genome sequencing of the bacteriophage $\phi X174$ were performed by $F. Sanger$ and his colleagues in $1977$. This was the first $DNA$-based genome to be fully sequenced.

(A) An $Auxotroph$ is a mutant strain of a microorganism that has lost the ability to synthesize a specific nutrient required for its growth due to a mutation.
It can only grow on a medium supplemented with that specific nutrient.
In contrast,a $Prototroph$ is a wild-type strain capable of synthesizing all its required nutrients from simple precursors.
Therefore,the correct term for the described organism is $Auxotroph$.

(B) The correct answer is $B$.
Cytosine,thymine,and uracil are pyrimidines,which are single-ring nitrogenous bases.
Adenine and guanine are purines,which are double-ring nitrogenous bases.
Therefore,the pair 'Purine Nitrogenous bases - Cytosine,thymine and uracil' is incorrectly matched.

(B) The correct answer is $B$. $DNA$ is directly involved in the process of replication (synthesis of another $DNA$ strand) and transcription (synthesis of $RNA$ molecules like $mRNA$,$tRNA$,and $rRNA$). Protein synthesis (translation) is the process where $mRNA$ is translated into a polypeptide chain by ribosomes and $tRNA$. Therefore,$DNA$ is not directly involved in the synthesis of proteins.

(D) Genetic research has utilized various model organisms based on the specific biological questions being addressed.
$1$. $Pisum$ $sativum$ (Garden pea) was used by Gregor Mendel to establish the fundamental laws of inheritance.
$2$. $Neurospora$ $crassa$ (a bread mold) is a classic model organism in genetics,particularly for studying biochemical pathways and gene function due to its haploid life cycle.
$3$. $Escherichia$ $coli$ ($E.$ $coli$) is the most widely used prokaryotic model organism for studying molecular genetics,$DNA$ replication,and gene regulation.
Since all three organisms have historically and currently been significant in genetic research,the correct answer is $D$.

(C) The chronological order of the events is as follows:
$1.$ Mendel's laws of inheritance were discovered in $1865$ (published in $1866$).
$2.$ The chromosome theory of heredity was proposed by Sutton and Boveri in $1902$.
$3.$ The experiments proving $DNA$ as the hereditary material (Hershey-Chase experiment) were conducted in $1952$.
Therefore,the correct sequence is $3, 1, 2$.

(A) In genetics,point mutations are classified based on the chemical nature of the base change:
$1$. Transition: $A$ purine ($A$ or $G$) is replaced by another purine,or a pyrimidine ($C$ or $T$) is replaced by another pyrimidine.
$2$. Transversion: $A$ purine is replaced by a pyrimidine,or a pyrimidine is replaced by a purine.
Analysis of the given mutations:
- $A$ to $G$: Purine to Purine $\rightarrow$ Transition
- $C$ to $T$: Pyrimidine to Pyrimidine $\rightarrow$ Transition
- $C$ to $G$: Pyrimidine to Purine $\rightarrow$ Transversion
- $T$ to $A$: Pyrimidine to Purine $\rightarrow$ Transversion
Therefore,the sequence is Transition,Transition,Transversion,and Transversion.

(D) $Escherichia \ coli$ $(E. \ coli)$ is widely used in genetic experiments for several reasons:
$1$. It is a haploid organism,which makes it easier to observe the effects of mutations as there is no dominant allele to mask recessive traits.
$2$. It has a very short generation time and can be easily cultured in a laboratory setting using simple nutrient media.
$3$. Its genome is relatively small and well-mapped,facilitating genetic manipulation and analysis.
Therefore,both $(b)$ and $(c)$ are correct reasons.

(D) $DNA$ acts as a template for the synthesis of $DNA$ during the process of replication,where the entire genome is copied.
Additionally,$DNA$ acts as a template for the synthesis of $RNA$ during the process of transcription,where specific segments of $DNA$ are copied into $RNA$ molecules.
Therefore,$DNA$ serves as a template for both $DNA$ and $RNA$ synthesis.

