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Molecular Biology MCQ Set 1

1. Transformation involves the uptake of _____________
a) Free DNA
b) Free RNA
c) Free tRNA
d) Free mRNA

View Answer

Answer: a [Reason:] Transformation involves the uptake of the free DNA molecules. These DNA are released from one bacterium and taken up by the other to complete the process of transformation.

2. What is conjugation?
a) Transformation of DNA from a donor to recipient cell
b) Interchanging of two types of DNA
c) Transfer of DNA from recipient to donor
d) Uptake of DNA from hosts

View Answer

Answer: a [Reason:] Conjugation occurs between two bacteria, one donor and one recipient cell. It involves the direct transfer of DNA from a donor cell to a recipient cell containing the F plasmid.

3. What is transduction?
a) Bacterial genes are transferred from donor to recipient cells
b) Bacterial genes are transferred from recipient to donor cells
c) Bacterial genes are carried from a donor cell to a recipient cell via bacteriophage
d) Uptake of naked DNA by hosts

View Answer

Answer: c [Reason:] In transduction process bacteria is lysed. Thus the bacterial genes are carried from a donor cell to a recipient cell by a bacteriophage which undergoes a lytic lifecycle inside the host.

4. In which year Frederic Griffith discovered transformation?
a) 1920
b) 1928
c) 1936
d) 1944

View Answer

Answer: b [Reason:] Frederick Griffith discovered transformation in Streptococcus pneumoniae in 1928. Pneumococci exhibited genetic variability in his experiments of causing pneumonia in rats.

5. During Frederick Griffith’s experiments what type of colonies were generated by the capsulated pneumococci?
a) Type R
b) Type L
c) Type S
d) Type B

View Answer

Answer: c [Reason:] During the experiment, when grown in blood agar media in petridishes. Pneumococci with capsules form large, smooth colonies. These colonies are designated as type S colonies.

6. What is the function of the capsule of the bacterial cells?
a) Virulence
b) Protection
c) Adherence
d) Both A and B

View Answer

Answer: d [Reason:] The polysaccharide capsule of the bacterial cell is required for virulence. It also provides protection to the bacterial cell from destruction by the host immune mechanism.

7. How many types of bacterial capsule are present for pneumococcus depending on the genotype of the cell?
a) 1
b) 2
c) 3
d) Infinite

View Answer

Answer: d [Reason:] The polysaccharide capsule may be of several different antigenic types. This depends on the specific molecular composition of the polysaccharide and it also depends on the genotype of the cell.

8. The genetic information is stored in DNA rather than protein.
a) True
b) False

View Answer

Answer: a [Reason:] The genetic information is stored in DNA. This was first discovered by Oswald Avery, Colin Macleod and Maclyn McCarty in 1944.

9. In the following which bacteria only take up their own DNA?
a) S. pneumoniae
b) B. subtilis
c) N. gonorrbacae
d) Mycoplasma

View Answer

Answer: c [Reason:] S. pneumoniae and B. subtilis take up DNA from any source. Only H. influenzae and N. gonorrbacae takes up DNA from their own or closely related species.

10. How many copies of special short nucleotide pair sequences for identification are present in H. influenzae and N. gonorrbacae?
a) 100 copies
b) 600 copies
c) 500 copies
d) 1000 copies

View Answer

Answer: b [Reason:] H. influenzae and N. gonorrbacae only takes up DNA from their own or closely related species. This is facilitated by the presence of a special short nucleotide pair sequences for identification which is present roughly about 600 copies in their respective genomes.

11. What do you mean by competent bacteria?
a) Able to take up genetic material from surrounding
b) Able to transfer genetic material to the surrounding
c) Takes up only its own genetic material
d) Can transfer genetic material from one bacterium to another

View Answer

Answer: a [Reason:] There are some bacterial species that have the ability to take up DNA from their surroundings. These bacteria encode certain proteins that facilitate the process of this type of transformation. Such bacteria are known as competent bacteria.

12. Competence in a bacteria ___________
a) DNA
b) RNA
c) Protein
d) Polysaccharides

View Answer

Answer: c [Reason:] The protein that mediates the transformation of the bacteria is known as competence and the bacteria are known as competent. This protein contains com segments.

Molecular Biology MCQ Set 2

1. In which phase bacteria develop competence?
a) Late phase
b) Log phase
c) Metaphase
d) Lag phase

View Answer

Answer: a [Reason:] Bacteria develop competence during the late phase of their growth cycle. In this cycle the density of bacteria is high but the cell division stops.

