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Which of the following statement is correct regarding the process of replication in E.coli? [NEET 2024]
Answer (4)
Sol. In Prokaryotes, like E.coli during replication, the DNA dependent DNA polymerase catalyse polymerization only in one direction, that is \(5^{\prime} \rightarrow 3^{\prime}\)
The lactose present in the growth medium of bacteria is transported to the cell by the action of [Neet 2024]
Answer (3)
Sol. The \(y\) gene lac operon codes for permease enzyme, which increase the permeability of cell to \(\beta\)-galactosides.
So, the lactose present in the growth medium of bacteria is transported into the cell by the action of permease.
What is the fate of a piece of DNA carrying only gene of interest which is transferred into an alien organism?Â
A. The piece of DNA would be able to multiply itself independently in the progeny cells of the organism.
B. It may get integrated into the genome of the recipient.
C. It may multiply and be inherited along with the host DNA.
D. The alien piece of DNA is not an integral part of chromosome.
E. It shows ability to replicate.
Choose the correct answer from the options given below: [NEET 2024]
Answer (3)
Sol. Correct answer is option (3) because
The fate of a piece of DNA carrying only gene of interest which is transferred into an alien organism are:
(B) It may get integrated into the genome of the recipient
(C) It may multiply and be inherited along with the host DNA
\(\Rightarrow\) This piece of DNA would not be able to multiply itself in the progeny cells of the organism but when gets integrated into the genome of the recipient, it may multiply and be inherited along with the host DNA.
A transcription unit in DNA is defined primarily by the three regions in DNA and these are with respect to upstream and down stream end; [NEET 2024]
Answer (4)
Sol. A transcription unit of DNA is defined primarily by the three regions in the DNA:
(i) A promoter
(ii) The structural gene
(iii) A terminator
The promoter is said to be located towards \(5^{\prime}\)-end (upstream) of the structural gene (the reference is made with respect to the polarity of coding strand)
The terminator is located towards \(3^{\prime}\)-end (downstream) of the coding strand.
Unequivocal proof that DNA is the genetic material was first proposed by [NEET 2023]
Answer (2)
Sol. The unequivocal proof that DNA is the genetic material came from the experiment of Alfred Hershey and Martha Chase.
Avery, Macleoid and McCarty gave the biochemical characterisation of Transforming Principle. The transformation experiments by using Pneumococcus was conducted by Frederick Griffith. Wilkins and Franklin produced X-ray diffraction data of DNA.
Main steps in the formation of Recombinant DNA are given below. Arrange these steps in a correct sequence.
A. Insertion of recombinant DNA into the host cell
B. Cutting of DNA at specific location by restriction enzyme
C. Isolation of desired DNA fragment
D. Amplification of gene of interest using PCR
Choose the correct answer from the options given below : [NEET 2023]
Answer (1)
Sol. The correct answer is option (1) because recombinant DNA technology involves several steps in specific sequence such as isolation of DNA, fragmentation of DNA by restriction endonucleases, isolation of desired DNA fragment, ligation of the DNA fragment into a vector, transferring the recombinant DNA into the host, culturing the host cells in a medium at large scale and extraction of the desired product.Main steps in the formation of Recombinant DNA are given below. Arrange these steps in a correct sequence.
A. Insertion of recombinant DNA into the host cell
B. Cutting of DNA at specific location by restriction enzyme
C. Isolation of desired DNA fragment
D. Amplification of gene of interest using PCR
\(
\begin{aligned}
&\text { Match List I with List II. }\\
&\begin{array}{lll}
\quad \text { List I } & & \text { List II } \\
\text { A. Gene ‘a’ } & \text { I. } & \beta \text {-galactosidase } \\
\text { B. Gene ‘y’ } & \text { II. } & \text { Transacetylase } \\
\text { C. Gene ‘i’ } & \text { III. } & \text { Permease } \\
\text { D. Gene ‘z’ } & \text { IV. } & \text { Repressor protein } \\
\text { Choose the correct answer from the options given below: }
\end{array}
\end{aligned}
\) [NEET 2023]
Answer (2)
Sol. In a lac operon,
Gene a codes for enzyme transacetylase.
Gene y codes for enzyme permease.
Gene i codes for repressor protein
Gene \(z\) codes for enzyme \(\beta\)-galactosidase.
Given below are two statements:
Statement I: In prokaryotes, the positively charged DNA is held with some negatively charged proteins in a region called nucleoid.
Statement II: In eukaryotes, the negatively charged DNA is wrapped around the positively charged histone octamer to form nucleosome.
In the light of the above statements, choose the correct answer from the options given below: [NEET 2023]
Answer (4)
Sol. In prokaryotes, the negatively charged DNA is held with some positively charged proteins in a region termed as nucleoid.
In eukaryotes, the negatively charged DNA is wrapped around the positively charged histone octamer to form a structure called nucleosome.
Given below are two statements:
Statement I: RNA mutates at a faster rate.
Statement II: Viruses having RNA genome and shorter life span mutate and evolve faster.
In the light of the above statements, choose the correct answer from the options given below: [NEET 2023]
Answer (1)
Sol. RNA being unstable, mutate at a faster rate. Consequently, viruses having RNA genome and having shorter life span mutate and evolve faster.
Which one is the correct product of DNA dependent RNA polymerase to the given template? 3’TACATGGCAAATATCCATTCA5′
[NEET 2024]
Answer (1)
Sol. Template DNA is :
3’TACATGGCAAATATCCATTCA5′
5’AUGUACCGUUUAUAGGUAAGU3′ m-RNA
\(
\begin{aligned}
&\text { Match List I with List II. }\\
&\begin{array}{lll}
\text { List I } & & \text { List II } \\
\text { A. Gene ‘a’ } & \text { I. } & \beta \text {-galactosidase } \\
\text { B. Gene ‘ } y \text { ‘ } & \text { II. } & \text { Transacetylase } \\
\text { C. Gene ‘i’ } & \text { III. } & \text { Permease } \\
\text { D. Gene ‘ } z \text { ‘ } & \text { IV.     Repressor protein }
\end{array}
\end{aligned}
\)
Choose the correct answer from the options given below: [NEET 2023]
Answer (4)
Sol. – In eukaryotes, RNA polymerase III codes for snRNAs, tRNA and 5s rRNA.
– Splicing of exons is performed by snRNPs.
– TATA box is present in promoter region of transcription unit.
– Rho factor is responsible for termination of transcription.
The process of translation of mRNA to proteins begins as soon as : [NEET 2022]
Answer (1)
Sol. When the small subunit of ribosome encounters an mRNA, the process of translation of the mRNA to protein begins. This process is followed by the binding of bigger/larger subunit.
\(\mathrm{t}-\mathrm{RNA}\) is activated by the addition of amino acid prior to the attachment of ribosome, in the first phase.
DNA polymorphism forms the basis of : [NEET 2022]
Answer (3)
Sol. Polymorphism in DNA sequence is the basis of genetic mapping of human genome as well as of DNA fingerprinting.
Read the following statements and choose the set of correct statements :
(a) Euchromatin is loosely packed chromatin
(b) Heterochromatin is transcriptionally active
(c) Histone octomer is wrapped by negatively charged DNA in nucleosome
(d) Histones are rich in lysine and arginine
(e) A typical nucleosome contains 400 bp of DNA helix
Choose the correct answer from the options given below : [NEET 2022]
Answer (2)
Sol. Heterochromatin is transcriptionally inactive. A typical nucleosome contains \(200 \mathrm{bp}\) of DNA helix.
Euchromatin is the loosely packed chromatin region.
The negatively charged DNA is wrapped around the positively charged histone octamer to form a structure called nucleosome. Histones are rich in basic amino acid residues lysine and arginine.
If a geneticist uses the blind approach for sequencing the whole genome of an organism, followed by assignment of function to different segments, the methodology adopted by him is called as : [NEET 2022]
Answer (1)
Sol. Sequencing the whole set of genome that contained all the coding and non-coding sequences and later assigning different regions in the sequence with fuctions is called sequence annotation.
If the length of a DNA molecule is 1.1 metres, what will be the approximate number of base pairs? [NEET 2022]
Answer (1)
Sol. Number of base pairs \(\times\) distance between 2 consecutive base pairs \(=\) Length of DNA molecule
\(
\begin{aligned}
& x \cdot 0.34 \times 10^{-9} \mathrm{~m}=1.1 \mathrm{~m} \\
& x=\frac{1.1}{0.3 \times 10^{-9}} \\
& =3.6 \times 10^9 \\
& \simeq 3.3 \times 10^9 \mathrm{bp}
\end{aligned}
\)
In an \(E\). Coli strain \(i\) gene gets mutated and its product can not bind the inducer molecule. If growth medium is provided with lactose, what will be the outcome? [NEET 2022]
Answer (3)
As the product of ‘ \(r\) ‘ gene binds with the operator region and blocks the transcription and translation of \(z, y\) and \(a\) genes.