(D) $DNA$ replication is semi-conservative and proceeds in the $5' \to 3'$ direction on both strands.
Since the two strands of the $DNA$ double helix are antiparallel, one strand (leading strand) is synthesized continuously in the $5' \to 3'$ direction towards the replication fork.
The other strand (lagging strand) is synthesized discontinuously in the $5' \to 3'$ direction away from the replication fork, using $RNA$ primers.
Option $D$ correctly depicts the antiparallel nature of the template strands ($3' \to 5'$ and $5' \to 3'$) and the synthesis of new strands in the $5' \to 3'$ direction.

(D) The clover leaf model of $tRNA$ is a secondary structure that includes several key components:
$1$. Acceptor arm: This is the site where the specific amino acid is attached.
$2$. $T\psi C$ arm (or $T$ arm): This arm is involved in binding the $tRNA$ to the ribosome.
$3$. Anticodon arm: This arm contains the anticodon sequence that recognizes and base-pairs with the complementary codon on $mRNA$.
$4$. $D$ arm (or $DHU$ arm): This arm contains the modified base dihydrouridine and is important for the recognition of aminoacyl-$tRNA$ synthetase.
Since all these structures are present in the clover leaf model of $tRNA$, the correct answer is $(d)$.

(B) In prokaryotes like $E. coli$,the $mRNA$ is highly unstable and has a very short half-life. The average life span of $mRNA$ in $E. coli$ is approximately $2$ minutes. This rapid turnover allows the bacteria to quickly adapt to changing environmental conditions by synthesizing new proteins as needed.

(D) $DNA$ (Deoxyribonucleic acid) and $RNA$ (Ribonucleic acid) differ in several chemical aspects:
$1$. Sugar: $DNA$ contains $2'$-deoxyribose sugar,whereas $RNA$ contains ribose sugar.
$2$. Nitrogenous base: $DNA$ contains thymine $(T)$,while $RNA$ contains uracil $(U)$ in place of thymine.
$3$. Structure: $DNA$ is typically double-stranded,whereas $RNA$ is typically single-stranded.
Therefore,all the given options describe chemical or structural differences between $DNA$ and $RNA$.

(C) The correct matches are as follows:
$(i)$ Oncogenes: These are genes that,when activated or mutated,can cause cancer. Thus,$(i) - (5)$.
(ii) Single circular $DNA$: This is a characteristic feature of the bacterial chromosome. Thus,$(ii) - (3)$.
(iii) Splicing: This is the process of removing introns and joining exons during the processing of $mRNA$ molecules in eukaryotes. Thus,$(iii) - (4)$.
(iv) Jacob and Monod: They proposed the $Lac$-Operon concept to explain gene regulation in bacteria. Thus,$(iv) - (2)$.
Therefore,the correct sequence is $(5), (3), (4), (2)$.

(B) Informosomes are complexes consisting of eukaryotic $mRNA$ molecules associated with specific proteins. These complexes are more stable than free $mRNA$ molecules and are involved in the transport and storage of $mRNA$ in the cytoplasm of eukaryotic cells.

(A) Eukaryotic genomes are characterized by the presence of large amounts of non-coding $DNA$ and repetitive sequences,which are largely absent or rare in prokaryotic genomes.
Prokaryotic genes are often organized into operons (a cluster of genes under the control of a single promoter),whereas eukaryotic genes are typically monocistronic.
In eukaryotes,$DNA$ is wrapped around histone proteins to form nucleosomes,whereas prokaryotes lack true histones.
Prokaryotic $DNA$ is typically circular and double-stranded,not single-stranded.

(A) The genetic code is read in triplets (codons). $A$ mutation in the first base of the $DNA$ segment $ATGACCAGGACCCCAACA$ will alter the first codon ($ATG$ to a different codon). Since the reading frame of the subsequent codons is determined by the sequence of bases,a change in the first base will lead to a change in the first codon,which subsequently alters the entire sequence of amino acids translated from this $DNA$ segment. This is known as a frameshift mutation or a missense mutation depending on the nature of the base change,but in this specific context,it results in a complete change in the type and sequence of amino acids.