2. In which bacteria competence pheromones was first identified?
a) S. pneumoniae
b) B. subtilis
c) H. influenzae
d) N. gonorrbacae

View Answer

Answer: b [Reason:] Bacillus subtilis contain small peptides called competence pheromones that are secreted by cells and accumulate at high cell density. High concentration of these pheromones induces the expression of genes encoding proteins required for facilitating the process of transformation.

3. What is the percentage of DNA molecules taken up by competent cells during transformation?
a) 0.1-0.5
b) 0.3-0.4
c) 0.5-0.6
d) 0.2-0.5

View Answer

Answer: d [Reason:] The DNA molecules taken up by competent cells during transformation are usually only about 0.2 – 0.5% of the complete genetic material. Thus plasmids for recombination are thus prepared to maintain the size of the adequate molecule to be taken up by the bacterium.

4. Transformation of E. coli does not occur.
a) True
b) False

View Answer

Answer: b [Reason:] E. coli is the most intensely studied bacterial species under natural conditions. In 1946, Joshua Lederberg and Edward Tatum discovered that E. coli cells transfer genes by conjugation thus exhibiting transformation.

5. During conjugation the Donor cell have cell surface appendages known as ____________
a) F pili
b) B pili
c) A pili
d) D pili

View Answer

Answer: a [Reason:] During conjugation Donor cells have cell surface appendages known as F pili. The synthesis of these F pili is controlled by gene present on a plasmid called the F factor or the fertility factor.

6. What is the size of the F factor that is involved in conjugation?
a) 106 base pairs
b) 105 base pairs
c) 10-12 base pairs
d) 109 base pairs

View Answer

Answer: b [Reason:] Most F factors are approximately 105 base pairs in size. This contains the gene that facilitates the growth if the F pili through which the bacteria is able to transfer the F factor to another bacteria.

7. In how many states the F plasmid can occur within bacteria?
a) 1
b) 2
c) 3
d) 4

View Answer

Answer: b [Reason:] The F plasmid can occur within bacteria in two states. They are the autonomous or plasmid state or the integrated state within the bacterial genome.

8. By which of the following methods does the F plasmid integrates into the bacterial genome?
a) Transformation
b) Conjugation
c) Recombination
d) Mutation

View Answer

Answer: c [Reason:] The F-factor can integrate into the bacterial chromosome by site specific recombination. This recombination is mediated by short stretches of homologous sequence in the genome and plasmid of the bacteria.

9. High frequency recombination cell contain an integrated F factor. What is the name of such a cell?
a) Har cell
b) Hra cell
c) Hfr cell
d) Hrf cell

View Answer

Answer: c [Reason:] High frequency recombination cell contain an integrated F factor. These cells are known as Hfr cell. F factor mediates the transfer of chromosome from the Hfr+ cell to an Hfr- cell transforming it into an Hfr+ cell.

10. What is the strain of the Hfr+ cell where the F factor is integrated near the tbr and leu loci?
a) Hfr H
b) Hfr A
c) Hfr B
d) Hfr K

View Answer

Answer: a [Reason:] The particular strain of the Hfr+ cell where the F factor is integrated near the tbr and leu loci is the Hfr H cell. In this strain the transfer of F factor induces the transfer of the other two genes with it.

11. Transformation, transduction and conjugation occurs in every bacterial cell.
a) True
b) False

View Answer

Answer: b [Reason:] Transduction is the only process that occurs in all bacteria. The occurrence of conjugation and transformation depend on whether the required genes and metabolic machinery have evolved in that species of the bacterium or not.

Molecular Biology MCQ Set 3

1. Which of the following is not a type of post translational modification?
a) Proteolysis
b) Protein folding
c) Glycosylation
d) Lipid addition

View Answer

Answer: b [Reason:] Post translational modifications include three types of modification that is proteolysis, glycosylation and lipid addition. Protein folding is the process taking place after the translational modification and that is used to produce the cognate protein by folding the polypeptide in proper 3-D form with the help of chaperons.

2. Proteolytic modifications of the polypeptide are an important process in the mechanism for protein sorting and transport.
a) True
b) False

View Answer

Answer: a [Reason:] Proteolytic modifications of the amino terminus play a part in the translocation of many proteins across the membranes, including secretory proteins in both bacteria and eukaryotes as well as proteins destined for incorporation into the plasma membrane, lysozomes, mitochondria and chloroplasts of eukaryotic cells. These proteins are targeted for transport to their destinations by amino terminal sequences that are removed by proteolytic cleavage as the protein crosses the membrane.

