Molecular Biology MCQ Set 1
1. How many kinds of mutation are found in DNA which includes mutation of only one base?
Answer: b [Reason:] There are two kinds of mutation which is observed in the In the DNA that include only one base and are also known as point mutation. These mutations are transition where mutation occurs chancing a purine to purine and pyrimidine to pyrimidine, and transversion where there purine is converted to purine and vice versa.
2. What is the overall rate at which new mutations arise spontaneously at any given site on the chromosome per round of replication?
a) ≈ 10-8 – 10-12
b) ≈ 10-7 – 10-9
c) ≈ 10-6 – 10-11
d) ≈ 10-5 – 10-10
Answer: c [Reason:] The overall rate at which new mutations arise spontaneously at any given site on the chromosome ranges from ≈ 10-6 – 10-11 per round of DNA replication. With some sites on the chromosomes being hot spots, mutations arise at a high frequency in these sites.
3. The mutation occurs at a random basis within a genome.
Answer: b [Reason:] The mutation occurs at a particular mutation prone region known as the hot spots. The hot spots are rich in di- or tri- nucleotide repeat sequences known as microsatellites.
4. What is the dinucleotide sequence of microsatellites?
Answer: a [Reason:] Microsatellites involves the repeats of the dinucleotide sequence of CA. the CA repeat is found at many widely scattered sites in the genome of humans and other eukaryotes.
5. By which process miss-incorporated base can change into a permanent mutation?
Answer: a [Reason:] A potential mutation may be introduced by misincorporation in any round of replication. In the next round of replication if the mutation is not repaired it gets permanently incorporated in the DNA sequence.
6. Detection of mismatches and fidelity of replication is maintained by mutation repair system.
Answer: a [Reason:] Proofreading by the polymerase is not always perfect and some mismatches may escape the detection which can become a permanent mutation if not corrected. This fidelity check is done by the mutation repair system; more precisely mismatch repair system, of the cell itself ensuring that perfect matches occur in the complementary strands.
7. How many steps are required to attain mismatch repair?
Answer: c [Reason:] The mismatch repair system involves two steps. The first step involves the scanning of the genome for mismatches. The second step ensures the correction of mismatch that has occurred in the genome.
8. In E. coli mismatches are detected by which repair protein?
a) Mut H
b) Mut L
c) Mut S
d) Mut D
Answer: c [Reason:] Mut S scans the DNA and recognizes mismatches from the distortion formed by the unpaired bases. Mut S then embraces the mismatched region introducing a pronounced kink in the DNA and a conformational change in the enzyme itself.
9. Mut S recruits how many component(s) to the mismatched site?
Answer: b [Reason:] Mut S on binding with the mismatched region recruits two more enzymes to form a complex. They are Mut L and Mut H. Mut L activates Mut H, the enzyme that causes an incision or nick on the newly synthesized strand near the site of mismatch.
10. The nicking of DNA is followed by the adherence of a helicase known as __________
a) Uvr D
b) Uvr A
c) Uvr B
d) Uvr C
Answer: a [Reason:] Nicking of the DNA is followed by the addition of a specific helicase for the repair system known as Uvr D. This helicase unwinds the DNA from the site of mismatch to the mismatched site so that the endonuclease could excise the mismatch.
11. Mismatch repair system is ATP dependent.
Answer: a [Reason:] Mismatch repair system is an ATP dependent process. The activation of Mut S enzyme is ATP dependent and even the recruitment of Mut L and Mut H requires the assistance of ATP hydrolysis to bind the mismatched sequence.
12. By which enzyme does E. coli tags its parental DNA strand?
Answer: a [Reason:] The E. coli enzyme Dam methylase tags the parental DNA strand by methylation. It methylates adenine residues on both the strands of DNA in a specific sequence of 5’….GATC….3’ which is frequently distributed along the whole genome.
13. If the Mut H cuts the DNA at the 5’ side of the mismatch then which nuclease is activated?
a) Exonuclease VII
b) Exonuclease VIII
c) Exonuclease I
d) Exonuclease IX
Answer: a [Reason:] When the Mut H cut the DNA at 5’ side then exonuclease VII or Rec J is activated. Rce J degrades the DNA strand in 5’ – 3’ direction.