It’s product is prevented from binding to the operator by attaching it with the inducer. As the inducer can now no more capable of binding with the repressor, thus, in all the cases, operator always gets attached with the repressor thereby preventing the transcription and transmission of \(z, y\) and \(a\).
Even in the presence of lactose, transcription and translation of \(z, y\) and a would not occur.
Ten E.coli cells with \({ }^{15} \mathrm{~N}-\) dsDNA are incubated in medium containing \({ }^{14} \mathrm{~N}\) nucleotide. After 60 minutes, how many E.coli cells will have DNA totally free from \({ }^{15} \mathrm{~N}\) ? [NEET 2022]
Answer (3)
Sol. From 10 parent E.coli cells
\(3^{\text {rd }}\) generation
\(
\begin{aligned}
& 40 \text { cells } \xrightarrow{20 \mathrm{~min}} 20^{14} \mathrm{~N} \text { containing cells }+20 \text { hybrid } \\
& \left(20 \text { cells hybrid }+20^{14} \mathrm{~N}\right. \\
& \text { and } 40{ }^{14} \mathrm{~N} \text { containing cells } \\
& \text { containing cells) } \\
&
\end{aligned}
\)
Therefore, after 60 minutes, 60 E.coli cells will have DNA totally free from \({ }^{15} \mathrm{~N}\).
The final proof for DNA as the genetic material came from the experiments of (NEET 2017)
(a)
If there are 999 bases in an RNA that code for a protein with 333 amino acids, and the base at position 901 is deleted such that the length of the RNA becomes 998 bases, how many codons will be altered?
(NEET 2017)
(b) : 1 codon consists of 3 bases. Therefore, a deletion on 901 position will affect 33 codons.
During DNA replication, Okazaki fragments used to elongate (NEET 2017)
(c) : Lagging strand is a replicated strand of DNA which is formed in short segments called Okazaki fragments. Its growth is discontinuous. The direction of growth of the lagging strand is \(3^{\prime} \rightarrow 5^{\prime}\) though in each Okazaki fragment it is \(5^{\prime} \rightarrow 3^{\prime}\).
Which of the following RNAs should be most abundant in animal cell? (NEET 2017)
(d) : \(r\) RNA (ribosomal RNA) is the most abundant of all types of RNA (70-88\%). Hence, it will be present in highest amount. Percentage of \(t\) RNA and \(m\) RNA is \(15 \%\) and \(2-5 \%\) respectively. miRNA (micro RNA) are 21-22 bp long RNA that bring degeneration of \(m\) RNA.
Spliceosomes are not found in cells of (NEET 2017)
(c) : Spliceosomes helps in removal of introns. They will not occur in prokaryotes because prokaryotes do not have introns and thus, processing does not require splicing of \(m\) RNA.
The association of histone \(\mathrm{H}_1\) with a nucleosome indicates that (NEET 2017)
(b) : Histones help in packaging of DNA. In eukaryotes, DNA packaging is carried out with the help of positively charged basic proteins called histones. Histones are of five types \(-\mathrm{H}_1, \mathrm{H}_2 \mathrm{~A}, \mathrm{H}_2 \mathrm{~B}\), \(\mathrm{H}_3\) and \(\mathrm{H}_4 \cdot \mathrm{H}_1\) is attached over the linker DNA. Histone contains a large proportion of the positively charged (basic) amino acids, lysine and arginine in their structure. DNA is negatively charged due to the phosphate groups on its backbone. The result of these opposite charges is strong attraction and therefore, high binding affinity between histones and DNA.
Taylor conducted the experiments to prove semiconservative mode of chromosome replication on
(NEET-II 2016)
(b) : Taylor et al (1957) conducted experiment on Vicia faba (broad bean) to prove semi-conservative replication of DNA. He fed dividing cells of root tips of Vicia faba with radioactive \({ }^3 \mathrm{H}\) containing thymine instead of normal thymine and found that all the chromosomes became radioactive. Labelled thymine was then replaced with normal one. Next generation came to have radioactivity in one of the two chromatids of each chromosome while in subsequent generation radioactivity was present in \(\)50 \%\(\) of the chromosomes. This is possible only if out of the two strands of a chromosome, one is formed a fresh while the other is conserved at each replication.Â
Â
The equivalent of a structural gene is (NEET-II 2016)
(b) Cistron (or gene) is a length of DNA that contains the information for coding a specific polypeptide chain or a functional RNA molecule (i.e., transfer RNA or ribosomal RNA). Hence, cistron is a unit of function. Currently such a gene is called structural gene.
Which of the following \(r\) RNAs acts as structural RNA as well as ribozyme in bacteria? (NEET-II 2016)
(c) : \(23 \mathrm{~S} r\) RNA acts as structural RNA as well as ribozyme in bacteria.
A molecule that can act as a genetic material must fulfill the traits given below, except (NEET-II 2016)
(c) : Genetic material should be structurally and chemically stable otherwise its expression will change and leading to loss of several metabolic functions, etc.
(a): The strand of DNA on which RNA polymerase binds to catalyse transcription is called template strand. It is also known as master or antisense strand. It has the polarity of \(3^{\prime} \rightarrow 5^{\prime}\).
Which one of the following is the starter codon? (NEET-I 2016)
(c) : Polypeptide synthesis is signalled by two initiator codons or start codons i.e., AUG (methionine codon) and rarely by GUG (valine codon).
Which of the following is required as inducer (s) for the expression of Lac operon? (NEET-I 2016)
(a): In Lac operon, lactose is an inducer. It binds with suppressor and inactivates it. It allows RNA polymers access to the promoter and transcription proceeds.
A complex of ribosomes attached to a single strand of RNA is known as (NEET-I 2016)
(c)
Which one of the following is not applicable to RNA? (NEET 2015)
(b): Chargaff’s rules are applicable only for double stranded DNA molecule. These are not applicable for single stranded DNA or RNA molecules. Chargaff’s rules state that DNA helices contain equal molar ratios of \(\mathrm{A}\) and \(\mathrm{T}, \mathrm{G}\) and \(\mathrm{C}\). This is because in a ds DNA molecule, complementary base pairing occurs between \(\mathrm{A}\) and \(\mathrm{T}\), and \(\mathrm{C}\) and \(\mathrm{G}\) base pairs. This complementary base pairing is not possible in case of single stranded RNA molecule. Thus, Chargaff’s rules are not applicable to RNA.
Balbiani rings are sites of (NEET 2015)
(b): In certain development stages the polytene chromosomes bear conspicuous swellings called chromosome puffs. The larger swellings are called Balbiani rings. In the region of a puff or Balbiani ring, the DNA strands uncoil, become active and produce number of copies of \(m\) RNA. The \(m\) RNAs may remain temporarily stored in the puff and they may undergo transcription to form proteins. Thus, Balbiani rings are the sites of RNA and protein synthesis.
Identify the correct order of organisation of genetic material from largest to smallest.(NEET 2015)
(a): In genome all the genes are contained in a single set of chromosomes. The instructions in our genome are present in the form of DNA. DNA has a complicated structure in the form of a double helix. Single strands of DNA are coiled up into structures called chromosomes. Within the chromosomes, segments of DNA are “read” together to form genes. Thus, a gene is a segment of DNA or chromosome situated at a specific locus (gene locus) which carries coded information associated with a specific functionand can undergo crossing over as well as mutation. A nucleotide is the basic unit of DNA made up of a pentose sugar, phosphoric acid and a nitrogenous base.
Satellite DNA is important because it (NEET 2015)
(d) : Satellite DNA is that part of repetitive DNA which has long repetitive nucleotide sequences in tandem that forms a separate fraction on density ultracentrifugation. DNA fingerprinting involves identifying differences in some specific regions in DNA sequence called as repetitive DNA, because in these sequences, a small stretch of DNA is repeated many times. These repetitive DNA sequences are separated from bulk genomic DNA as different peaks during density gradient centrifugation. The bulk DNA forms a major peak and the other small peaks are referred to as satellite DNA. Depending on base composition (A:T rich or \(\)\mathrm{G}: \mathrm{C}\(\) rich), length of segment, and number of repetitive units, the satellite DNA is classified into many categories, such as microsatellites, mini-satellites etc. These sequences normally do not code for any proteins, but they form a large portion of human genome. These sequences show high degree of polymorphism and form the basis of DNA fingerprinting. Since DNA from every tissue (such as blood, hair-follicle, skin, bone, saliva, sperm etc.) from an individual show the same degree of polymorphism, they become very useful identification tool in forensic applications.
