Molecular Biology MCQ Set 1
1. Which of the following is not a feature of the genetic code?
c) Non – overlapping
Answer: d [Reason:] The genetic code is non – ambiguous. This means that there is no ambiguity about a particular codon. A particular code will always code for the same amino acid wherever it is found.
2. The codon is a ____________
Answer: c [Reason:] The codon is a triplet. Singlet and doublet codes are not enough to code for 20 amino acids. Again in case of a quadruplet codon there will be 256 possible codons which are highly excessive. Triplet codon thus is the minimum requisite having 64 possible codons.
3. There is one amino acid for one genetic code.
Answer: b [Reason:] In a triplet code for a particular amino acid more than one word can be used. This phenomenon is described by saying that the code is degenerate. A non – degenerate could be one where there is one to one relationship between amino acids and the codons so that 44 codons out of 64 will be useless or nonsense codons.
4. One base in an mRNA can be used for more than one codon.
Answer: b [Reason:] One base in an mRNA cannot be used for more than one codon. This is because the codes are non – overlapping.
5. Which of the following is not a termination codon?
Answer: d [Reason:] UGA, AGA and AGG are termination codons of which UGA is the universal termination codon and AGA and AGG are mitochondrial termination codons. But UAC is the universal codon for tyrosine.
6. In case of mitochondrial genetic code UGA is a ____________ codon.
Answer: a [Reason:] In case of mitochondrial genetic code UGA is a tryptophan codon. But UGA is a stop codon in the universal genetic code.
7. The wobble hypothesis was devised by ____________
a) Arthur Kornberg
b) Francis Crick
c) James Watson
d) William Asbury
Answer: b [Reason:] In 1966, Francis Crick devised the wobble hypothesis to explain the observations regarding base pairing. It states that the base at the 5’ end of the anticodon is not as spatially confined as the other two, allowing it to form hydrogen bonds with any of the several bases located at the 3’ end of a codon.
8. The distribution of codon is made in such a way to minimize mutation effect.
Answer: a [Reason:] Inspection of the distribution of codons in the genetic code suggests that the code evolved in such a way as to minimize the deleterious effects of mutations. For instance, mutations in the first position of a codon will often give a similar, if not same, amino acid.
9. Point mutation of the second place mutating “A” with “T” will result in a similar if not same amino acid.
Answer: b [Reason:] Codons with pyrimidines in the second position specify mostly hydrophobic amino acids, whereas those with purines in the second position correspond mostly to the polar amino acids. Thus when purine “A” is replaced by pyrimidine “T” it results in an amino acid swap between hydrophobic to polar amino acids.
10. Which of the following genetic code shows ambiguity?
Answer: d [Reason:] UGA is one of the universal termination codon. But UGA codes for tryptophan in mitochondrial genetic code.
Molecular Biology MCQ Set 2
1. In the beads on a string model, the bead is made up of __________
a) 6 histone proteins
b) 8 histone proteins
c) 6 histone proteins and DNA
d) 8 histone proteins and DNA
Answer: b [Reason:] The “beads on a string” model is for the nucleosome. It consists of the 8 histone protein core or the bead and the DNA wound around imitating a string.
2. The unpacked stretches of DNA is the extra chromosomal load found in eukaryotic genome ___________
Answer: b [Reason:] Linker DNA is the stretches of DNA that are not packed into a nucleosome. Typically these are the regions engaged in gene expression, replication and recombination and are generally associated with non – histone proteins.
3. How many types of histone molecules are found in nature?
Answer: c [Reason:] Eukaryotic cells commonly contain five abundant histone molecules. They are named as H1, H2A, H2B, H3 and H4.
4. Nucleosome is made up of __________
a) DNA, histone core protein
b) DNA, histone core protein, linker H1
c) RNA, histone core protein
d) RNA, histone core protein, linker H1
Answer: b [Reason:] The core histone proteins are H2A, H2B, H3 and H4, over which the DNA is wrapped. Histone H1 is not a part of nucleosome core particle, instead it binds to the linker DNA and thus is referred to as linker histone. Thus, the histone core, linker histone and DNA are the components of the nucleosome.
5. Histones have a high content of negatively charged amino acids.
Answer: b [Reason:] As histones maintain a constant association with negatively charged DNA thus histone molecules are made up of high content of positively charged amino acid. Greater than 20% of the residues in each histone molecules are either lysine or arginine.