(D) $DNA$ is not directly involved in protein synthesis.
$DNA$ acts as a template for the synthesis of $mRNA$ (transcription),$tRNA$,and $rRNA$.
$DNA$ also acts as a template for its own replication (synthesis of another $DNA$ strand).
However,$DNA$ does not move to the site of protein synthesis (ribosomes) to directly guide the process.
Instead,it transfers its genetic information to $mRNA$ molecules,which then move to the ribosomes to direct the synthesis of proteins.

(A) The concept of 'Central Dogma' was proposed by Francis Crick in $1958$.
It states that the genetic information flows from $DNA \rightarrow RNA \rightarrow \text{Protein}$.

(C) mutation at a genetic locus alters the nucleotide sequence of the $DNA$.
This change is transcribed into an altered $mRNA$ sequence.
The abnormal $mRNA$ sequence leads to the incorporation of different amino acids during translation,resulting in a change in the primary structure of the protein.
Since proteins (enzymes/structural proteins) determine the phenotype of an organism,a change in protein structure leads to a change in the organism's character.

(A) The concept of the circular flow of genetic information,which suggests that genetic information can flow from $DNA$ to $RNA$ to protein and potentially back,was first proposed by $Barry$ $Commoner$ in $1968$.
This challenged the strictly linear 'Central Dogma' proposed by $Francis$ $Crick$.

(D) Jumping genes,also known as transposons,are $DNA$ sequences that can move from one location to another within the genome.
These elements are found in both prokaryotes (such as bacteria) and eukaryotes (such as maize,humans,etc.).
They were first discovered by Barbara McClintock in maize $(Zea \ mays)$.

(C) Genes control protein synthesis by directing the synthesis of specific proteins (polypeptides) through the processes of transcription and translation. Additionally,genes control heredity by acting as the units of inheritance,ensuring the transmission of hereditary traits from one generation to the next via $DNA$ replication and gamete formation. Therefore,genes are responsible for both functions.

(D) $Cistron$ is defined as a segment of $DNA$ that codes for a single polypeptide chain.
It represents the functional unit of inheritance in terms of gene expression.
$Recon$ is the unit of recombination,$Muton$ is the unit of mutation,and $Codon$ is the unit of the genetic code consisting of three nucleotides.

(D) . Karl Landsteiner is credited with the discovery of human blood groups ($ABO$ blood group system).
$B$. Erwin Chargaff proposed the base pairing rules for $DNA$,stating that the amount of Adenine $(A)$ equals Thymine $(T)$ and Guanine $(G)$ equals Cytosine $(C)$. This is known as Chargaff's rule.
$C$. James Watson and Francis Crick proposed the double-helical structure of $DNA$ in $1953$.
Since all the given combinations are scientifically correct,the correct answer is $D$.

(C) In biological systems,$DNA$ serves as a template for two primary processes:
$1$. Replication: $DNA$ acts as a template for the synthesis of new $DNA$ strands,ensuring genetic continuity.
$2$. Transcription: $DNA$ acts as a template for the synthesis of $RNA$ molecules (mRNA,tRNA,rRNA).
Therefore,$DNA$ is a template for both $DNA$ and $RNA$ synthesis.

(D) The process of transferring genetic information from $RNA$ to $DNA$ is known as $Reverse \text{ } transcription$.
This process is mediated by the enzyme $Reverse \text{ } transcriptase$.
$Replication$ refers to the synthesis of $DNA$ from $DNA$.
$Transcription$ refers to the synthesis of $RNA$ from $DNA$.
$Translation$ refers to the synthesis of proteins from $mRNA$.