Gene regulation governing lactose operon of E.coli that involves the lac I gene product is
(NEET 2015)
(d): The control of expression of lac operon is negative (as it is turned off normally) and inducible. Inducible operon is an operon which remains switched off normally but becomes operational in the presence of an inducer (lactose, actually allolactose a metabolite of lactose, in case of lac operon). The inducible operon generally functions in catabolic pathways. In the presence of an inducer, the repressor has a higher affinity for the inducer than for the operator gene. When lactose is added, a few lactose molecules are carried into the cell by the enzyme lactose permease as small amount of this enzyme is present in the cell even when the operon is not working. These few lactose molecules are converted into allolactose molecules which act as an inducer and bind to the repressor (a product of regulator gene). The repressor-inducer complex fails to join with the operator gene, thus it is turned on.
In sea urchin DNA, which is double stranded. \(17 \%\) of the bases were shown to be cytosine.
The percentages of the other three bases expected to be present in this DNA are (2015 Cancelled)
(a) : According to Chargaff’s rule, the amount of adenine is always equal to that of thymine and the amount of guanine is always equal to that of cytosine, i.e., \(\mathrm{A}=\mathrm{T}\) and \(\mathrm{G}=\mathrm{C}\). Also, the purines and pyrimidines are always in equal amounts, i.e., \(\mathrm{A}+\mathrm{G}\)\(=\mathrm{T}+\mathrm{C}\). Now, given dsDNA has \(17 \%\) cytosine and hence guanine will be also \(17 \%\). So, A + T must be \(66 \%\), therefore, percentage of adenine or thymine would be \(66 / 2=33 \%\).
Which one of the following is wrongly matched? (NEET 2014)
(a, d) : Transcription is the process in living cells in which the genetic information of DNA is transferred to \(m\) RNA as first step of gene expression. An operon consists of structural genes, promoter, operator and regulator gene.
Transformation was discovered by (NEET 2014)
(c) : Transformation was first studied by S.F. Griffith in 1928 while studying Streptococcus pneumoniae. He found that R-Type non virulent bacteria pick up virulence from heat killed virulent S-type bacteria and transform into virulent forms. It was this experiment which indicated presence of a ‘transforming principle’ which was later found out to be DNA, by Avery et al.
Select the correct option.
\(
\begin{array}{|l|l|l|}
\hline & \begin{array}{l}
\text { Direction of } \\
\text { RNA synthesis }
\end{array} & \begin{array}{l}
\text { Direction of reading of the } \\
\text { template DNA strand }
\end{array} \\
\hline 1 . & 5^{\prime}-3^{\prime} & 3^{\prime}-5^{\prime} \\
\hline 2 . & 3^{\prime}-5^{\prime} & 5^{\prime}-3^{\prime} \\
\hline 3 . & 5^{\prime}-3^{\prime} & 5^{\prime}-3^{\prime} \\
\hline 4 . & 3^{\prime}-5^{\prime} & 3^{\prime}-5^{\prime} \\
\hline
\end{array}
\)Â (NEET 2014)
(a) RNA polymerase initiates and extends the RNA (chain elongation) and functions always in \(5^{\prime}\) to \(3^{\prime}\) direction. The structural component of DNA has \(3^{\prime}\) to \(5^{\prime}\) polarity. It is also called template DNA strand or antisense (-) strand.
Which of the following statements is not true of two genes that show \(50 \%\) recombination frequency?
(NEET 2013)
(d)
The diagram shows an important concept in the genetic implication of DNA. Fill in the blanks A to \(\mathrm{C}\).
\(
\xrightarrow[\rightarrow]{\text { DNA }} \xrightarrow{\mathrm{A}} m \mathrm{RNA} \xrightarrow{\mathrm{B}} \text { Protein } \xrightarrow{\text { Proposed by }} \mathrm{C}
\) (NEET 2013)
(a) : The expression of the genetic material occurs normally through the production of proteins. This involves two consecutive steps. These are transcription and translation. The DNA codes for the production of messenger RNA ( \(m\) RNA) during transcription. Messenger RNA carries coded information to ribosomes. The ribosomes read this information and use it for protein synthesis. This process is called translation. F.H.C. Crick described this undirectional flow of information in 1958 as the ‘central dogma of molecular biology’.
Which enzyme will be produced in a cell if there is a nonsense mutation in the lac \(\mathrm{Y}\) gene?
(NEET 2013)
(c) : A nonsense mutation is the one which stops polypeptide synthesis due to the formation of termination or non-sense codon. In lac operon, sequence of structural genes is \(\mathrm{Z}, \mathrm{Y}\), and \(\mathrm{A}\), which respectively code for \(\beta\)-galactosidase, lactose permease and transacetylase. If the gene \(\mathrm{Y}\) has nonsense mutation, gene expression will stop at it, resulting in non-expression of both gene \(\mathrm{Y}\) and successive gene \(\mathrm{A}\). Thus, only \(\beta\)-galactosidase enzyme will be produced.
\(
\xrightarrow[\rightarrow]{\text { DNA }} \xrightarrow{\mathrm{C}} m \mathrm{RNA} \xrightarrow{\mathrm{B}} \text { Protein } \xrightarrow{\text { Proposed by }} \mathrm{A}
\)
The figure gives an important concept in the genetic implication of DNA. Fill the blanks \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\).
(NEET 2013)
(d) :
Satellite RNAs are present in some (NEET 2013)
(d) : Plant viruses often contain parasites of their own, referred to as satellites. Satellite RNAs are highly dependent on their helper virus for both replication and encapsidation. Their size vary from 194 to 1500 nucleotides (approx.) The larger satellitescontain open reading frame and express proteins, whereas smaller satellites do not produce functional proteins.
Which of the following is not a property of the genetic code? (Karnataka NEET 2013)
(b) : Genetic code is non-ambiguous. Nonambiguous code means that there is no ambiguity about a particular code. One codon specifies only one amino acid and not any other. There are 64 codons. Out of 64,3 are stop codons or nonsense codons, i.e., these do not code for any amino acid and rest 61 code for one of the 20 amino acids. Neither of them code for more than one amino acids except GUG which normally code for valine but in certain conditions it also codes for \(\mathrm{N}\)-formyl methionine as initiation codon.
Genes of interest can be selected from a genomic library by using (Karnataka NEET 2013)
(b) : Gene bank or genomic library is a complete collection of cloned DNA fragments which comprises the entire genome of an organism. Molecular probes are small DNA segments that are used to detect the presence of complementary sequences in nucleic acid samples in genomic library. These are usually formed of 200-500 nucleotide sequences. These segments or probes are labelled either with radioactive or with nonradioactive compound. Probes with DNA sequence complementary to the gene to be isolated are used. They bind with the desired gene, making it visible and help in isolating it from the library.
In an inducible operon, the genes are (Karnataka NEET 2013)
(a):
One of the most frequently used techniques in DNA fingerprinting is (Karnataka NEET 2013)
(a)
Removal of introns and joining of exons in a defined order during transcription is called (NEET 2012)
(d): Introns, which occur principally in eukaryotes, are transcribed into messenger RNA ( \(m\) RNA) but are subsequently removed from the transcription before translation. In certain cases, removal of the introns is an autocatalytic process (selfsplicing) whereby the RNA itself has the properties of an enzyme.
If one strand of DNA has the nitrogenous base sequence as ATCTG, what would be the complementary RNA strand sequence?
(NEET 2012)
(b) : In RNA, thymine is substituted with uracil thus, the RNA strand complementary to DNA strand ATCTG will be UAGAC.
Ribosomal RNA is actively synthesized in (NEET 2012)
(b) : Nucleolus is the centre for synthesis of ribosomal RNA \((r \mathrm{RNA})\) that form ribosomal subunits. Ribosomal proteins migrate to the nucleolus from their assembly sites in the cytoplasm and are packaged into ribonucleoproteins. These return to the cytoplasm where they become mature ribosome particles.
Which one of the following is not a part of a transcription unit in DNA? (NEET 2012)
(a) : A transcription unit is a part of DNA that is able to transcribe a complete RNA. It consists of a promoter region (where RNA polymerase binds to start transcription), the structural gene (coding region) and the terminator region (that signals release of RNA polymerase and newly formed RNA strand).
Removal of RNA polymerase III from nucleoplasm will affect the synthesis of (NEET 2012)
(a) : In eukaryotes, RNA polymerase enzymes (Type I, II, III) catalyze the synthesis of RNA using as a template either an existing DNA strand or an RNA strand. Type I is responsible for synthesis of \(r\) RNA, type II for \(m\) RNA and type III for \(t\) RNA synthesis.
