Molecular Biology MCQ Set 1
1. How many types of specific domains are present in the SR protein?
Answer: b [Reason:] The SR protein has 2 special RNA binding domains. They are RNA-recognition motif (RRM) and the other is RS domain.
2. Which of the following components of the splicosome machinery is recruited by the SR protein?
Answer: c [Reason:] The SR proteins help efficient recruitment of the components to the different splicosome components to the nearby splice site. Specifically the SR proteins recruit the U2AF proteins to the 3’ splice site and U1 to the 5’ splice site.
3. SR proteins are not essential for RNA splicing.
Answer: b [Reason:] SR proteins are essential for RNA splicing. They not only ensure the accuracy and efficiency of constitutive splicing but also regulate alternative splicing. They come in many varieties, some controlled by physiological signals, others constitutively active.
4. The presence and activity of SR determines the ___________
a) Product formed after splicing
b) Mechanism of splicing
c) Usage of particular splice site
d) Repression of splicing mechanism
Answer: c [Reason:] The SR family protein is large and diverse that has specific roles in regulated alternative splicing by directing the splicing machinery to different splice sites under different conditions. Thus, the presence of activity of a given SR protein can determine whether a particular splice site is used in a particular cell type, or at a particular stage of development.
5. Alternative splicing is regulated by activators and repressors.
Answer: a [Reason:] Proteins that regulate splicing bind to the specific sites called exonic or intronic splicing enhancers, or exonic or intronic splicing silencers. The former enhance and the latter repress splicing in nearby splice sites.
6. The Drosophila half-pint protein is _____________
a) Positive regulator
b) Negative regulator
c) Non regulator
d) Not related to fly ovarian protein
Answer: a [Reason:] An activator that promotes a particular alternative splicing event in a specific tissue type is the Drosophila half-pint protein. This activator regulates the alternative splicing of a set of pre-mRNA in the fly ovary.
7. Most silencer proteins are able to bind to pre-mRNA but cannot recruit the splicing machinery.
Answer: a [Reason:] Most silencers are recognized by members of the heterogeneous nuclear ribonucleoprotein (hnRNP) family. These bind RNA but lack the RS domains and so cannot recruit the splicing machinery.
8. hrRNPA1 acts as the repressor for the final mRNA form of ____________
Answer: d [Reason:] The hrRNPA1 binds to an exonic silencer element within an exon of the HIV at pre-mRNA and represses the inclusion of that exon in the final mRNA. Bi binding to its site, the repressor blocks binding of the activator SC35 to a nearby enhancer element.
9. The pre-mRNA binding minor splicosome is also known as ___________
a) AG – CT splicosome
b) AT – GC splicosome
c) AC – AG splicosome
d) AT – AC splicosome
Answer: d [Reason:] The minor splicosome recognizes rarely occurring introns having consensus sequences distinct from the sequences of most pre-mRNA introns. This recently discovered form is known as the AT – AC splicosome, because the termini of the originally identified rare introns contain AU at the 5’ splice site and AC at the 3’ site in RNA or AT and TC in DNA.
10. hnRNPI binds to which segment of the mRNA?
a) 5’ splice site
b) Branch point site
c) 3’ splice site
d) AT – AC splicosome
Answer: c [Reason:] The hnRNPI binds to the pyrimidine tract of the intron. Thus it is also known as the polypyrimidine tract-binding protein. This is a repressor protein which when binds to the mRNA blocks the binding of the splicosome machinery.
11. Alternative splicing of the Troponin T gene produces ____________ alternative forms of mRNA.
Answer: b [Reason:] A region of the Troponin T gene encodes 5 exons. After splicing they recombine in such a way that only two forms of mRNA are generated. One contains exon 1, 2, 4 and 5 known as the β-tropoinin T mRNA whereas the other containing exon 1, 2, 3 and 5 known as the α-tropoinin T mRNA.
12. Which of the following mRNA is an example of having an extended exon?
a) Human slo gene
b) Mammalian muscle Troponin T mRNA
c) T-antigen mRNA of monkey virus SV40
d) Drosophila Dscam mRNA
Answer: c [Reason:] The pre-mRNA of the T-antigen of monkey virus SV40 contains 3 exons interrupted by 2 introns. These are spliced in 5 different alternative pathways – normal, exon skipped, exon extended, intron retained and alternative exons.