6. With respect to assembly of every core histone which of the following is wrong?
a) A conserved region
b) Histone fold domain
c) Disc shaped structure
d) 2 α helices and an unstructured loop
Answer: d [Reason:] The histone fold is composed of 3 α helices and two unstructured loops. In each of these cases the histone fold mediates the formation of head to tail heterodimers of specific pairs of histone.
7. Which of the following histone pairs forms tetramers in solution?
a) H1, H2A
b) H2A, H2B
c) H2B, H3
d) H3, H4
Answer: d [Reason:] H3 and H4 histone first forms heterodimers then they come together to form a tetramer with two molecules of each. In contrast, H2A and H2B form heterodimers only in the solution and histone H1 only acts as the linker histone.
8. With respect to the “tails” of the histone molecules which of the following is not true?
a) N – terminal extension
b) Lacks defined structure
c) Required for the association of nucleosome
d) Sites for extensive modifications
Answer: c [Reason:] The “tail” of the histone is not required for the association for the DNA with the histone octamer into a nucleosome. This is proved when the nucleosome is treated with the protease, trypsin. Trypsin is known to cleave proteins after positively charged amino acid thus , when the N – terminal tail is removed no structural variation is observable in the nucleosome.
9. How many contacts are observed between the DNA and the histone core protein?
Answer: a [Reason:] 14 distinct sites of contact are observed, one for each time the minor grove of the DNA faces the histone octamer. 147 base pairs of DNA is wound around the histone octamer and each minor grove occurs after 10 base pairs thus, 14 contacts are observable.
10. Association of DNA and histone is mediated by_________
a) Covalent bonding
b) Hydrogen bonding
c) Hydrophobic bonding
d) Vander Waals interactions
Answer: b [Reason:] Association of DNA and histone is mediated by a large number of hydrogen bonds, that is, ≈140 bonds. The majority forms between the protein and the oxygen of the phosphodiester backbone near the minor grove. Only 7 hydrogen bonds are made between the protein side chains and the bases in the minor groves of the DNA.
Molecular Biology MCQ Set 3
1. Which of the following does not affect the formation of hybrid DNA?
a) Ionic strength
d) Homologous DNA
Answer: b [Reason:] Hybrid molecules are formed when homologous, denatured DNAs from two different sources are mixed with each other. This occurs due to the presence of complementary bases and under appropriate ionic strength and temperature.
2. Which of the following is not termed as hybridization?
a) DNA and cDNA
b) DNA and mRNA
c) DNA from different species
d) DNA from male and female of same species
Answer: d [Reason:] The process of base – pairing between complementary DNA strands from different sources of species is termed as hybridization. Thus DNA from male and female of same species is not known as hybridization.
3. What is a probe?
a) Chemically synthesized DNA
b) Purified DNA
c) Fragmented DNA duplex
d) Either purified or synthesized single stranded DNA
Answer: d [Reason:] A probe is a known DNA oligonucleotide used to search for specific DNA sequences containing its complementary sequence. These probes can be both, purified DNA or chemically synthesized single stranded DNA.
4. Why is a probe labeled?
a) Improve visibility
b) Improve stability
c) Improve location identification
d) Improve binding capability
Answer: c [Reason:] The probe is labeled because to improve its location identification. This means, the probe DNA must be labeled so that it can be readily located once it has found its target sequence.
5. In how many ways a DNA can be labeled?
Answer: a [Reason:] There are two basic methods for DNA labeling. The first involves the synthesis of the probe DNA in the presence of a labeled precursor. The other involves adding a label to the end of an intact DNA molecule.
6. What is the function of polynucleotide kinase?
a) Addition of γ – phosphate at 3’ – OH.
b) Addition of γ – phosphate at 5’ – OH.
c) Removal of γ – phosphate at 3’ – OH.
d) Removal of γ – phosphate at 5’ – OH.
Answer: b [Reason:] The enzyme polynucleotide kinase is known to add γ – phosphate at 5’ – OH end of the DNA strand. This reaction occurs at the cost of an ATP molecule which id hydrolyzed to extract energy and a phosphate.
7. Fluorescent labeling is an important phenomenon in case of hybridization and manipulations in molecular biology____________
Answer: b [Reason:] DNA labeled with fluorescent precursors can be easily detected by irradiating the DNA sample with appropriate wavelength of UV light. The results can be monitored by visualizing the longer wavelength light emitted by the labeled DNA. This is thus very important in localization of probes and thus identifying the location of the target DNAs.