What are the structures called that give an appearance as ‘beads-on-string’ in the chromosomes when viewed under electron microscope?(NEET 2011)
(c) Nucleosomes appear as ‘beadsÂonÂstring’ in
the chromosome when viewed under electron
microscope. The beads in ‘beadsÂonÂstring’
arrangement are complexes of histones and DNA.
The bead plus the connecting DNA that leads to the
next bead from the nucleosome. Nucleosome is the
fundamental unit of organization on which the higherÂ
order packaging of chromatin is built. The bead of
each nucleosome contains eight histone molecules in
which two copies each of H 2 A, H 2 B, H 3 and H 4 are
found.
The unequivocal proof of DNA as the genetic material came from the studies on a (Main 2011)
 (d): The unequivocal proof that DNA is the genetic material came from the experiments of Alfred Hershey and Martha Chase (1952). They worked with viruses that infect bacteria called bacteriophages.
Which one of the following does not follow the central dogma of molecular biology? \((2010)\)
(d) :
\(
\text { DNA } \xrightarrow{\text { Transcription }} m \text { RNA } \xrightarrow{\text { Translation }} \text { Protein }
\)
This one-way flow of information from DNA to \(m\) RNA and then to protein is called the central dogma of molecular biology by F.H.C. Crick (1958). But later on two American workers H. Temin and D. Baltimore reported that DNA is also formed from RNA in retroviruses, e.g., HIV. This is called reverse transcription or Teminism, i.e.,
This reverse transcription occurs under the influence of reverse transcriptase enzyme. So, HIV viruses does not follow central dogma.
Which one of the following palindromic base sequences in DNA can be easily cut at about the middle by some particular restriction enzyme [NEET 2010]?
(c) Palindromic nucleotide sequences in the DNA molecule are groups of bases that form the same sequence when read in both forward and backward direction. In the given question, only option (c) represents a palindromic sequence, that can be easily cut at about the middle by some particular restriction enzyme.
The one aspect which is not a silent feature of genetic code, is its being (NEET 2010)
(b): Genetic code is non-ambiguous. Nonambiguous code means that there is no ambiguity about a particular code. One codon specifies only one amino acid and not any other. There are 64 codons. Out of 64,3 are stop codons or nonsense codons, i.e., these do not code for any amino acid and rest 61 code for one of the 20 amino acids. Neither of them code for more than one amino acids except GUG which normally code for valine but in certain conditions it also codes for \(\mathrm{N}\)-formyl methionine as initiation codon.
Select the two correct statements out of the four (i -iv) statements given below about lac operon.
(i) Glucose or galactose may bind with the repressor and inactivate it.
(ii) In the absence of lactose the repressor binds with the operator region.
(iii) The \(z\)-gene codes for permease.
(iv) This was elucidated by Francois Jacob and Jacques Monod.
The correct statements are (NEET 2010)
(c) : The two French scientists, Jacob and Monod proposed the lac operon of E. coli. The lac operon (an inducible operon) contains a promoter, an operator, a regulator gene and three structural genes \(z, y\), and \(a\), coding for the enzyme \(\beta\)-galactosidase, \(\beta\)-galactoside permease, and \(\beta\)-galactosidetransacetylase, respectively. \(\beta\)-galactoside permease “pumps” lactose into the cell, where \(\beta\)-galactosidase cleaves it into glucose and galactose. The function of the transacetylase is still not clear. The lac regulator gene, designated the \(i\) gene, codes for a repressor. In the absence of the inducer (i.e., lactose, actually allolactose), the repressor binds to the lac operator sequence, preventing RNA polymerase from binding to the promoter and transcribing the structural genes. The inducer of the operon, allolactose, is derived from lactose in a reaction that is catalyzed by \(\beta\)-galactosidase. Once formed, allolactose binds to the repressor, causing it to be released from the operator; in doing so, it induces transcription of the \(z, y\) and \(a\) structural genes. CAP is activator called catabolic activator protein. It exerts a positive control in lac operon because in its absence RNA polymerase is unable to recognise promotor gene. CAP activates lac genes only when glucose is absent. Such enzymes whose synthesis can be induced by adding the substrate are known as inducible enzymes and the genetic systems responsile for the synthesis of such an enzyme are known as inducible operons.
The \(3^{\prime}-5^{\prime}\) phosphodiester linkages inside a polynucleotide chain serve to join (Main 2010)
(c) : The phosphodiester bonds is formed between the phosphate group, which is connected to carbon \(5^{\prime}\) of the sugar residue of one nucleotide, and carbon \(3^{\prime}\) of the sugar residue of the next nucleotide.
The lac operon consists of (Main 2010)
(b) : The two French scientists, Jacob and Monod proposed the lac operon of E. coli. The lac operon (an inducible operon) contains a promoter, an operator, a regulator gene and three structural genes \(z, y\), and \(a\), coding for the enzyme \(\beta\)-galactosidase, \(\beta\)-galactoside permease, and \(\beta\)-galactosidetransacetylase, respectively. \(\beta\)-galactoside permease “pumps” lactose into the cell, where \(\beta\)-galactosidase cleaves it into glucose and galactose. The function of the transacetylase is still not clear. The lac regulator gene, designated the \(i\) gene, codes for a repressor. In the absence of the inducer (i.e., lactose, actually allolactose), the repressor binds to the lac operator sequence, preventing RNA polymerase from binding to the promoter and transcribing the structural genes. The inducer of the operon, allolactose, is derived from lactose in a reaction that is catalyzed by \(\beta\)-galactosidase. Once formed, allolactose binds to the repressor, causing it to be released from the operator; in doing so, it induces transcription of the \(z, y\) and \(a\) structural genes. CAP is activator called catabolic activator protein. It exerts a positive control in lac operon because in its absence RNA polymerase is unable to recognise promotor gene. CAP activates lac genes only when glucose is absent. Such enzymes whose synthesis can be induced by adding the substrate are known as inducible enzymes and the genetic systems responsile for the synthesis of such an enzyme are known as inducible operons.
In eukaryotic cell transcription, RNA splicing and RNA capping take place inside the (Main 2010)
(b) : Unlike in prokaryotes where transcription and translation take place in the same compartment, in eukaryotes primary transcript is first processed in the nucleus and then transported outside of the nucleus. Since the primary transcripts of the eukaryotes contains both the expressing genes (exons) and nonexpressing genes (introns), it undergoes splicing of introns and later capping and tailing at \(5^{\prime}\)-end and \(3^{\prime}\)-end respectively.
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Which one of the following statements about the particular entity is true? (Main 2010)
(b) : Insulin gene is found in every body cell but is not expressed in all cells. It is nucleosome which consists of a core of eight histones. DNA is composed of nucleotides. Centriole is found in animal cells, which produces aster during cell division.
Whose experiments cracked the DNA and discovered unequivocally that a genetic code is a “triplet”? (NEET 2009)
(d) : Genetic code was deciphered in 1960 ‘s by Crick, Ochoa, Nirenberg, Mathaei and Khorana.
Semi-conservative replication of DNA was first demonstrated in (NEET 2009)
(a) : Mathew Meselson and Franklin Stahl (1958) conducted various experiments using isotopically labelled DNA of Escherichia coli to provide evidence in favour of semi-conservative mode of DNA replication.
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Removal of introns and joining the exons in a defined order in a transcription unit is called (NEET 2009)
(d) : \(m\) RNA is not made directly in a eukaryotic cell. It is transcribed as heterogeneous nuclear RNA (hnRNA) in the nucleus. hnRNA contains introns and exons. The introns are removed by RNA splicing leaving behind the exons, which contain the information. The exonic regions of RNA are joined together to produce a single chain RNA required for functioning as translational template.
What is not true for genetic code? (NEET 2009)
(d) : The relationship between the sequence of amino acids in a polypeptide and nucleotide sequence of DNA or \(m\) RNA is called genetic code. The genetic code is continuous and does not possess pause after the triplets. So a codon in \(m\) RNA is a non-contiguous fashion. If a nucleotide is deleted or added, the whole genetic code will read differently.
In the DNA molecule, (NEET 2008)
(d) : The relationship between the sequence of amino acids in a polypeptide and nucleotide sequence of DNA or \(m\) RNA is called genetic code. The genetic code is continuous and does not possess pause after the triplets. So a codon in \(m\) RNA is a non-contiguous fashion. If a nucleotide is deleted or added, the whole genetic code will read differently.
Which one of the following pairs of nitrogenous bases of nucleic acids, is wrongly matched with the category mentioned against it?
