Molecular Biology MCQ Set 1
1. Who were the first to suggest that one strand of DNA might act as a template for the synthesis of its complementary strand?
a) Meselson and Stahl
b) Watson and crick
c) Walter Flemming
d) Rosalind Franklin and Maurice Wilkins
Answer: b [Reason:] In Watson and Crick’s paper on the model of DNA double helix they ended with a statement that it had not escaped their notice that the specific pairing they had postulated immediately suggested that one strand might be the template for the complementary strand synthesis.
2. Semiconservative nature of replication of eukaryotic chromosome was first demonstrated by _______
a) Walter Flemming on root tip cells of Vicia faba
b) J. Herbert Taylor, Philip Wood and Walter Hughes on root tip cells of Vicia faba
c) Walter Flemming on root tip cells of Phaseolus vulgaris
d) J. Herbert Taylor, Philip Wood and Walter Hughes on root tip cells of Phaseolus vulgaris
Answer: b [Reason:] as Walter Flemming discovered cell division not semi conservative nature of replication. Semiconservative nature of replication was first demonstrated by J. Herbert Taylor, Philip Wood and Walter Hughes on root tip cells of broad bean which is the common name for Vicia faba.
3. Watson and Crick’s suggestion of the complementary strand synthesis taking one of the parent strand as template was proposed in _________
Answer: c [Reason:] As Watson and Crick’s paper on structure of DNA was published in 1953 they ended their journal suggesting the possibility of the synthesis of complementary strand taking one of the parent strands as the template.
4. Which of the following statements is true with respect to the DNA double helix?
a) Composed of two or more polynucleotide chains
b) The base pairs have opposite polarity
c) Covalent bond exists in base pairing
d) T is transcribed as U
Answer: b [Reason:] The two strands having complementary base pairing have the same helical geometry but have opposite polarity. This is because the strands are held together are antiparallel in nature.
5. Who, in which year showed firm enzymological proof that DNA alone functions as the template for the synthesis of new DNA strands?
a) Watson and Crick in 1953
b) Avery, MacCarty and Macleod in 1944
c) Meselson and Stahl in 1958
d) Hershey and Chase in 1952
Answer: c [Reason:] Within five years of the discovery of the double helix, evidence emerged for the separation of the complementary strands during DNA replication as suggested by Watson and Crick. This firm enzymological proof was first provided by Meselson and Stahl’s experiment of Semiconservative way of replication of DNA in 1958.
6. Replication of DNA ends are carried out by polymerase.
Answer: b [Reason:] Replication of DNA ends, that is, the telomeres cannot be done using polymerases. Cell has a unique enzyme, telomerase which uses its own unique mechanism for telomere replication.
7. Which of the following regarding the basic mechanism of gene expression is correct?
a) DNA ―> tRNA―> protein
b) RNA ―> cDNA ―> mRNA―> protein
c) RNA ―> DNA―> mRNA ―> protein
d) DNA ―> protein
Answer: b [Reason:] Gene expression is carried forward from DNA to mRNA (transcription) and mRNA to protein (translation). To this RNA can be converted to cDNA by reverse transcription. Then the process is carry forwarded in the same manner.
8. Which of the following does not take part in gene expression?
c) RNA processing
Answer: a [Reason:] Replication is the coping of the gene in double but plays no role in the production of protein thus, expression of gene. Transcription is the production of RNA from DNA and RNA processing is used for stabilizing the RNA in cytosol ad excision of non-coding regions. Translation is the major process which leads to the formation of polypeptide chain.
9. Multiple copies of RNA could be formed at the same time.
Answer: a [Reason:] The RNA produced does not remain base paired to the template DNA strand and is displaced only by a few nucleotides behind the transcription site. Thus another RNA polymerase can attach itself to the DNA template facilitating multiple RNA production.
10. Amino acids prior to their incorporation into polypeptide must be attached to a special adaptor molecule. Who proved this and when?
a) Francis H. Crick in 1955
b) Paul C. Zamecnic and Maholon B. Hoagland in 1957
c) James Watson and Francis H. Crick in 1953
d) Linus Pauling in 1950
Answer: b [Reason:] In 1955 Francis H. Crick proposed that prior to the incorporation of amino acids in polypeptide chain they must be attached to an adapter molecule which can interact with the three-nucleotide long coding units of mRNA. This was finally proved in 1957 by Paul C. Zamecnic and Maholon B. Hoagland. They demonstrated that amino acids were carried by the transfer RNAs (tRNA) to the ribosome for its transfer to the growing polypeptide chain.