3. The amino acid is the signal sequence in any polypeptide chain for ____________
a) Protein activity
b) Glycosylation site
c) Proteolytic site
d) Site for lipid addition

View Answer

Answer: c [Reason:] The amino acid is the signal sequence is usually about 20 amino acids long and targets many secretory proteins to the plasma membrane of bacteria or the ER of eukaryotic cells while translation is still in progress. The signal sequence, which consists predominantly of hydrophobic amino acids, is inserted into a membrane channel as it emerges from the ribosome. This signal is cleaved off to produce the mature polypeptide.

4. The first step in protein targeting is ___________
a) Synthesis of protein
b) Translocation to Golgi body
c) Translocation to nucleus
d) Translocation to ER

View Answer

Answer: d [Reason:] The first step in protein targeting is its translocation to the ER after its production from the ribosome. Only after this translocation proteolytic cleavage occurs producing the mature protein which undergoes many modifications before it is targeted to its destination.

5. How many Proteolytic cleavages produce the mature insulin?
a) 1
b) 2
c) 3
d) 4

View Answer

Answer: b [Reason:] Insulin forms by two cleavages. The initial precursor, preproinsulin, contains an amino terminal signal sequence that targets the polypeptide chain to the ER. Removal of the signal sequence during transfer to the ER yields a second precursor, called proinsulin. This precursor is then converted to insulin by proteolytic removal of an internal peptide.

6. Glycosylation is the addition of ___________ to the protein.
a) Carbohydrate
b) Lipid
c) Fat
d) Minerals

View Answer

Answer: a [Reason:] Glycosylation is the addition of carbohydrate moieties to the protein. The proteins to which the carbohydrate groups are added are called glycoproteins. The can be added in but amino and carboxy terminals of the amino acids depending on the type of amino acid.

7. In the N-linked glycoprotein the carbohydrates are attached to which of the following bases?
a) Valine
b) Threonine
c) Asparagine
d) Serine

View Answer

Answer: c [Reason:] In the N-linked glycoprotein the carbohydrates are attached to the amino terminus of the amino acid. An example of such an amino acid to which N-linked glycosylation occurs is the nitrogen atom in the side chain of Asparagine.

8. In O-linked glycoproteins the carbohydrates may be attached to how many amino acids?
a) 1
b) 2
c) 3
d) 4

View Answer

Answer: b [Reason:] In O-linked glycoproteins the carbohydrates may be attached to two of the 20 amino acids found generally. These carbohydrate moieties are added to the oxygen atom in the side chain of serine or Threonine.

9. Initiation of glycosylation occurs in the Golgi body.
a) True
b) False

View Answer

Answer: b [Reason:] The proteins are usually transferred to the ER as soon as they are produced by the ribosome. Thus most of the modification such as proteolysis and glycosylation occurs within the lumen of the ER.

10. How many types of modification are possible in eukaryotes by addition of lipids?
a) 1
b) 2
c) 3
d) 4

View Answer

Answer: d [Reason:] Three general types of lipid additions – N-myristoylation, prenylation and palmitoylation – are common in eukaryotic proteins associated with the cytosolic face of the plasma membrane. A fourth type of modification, the addition of glycolipids, plays an important role in anchoring some cell surface proteins to the extracellular face of the plasma membrane.

11. Prenylation adds prenyl groups to the ___________ amino acid residues.
a) Methionine
b) Cystine
c) Threonine
d) Arginine

View Answer

Answer: b [Reason:] Prenylation adds prenyl groups to the Cystine residue. The prenyl group is added to the sulphur atom in the side chains of cystine residues located near the C-terminus of the polypeptide chain.

12. Frame shift mutation can increase or decrease the length of a polypeptide.
a) True
b) False

View Answer

Answer: a [Reason:] Frame shift mutation is a type of point mutation that introduces alterations in the genetic code. Thus frame shift mutation introduces insertion or deletions of one or a small number of base pairs that alter the reading frame.

13. Consider a tandem repeat of the sequence GCU. Now if an insertion of an A residue occurs in the message as follows
molecular-biology-questions-answers-post-translational-modifications-proteins-mistranslation-protein-synthesis-q13
What type of mutation has taken place in this case?
a) Missense mutation
b) Stop mutation
c) Frame shift mutation
d) Reverse mutation

View Answer

Answer: c [Reason:] GCU generally codes for alanine but the insertion of A in the genetic message changes the reading frame thus generates a serine codon (AGC) at the site of insertion. This results in the frame shift in the open reading frame from alanine to cystine in the downstream of the insertion. Thus the insertion of a single base alters the coding capacity of the message.