(NEET 2008)
(b) : The two DNA chains are held together by hydrogen bonds between their nitrogenous bases. Adenine (A), a purine of one chain lies exactly opposite thymine (T), a pyrimidine of the other chain. Similarly, cytosine (C), a pyrimidine lies opposite guanine \((\mathrm{G})\), a purine. Three hydrogen bonds occur between cytosine and guanine \((\mathrm{C} \equiv \mathrm{G})\) at positions \(1^{\prime}, 2^{\prime}\), and \(6^{\prime}\) and two hydrogen bonds between adenine and thymine \((\mathrm{A}=\mathrm{T})\) at positions \(1^{\prime}\) and \(6^{\prime}\).
Polysome is formed by (NEET 2008)
(c): Ribosomes may occur in rosettes or helical groups called polyribosomes or polysomes (Gk. Poly – many, soma – body). The different ribosomes of a polyribosome are connected with a \(10-20 \AA\) thick strand of messenger or \(m\) RNA and its maintenance requires energy. Polyribosomes are formed during periods of active protein synthesis when a number of copies of the same polypeptide are required.
Which one of the following pairs of codons is correctly matched with their function or the signal for the particular amino acid?
(NEET 2008)
(d) : AUG codes for methionine and is initiation or start codon which starts the synthesis of polypeptide. UAA (ochre), UAG (amber) and UGA (opal) do not specify any amino acid so they are called termination codons. CUU, CUC, CUA and CUG codes for leucine whereas GCU, GCC, GCA and GCG codes for alanine.
One turn of the helix in a B-form DNA is approximately (NEET 2006)
(d) : DNA or deoxyribose nucleic acid is the largest macromolecule made of the helically twisted two antiparallel polydeoxyribonucleotide strands held together by hydrogen bonds. The two strands of DNA are together called DNA duplex. It has a diameter of 20 Angstrom . One turn spiral has a distance of 34 Angstrom . It contains 10 deoxyribonucleotides in each strand so that the distance between two adjacent nucleotides is 3.4 Angstrom.
One gene-one enzyme hypothesis was postulated by (NEET 2006)
(a) : In 1948, Beadle and Tatum proposed onegene one-enzyme hypothesis which states that a gene controls metabolic machinery of the organism through synthesis of an enzyme. This laid the foundation of biochemical genetics. Beadle and Tatum were awarded Nobel Prize in 1958. This one gene one enzyme theory has been changed to one gene one polypeptide hypothesis proposed by Yanofsky. i.e., one gene synthesizes one polypeptide and many polypeptides form one enzyme.
Antiparallel strands of a DNA molecule means that (NEET 2006)
(d) : DNA is a type of nucleic acid that forms genetic material in many organisms. It consists of a long polymer of nucleotides which transcribes the coded information in the form of a triplet code of nucleotides in \(m\) RNA. It is a double helical molecule. The two strands of DNA run in opposite directions to one another with the hydrogen bonds between them. One strand of DNA has \(5^{\prime}-3^{\prime}\) direction and the other strand has \(3^{\prime}-5^{\prime}\) direction. So they are antiparallel. This direction is determined by the presence of a free phosphate or \(\mathrm{OH}\) group at the end of the strand. If the strand has phosphate group at the the \(5^{\prime}\) end and with a free \(\mathrm{OH}\) group at the \(3^{\prime}\) end.
Which antibiotic inhibits interaction between \(t\) RNA and \(m\) RNA during bacterial protein synthesis?
(NEET 2006)
(c) : Neomycin is a broad spectrum antibiotic which was first isolated from a strain of Streptomyces feadiae. It is effective against Gram positive as well as Gram negative bacteria. Its mechanism of action is by selective inhibition of protein synthesis on the \(70 \mathrm{~S}\) (prokaryotic) ribosome by inhibiting the interaction of \(m\) RNA and \(t\) RNA during translation process.
Amino acid sequence, in protein synthesis is decided by the sequence of (NEET 2006)
(c) : Messenger RNA or \(m\) RNA has been named so because it carries the coded information from DNA for the synthesis of proteins. It carries the coded information in a number of base triplets called codons. It is transcribed on DNA by the enzyme RNA polymerase. Hence, its base sequence is complementary to DNA on which it has been synthesized. In eukaryotes each gene transcribes its own \(m\) RNA, therefore the number of \(m\) RNAs corresponds to the number of genes. \(r\) RNA is a type of RNA that forms structural and functional components of ribosomes. \(t \mathrm{RNA}\) is a class of RNAhaving structures with triplet nucleotide sequences that are complementary to the triplet nucleotide coding sequences of \(m\) RNA. It binds with amino acids and transfers them to ribosomes.
E.coli cells with a mustard \(z\) gene of the lac operon cannot grow in medium containing only lactose as the source of energy because (NEET 2005)
(b)
Telomerase is an enzyme which is a (NEET 2005)
(c) : Telomerase is a ribonucleoprotein molecule that is enzymatic in nature. It uses a special mechanism for the synthesis of DNA at telomeric ends. The DNA repeat sequence of telomere has one \(\mathrm{G}\) rich strand and other \(\mathrm{C}\) rich strand. The \(\mathrm{G}\) rich strand has a single stranded overhand. This overhand works as a primer and for its elongation uses as template the RNA component of telomerase enzyme. Thus telomerase synthesizes only the G rich strand of telomeres.
Using imprints from a plate with complete medium and carrying bacterial colonies, you can select streptomycin resistant mutants and prove that such mutations do not originate as adaptation. These imprints need to be used
(NEET 2005)
(c) : Streptomycin is broad spectrum (active against both Gram-positive and Gram-negative bacteria) and was the first really effective drug against tuberculosis, but its use is limited by the development of resistant strains and by toxic side-effects. The bactericidal action of streptomycin, as with other aminoglycoside antibiotics (e.g., neomycin) is through selective inhibition of protein synthesis on \(70 \mathrm{~S}\) ribosomes.
To check resistance of mutants against streptomycin they must be grown on plates with streptomycin.
Only those bacterial colonies will propagate from the master that are resistant to the antibiotic.
Protein synthesis in an animal cell occurs (NEET 2005)
(d) : The mitochondria contains its own set of ribosomes which synthesize proteins, so protein synthesis occures both in mitochondria and cytoplasm.
Which one of the following makes use of RNA template to synthesize DNA? (NEET 2005)
(c)Â This one-way flow of information from DNA to \(m\) RNA and then to protein is called the central dogma of molecular biology by F.H.C. Crick (1958). But later on two American workers H. Temin and D. Baltimore reported that DNA is also formed from RNA in retroviruses, e.g., HIV. This is called reverse transcription or Teminism.
Which one of the following hydrolyses internal phosphodiester bonds in a polynucleotide chain? (NEET 2005)
(c) : Endonucleases hydrolyse the internal phosphodiester bond. Exonucleases cleave the terminal nucleotides. Lipase digest fats and proteases break down proteins.
During transcription holoenzyme RNA polymerase binds to a DNA sequence and the DNA assumes a saddle like structure at that point. What is that sequence called?
(NEET 2005)
(b) : After 25 bases from start of transcription point are TATA boxes. After 40 bases from TATA boxes appears CAAT boxes. Both of these sequences serve as recognitions sites in eukaryotic promoters. Transcription in eukaryotic genes is a far more complicated process than in prokaryotes.
After a mutation at a genetic locus the character of an organism changes due to change in (NEET 2004)
(a) : A mutation involves a change in the sequence of nucleotides in a nucleic acid molecule. This change will express itself in the form of a change in the sequence of aminoacids in the protein molecule synthesized through the information, encoded in nucleic acid segment. Therefore mutations at molecule level can be studied both by the study ofthe sequence of amino acids in a protein and also by the study of sequence of nucleotides in a segment of nucleic acid.
Which form of RNA has a structure resembling clover leaf? (NEET 2004)
(d) : Transfer RNA ( \(t\) RNA) are species of RNA responsible for the transfer of specific amino acids to the growing end of a polypeptide chain during translation. R.Holly in 1965 gave clover leaf model of \(t\) RNA for yeast alanyl \(t\) RNA. It has four major sites -AAbinding site, anticodon, site, TUC loop and DHU loop. The chain is having unpaired base sequence \(\mathrm{CCA}\) at \(3^{\prime}\) end and \(\mathrm{G}\) at \(5^{\prime}\) end.
During transcription, if the nucleotide sequence of the DNA strand that is being coded is ATACG then the nucleotide sequence in the \(m\) RNA would be
(NEET 2004)
(c) : During transcription RNA synthesis from a DNA template takes place. It involves rewriting of the code without a change in its language. In \(m\) RNA, adenine pairs with uracil because thymine is not present \(m\) RNA.