8. Radiolabelling is generally brought about by addition of radioactive phosphorus ___________
Answer: a [Reason:] Radioactively labeled precursors typically have radioactive 32P or 35S incorporated into α – phosphate of one of the four nucleotides. Radioactive DNA can be detected by exposing the sample of interest to X – ray film or by photomultipliers that emit light in response to excitation by the β – particles emitted by from 32P and 35S.
9. During electrophoresis denaturation of the double stranded DNA is brought about by __________
a) Treatment with alkali
b) Application of current
c) Treatment with EtBr
d) Application of heat
Answer: a [Reason:] During electrophoresis denaturation of the double stranded DNA is brought about by its treatment with alkali. This step is done once the sample is separated by electrophoresis and taken up in the nitrocellulose membrane.
10. In case of northern blotting the hybridization occurs between two RNA molecules _________
Answer: b [Reason:] The electrophoretically separated RNAs are transferred into a positively charged membrane. Then probe of radioactive DNA of choice is used thus in case of northern blotting the hybridization occurs between a DNA and a RNA molecule.
11. Northern blotting is performed for ____________
a) Determining the size of DNA
b) Determining the size of RNA
c) Quantification of RNA
d) Sequencing of RNA
Answer: c [Reason:] Northern blotting is performed by an experimenter to ascertain the amount of a particular mRNA present in the sample rather than its size. This measure is a reflection of the level of expression of the gene that encodes it.
12. The process by which a probe is used to screen a library is known as ___________
b) Southern blotting
c) Colony hybridization
d) Western blotting
Answer: c [Reason:] The process by which a labeled DNA probe is used to screen a library is known as colony hybridization. This process uses the technique of replica plating.
Molecular Biology MCQ Set 4
1. The splice site is found in __________
a) 3’ end of exon
b) 5’ end of intron
c) Within the exon
d) Within the intron
Answer: b [Reason:] The splice site is found in the borders between intron and exon. They are marked by specific nucleotide sequences and are found on the borders of the intron that is the 3’ and 5’ ends. They are known as 3’ splice site and 5’ splice site respectively.
2. During splicing which of the following sites is not recognized by the splicosome?
b) Branch point
d) G-rich site
Answer: d [Reason:] The splicosome splice site recognizes three specific sites. They are 5’ splice site or the donor, the branch point site and the 3’ splice site known as the acceptor.
3. The 3’ end of an intron is marked by __________
a) Donor site
b) Acceptor site
c) AT rich site
d) Branch point site
Answer: b [Reason:] The 3’ end of an intron is marked by the acceptor site also known as the 3’ splice site. It is a track of pyrimidines that is recognized by the splicosome complex.
4. The highly conserved base at the branch point site is __________
Answer: a [Reason:] The highly conserved base at the branch point site is A whereas the 3’ splice site is AG rich and the 5’ splice site is GU rich. These highly conserved sequences are found within the intron itself as the bases in the exons are constrained by the need of encoding specific amino acid and protein products.
5. How many consensus sequences for splicing are found in an exon?
Answer: d [Reason:] None of the consensus sequences for splicing are found in an exon. The highly conserved base at the branch point site is A whereas the 3’ splice site is AG rich and the 5’ splice site is GU rich. These highly conserved sequences are found within the intron itself.
6. To achieve splicing of an intron how many phosphodiester bonds are formed __________
Answer: b [Reason:] The splicing is achieved by the two successive transesterification reactions. For the formation of these two new phosphodiester bonds, two existing phosphodiester bonds are broken thus no net change in the number of bond is observed.
7. The first transesterification reaction between the branch point site and the 5’ splice site is a __________ reaction.
Answer: b [Reason:] The first transesterification reaction is triggered by the 2’-OH of the conserved A at the branch site. This group acts as a nucleophile to attack the phosphoryl group of the conserved G in the 5’ splice site. This is an SN2 reaction that proceeds through pentavalent phosphorus intermediate.
8. The second transesterification reaction occurs between __________
a) The 3’ splicing site of intron and 5’ splice site of intron
b) The 3’ splicing site of intron and the branch point site
c) The 5’ splicing site of intron and 3’ end of exon
d) The 5’ end of exon and the 3’ splice site
Answer: d [Reason:] The second transesterification reaction occurs between the 5’ end of exon and the 3’ splice site. In this case the newly liberated 3’-OH of the 5’ exon acts as a nucleophile. It attacks the phosphoryl group at the 3’ splice site thus joining the two exons by a phosphodiester bond.