14. If the insertions of three nucleotides occur in a consecutive manner it causes which of the following type of mutation?
a) Missense mutation
b) Stop mutation
c) Frame shift mutation
d) Reverse mutation

View Answer

Answer: c [Reason:] If the insertions of three nucleotides occur in a consecutive manner it causes frame shift mutation. This is because frame shift mutation introduces insertion or deletions of one or a small number of base pairs that alter the reading frame.

Molecular Biology MCQ Set 4

1. How many pre – rRNA transcript in eukaryotes are used to produce all the different mature rRNAs?
a) 1
b) 2
c) 3
d) 4

View Answer

Answer: b [Reason:] Eukaryotes have 4 species of ribosomal RNA molecules. Three of these rRNAs are derived by the cleavage of a single precursor whereas the last rRNA is synthesized as a single molecule.

2. In bacterial cell the 3 rRNAs 23S, 16S and 5S are derived from different transcripts.
a) True
b) False

View Answer

Answer: b [Reason:] No in bacterial cell the 3 rRNAs 23S, 16S and 5S are not derived from different transcripts. As in eukaryotes, the bacterial rRNAs 23S, 16S and 5S are derived by the cleavage of one single pre – rRNA transcript.

3. Which one of the following rRNA undergoes least post – transcriptional processing?
a) 28S
b) 18S
c) 5.8S
d) 5S

View Answer

Answer: d [Reason:] The rRNA undergoes least post – transcriptional processing is the 5S rRNA. This is because it is synthesized separately and thus do not undergo any kind of cleavage. The rest three are derived by the cleavage of one single pre – rRNA transcript.

4. The bacterial pre – rRNA undergo __________ cleavage.
a) 1
b) 2
c) 3
d) 4

View Answer

Answer: b [Reason:] The bacterial pre – rRNA undergo 2 cleavages. As three rRNA molecules are flanked together the first cleavage separates the precursors of the three rRNAs. A further cleavage known as the secondary cleavage ensures the production of the mature rRNA molecules.

5. Which of the following types of processing is not used to prepare mature rRNA?
a) Splicing
b) Methylation
c) Glycosylation
d) Conversion of uridine

View Answer

Answer: a [Reason:] The preparation of mature rRNA does not include splicing in most of the cases. They only include methylation of specific bases, and some specific sugar moieties and conversion of uridines to pseudouridines.

6. Each tRNA is produced from different transcripts .
a) True
b) False

View Answer

Answer: b [Reason:] Like rRNAs, tRNAs in both bacteria and eukaryotes are synthesized as a longer precursor molecule. Some of these pre – tRNAs contain several individual tRNA sequences. Also in bacteria some of the tRNAs are included in pre – rRNA transcripts.

7. The pre – tRNA involves cleavage by __________
a) tRNA itself
b) RNase H
c) RNase P
d) RNA polymerase

View Answer

Answer: c [Reason:] The processing of the 5’ end of the pre – tRNAs involves cleavage by an enzyme called RNase P. RNase P consists of RNA and in molecules, both of which are required for maximal activity.

8. RNase P is a ribozyme.
a) True
b) False

View Answer

Answer: a [Reason:] In 1983 Sidney Altman and his colleagues demonstrated that the isolated RNA component of the RNase P is itself capable of catalyzing pre – tRNA cleavage. These experiments established that RNase P is a ribozyme – an enzyme in which RNA rather than protein is responsible for catalytic activity.

9. The common sequence for all tRNAs at their 3’ end is __________
a) CCC
b) CCT
c) GGA
d) CCA

View Answer

Answer: d [Reason:] All tRNAs have the common sequence CCA at their 3’ end. This sequence is the site of amino acid attachment, so it is required for tRNA function during protein synthesis.

10. The 3’ CCA terminus is exceptionally added by the RNA polymerase during transcription.
a) True
b) False

View Answer

Answer: b [Reason:] The 3’ CCA terminus is encoded in the DNA of some tRNA genes, but in others it is not. Thus the 3’ CCA terminus is added during RNA processing by an enzyme which recognizes and adds CCA to the 3’ end of all tRNAs that lack this sequence.