14. What are the eukaryotic components of Mut S and Mut L of E. coli?
a) MSH, MLH
b) MHS, MHL
c) MLS, PMS
d) PMS, MHL
Answer: a [Reason:] The homologous component of Mut S in eukaryotes is called MSH. The other homologous component of Mut L in eukaryotic cells is called MLH or PMS.
15. High level of mutation in the germ cells is acceptable but in somatic cells can be catastrophic.
Answer: b [Reason:] Mutations in both somatic and germ cells can be catastrophic. High rates of mutation in the germ line would destroy the species and high rate of mutation in the somatic cells would destroy the individual.
Molecular Biology MCQ Set 2
1. Which of the following is a character of ORF?
b) 3 – nucleotide codons
d) Non – overlapping
Answer: c [Reason:] Exons are the part of nascent RNA that are connected together to form RNA after the removal of introns by the splicosome. The splicosome in the complex formed in the nucleus with nascent RNA being the substrate for the formation of mature RNA after its splicing and other post transcriptional mechanism.
2. With respect to the open reading frame (ORF) which of the following is wrong?
a) Starts at 5’ end
b) ‘AUG’ is the only start codon for both prokaryotes and eukaryotes
c) The start codon defines the reading frame for all the subsequent codons
d) The stop codon is distinct from the ends of the mRNA
Answer: b [Reason:] ‘AUG’ is the only start codon for eukaryotes. For prokaryotic system ‘AUG’, ‘GUG’ and sometimes even ‘UUG’ act as the start codon.
3. With respect to polycistronic mRNAs which of the following is wrong?
a) Multiple ORFs
b) Found in eukaryotes
c) Encodes proteins with related functions
d) Multiple polypeptide chain
Answer: b [Reason:] Eukaryotes only contain 1 ORF per mRNA and are thus monocistronic. Polycistronic mRNAs are generally found in prokaryotes with 2 or more ORFs.
4. What was the name of ribosome binding site?
c) A site
d) Shine – Dalgarno sequence
Answer: c [Reason:] Upstream the ORF a 3 – 9 base pair sequence on the 5’ side of the sequence is identified as the ribosome binding site (RBS). This element is referred to as the Shine – Dalgarno sequence, named after the scientists who discovered it by comparing the sequences of multiple mRNAs.
5. Which part of the ribosome identifies the Shine – Dalgarno sequence?
b) 16S rRNA
c) 23S rRNA
d) 5S rRNA
Answer: b [Reason:] The Shine – Dalgarno sequence is identified by the 16S rRNA. The core of the 16S rRNA has the sequence of 5’…..CCUCCU…..3’ and is located near the 3’ end of the rRNA. Not surprisingly the prokaryotic RBS are most often the subset of sequence 5’…..AGGAGG…..3’. Thus, 16S rRNA is the one that aligns the ribosome with the mRNA.
6. Eukaryotic mRNAs recruit ribosomes using the Shine – Dalgarno sequence.
Answer: b [Reason:] Eukaryotic mRNAs recruit ribosomes using specific chemical modifications called 5’ cap. The 5’ end of the mRNA is capped with methylated Guanine nucleotide to the mRNA via an unusual 5’ to 5’ linkage. To this methylated Guanine three phosphates are added. This cap binds to the ribosome which then slides along the mRNA length to find the ‘AUG’ for start of translation.
7. Eukaryotic mRNA is read in the 3’ to 5’ direction.
Answer: b [Reason:] The eukaryotic mRNA is read in the 5’ to 3’ direction. This is because capping of mRNA occurs at the 5’ end, thus the ribosome attaches to the 5’ end and starts translating in the 5’ to 3’ direction.
8. Which one of the following is the following is known as the Kozak sequence?
Answer: b [Reason:] The Kozak sequence is named after its identification by Marilyn Kozak. This sequence contains purines (A/G) three bases upstream of ‘AUG’ and a ‘G’ immediately following it. Thus the sequence stands up to 5’…..A/GNNAUGG…..3’. this sequence is known to increase translation efficiency.