13. Alternative splicing is always used to produce multiple isoforms of proteins.
Answer: b [Reason:] Some genes that encode only a single functional protein show alternative splicing. In those cases, alternative splicing is used simply as a way of switching expression of the gene on and off.
14. Alternative splicing can be used to switch off a gene’s expression. How many ways are there to achieve it?
Answer: b [Reason:] There are 2 alternative splicing can be used to switch off a gene’s expression. Most straightforwardly, alternative splicing determines the presence of a stop codon to be included with the exons as stop codons are extensively found in the intronic regions. The second way is by regulating the use of an intron, which, when retained in the mRNA, ensures that species is not transported out of the nucleus and so is never translated.
15. Alternative splicing was discovered in the year ______________
Answer: a [Reason:] Alternative splicing was discovered in the studies of gene expression in the mammalian adenovirus, where mRNAs are alternatively spliced. It was discovered in the year 1977.
16. Which of the following is not a part of the minor splicosome machinery?
Answer: a [Reason:] U2 is not a part of the minor splicosome machinery and is only found in the major splicosome machinery. Instead U12 is found in the minor splicosome machinery and is known to recognize different sites from the U2 snRNP although carrying out the same function as U2.
Molecular Biology MCQ Set 2
1. Eukaryotic ribosome is recruited to the mRNA by _____________
b) Shine – Dalgarno sequence
c) 5’ capping
d) 3’ tailing
Answer: c [Reason:] Prokaryotic ribosome is recruited to the mRNA by the Shine – Dalgarno sequence. In case of eukaryotes ribosome is recruited to the mRNA by the 5’ cap.
2. After bounding to the mRNA eukaryotic ribosome scans the entire mRNA until it encounters a start codon.
Answer: a [Reason:] The eukaryotic ribosome is recruited to the mRNA by the 5’ cap. The guanine residue of the 5’ cap is connected to the 5’ end of the mRNA through three phosphate groups that recognizes and recruits the ribosome. Once bound to the mRNA, the ribosome moves in a 5’→3’ direction until it encounters a 5’…….AUG…….3’ start codon by a process called scanning.
3. The end of all tRNAs is ___________
a) 5’ ACC 3’
b) 5’ CCA 3’
c) 3’ CAC 5’
d) 3’ GAG 5’
Answer: b [Reason:] All tRNAs end at the 3’ terminus with the sequence 5’ CCA 3’. This is the site that is attached to the cognate amino acid by the enzyme aminoacyl tRNA synthetase.
4. How many unusual bases are observed in a tRNA molecule?
Answer: d [Reason:] Five unusual bases, produced by enzymatic modification of the usual bases, may be observed in a tRNA molecule. These are pseudouridine, dihyrdouridine, hypoxanthine, thymine and methylguanine. Of these five, unusual bases pseudouridine and dihyrdouridine are most commonly found in the tRNA molecules.
5. In pseudouridine the attachment of Uracil base to ribose through ___________ of uracil.
a) N at position 3
b) C at position 6
c) C at position 5
d) N at position 1
Answer: c [Reason:] Pseudouridine is produced by the enzymatic modification of the regular base uridine through isomerization. In the modification process of uridine to pseudouridine the attachment of base to the ribose sugar is switched from nitrogen at ring position 1 to the carbon at ring position 5.
6. Dihydrouridine is the result of oxidation of the uracil base.
Answer: b [Reason:] Dihydrouridine is a derivative of uridine formed by enzymatic reduction. The process of reduction forms the double bond between the carbons at positions 5 and 6.
7. How many loops are seen in the tRNA?
Answer: d [Reason:] Four different loops are observable in the tRNA 3˚ motif. They are the ΨU loop, D loop, variable loop and the anticodon loop. The ΨU loop is named because of the presence of the pseudouridine base. The D loop contains the dihyrdouridine base. The anticodon loop contains the anticodon and the variable loop is named so because its length is variable.
8. How many double helix regions are observed in a tRNA molecule?
Answer: a [Reason:] The tRNA molecules consist of 4 double helix regions. They are the acceptor stem and the stem of the three loops, the ΨU loop, D loop and the anticodon loop.
9. The variable loop is found between the ___________
a) Acceptor stem and ΨU loop
b) Acceptor stem and D loop
c) Anticodon loop and ΨU loop
d) Anticodon loop and D loop
Answer: c [Reason:] The variable loop sits between the anticodon loop and ΨU loop. Also as the name implies the size of the variable loop varies from 3 to 21 bases.