In mutational event, when adenine is replaced by guanine, it is a case of (NEET 2004)
(c): Transition mutant is one in which a purine is substituted by a different purine, or a pyrimidine by a different pyrimidine. Such a change involves a base pair change between a G-C pair and an A-T pair in the DNA whereas transversion results when one nitrogen base is replaced by another different type e.g., C-G and A-T. Transcription is the formation of \(m\) RNA on DNA templete.
The following ratio is generally constant for a given species: (NEET 2004)
(c) :
What would happen if in a gene encoding a polypeptide of 50 amino acids, \(25^{\text {th }}\) codon (UAU) is mutated to UAA ?
(NEET 2003)
(a): UGA, UAG and UAA are three non sense (or termination) codon which do not code for any amino acid. If in a gene encoding a polypeptide of 50 amino acid, \(25^{\mathrm{hm}}\) codon is mutated to UAA or any of the termination codon, then the chain will be terminated at that place because it will become difficult for \(t\) RNA to bring amino acid from amino acid pool. So in that case a polypeptide of 24 amino acid will be formed.
What does “lac” refer to in what we call the lac operon? (NEET 2003)
(a) : In lac operon, lac refers to lactose. The lac operator is a part of the structural genes (lac \(\mathrm{Z}\), lac \(\mathrm{Y}\), lac A and lac I). It is responsible for the uptake and initial catabolism of lactose.
During translation initiation in prokaryotes, a GTP molecule is needed in (NEET 2003)
(c) : The initiation of polypeptide chain in prokaryotes is always brought about by the amino acid methionine but it has to be formylated to form \(t \mathrm{RNA}\) \(\mathrm{f}^{\text {net }}\). Then methionine binds with \(t \mathrm{RNA} \mathrm{f}^{\text {net }}\) to form \(\mathrm{f}\)-mot \(-t\) RNA \(\mathrm{f}^{\text {net. }}\). This \(\mathrm{f}^{\text {net }}-t \mathrm{RNA}\) fmet complex binds with the \(m\) RNA-30S subunit complex using initiation factors IF-2 and IF-1 and GTP.
Which one of the following triplet codes, is correctly matched with its specificity for an amino acid in protein synthesis or as ‘start’ or ‘stop’ codon?
(NEET 2003)
(d) : Codon UAC is correctly matched as it codes for amino acid tyrosine. UCG codes for serine, UUU codes for phenylalanine and UGU codes for cysteine. Start codon is AUG and stop codons are UAA, UAG and UGA.
During transcription, the DNA site at which RNA polymerase binds is called (NEET 2003)
Â
(a) : Promoter is region on a DNA molecule upstream from the coding sequence, area to which RNA polymerase initially binds prior to the initiation of transcription. The promoter,or at least part of it, determines the nature of the polymerase that associates with it. Certain consensus sequences within the promoter region seem to be particularly important in the binding of RNA polymerase, and these are known as CAAT and TATA boxes. The promoter region extends from some 40 nucleotides to about five nucleotides upstream from the start of the genecoding region, the CAAT and TATA boxes being located within the promoter region as short six or seven nucleotide sequence.
Degeneration of a genetic code is attributed to the (NEET 2003)
Â
(d) : In a triplet for a particular amino acid more than one word (synonyms) can be used. This phenomenon is described by saying that the code is degenerate. A degenerate code would be one where there is one to one relation between aminoacids and the codons that 44 codons out of 64 will be useless or nonsense codons. A code is degenerate because of the third base of the codon. It has been shown that the same \(t\) RNA can recognize more than one codons differing only at the third position. For example GCU, GCC and GCA all code for alanine amino acids.
In the genetic code dictionary, how many codons are used to code for all the 20 essential amino acids?
(NEET 2003)
(b) :
In a DNA percentage of thymine is \(20 \%\) then what will be percentage of guanine? (NEET 2002)
(c) : In a DNA, the percentage of thymine is \(20 \%\). So, as it pairs with adenine, it is also \(20 \%\). So the guanine and cytosine together forms \(60 \%\) of DNA and hence, guanine is \(30 \%\).
Transformation experiment was first performed on which bacteria? (NEET 2002)
(b) : Transformation involves transfer of genetic material of one bacterial cell into another bacterialcell by some unknown mechanism and it converts one type of bacterium into another type.
This was first studied by Griffith (1928) in Diplococcus pneumoniae and hence is known as Griffith effect.
Jacob and Monod studied lactose metabolism in E. coli and proposed operon concept. Operon concept is applicable for:
(NEET 2002)
(c) : Operon model was given by Jacob and Monod (1961) for regulation of protein synthesis in prokaryotes. In bacteria, the genes that contain the information for assembling the enzymes for a metabolic pathway are usually clustered together on the chromosome in a functional complex called an operon.
Regulation of protein synthesis in eukaryotes is explained by gene battery model given by Britten and Davidson.
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In \(E\). coli, during lactose metabolism repressor binds to: (NEET 2002)
(b) : In the lac operon of E.coli due to the activity of regulator gene synthesis of repressor molecules occurs. These repressor molecules get attached to the operator gene and thus check \(m\) RNA synthesis and because of this no protein synthesis occurs.
Out of 64 codons, 61 codons code for 20 types of amino acid. It is called (NEET 2002)
(a):
Which of the following reunites the exon segments after RNA splicing? (NEET 2002)
(c) : RNA polymerase enzyme catalyses the synthesis of RNA. It is single in prokaryotes. There are three types of RNA polymerases in eukaryotes-I for \(28 \mathrm{~S}, 18 \mathrm{~S}\) and \(5.8 \mathrm{~S}\) RNA, II for \(m\) RNA and snRNA and III for \(t\) RNA, 5SRNA and scRNA.
Primase is an RNA polymerase that is used to initiate DNA synthesis. RNA ligase reunites the exon segment after RNA splicing.
Which of the following enzymes are used to join bits of DNA? (NEET 2002)
(a) : Ligases are used to join bits of DNA. Primase is an RNA polymerase, used to initiate DNA synthesis. DNA polymerase enzyme catalyses the synthesis of DNA. Endonuclease, causes the splicing of the intron carrying the coding sequence of the same endonuclease.
Exon part of \(m\) RNAs have code for (NEET 2002)
(a): DNA transcribes to form \(m\) RNA. Its function is to carry coded information from DNA for the synthesis of proteins. The RNA consists of a coding region called exon and non-coding region called introns. The exons are thus the functional part that have code for proteins.
Change in sequence of nucleotide in DNA is called as (NEET 2002)
(b) : A mutation involves a change in the sequence of nucleotides in a nucleic acid molecule. This change will express itself in the form of a change in the sequence of aminoacids in the protein molecule synthesized through the information, encoded in nucleic acid segment. Therefore mutations at molecule level can be studied both by the study ofthe sequence of amino acids in a protein and also by the study of sequence of nucleotides in a segment of nucleic acid.Â
\(m\) RNA is synthesised on DNA template in which direction (NEET 2001)
(a) : \(m \mathrm{RNA}\) is synthesized on DNA template in \(5^{\prime}-3^{\prime}\) direction. Synthesis of \(m\) RNA exhibits several features that are synonymous with DNA replication. RNA synthesis requires accurate and efficient initiation, elongation proceeds in the \(5^{\prime}-3^{\prime}\) direction (i.e., the polymerase oves along the template strand of DNA in the \(5^{\prime}-3^{\prime}\) direction), and RNA synthesis requires distinct and accurate termination. Transcription exhibits several features that are distinct from replication.
In negative operon, (NEET 2001)
(a): The tryptophan operon (trp operon) in bacteria is a repressible operon. Here, repressor is inactive and it becomes active as DNA binding protein only when complexed with a co-repressor (tryptophan). In absence of tryptophan, the operator site is open to binding by RNA polymerase, which transcribes the structural genes of tryptophan operon, leading to production of enzymes that synthesize tryptophan. When tryptophan becomes available, the enzymes of tryptophan synthetic pathway are no longer needed and tryptophan (co-repressor)-repressor complex blocks transcription. The regulation of this operon is also a negative control.
Gene and cistron words are sometimes used synonymously because (NEET 2001)
(c) : A gene is a hereditary unit consisting of a sequence of DNA and occupying a specific position or locus within the genome. Gene activity ultimately affects the phenotype of the organism possessing the gene. Thus gene is a physical and functional unit of genetic information. A cistron is a unit of genetic function. In prokaryotes there is one gene one enzyme correspondence. It means that in these organisms genes and cistrons are equivalent.
Types of RNA polymerase required in nucleus of eukaryotes for RNA synthesis (NEET 2001)
(c):
Method of DNA replication in which two strands of DNA separate and synthesize new strands (NEET 2000)
(c) : The method of DNA replication is semiconservative. According to the semi-conservative model proposed by Watson and Crick, each strand of the two double helices formed would have one old and one new strand. So, the parental identity is conserved upto half extent and hence DNA replication is semi-conservative.
