Molecular Biology MCQ Number 01216

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.

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.

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.

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

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.

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.

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.

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

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.