Molecular Biology MCQ Set 2
1. Nucleosome was first described in 1974 by ___________
a) William Asbury
b) Rosalind Franklin
c) Roger Kornberg
d) John Crick
Answer: c [Reason:] The basic structural unit of chromatin, the nucleosome, was described by Roger Kornberg in 1974.Two types of experiments led to Kornberg’s proposal of the nucleosome model. First, partial digestion of chromatin with micrococcal nuclease was found to yield DNA fragments approximately 200 bp long. In contrast, a similar digestion of naked DNA yielded a continuous smear of randomly sized fragments.
2. The extent of chromosome coiling in non – dividing cells is _________
Answer: b [Reason:] In non – dividing or interphase cells most of the chromatin is decondensed and distributed throughout the nucleus. This form of chromatin is known as Euchromatin. During this period of the cell cycle, genes are transcribed and the DNA is replicated in preparation for cell division.
3. The whole length of DNA is transcriptionally active ___________
Answer: b [Reason:] About 10% of interphase chromatin is in a very highly condensed state that resembles the chromatin of cells undergo mitosis. This type of condensed chromatin is known as the heterochromatin. Heterochromatin is transcriptionally inactive and contains highly repeated DNA sequences, such as those present at centromeres and telomeres.
4. Why are chromosomes condensed?
a) To facilitate accommodation
b) Always condensed
c) To facilitate cell division
d) To facilitate distribution in daughter cells
Answer: d [Reason:] As the cells enter mitosis, there chromosomes become highly condensed so that they can be distributed to daughter cells. The chromatin in interphase nuclei are organized in loops, which are thought to fold upon themselves to form the compact metaphase chromosomes of mitotic cells in which the DNA has been condensed nearly ten – thousand folds.
5. At which phase do transcription ceases?
c) S phase
d) G phase
Answer: b [Reason:] The chromatin in interphase nuclei are organized in loops that fold upon them to form the compact metaphase chromosomes of mitotic cells which are condensed nearly upto ten – thousand folds. Such condensed chromatin can no longer be used as a template for RNA synthesis and thus transcription ceases during mitosis. As prophase marks the beginning of mitosis thus it is the right option.
6. The part that plays a critical role in even distribution of parental DNA during division is ___________
c) Spindle fibre
Answer: b [Reason:] The centromere plays a critical role in even distribution of parental DNA during cell division. It is a specialized region on the chromosome known as the constricted chromosomal region that holds the sister chromatids together and attaches the chromosomes to the spindle fibres during metaphase of division.
7. With respect to centromere which of the following is wrong?
a) Constricted chromosomal region
b) Holds the sister chromatids together
c) Attaches to spindle fibres
d) Facilitates even distribution
Answer: d [Reason:] As the cell enters mitosis, chromatin condensation leads to the formation of metaphase chromosome consisting of two identical sister chromatids. These sister chromatids are held together at the centromere, which is known as the constricted chromosomal region. As mitosis precedes the microtubules of mitotic spindles attaches to the centromere, and the two sister chromatids separate and move to the opposite poles of the spindle.
8. The protein that binds to the spindle fibres are known as centromeric proteins _________
Answer: b [Reason:] The proteins associated with centromeres form a specialized structure called the kinetochore. The spindle fibres thus formed attaches to these proteins which finally lead to the segregation of the chromosomes to the daughter cells.
9. Centromeric DNA was initially defined in ___________
Answer: c [Reason:] Centromeric DNA sequences were initially identified in yeasts, where their functions were assayed by following the segregation of plasmids at mitosis. Plasmids that contain functional centromeres segregate like chromosomes and are equally distributed to daughter cells following mitosis.
10. How many satellite sequence elements are present in yeast centromere?
Answer: b [Reason:] The centromere sequences of the well studied yeast are contained in approximately 125 base pairs. It has been found to consist of three sequence elements: two short sequences of 8 to 25 bp separated by 78 to 86 bp of very A/T rich DNA.