(D) Option $D$ is incorrect. The sweetness index of saccharin is approximately $300$ to $500$ times that of sucrose,not $10,000$.
$DNA$ is chemically less reactive and structurally more stable than $RNA$ due to the absence of the $2'-OH$ group.
$RNA$ is more labile and prone to mutations,making it evolve faster.
Guanylyl transferase is indeed the enzyme responsible for adding the $7-methylguanosine$ cap to the $5'$ end of $hn-RNA$ during processing.

(B) The Central Dogma of molecular biology states that genetic information flows from $DNA \rightarrow RNA \rightarrow \text{Protein}$.
$HIV$ (Human Immunodeficiency Virus) is a retrovirus that contains $RNA$ as its genetic material.
Upon infecting a host cell, $HIV$ uses the enzyme reverse transcriptase to synthesize $DNA$ from its $RNA$ template $(RNA \rightarrow DNA)$.
This process of reverse transcription violates the unidirectional flow of information proposed in the standard Central Dogma, making $HIV$ an exception.

(A) Genes that maintain their position on a chromosome are stable and do not change their location. In contrast,$Jumping$ $genes$ (also known as $Transposons$) are $DNA$ sequences that can change their position within the genome. Since the question asks for genes that maintain their position,the term $Jumping$ $genes$ is the correct category to distinguish from,but in the context of standard genetics terminology,genes that do not move are simply stable genes. However,among the given options,$Jumping$ $genes$ are the ones known for changing positions,and the question implies identifying the phenomenon. If the question implies genes that are $NOT$ jumping,they are stable. Given the options,$Jumping$ $genes$ is the specific term for mobile genetic elements. If the question asks for the name of genes that move,the answer is $Jumping$ $genes$. If the question implies the opposite,there might be a phrasing issue,but $Jumping$ $genes$ is the standard biological term for mobile elements.

(C) Western blotting is a laboratory technique used to detect specific protein molecules from among a mixture of proteins.
$1$. Southern blotting is used for the detection of $DNA$.
$2$. Northern blotting is used for the detection of $RNA$.
$3$. Western blotting is used for the detection of proteins.
Therefore,the correct option is $C$.

(A) The one-gene-one-enzyme hypothesis was proposed by George Beadle and Edward Tatum in $1941$.
They conducted experiments on the bread mold $Neurospora$ $crassa$.
By inducing mutations using $X$-rays, they identified nutritional mutants that could not synthesize specific amino acids or vitamins.
This led to the conclusion that each gene is responsible for the synthesis of a specific enzyme, thus establishing the one-gene-one-enzyme relationship.

(C) Let us analyze each statement:
$1$. In transcription,$DNA$ is transcribed into $RNA$. Adenine $(A)$ in $DNA$ pairs with Uracil $(U)$ in $RNA$. This statement is correct.
$2$. Regulation of $lac$ operon by a repressor is called negative regulation,not positive regulation. Positive regulation involves activators like $CAP$. This statement is incorrect.
$3$. The human genome is estimated to contain approximately $20,000$ to $25,000$ genes,not $50,000$. This statement is incorrect.
$4$. Haemophilia is a well-known sex-linked recessive disorder caused by a defect in blood clotting factors. This statement is correct.
Thus,statements $1$ and $4$ are correct. The total number of correct statements is $2$.

(A) The smallest unit of genetic material that can undergo a mutation is called a $Muton$.
$1$. $A$ $Muton$ is defined as the smallest segment of $DNA$ that,when changed,results in a mutation.
$2$. $A$ $Cistron$ is the unit of function (gene).
$3$. $A$ $Recon$ is the smallest unit of genetic material capable of recombination.
Therefore,the correct option is $A$.

(C) $Prototroph$ is an organism that has the same nutritional requirements as the wild type. It can synthesize all its essential nutrients,such as amino acids and vitamins,from simple inorganic sources. In contrast,an $Auxotroph$ is a mutant organism that has lost the ability to synthesize a specific nutrient and therefore requires that nutrient to be added to its growth medium.