9. With respect to the composition of ribosome which of the following is correct.
a) Ribosome is composed of 60S and 30S subunit
b) Eukaryotic ribosome small subunit contains only one 16S rRNA
c) 60S subunit consists of 5S rRNA and 23S rRNA
d) 60S and 40S makes up the 80S ribosome
Answer: d [Reason:] There is a discrepancy in the sedimentation velocity of the subunits separately and as a whole. This is because of the fact that the sedimentation velocity is determined both by shape and size and hence, it is not an exact measure of mass. Prokaryotic ribosome consists of 50S and 30S subunit and the 30S subunit contains only one 16S rRNA. The 50S subunit consists of 5S and 23S rRNA.
10. Which element of the ribosome plays the key role in mRNA translation?
a) rRNA of the large subunit
b) Proteins of the large subunit
c) rRNA of the small subunit
d) Proteins of the small subunit
Answer: c [Reason:] The anticodon loops of the charged tRNA and the codons of the mRNA contacts the 16S rRNA. This 16S rRNA is present in the small subunit and plays the key role in mRNA translation even though all the rest needed without this 16S rRNA no translation would occur.
Molecular Biology MCQ Set 3
1. X-chromosome inactivation is known as ______________
c) Dosage compensation
Answer: c [Reason:] In placental mammals, dosage compensation of X-linked genes is achieved by the inactivation of one of the female X-chromosomes. This mechanism was first proposed by Mary Lyon in 1961.
2. The proposal for the concept of X-chromosome inactivation in female XX-chromosomes was experimentally shown in ___________
Answer: b [Reason:] The mechanism of X-chromosome inactivation was proposed by Mary Lyon in 1961 who inferred it from the studies on mice. Subsequent research by Lyon and others has shown that the inactivation event occurs when the mouse embryo consists of a few thousand cells.
3. The maternal X-chromosome is inactivated during Dosage compensation.
Answer: b [Reason:] At a few thousand cell stage of the embryo the inactivation of one X chromosome occurs. The X-chromosome is chosen at random, though once chosen it remains inactivated in all the descendents of that cell.
4. Female mammals are called genetic mosaics.
Answer: a [Reason:] The female mammals are called genetic mosaics as they contain two types of cell lineages. The maternally inherited X-chromosome is inactivated in roughly half of these cells and the paternal inherited X-chromosome is inactivated in the other half.
5. The best genetic mosaicism is observed in which of the following?
Answer: d [Reason:] The best genetic mosaicism is observed in the study of fur coloration of cats and mice. A female that is heterozygous for an X-linked gene is therefore able to show two different phenotypes.
6. Barr body is the ___________
a) Active X chromosome
b) Inactive X chromosome
c) Active Y chromosome
d) Inactive Y chromosome
Answer: b [Reason:] Chemical analysis shows that the inactivated X chromosome condenses into a dark staining structure. This structure was called the Barr body after the scientist Murray Barr who discovered it.
7. What changes occur in the chromosome to make it inactive?
Answer: a [Reason:] An X chromosome that has been inactivated does not look or acts like other chromosomes. Chemical analysis shows that the DNA is modified by the addition of numerous methyl groups. This provides the structure a highly condensed darkly stained structure called the Barr body.
8. The X-chromosome once inactivated remains inactive throughout the life in all the cells of the body.
Answer: b [Reason:] The inactivated X chromosome remains in this altered state in all the somatic cells. However in germ cells it is reactivated because two copies of some X linked genes are needed for the successful completion of oogenesis.
9. All genes on the inactivated X chromosome are transcriptionally silent.
Answer: b [Reason:] All genes on the inactivated X chromosome are not transcriptionally silent. One that remains active is called X-inactive specific transcript or XIST.
10. The functional product of the Xist gene is ___________
c) RNA devoid of ORF
Answer: c [Reason:] The Xist gene in human beings encodes a 17 kb transcript devoid of any significant open reading frame. It therefore seems unlikely to code protein. Instead the RNA itself is probably the functional product of the Xist gene.
11. The one which is able to repress the production of Xist transcript is the active X chromosome.
Answer: a [Reason:] In mammalian dosage compensation system the X chromosome that is able to repress the production of Xist transcript remains active. This is because the production of high amount of Xist RNA leads to the higher rate of methylation of DNA leading it to condense into the inactivated Barr body.