11. The centromere is A/T rich region ____
Answer: a [Reason:] Yes the centromere is A/T rich region. For example, the centromere of Drosophila is about 420 kb consisting of two highly repetitive satellite DNAs, AATAT and AAGAG, in addition to a non-repetitive region of A/T rich DNA. Again in Arabidopsis, centromeres consist of 3 million bp of an A/T rich 178 bp satellite DNA. In humans also, this feature is observable which is a 171 bp A/T sequence arranged in tandem repeats spanning around 1–5 million base pairs.
12. Which alternate form of histone is seen in centromeric histones of humans?
b) SMC protein
Answer: d [Reason:] It has been observed that the human chromatin near the centromere has a unique structure. In particular, histone H3 is replaced by its variant CENP-A. CENP-A is uniformly present in centromere of all organisms that have been studied and thus are thought to be of prime requisite for the assembly of the kinetochore proteins needed for centromere function.
13. Telomere is not related to __________
b) Chromosome degradation
Answer: c [Reason:] The sequences at the ends of eukaryotic chromosomes are called the telomeres. They play a critical role in maintenance and replication of the DNA and also play a part in chromosomal degradation. Degradation of the telomeres leads to ageing.
14. Which of the following nucleotides is rich I telomere of an organism?
a) A, T
b) T, G
c) G, C
d) C, A
Answer: c [Reason:] The telomere DNA sequences of a variety of eukaryotes are similar, consisting of repeats of a simple sequence DNA containing a cluster of G residues on one strand. For example the repeat sequence of telomere in humans and other mammals is TTAGGG, and in Tetrahymena is TTGGGG. This sequence is repeated over hundreds and thousands of times and is terminated with a 3’ overhanging DNA.
15. How does telomerase activity depend on age?
c) Remains the same
d) Does not occur
Answer: d [Reason:] Telomerase activity does not occur in normal somatic cell as it is present in inactive form. Thus the ends of the DNA does not get replicated every time a cell divides which leads to aging.
16. In cancer telomerase activity__________
c) Remains constant
d) Plays no role
Answer: a [Reason:] Cancer cells have high levels of telomerase activity allowing them to maintain the ends of their chromosomes through indefinite divisions. Since normal somatic cells lacks active telomerase activity they do not divide indefinitely.
Molecular Biology MCQ Set 3
1. The origin of replication is rich in __________
a) A, T
b) G, C
c) A, G
d) C, T
Answer: a [Reason:] The origin of replication is an A, T rich segment of DNA which unwinds readily but not spontaneously. Unwinding of DNA at this region is controlled by the replication initiation proteins.
2. How many origin of replication are present in the E. coli genome ___________
Answer: a [Reason:] The E. coli genome has only one origin of replication, thus only one replicon. The eukaryotic genome has multiple origin of replication sites, thus have a multiple replicon system. The origin of replication in E. coli genome is known as the ori C.
3. There are two repeated motifs critical for ori C function. They are ___________
a) 9 mer and 18 mer
b) 6 mer and 9 mer
c) 9 mer and 13 mer
d) 13 mer and 18 mer
Answer: c [Reason:] The two repeated motifs critical for ori C functions are 9 mer and 13 mer. 9 mer is present in 5 copies and 13 mer is present in 3 copies in the origin of replication site.
4. With respect to the ori C motif choose the correctly paired option.
a) 9 mer = 9 times
b) 9 mer = 3 times
c) 13 mer = 5 times
d) 13 mer = 3 times
Answer: d [Reason:] 9 mer is present in 5 copies and 13 mer is present in 3 copies in the origin of replication site. 9 mer is the site of binding of Dna A and 13 mer is the unwinding site of DNA to facilitate replication.
5. Which of the following is not paired with its correct function?
a) 9 mer = binding site for Dna A
b) 13 mer = binding site for replication initiator protein
c) AT rich DNA = unwinding site
d) Dna A = replication initiator protein
Answer: b [Reason:] The replication initiator protein is known as Dna A which binds in the 9 mer region. 13 mer regions is the initial site for the formation of single stranded DNA formation during initiation.