(B) Lederberg's replica plating experiment was designed to demonstrate the genetic basis of antibiotic resistance in bacteria.
$1$. Initially,a bacterial colony is grown on a complete medium (master plate).
$2$. $A$ velvet-covered block is used to transfer the pattern of colonies onto a new plate containing a selective medium.
$3$. To isolate streptomycin-resistant strains,the replica plate must contain the antibiotic streptomycin.
$4$. Only those colonies that possess the mutation for streptomycin resistance will survive and grow on the medium containing streptomycin.
$5$. Therefore,a complete medium supplemented with streptomycin is used to identify these resistant strains.

(D) Genetic mutations are permanent alterations in the $DNA$ sequence that makes up a gene.
Since genes are located on chromosomes,mutations can also occur at the chromosomal level (chromosomal aberrations).
In some viruses,the genetic material is $RNA$,and mutations can occur in $RNA$ as well.
Therefore,genetic mutations can affect $DNA$,$RNA$,and chromosomes.

(B) $Silent$ $mutation$ is a type of genetic mutation where the change in the nucleotide sequence of the $DNA$ does not alter the amino acid sequence of the resulting protein. This occurs because the genetic code is degenerate, meaning multiple codons can code for the same amino acid. As a result, there is no change in the phenotype of the organism.

(A) Base analogs are chemical compounds that are structurally similar to normal nitrogenous bases of $DNA$.
Because of their structural similarity,they can be incorporated into the $DNA$ molecule during replication in place of the normal bases.
For example,$5$-bromouracil ($5$-$BU$) acts as an analog of thymine and can pair with guanine instead of adenine.
This leads to a point mutation where one purine is replaced by another purine or one pyrimidine is replaced by another pyrimidine.
This specific type of base substitution is known as $Transition$.

(B) frame-shift mutation occurs due to the insertion or deletion of one or more nucleotides in a $DNA$ sequence,which alters the reading frame of the genetic code.
Original sequence: $AAG, GAG, GAC, CAA, CCA$ ($15$ bases).
In option $B$: $AAG, AGG, ACC, AAC, CAA$,the reading frame has shifted significantly compared to the original sequence,indicating an insertion or deletion event occurred. Options $C$ and $D$ represent point mutations (substitutions),where only a single base is changed without altering the entire reading frame downstream. Therefore,option $B$ is the correct representation of a frame-shift mutation.

(D) The replica plating experiment,famously conducted by Joshua and Esther Lederberg,was used to demonstrate the genetic basis of antibiotic resistance in bacteria.
In this experiment,a master plate containing bacterial colonies was pressed onto a velvet-covered block and then transferred to new plates containing a medium supplemented with the antibiotic streptomycin.
The streptomycin acts as a screening agent (or selective agent) to identify colonies that possess the genetic mutation for antibiotic resistance.
Only those colonies that have the resistance gene can grow on the streptomycin-supplemented medium,while others are inhibited.

(A) An $Auxotroph$ is a mutant microorganism that has lost the ability to synthesize a specific nutrient required for its growth due to a mutation in a biosynthetic pathway.
Consequently,it cannot grow on minimal media unless the specific compound it cannot synthesize is added to the medium.
In contrast,a $Prototroph$ is a wild-type organism that can synthesize all its required nutrients from simple inorganic sources.
Therefore,the correct term for such a mutant is an $Auxotroph$.

(A) In genetics,a point mutation is a change in a single nucleotide base pair.
Transitions are a specific type of point mutation where a purine is replaced by another purine (e.g.,$A \leftrightarrow G$) or a pyrimidine is replaced by another pyrimidine (e.g.,$C \leftrightarrow T$).
Transversions,on the other hand,involve the substitution of a purine for a pyrimidine or vice versa.
Therefore,the correct term for the substitution of a purine with another purine or a pyrimidine with another pyrimidine is Transition.