9. The transesterification reaction is ATP independent.
Answer: a [Reason:] In the two reaction steps, there is no net gain in the number of chemical bonds – two phosphodiester bonds are broken, and two new ones made. As it is just a question of shuffling of bonds, no energy input is demanded by the chemistry of this process.
10. In trans-splicing the excised intron forms a lariat structure.
Answer: b [Reason:] The lariat structure of intron is formed in case of general splicing mechanism. In case of trans-splicing the excised intron forms a Y – shaped branch structure.
Molecular Biology MCQ Set 5
1. The type of DNA amplification where the region of DNA amplified lies on either side of a known segment __________
a) RT – PCR
b) Anchored – PCR
c) Inverse – PCR
d) Nested – PCR
Answer: c [Reason:] PCR can be used to amplify the sequences flanking on either sides of a known DNA segment. The process uses primers complementary to the 5’ ends of the interested segment. This is a reverse process of the general PCR mechanism thus is known as the inverse PCR.
2. Primer complementary to regions used in inverse PCR is _____________
a) 3’ end of unknown region
b) 5’ end of unknown region
c) 3’ end of known region
d) 5’ end of known region
Answer: c [Reason:] Inverse PCR uses primers complementary to the 3’ ends of the known segment of DNA. this thus, amplifies the unknown regions on the either sides of the known region and thus is known as inverse PCR.
3. With respect to target DNA used in inverted PCR which of the following is not true?
a) Restricted segment
b) Blunt ended segment
c) Intact known segment
d) Flanked unknown segment flanked on either side
Answer: b [Reason:] The target DNA is cut with a restriction enzyme that produces sticky ends and that does not cut within the region of which the sequence is known. The cutter instead cuts at unknown sites on the either sides of the known sequence. This lets the DNA circularize later, which is a very important step for achieving our desired results.
4. The DNA concentration in inverse PCR is kept low _______
Answer: a [Reason:] The cut made by the restriction enzymes lie beyond the unknown region of our interest. As the restricted fragment thus produced has sticky ends it is left to circularize. The DNA taken for this process is in small quantity or else all the restricted fragments would bind among themselves and we will not get the desired circularized DNA.
5. The linear fragment used for inverted PCR has known ends _____________
Answer: a [Reason:] The molecular cutters used are rare and the cut producing the fragment involves the two regions of interest flanked to either sides of a known region. This fragment is left to circularize. This circular DNA is then restricted within the known DNA segment to produce a linear fragment having known ends on either sides.
6. With respect to RAPD which of the following is false?
a) 10 bases long
b) G/C rich
c) Has inverted repeats
d) Radom sequences are used
Answer: c [Reason:] RAPDs are generated by using random sequences of ordinarily 10 bases long oligonucleotides. These oligonucleotides are generally G + C rich and do not contain any repeated sequences.
7. Polymorphism in RAPD is observed because ______________
a) DNA used is from different chromosomes of same species
b) DNA used is from same chromosomes of same species
c) DNA used is from different chromosomes of different species
d) DNA used is from complementary chromosomes of same species
Answer: c [Reason:] In RAPD, primers for PCR amplification of genomic DNAs are from different species/strains. Thus polymorphism is produced due to the complementary sequence for the primer used being present in one strain and absent in another.
8. RAPDs cannot be used for PCR amplification.
Answer: b [Reason:] As the RAPDs behave as dominant markers, they are easily recognized by the primers for amplification. Thus, RAPDs turn out to be great templates for the primers for PCR amplification.
9. RAPDs are much more convenient markers than RFLPs .
Answer: a [Reason:] RAPDs have similar applications as RFLPs, but they are considerably faster and more convenient, particularly with such species where little previous work has been done. In some cases, RAPDs may generate such an amount of information in 4 weeks, which would have taken about 2 years to obtain using RFLPs.
10. The inheritance pattern of RAPD is _______________
Answer: a [Reason:] The inheritance pattern of RAPD is dominant type. The inheritance pattern of RFLP is Codominant type.
11. RAPDs can be used to detect multiple alleles of a marker.
Answer: b [Reason:] RAPDs can never be used to detect multiple alleles of a marker. Only RFLPs are able to detect multiple alleles of a marker within the samples of DNA.
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