6. Which of the following is not a function of Dna A?
a) DNA binding
b) DNA strand separation
c) Initiates DNA melting
d) Replication protein recruitment
Answer: c [Reason:] Dna A does not initiate DNA melting, that is, DNA strand separation. DNA strand separation takes place at the AT rich region which unwinds readily. To this readily unwound DNA region at the 13 mer site, the Dna A protein recruits different proteins required for replication initiation.
7. The replication terminus is a segment of a number of ____________ bp.
Answer: b [Reason:] The replication terminus is a segment of 350 kb region. 350 kb is equivalent to 350000 base pairs and has 6 nearly identical non palindromic sites.
8. The replication terminus involves ____________ identical non palindromic termination site.
Answer: c [Reason:] The replication terminus involves 6 identical non palindromic termination sites. They are Ter E, Ter D and Ter A on one side and Ter F, Ter B and Ter C on the other.
9. The termination sites are one way valves known as non-polar valves ____________
Answer: b [Reason:] A replication fork travelling counterclockwise passed through Ter F, Ter B and Ter C but stops upon encountering Ter E, Ter D or Ter A. Similarly a clockwise travelling replication fork transits Ter E, Ter D and Ter A but halts at Ter F, Ter B or Ter C. Thus, these termination sites are polar as they acts as one way valves that allow replication fork to enter the terminus region but not to leave it.
10. The arrest of replication fork motion at Ter sites require the action of ____________ protein.
Answer: a [Reason:] The arrest of replication fork is motioned at Ter sites require the action of Tus protein. It is a 309 – residue monomer that is the product of the tus gene.
11. The topological unlinking of DNA in prokaryotes is promoted by ______________
d) Dna C
Answer: b [Reason:] The final step in prokaryotic DNA replication is the topological unlinking of the parental DNA strands. This process is catalyzed by topoisomerase.
12. Helicase attaches to the leading strand during replication.
Answer: b [Reason:] Helicase is used to unwind the DNA during replication. It attaches to the lagging strand to unwind the DNA helix.
13. The primer used for lagging strand synthesis in prokaryotes is a RNA primer.
Answer: b [Reason:] The primer used for leading strand synthesis in prokaryotes is a RNA primer. The primer used for lagging strand synthesis in prokaryotes is a DNA primer.
14. Which enzyme is used to remove the primer from the Okazaki fragment?
b) RNase H
c) 5’ exonuclease
Answer: d [Reason:] Primer used for prokaryotic replication of lagging strand is a DNA primer thus RNase H and 5’ exonuclease is not used. Endonuclease is used for producing restrictions within the strand. Thus to remove DNA primer polymerase is used in the prokaryotic organisms.
Molecular Biology MCQ Set 4
1. Non – coding sequence in mRNA is known as __________
b) Non – template
Answer: c [Reason:] Long stretches of non – coding DNA are found within most eukaryotic genes. These sequences are known as introns which are removed from the pre-mature RNA to produce mature RNAs.
2. Which of the following organisms have monocistronic DNA?
Answer: b [Reason:] The bacterial genome is monocistronic. This means that the genes are not disrupted by the presence of introns.
3. Which of the following organisms have overlapping genes?
Answer: a [Reason:] The viruses have overlapping genes to accommodate the genome in such a small cell. The bacterial cell is monocistronic whereas all eukaryotic cells are polycistronic.
4. Introns were first discovered in __________
Answer: d [Reason:] Introns were first discovered in 1977 during the studies of replication of adenovirus in cultured human cells. It was discovered independently in the laboratories of Phillip Sharp and Richard Roberts.
5. Adenovirus is a useful model of studying gene expression because __________
a) High infection rate
b) Assured transformation
c) High protein production
d) Short duplication time
Answer: c [Reason:] Adenovirus is a useful model of studying gene expression because of two reasons:
i) The size of the viral genome is small that is only 3.5*104 base pairs long.
ii) The adenovirus mRNAs are produced at high levels in infected cells thus yielding high amount of protein.
6. Mouse β – globin gene has __________ introns.
Answer: b [Reason:] Mouse β – globin gene has 2 introns. This gene encodes the β – subunit of haemoglobin. This is found by the observations produced by the electron microscopic analysis of RNA – DNA hybrids and subsequent nucleotide sequencing of cloned genomic DNA and cDNAs.