11. Though unusual, pre – tRNA also undergo splicing. It is achieved by the help of __________
a) tRNA itself
b) Splicosome
c) RNA polymerase exonuclease activity
d) Endonuclease

View Answer

Answer: d [Reason:] Though unusual, pre – tRNA and pre – rRNAs also undergo splicing. In contrast to other splicing reactions, which involve the activity of catalytic RNAs, tRNA splicing is mediated by conventional protein enzymes. An endonuclease cleaves the pre – tRNA at the splice site to excise the intron, followed by joining of the exons to form a mature tRNA molecule.

Molecular Biology MCQ Set 5

1. Which of the following types of DNA polymerase has 3’→5’ exonuclease activity?
a) DNA polymerase I
b) DNA polymerase II
c) DNA polymerase III
d) DNA polymerase IV

View Answer

Answer: a [Reason:] DNA polymerases I, II, III and IV all has 5’→3’ exonuclease activity. DNA polymerases I is the only polymerase to have the 3’→5’ exonuclease activity which is the proof reading activity of DNA polymerase.

2. Choose the odd one out with respect to DNA polymerase III.
a) dna E
b) dna N
c) dna Q
d) dna B

View Answer

Answer: d [Reason:] DNA polymerase III is the product of dna E, dna N, dna Q and 7 other structural genes. Only dna B is the odd one in the provided group as it is not the structural gene for any of the subunits of DNA polymerase III.

3. Which of the following is used in prokaryotic replication?
a) DNA polymerase I
b) DNA polymerase II
c) DNA polymerase III
d) DNA polymerase δ

View Answer

Answer: c [Reason:] Prokaryotic replication is polymerized by the enzyme DNA polymerase III. DNA polymerase I, DNA polymerase II and DNA polymerase δ all take part in the eukaryotic replication.

4. DNA polymerase II can polymerase upto ______________ nucleotides per minute at 37˚C.
a) 50
b) 500
c) 5000
d) 50000

View Answer

Answer: a [Reason:] DNA polymerase II has the lowest affinity for dNTPs and can polymerase upto 50 nucleotides per minute at 37˚C. DNA polymerase III has the highest affinity for dNTPs and can polymerase upto 15000 nucleotides per minute at 37˚C.

5. All mutation in polymerases is lethal except the mutation of polymerase III.
a) True
b) False

View Answer

Answer: b [Reason:] Mutations in both DNA polymerase I and DNA polymerase III are lethal. Only exception to this is polymerase III, mutation to which is not lethal.

6. Polymerase I has the highest affinity for dNTPs.
a) True
b) False

View Answer

Answer: b [Reason:] DNA polymerase I have a low affinity for dNTPs and can polymerase upto 1000 nucleotides per minute at 37˚C. DNA polymerase III has the highest affinity for dNTPs and can polymerase upto 15000 nucleotides per minute at 37˚C.

7. Polymerase I has two fragments. The residue of larger fragment consists from ____________residues.
a) 1 – 234
b) 234 – 658
c) 324 – 928
d) 234 – 928

View Answer

Answer: c [Reason:] Polymerase I has two fragments. The residue of larger fragment consists from 324 – 928 residues is known as the klenow fragment which has the polymerase activity as well as the 5’→3’ exonuclease activity.

8. Which of the following is false about klenow fragment?
a) Polymerization activity
b) 3’→5’ exonuclease activity
c) 5’→3’ exonuclease activity
d) 324 – 928 residue of polymerase I

View Answer

Answer: b [Reason:] The residue of larger fragment consists from 324 – 928 residues is known as the klenow fragment which has the polymerase activity as well as the 5’→3’ exonuclease activity. The 3’→5’ exonuclease activity or the proofreading activity is present in the small fragment of DNA polymerase I.

9. DNA polymerase III has ___________ subunits.
a) 4
b) 6
c) 8
d) 10

View Answer

Answer: d [Reason:] DNA polymerase II has 10 subunits. They are αa, ϵa, θa, τ, γb, δb, δ’b, χb, Ψb and β the structural genes of which are pol C, dna Q, hot E, dna xc, dna χc, hot A, hot B, hot C, hot D AND dna N respectively.

10. Which of the following types of DNA polymerase does not take part in DNA repair ?
a) DNA polymerase I
b) DNA polymerase II
c) DNA polymerase III
d) DNA polymerase IV

View Answer

Answer: c [Reason:] DNA polymerase I take part in both replication and repair. DNA polymerase II and DNA polymerase IV take part in repair only. DNA polymerase III take part in chromosomal replication only.

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