Which of the following is initiation codon? (NEET 2000)
(c): AUG codes for methionine and is initiation or start codon which starts the synthesis of polypeptide. UAA (ochre), UAG (amber) and UGA (opal) do not specify any amino acid so they are called termination codons. CUU, CUC, CUA and CUG codes for leucine whereas GCU, GCC, GCA and GCG codes for alanine. Â
Anticodon occurs in (NEET 2000)
(a)
Length of one loop of B-DNA (NEET 2000)
(a) : B-DNA is an antiparallel double helix. The double strand or duplex is coiled plectonemically in right handed fashion around a common axis like a rope stair case twisted in a spiral. The coiling produces alternate major and minor grooves. One turn of spiral has a distance between two adjacent nucleotides is 3.4 Angstrom .
In three dimensional view the molecule of \(t \mathrm{RNA}\) is (NEET 2000)
(a) : 3-D model of \(t\) RNA looks like flattened L-shaped molecule.\(t\) RNA acts as adoptor molecule which carries amino acids to the site of protein synthesis (i.e., ribosomes). Most accepted model for \(t\) RNA structure is ‘clover leaf model.
Similarity in DNA and RNA is that (NEET 2000)
(a) : Deoxyribonucleic acid and ribonucleic acid as the name suggests are made up of several nucleotide monomers. Each nucleotide consists of pentose sugar, phosphate group and nitrogenous bases. DNA has deoxyribose sugar whereas RNA has ribose sugar. The bases in DNA molecule are A, T, G and \(\mathrm{C}\) whereas in RNA, thymine is absent and instead uracil is found.
The Pneumococcus experiment proves that (AIPMT 1999)
(c) : Transformation was first discovered by Griffith (1928), in Pneumococcus (Streptococcus pneumoniae), that causes pneumonia.
Griffith injected a group of mice with nonencapsulated, rough (R), pneumococci; a second group with heat-killed encapsulated pneumococci cells, and a third group a mixture consisting of a few living nonencapsulated, rough pneumococci derived from a type \(S\) culture, and heat-killed encapsulated cells (S type). Griffith observed that the mice in the first two groups were not infected, and the mice in the third group died within a few days. The mice of the third group should have survived as the organisms which could kill them had been killed, and the cell of \(\mathrm{R}\) type were incapable of causing disease. However, the mice died, and living virulent encapsulated cells of the type \(\mathrm{S}\) were recovered from their dead bodies. It was observed by Griffith, that killed encapsulated pneumococci had liberated some substance which favoured non-capsulated cells (R type) to produce a capsular substance.
This substance in later experiments was proved to be DNA. These experiments showed that DNA is the genetic material.
In operon concept, regulator gene functions as (AIPMT 1999)
(b) : Regulator gene is a gene whose function is to control the transcriptional activity of other genes, either adjacent or distant in the genome. In the case of the lac operon of E.coli the regulator gene lac \(i\) produces a protein product that represses the operator gene of the operon. In bacteria the same regulator gene may affect a series of non-adjacent operons.
Initiation codon in eukaryotes is (AIPMT 1999, 1994)
(c): AUG codes for methionine and is initiation or start codon which starts the synthesis of polypeptide. UAA (ochre), UAG (amber) and UGA (opal) do not specify any amino acid so they are called termination codons. CUU, CUC, CUA and CUG codes for leucine whereas GCU, GCC, GCA and GCG codes for alanine.Â
DNA is mainly found in (AIPMT 1999)
(b) : DNA is mainly found in nucleus. It is associated with RNA and proteins to form compact chromosomes. But some amount of DNA is also found in chloroplasts and mitochondria. This DNA is called extra-chromosomal DNA.
In prokaryotes, the genetic material is (AIPMT 1999)
(b): The genetic material of prokaryotes is circular and single stranded DNA. It has no association of histones. The eukaryotic geneticmaterial is linear and double stranded DNA. It is associated with histone proteins to form nucleosome unit.
In DNA, when AGCT occurs, their association is as per which of the following pair? (AIPMT 1999)
(a) : DNA molecule has four bases – adenine, guanine, cytosine and thymine. Adenine always pairs with thymine and guanine pairs with cytosine. Their association is A-T and G-C.
The eukaryotic genome differs from the prokaryotic genome because (AIPMT 1998)
(b) : Genome refers to the total sets of chromosomes carried by each cell of the organism. In prokaryotes the genetic material is circular and single stranded DNA. It has no association of histones. The eukaryotic genetic material is linear and double stranded DNA. It is associated with histone proteins to form nucleosome unit.
What base is responsible for hot spots for spontaneous point mutations? (AIPMT 1998)
(c): Mutations are rare events in nature and are then described as spontaneous mutations. Some of these mutations originate from mistakes in normal duplication of DNA. Transitions may be produced by tautomeric shift or ionization of bases which leads to mistaken, \(\mathrm{A}-\mathrm{C}\) base pairing and more frequently mistaken \(\mathrm{G}-\mathrm{T}\) base pairing. Guanine pairs with the rare enol form of thymine and is thus considered as hot spot for spontaneous point mutations.
Genes that are involved in turning on or off the transcription of a set of structural genes are called
(AIPMT 1998)
(d): Operator genes are a region of DNA sequence capable of interacting with a specific repressor molecule and in doing so it affects the activity of other genes downstream from it.
DNA elements, which can switch their position, are called (AIPMT 1998)
(b) : Transposons are portable genetic elements which can insert themselves at random into a plasmid or any chromosome independently of the host cell recombination system. It was discovered by Barbara Mc Clintock (1940) in maize and termed as jumping genes. Later Headges and Jacob termed them as transposons.
Introns are nontranslated sequences within the coding sequence of a gene. Such sequences are transcribed into hnRNA but are then spliced out and are not represented in the message. The non-intron sequences of the gene are referred to as exons.
Cistron sequence of nucleotides in a DNA molecule code for one particular polypeptide chain.
The codons causing chain termination are (AIPMT 1997)
(b) : AUG codes for methionine and is initiation or start codon which starts the synthesis of polypeptide. UAA (ochre), UAG (amber) and UGA (opal) do not specify any amino acid so they are called termination codons. CUU, CUC, CUA and CUG codes for leucine whereas GCU, GCC, GCA and GCG codes for alanine.Â
DNA synthesis can be specifically measured by estimating the incorporation of radio-labelled (AIPMT 1997)
(a) : Autoradiography is the study of labelled precursors like \({ }^3 \mathrm{H}\) by knowing the movement of radioactivity with the help of photographic films and emulsions at short intervals.
Radioactive material like tritiated thymidine which is formed by replacing normal hydrogen of thymidine with \(\mathrm{H}^3\) (heavy isotope of hydrogen). Thymidine only is used for this purpose because RNA will not be labelled by this.
The RNA that pick up specific amino acid from amino acid pool in the cytoplasm to ribosome during protein synthesis is called
(AIPMT 1997)
(d) : Transfer RNA or \(t\) RNA help in transfer of amino acids to ribosomes \(m\) RNA complex to form the polypeptide chain. It has four key regions a carrier and recognition end, enzyme site and ribosome site. This recognition end has three anticodons with the help of which amino acids are identified. \(r\) RNA forms \(67 \%\) of \(70 \mathrm{~S}\) ribosomes and \(50 \%\) of \(80 \mathrm{~S}\) ribosomes. \(m\) RNA carries the coded information from DNA for the synthesis of proteins.
Which of the following step of translation does not consume a high energy phosphate bond? (AIPMT 1997)
(a): Protein synthesis or translation consists of ribosomes, amino acids, \(m\) RNA, \(t\) RNAs and aminoacyl \(t\) RNA synthetases. The ribosomes have two binding sites namely aminoacyl site or A-site and peptide site or \(\mathrm{P}\)-site. The starting amino acid methionine lies at the P-site of the ribosome. The next incoming \(t \mathrm{RNA}\) is called amino acyl \(t \mathrm{RNA}\), it is bound to A-site. A peptide bond is formed between \(\mathrm{COOH}\) group of the \(t \mathrm{RNA}\) at \(\mathrm{P}\)-site and \(\mathrm{NH}_2\) group of aminoacyl \(t\) RNA. This is facilitated by the enzyme peptidyl transferase and does not require high energy phosphate bonds.
Which of the following serves as a terminal codon? (AIPMT 1996)
(a) : AUG codes for methionine and is initiation or start codon which starts the synthesis of polypeptide. UAA (ochre), UAG (amber) and UGA (opal) do not specify any amino acid so they are called termination codons. CUU, CUC, CUA and CUG codes for leucine whereas GCU, GCC, GCA and GCG codes for alanine.