7. The amount of exon in the eukaryotic genome is higher than that of the introns.
Answer: b [Reason:] The intron – exon structure of many eukaryotic genes is quite complicated, and the amount of DNA in the intron sequences is often greater than that in the exons. For example, an average human gene contains about 10 exons, interrupted by introns distributed over approximately 56 kb of genomic DNA. The exons generally total only about 4.1 kb including both the 5’ and 3’ ends of the mRNA that are not translated into proteins.
8. The mRNA of which eukaryotic protein lacks introns?
Answer: c [Reason:] Introns are present in almost all genes of complex eukaryotes, although they are not universal. Almost all histone genes for example, lacks introns, so introns are not clearly not required for gene function in eukaryotes.
9. Introns are nothing more than genetic load in the eukaryotic genome.
Answer: b [Reason:] Although most introns do not specify the synthesis of protein product, they have other cellular activities. Many introns encode functional RNAs, including the small nucleolar RNAs that function in ribosomal RNA processing and the microRNAs, which are major regulators of gene expression in eukaryotic cells. Introns also contain regulatory sequences that control transcription and mRNA processing.
10. Presence of introns facilitates the formation of several different mRNAs, thus increasing protein yield of different types but from same gene.
Answer: a [Reason:] Presence of introns facilitates the formation of several different mRNAs by the process of alternative splicing. In this the exons of the gene join in different combinations resulting in the synthesis of different proteins from the same gene.
Molecular Biology MCQ Set 5
1. Who of the following was not one of the scientists explaining the replicon model?
a) Francois Jacob
b) Sydney Brenner
c) Jacques Cuzin
d) Arthur Kornberg
Answer: d [Reason:] Arthur Kornberg was the first to isolate DNA polymerase from E. coli. Francois Jacob, Sydney Brenner and Jacques Cuzin proposed a model explaining the events controlling the initiation of replication in bacteria known as the replicon model in 1963.
2. The replicon model was first explained in the year ____________
Answer: c [Reason:] In 1963 Francois Jacob, Sydney Brenner and Jacques Cuzin proposed a model explaining the events controlling the initiation of replication in bacteria. They defined that the entire DNA of bacteria is replicated from a particular origin and thus termed the model a replicon.
3. The eukaryotic chromosome has a single replicon.
Answer: b [Reason:] A replicon is a stretch of DNA that is replicated from a single origin. The E. coli is so small that it has only one origin of replication thus have only one replicon. The eukaryotic chromosome is large and the replication of the whole genome is complicated thus it has multiple origin of replications giving rise to multiple replicons.
4. The replicon model comprises of an initiator and _______________
b) Origin of replication
c) Dna A protein
d) AT rich DNA
Answer: a [Reason:] The replicon model comprises of an initiator and a replicator. The initiator protein helps in the recognition of the DNA element in the replicator and activates the process of initiation. The replicator is the cis – acting DNA segment that directs the initiation of DNA replication.
5. Replicator is the synonym for origin of replication.
Answer: b [Reason:] The replicator is the cis – acting DNA segment that directs the initiation of DNA replication. But the origin of replication is the site on the DNA where the DNA is unwound and the replication initiates.
6. The initiator protein binds to a segment ____________ of the replicator sequence.
c) At the end
Answer: d [Reason:] The initiator protein binds to a segment within the replicator sequence. This protein finally attaching to its specific sequence activates the initiation of replication by recruiting different replication enzymes and signaling the unwinding of DNA.
7. Binding of the Dna A protein to the DNA is energy dependent _______________
Answer: a [Reason:] The Dna A protein binds to the repeated 9 mer region of the ori C and is regulated by ATP. When bound to ATP, Dna A interacts with DNA in the region of the repeated 13 mer repeats of ori C.
8. Assembly of how many proteins make up the origin replication complex (ORC) in eukaryotes?
Answer: c [Reason:] In eukaryotes the initiator is a 6 protein complex called the origin replication complex (ORC). The function of ORC is best understood in yeast cells.
9. Which of the following proteins is not recruited by Dna A protein?
a) Dna A protein
c) Dna C protein
d) SSB protein
Answer: d [Reason:] Dna A recruits a complex of two proteins, the helicase, Dna B protein and the helicase loader (Dna C). Both proteins are present in 6 copies within the complex. The DNA helicase is maintained in an inactive state in the helicase – helicase loader complex.
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