10. The 3 dimension structure of an adaptor molecule of translation is revealed to be ___________
a) Y shaped
b) L shaped
c) K shaped
d) X shaped
Answer: b [Reason:] The adaptor molecule of translation is the tRNA molecule which acts as the bridge between the mRNA code and the respective amino acid. X-ray crystallography reveals an L-shaped tertiary structure in which the terminus of the acceptor stem is at one end of the molecule and the anticodon loop is about is about 70 Å away at the other end.
11. Which of the following does not take part in stabilizing the cloverleaf model of the tRNA?
a) Base stacking
b) Base and sugar-phosphate backbone interaction
c) Ionic bond
d) Hydrogen bond
Answer: c [Reason:] Three types of interactions stabilize the L-shaped structure of the tRNA. The first is hydrogen bonds between the bases to form the helical parts of the tertiary structure of the tRNA molecule. Second the interaction between the bases and the respective sugar phosphate backbone. Finally the additional stabilization is provided by the base stacking between the two extended regions of base pairing.
Molecular Biology MCQ Set 3
1. Automated sequencing is defined as _________
a) Chain termination sequencing
b) Radio labeled sequencing
c) Real time fluorescence sequencing
Answer: c [Reason:] To automate the process of sequencing DNA it is desirable to acquire sequence data in real time. This is done by attaching fluorescent tags to the chain terminating nucleotides and then detecting the bands within the gel during the electrophoretic separation.
2. Automated sequencing uses _____________
a) Enzymatic process
c) Chain termination
d) Fluorescent tagging
Answer: d [Reason:] Automated sequencing uses fluorescent tagging method for the detection of bands. These tags are attached to the chain terminating nucleotides. Each of the four dideoxynucleotide carries a different flurophore.
3. No chain termination takes place in automated sequencing ____________
Answer: b [Reason:] As the tags are attached to the four dideoxynucleotide polymerase cannot synthesize after that. Thus incorporation of the tags ensures the termination of the synthesis.
4. As different fluorescent tags affect the mobility of fragments thus only one type of tag is used.
Answer: b [Reason:] The four dideoxynucleotide carries a different flurophore to simplify the base identification. As different fluorescent tags affect the mobility of fragments thus sophisticated software is incorporated into the scanning process to ensure that the bands are read in the correct order.
5. With respect to automated sequencing which of the following is true?
a) Detected by radioactive tags
b) Multiple plane sequencing
c) Horizontal scanning one sequence per lane
d) Vertical reading software incorporation
Answer: c [Reason:] By using four different fluorescent dyes, it is possible to electrophorese all four chain terminating reactions together in one plane of a sequencing gel. The DNA bands are detected by their fluorescence as they electrophorese past the detector. In the detector it is made to scan horizontally across the base of a slab gel.
6. Which of the following feature is required for the high sensitivity of the tags used for autosequencing?
a) Strong absorption at UV spectrum
b) Maximum emittion at different wavelengths
c) Different mobility shifts
d) Induce circularization of DNA fragments
Answer: b [Reason:] For high sensitivity DNA detection in four colours sequencing the following criteria must be met:
i) Each of the four dyes must exhibit strong absorption at the common laser wavelength.
ii) To have an emission maximum at a distinctly different wavelength.
iii) To introduce the same relative mobility shift of the sequencing fragments.
7. Automated sequencing is a real time one trial method of sequencing and is highly handy.
Answer: b [Reason:] The use of single gel track for all four dideoxy reactions means that the problem is less acute in automated sequencing than manual sequencing. It is desirable to sequence a piece of DNA several times and on both strands, to eliminate caused by technical problems.
8. A third generation sequencing can sequence upto ____________nucleotides per day.
Answer: a [Reason:] The third generation sequencing can sequence upto 750000 nucleotides per day. These new generation sequencers uses a 96 capillary sequencer filled with the gel matrix.
Molecular Biology MCQ Set 4
1. What is the consensus sequence of the Pribnow box?
Answer: b [Reason:] The Pribnow box is also known as the –10 promoter site. It was first recognized by Pribnow in 1975. It has a consensus sequence of TATAAT.
2. The Pribnow box is present on the coding strand of the DNA template.
Answer: b [Reason:] The Pribnow box or the –10 sequence is present in the sense strand of DNA duplex. The sense strand is also known as the non – coding strand. It also harbors the other two conserved sequences for polymerase binding.