12. In which stage of the cell cycle the Barr body can be identified?
Answer: a [Reason:] Inactivated X chromosome or the Barr body is readily identified in mammalian cells. During interphase, they condense into a darkly stained mass associated with the nuclear membrane.
Molecular Biology MCQ Set 4
1. What does the diagram denote?
b) Alternative splicing
d) Exon shuffling
Answer: c [Reason:] The above structure denotes the mechanism of trans-splicing. In trans-splicing, the two exons are found in two separate RNA molecules that are spliced together into a single mRNA. The intron is thus spliced is a Y – shaped structure.
2. Splicosome comprises of ______________
a) Only proteins
b) Only RNA
c) RNA and protein
d) RNA and lipoprotein
Answer: c [Reason:] The transesterification reaction is mediated by a molecular complex known as the splicosome. It comprises of about 150 proteins and 5 RNAs and is similar in size to a ribosome.
3. Splicosome is a proteinaceous enzyme.
Answer: b [Reason:] As many of the splicosome functions is mediated by its RNA components rather than proteins thus it is a ribozyme. RNAs located in the intron – exon borders participate in the splicing mechanism along with the RNAs of the splicosome to promote splicing.
4. “snurps” are made of ____________
c) DNA and protein
d) RNA and protein
Answer: d [Reason:] The five RNAs of the splicosome are collectively known as the small nuclear RNAs. Each of these RNAs is between 100 – 300 nucleotides long and is complexed with several proteins. These RNA – protein complexes are called small nuclear ribonuclear proteins (snRNPs- pronounced as “snurps”).
5. Which of the following is not a function of snRNPs?
a) 5’ splice site reorganization
b) 3’ splice site reorganization
c) Branch point site
d) Catalyze 5’ splice site cleavage
Answer: b [Reason:] The snRNPs have three roles in splicing:
i) 5’ splice site reorganization,
ii) Branch point site,
iii) Catalysis of 5’ splice site cleavage by bringing the 5’ splice site and the branch point site close together.
6. 5’ splice site is recognized by which of the following type of RNA present in the splicosome?
Answer: d [Reason:] 5’ splice site is primarily recognized by U1 snRNP in the pre – mRNA. Later in the reaction, this splice site is further recognized by U6 snRNP of the splicosome complex.
7. Which of the following splicosomal RNAs does not take part in the 5’ site splicing of the intron?
Answer: c [Reason:] 5’ splice site is primarily recognized by U1 snRNP in the pre – mRNA. Later in the reaction, this splice site is further recognized by U6 snRNP of the splicosome complex. The branch point site is recognized by U2 and thus these three snRNPs take part in the splicing of the 5’ splice site.
8. Which of the following component of splicosome recognizes the 3’ splice site?
Answer: a [Reason:] The 3’ splice site is recognized by U2AF or U2 auxillary factor. This recognizes the polypyrimidine tract of the 3’ splice site. This U2AF is an RNA – free protein element of the splicosome complex.
9. Which of the following proteins is correctly paired to its function?
a) U2AF – binds to branch point site
b) DEAD-box – recognizes pyrimidine tract
c) RNA annealing factor – loading snRNPs onto mRNA
d) BBP – dissociation of RNA–RNA interactions
Answer: c [Reason:] The correct functions for each of the enzymes are:
i) U2AF – recognizes pyrimidine tract
ii) DEAD-box – dissociation of RNA–RNA interactions
iii) RNA annealing factor – loading snRNPs onto mRNA
iv) BBP – binds to branch point site.
10. Which of the following is not involved in creating the early splicosome complex?
a) Base pairing of U1 and splice site
b) U2AF binding to 3’ splicing site
c) U2 binding to branch point site
d) BBP interaction with U2AF
Answer: c [Reason:] The early splicosome complex or the A complex consists of the following components:
i) 5’ splice site is recognized by the U1 snRNP.
ii) One subunit of U2AF binds to the 3’ subunit and the other to the pyrimidine tract.
iii) The BBP then interaction with U2AF thus finally binding to the branch point site.
This finally gives rise to the to the early complex of splicosome.
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