(A) The correct answer is $A$. Ribosomal $RNA$ $(r-RNA)$ acts as a structural and catalytic component of the ribosome. Specifically,the $23S$ $r-RNA$ in prokaryotes (and $28S$ $r-RNA$ in eukaryotes) functions as a ribozyme,which is an enzyme made of $RNA$ that catalyzes the formation of peptide bonds between amino acids during protein synthesis.

(A) Statement $P$ is incorrect because the length of $m-RNA$ is not fixed at $75$ nucleotides; it varies depending on the length of the polypeptide chain it encodes.
Statement $Q$ is incorrect because there are generally $61$ codons that code for amino acids,and there are typically $61$ types of $t-RNA$ molecules (excluding stop codons),not $20$. While there are $20$ types of amino acids,the number of $t-RNA$ types corresponds to the number of sense codons.

(C) The correct matching is as follows:
$(a)$ $m-RNA$ (messenger $RNA$) carries the genetic information from $DNA$ to the cytoplasm to serve as a template for protein synthesis. Thus,$(a-ii)$.
$(b)$ $t-RNA$ (transfer $RNA$) is a small molecule consisting of approximately $75$ to $90$ nucleotides that brings specific amino acids to the ribosome. Thus,$(b-iii)$.
$(c)$ $r-RNA$ (ribosomal $RNA$) is a structural component of ribosomes,the site of protein synthesis. Thus,$(c-i)$.
$(d)$ $RNA$ is a polynucleotide chain that contains ribose sugar and the nitrogenous base uracil instead of thymine,and it exists in three main types $(m-RNA, t-RNA, r-RNA)$. Thus,$(d-iv)$.
Therefore,the correct sequence is $(a-ii), (b-iii), (c-i), (d-iv)$.

(B) The pentose sugar present in $DNA$ is $2$-deoxyribose.
In a ribose sugar,there is an $-OH$ group at the $2'$ carbon position.
In $2$-deoxyribose,the oxygen atom is removed from the $2'$ position,leaving only a hydrogen atom $(-H)$ at that position.
The structure provided shows the $2'$ carbon position marked as '$X$'.
Since it is $DNA$,the sugar is $2$-deoxyribose,so the '$X$' position must be occupied by a hydrogen atom $(-H)$.
Looking at the options,option $B$ represents the $-H$ and $-H$ configuration at that carbon,which corresponds to the deoxy position.

(A) The correct matching is as follows:
$(1)$ $DNA$: $A$ specific part of $DNA$ acts as a template for transcription,so $(1-S)$.
$(2)$ $m-RNA$: It carries the genetic information from $DNA$ to the ribosome in the form of codons,so $(2-R)$.
$(3)$ $t-RNA$: It has anticodons that pair with codons on $m-RNA$,and there are different types of $t-RNA$ for different amino acids,so $(3-Q)$.
$(4)$ $r-RNA$: It is the most abundant type of $RNA$ in the cell,constituting about $80\%$ of the total $RNA$,so $(4-P)$.
Therefore,the correct sequence is $(1-S), (2-R), (3-Q), (4-P)$.

(A) The correct matches are as follows:
$(1)$ Chromosome contains the genetic material, which is the Gene $(Q)$.
$(2)$ $m-RNA$ (messenger $RNA$) carries the genetic code $(P)$ for protein synthesis.
$(3)$ $t-RNA$ (transfer $RNA$) contains the anticodon $(S)$ that pairs with the codon on $m-RNA$.
$(4)$ $r-RNA$ (ribosomal $RNA$) is a structural component of the Ribosome $(R)$.
Therefore, the correct sequence is $(1-Q), (2-P), (3-S), (4-R)$.

(D) The correct answer is $(d)$.
Transcription is the process in living cells in which the genetic information of $DNA$ is transferred to $mRNA$,not $tRNA$,as the first step of gene expression.
An operon consists of structural genes,a promoter,an operator,and a regulator gene. Therefore,the definition in $(c)$ is incomplete as it omits the regulator gene.
Since both $(a)$ and $(c)$ are incorrectly matched,$(d)$ is the correct choice.