3. The –35 sequence is highly conserved and has a consensus sequence of ___________
Answer: d [Reason:] The –35 sequence is highly conserved in efficient promoters and has a consensus sequence of TTGACA. The first three positions of this hexameric sequence are the mostly conserved.
4. The +1 site of conserved promoter sequence is generally a pyrimidine.
Answer: b [Reason:] The +1 site of conserved promoter sequence is the transcription start site and is 90% purine. “A” is more common than “G” and often have “C” and “T” on its either side.
5. –10, – 35 and +1 sites are the consensus promoter sites of sigma factor ______________
Answer: a [Reason:] Different sigma factors recognize different promoter sites. As σ70 sigma factor is the most common sigma factor the highly studied –10, – 35 and +1 sites of promoter belongs to it.
6. The RNA polymerase core enzyme has a high specificity for promoter sites and can perfectly work alone.
Answer: b [Reason:] The RNA polymerase core enzyme has a high non – specific affinity for DNA. This is referred to as ‘loose binding’ and is not stable. Addition of σ factor converts the RNA polymerase core protein into a holoenzyme and also increases its specificity by 100 folds.
7. The complex formed by the polymerase and the promoter DNA is known as the _____________
a) λ complex
b) γ complex
c) Open complex
d) Closed complex
Answer: d [Reason:] At the promoter the polymerase recognizes the –35 and the –10 sequence and tightly binds to it. The initial complex of the polymerase with the base paired promoter DNA is referred to as a closed complex.
8. Negative supercoiling enhances the rate of transcription.
Answer: a [Reason:] Negative supercoiling enhances the rate of transcription of many genes. This is because it facilitates the unwinding of the DNA duplex by the RNA polymerase.
9. The –10 promoter region is rich in ______________ bases.
a) A, G
b) C, T
c) A, T
d) C, G
Answer: c [Reason:] The –10 promoter region is rich in A, T bases. It has a consensus sequence of TATAAT and is also known as the Pribnow box. The presence of A and T is significant because the unwinding of DNA starts at this site and AT base parts melt easily than the GC base pairs.
Molecular Biology MCQ Set 5
1. The CpG islands generally consist of how many GC bases?
b) Less than 100
c) More than 1000
Answer: d [Reason:] At certain sites CpG dinucleotide occurs at higher frequency and is referred to as CpG islands. CpG islands are approximately 1000 bases long and show an elevated G+C base composition.
2. In humans CpG islands are generally located in the ____________
a) Operator region
b) Promoter region
c) Transcript region
d) Terminator region
Answer: b [Reason:] CpG islands are often associated with the transcription start site, that is, the promoter region. About 60% of human genes have CpG islands at their promoter regions.
3. The Intergenic CpGs are more commonly methylated than in the intragenetic region.
Answer: a [Reason:] CpG islands in the genome are located in both intragenic and intergenic regions. In vertebrates over 80% of the methylated cytosine residues are found in the intragenetic regions. In contrast CpGs within the CpG islands are generally either not methylated or have relatively low levels of methylation.
4. Major methylation of CpG islands does not include?
Answer: d [Reason:] The methylation of genome is persistent throughout and is missing only in regions such as CpG islands within the promoter and enhancer regions. Methylation in these two regions may lead to the silencing of the gene itself.
5. De-novo methylation leads to the phenomenon of genetic imprinting in the offspring.
Answer: b [Reason:] Maintenance methylation leads to the phenomenon of genetic imprinting in the offspring. This phenomenon controls the expression of certain genes involved in the development of mammalian embryos.
6. Both the imprinted genes inherited from either parent are expressed in an offspring.
Answer: b [Reason:] In an offspring only one of the imprinted genes inherited from either parent is expressed. This phenomena is known as allelic exclusion.
7. The phenomenon of allelic exclusion leads to the inactivation of one allele of a gene. The active allele is known as ____________
Answer: c [Reason:] In the phenomenon of allelic exclusion the active allele is known as hemizygote. And the phenomenon that is exhibited by it is known as functional hemizygosity.
8. Gene of which of the following protein/enzyme/hormone is an example of paternal Hemizygote?
Answer: b [Reason:] An example of an imprinted gene in mammals is the insulin like growth factor 2 or Igf-2 which encodes a growth factor. Only the paternal copy of the Igf-2 gene is transcribed and is not expressed from the maternally inherited chromosome.
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