The maximum formation of \(m\) RNA occurs in (AIPMT 1996)
(d) : Nucleolus is a plasmosome body that is formed around the nucleolus organizer and is located in the secondary constriction on that chromosome. It is made up of RNA and proteins. The associated nucleolar chromatin contains DNA. It forms \(m\) RNA that has low molecular weight. Ribosomes are mainly concerned with proteins synthesis. They are sites for synthesis of \(r\) RNA and \(t\) RNA is synthesized in the cytoplasm.
Radio-tracer technique shows that DNA is in (AIPMT 1996)
(c) : \({ }^{14} \mathrm{C}\) and \({ }^3 \mathrm{H}\) are incorporated in bases like thymidine, uridine and amino acids to study the structure of DNA and proteins. Radio tracer technique shows that DNA is in double helical form.
The wild type E.coli cells are growing in normal medium with glucose. They are transferred to a medium containing only lactose as sugar. Which of the following changes take place?
(AIPMT 1995)
(a): When E.coli bacteria are transfered to medium containing lactose, then the lac opeon is indueed. The lac opeaon consists of 3 structural gene (lac Z, lac Y and lac A). It involves the synthesis of \(\beta\)-galactosidase enzyme in E.coli, which hydrolyses lactose into glucose and galactose.
The lac operon is an example of (AIPMT 1995)
(d) :
An environmental agent, which triggers transcription from an operon, is a (AIPMT 1995)
(d) : Inducer is a metabolite (or analogue of similar chemical structure), usually of low molecular weight, which promotes the production of an enzyme. Inducers are often substrates for the enzymes they induce, e.g. lactose in case of the synthesis of \(\beta\) galoctosidase in lac operon.
If the sequence of bases in DNA is ATTCGATG, then the sequence of bases in its transcript will pe (AIPMT 1995)
(d) : In transcription, \(m\) RNA is formed from DNA template and thymine of DNA is replaced by uracil of RNA. Uracil pairs with adenine.
If the DNA codons are ATG ATG ATG and a cytosine base is inserted at the beginning, then which of the following will result?
(AIPMT 1995)
(a): Nonsense mutation is a mutation which interconverts a nonsense to or from a sense-coding triplet, resulting in an abnormally foreshortened or elongated polypeptide chain. But in this example cytosine is added at the beginning so CAT GAT GATG will result.
In split genes, the coding sequences are called (AIPMT 1995)
(a) : Split gene are those genes that consist of continuous sequence of nucleotide (coding sequence) interrupted by intervening sequences. Most eukaryotic genes are split as are genes of some animal viruses. The continuous coding sequences are called exons and the intervening non-coding sequence are called introns. These introns are not represented in \(m\) RNA transcribed from the gene and are not utilized for the synthesis of proteins.
Anticodon is an unpaired triplet of bases in an exposed position of (AIPMT 1995)
(a): Anticodon is the sequence of three nucleotides in a transfer RNA molecule that pairs with a complementary sequence of three nucleotides (codon) on a molecule of messenger RNA. \(t\) RNA has clove like shape or \(\mathrm{L}\) shape (three dimensional). It has \(\mathrm{G}\) at \(5^{\prime}\) end \(\mathrm{CCA}\) at \(3^{\prime}\) end. CCA at \(3^{\prime}\) end is meant for attaching to a specific amino acid (AA-binding site). On the opposite side lies an anticodon that is complementary to a specific codon of \(m\) RNA. The two are called recognition sites.
‘Lac operon’ in E. coli, is induced by (AIPMT 1994)
(c) : Lac operon in E. coli is induced by \(\beta\)-galactosidase an enzyme meant for hydrolysis of lactose in glucose and galactose.
\(
\text { Lactose } \xrightarrow{\beta \text {-galactosidase }} \text { Glucose }+ \text { Galactose }
\)
These enzymes are called as inducible enzymes, because the synthesis of such enzymes are induced by adding substrate such as lactose by 10,000 times.
Initiation codon in eukaryotes is (AIPMT 1994)
(c) : Â AUG codes for methionine and is initiation or start codon which starts the synthesis of polypeptide. UAA (ochre), UAG (amber) and UGA (opal) do not specify any amino acid so they are called termination codons. CUU, CUC, CUA and CUG codes for leucine whereas GCU, GCC, GCA and GCG codes for alanine.
There are special proteins that help to open up DNA double helix in front of the replication fork.Â
These proteins are (AIPMT 1994)
(b) : DNA is a double helical molecule and it opens to form a replication fork for its replication. The two strands of DNA are joined with the help of \(\mathrm{H}\)-bonds between the strands. Topoisomerases are specialized to cause nicks or breaks in the double helix and helps separate the DNA stands. Helicase unwindsthe DNA helix from that nick caused by the topoisomerase and this seperates the two strands.
DNA gyrase introduces negative supercoils in DNA strands of prokayotes.
DNA polymerase adds nucleotides units to the \(3^{\prime}\) end of a DNA chain. DNA ligase joins the ends of DNA.
In protein synthesis, the polymerization of amino acids involves three steps. Which one of the following is not involved in the polymerisation of protein ?
(AIPMT 1994)
(d) : Transcription is the mechanism of copying the message of DNA on RNA with the help of enzyme RNA polymerase. It is meant for taking the coded information from DNA to the site where it is required for protein synthesis.
Translation or protein synthesis is a complicated process involving several steps such as – activation of amino acid, transfer of amino acid to \(t\) RNA, initiation of polypeptide synthesis, elongation of polypeptide chain and, termination of polypeptide chain.
Nucleosome core is made of (AIPMT 1993)
(d): Nucleosome core is made up of \(\mathrm{H}_2 \mathrm{~A}, \mathrm{H}_2 \mathrm{~B}\), \(\mathrm{H}_3\) and \(\mathrm{H}_4\). It is about \(7-10 \mathrm{~nm}\) in diameter, consisting of histones around which a DNA strand, about 120 base pair long is wrapped in chromosomes.
Initiation codon of protein synthesis (in eukaryotes) is (AIPMT 1993)
(d): AUG codes for methionine and is initiation or start codon which starts the synthesis of polypeptide. UAA (ochre), UAG (amber) and UGA (opal) do not specify any amino acid so they are called termination codons. CUU, CUC, CUA and CUG codes for leucine whereas GCU, GCC, GCA and GCG codes for alanine.
The transforming principle of Pneumococcus as found out by Avery, MacLeod and McCarty was (AIPMT 1993)
(b): The transforming principle of Pneumococcus as found out by Avery, MacLeod and McCarty was DNA. In 1944, Avery, MacLeod and McCarty repeated Griffith’s experiment successfully. They separated the proteins, carbohydrates and DNA of S III strains and separately mixed them in the pure cultures of R II. Only DNA could bring about transformation of R II type into S III and not the proteins or the carbohydrates.
Who proved that DNA is basic genetic material? (AIPMT 1993)
(d): Hershey and Chase proved that DNA is a basic genetic material. Hershey and Chase, 1952, by using \(\mathrm{P}^{32}\) and \(\mathrm{S}^{35}\) with a \(\mathrm{T}-2\) type phage concluded that DNA is the genetic material.
Because most of the amino acids are represented by more than one codon, the genetic code is (AIPMT 1993)
(c) : Certain amino acids are identified by more than one codons. This phenomenon is called as degeneracy e.g., only AUG codes for methionine and UGG tryptophan.
The process of translation is (AIPMT 1993)
(b): The process of translation is protein synthesis. Emil Fischer, a German chemist established that the proteins are polymers of amino acids. There are some twenty amino acids involved in protein synthesis. In translation, the message coded by DNA on \(m\) RNA is translated into a specific protein.
A DNA with unequal nitrogen bases would most probably be (AIPMT 1993)
(a): A DNA with unequal nitrogen bases would most probably be single stranded. Nitrogenous bases are unequal in number in single stranded DNA, because they do not possess complementary base pairs.
During DNA replication, the strands separate by (AIPMT 1993)
(c) : During DNA replication, the strands separate by unwindase/helicase. The molecule is unwound by DNA unwinding proteins called helicases. The helicases II and III get attached to logging strand and protein to the leading strand. The formation of bands is avoided by single stranded DNA binding proteins (SSB).
Nucleotide arrangement in DNA can be seen by (AIPMT 1993)
(a) : Nucleotide arrangement in DNA can be seen by X-ray crystallography. Watson and Crick, 1953 proposed the double helical model for DNA. They were awarded Nobel prize in 1962. This model was developed by them on the basis of several previous observations including the \(d\)-helix of Pauling, 1951 and X-ray reflection studies of Franklin and Gosling, 1953.
In the genetic dictionary, there are 64 codons as (AIPMT 1990)
(d):
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