Molecular Biology MCQ Set 1
1. Which of the following statement is false about DNA?
a) Located in chromosomes
b) Carries genetic information from parent to offspring
c) Abundantly found in cytoplasm
d) There is a precise correlation between amount of DNA and number of sets of chromosome per cell
Answer: c [Reason:] In case of eukaryotes DNA is abundantly found in nucleoplasm which is surrounded by the nuclear membrane. This structure is known as nucleus which is found in the cytoplasmic matrix. In case of prokaryotes DNA is found in a less dense cytoplasmic matrix known as the nucleoid.
2. Which of the following function of DNA is necessary for the purpose of evolution?
Answer: d [Reason:] Mutation facilitates the change of bases within a DNA and if this change encodes for a viable amino acid which in turn may lead to the synthesis of a different protein. This protein exerts a phenotypic character to the organism which may be different from the wild type character or may generate a unique character itself, thus leading to evolution.
3. According to the phenotypic characters of pneumococcus considered in Griffith’s experiment of transformation, which of the following statements are correct?
i) Presence of slime layer
ii) Presence of capsule
iii) Absence of capsule
iv) Type of adhesion
v) Molecular makeup of capsule
Choose the correct option
a) i, ii, iii
b) ii, iii, iv
c) ii, iii, v
d) i, iv, v
Answer: c [Reason:] As the two phenotypic characteristics of importance in Griffith’s experiment of transformation are:-
i) Presence or absence of polysaccharide capsule that surrounds the bacterial cell which is responsible for its characteristic smooth or rough colonies respectively
ii) The type of capsule i.e., the specific molecular composition of the polysaccharide present in the capsule coded by its genome, thus triggering immune response of a specific type.
4. Which of the following combination is a correct observation for the transformation experiment performed by Griffith?
a) Type IIIS (living) + mouse = dead
b) Type IIIS (heat killed) + mouse = dead
c) Type IIR (living) + mouse = dead
d) Type IIIS (heat killed) + type IIR (living) + mouse = living
Answer: a [Reason:] Type IIIS strain means smooth strain of pneumococci which is virulent in nature. If this strain is injected alive in a mouse it is bound to develop the disease and thus fall dead.
5. Fredrick Griffith’s experiment involving Streptococcus pneumoniae lead to the discovery of____________
a) DNA as genetic material
b) RNA as genetic material
c) Protein as genetic material
d) Transforming principle
Answer: d [Reason:] When heat killed virulent (smooth) type bacteria were injected in the mouse along with the living avirulent (rough) type of bacteria the mouse developed the disease, which was an unlikely result. Moreover when the bacteria were isolated from the infected mouse they were found to be of the virulent type. Thus, it was believed that some factor from the heat killed virulent type bacteria transformed the avirulent to virulent type bacteria which was known to be as the transforming principle.
6. Definite results proving DNA to be genetic material was given by __________
a) Fredrick Griffith
b) Hershey and Chase
c) Avery, Macleod and MacCarty
d) Meselson and Stahl
Answer: c [Reason:] Avery, Macleod and MacCarty in their experiment with pneumococcus strains type IIIS and type IIR they performed three separate experimental setups:-
i) Using DNase to degrade DNA
ii) Using RNase to degrade RNA
iii) Using Protease to degrade proteins
Thus, when the separate combinations were injected into mice respectively the transforming principle was found to be DNA.
Note- Only these three enzymes were used as only their subsequent substrates were the main factors which were in conflict to be the genetic material.
7. Which of the following statements regarding the mechanism of transformation in Bacillus subtilis is false?
a) A competent bacteria contains a DNA receptor/translocation complex
b) While translocation of exogenous DNA, both strands gets passage into the cell
c) While translocation of exogenous DNA, only one strand gets passage into the cell
d) The exogenous DNA recombines and incorporates itself in the chromosome of the recipient cell.
Answer: b [Reason:] During the translocation of exogenous duplex DNA, one of its strands gets degraded by deoxyribonuclease (DNase) in the periplasmic space, thus allowing passage to only single stranded DNA molecule into the cell.
8. What were the main criteria taken under consideration for the experiment by Hershey and Chase?
a) DNA contains phosphorus, protein contains sulfur
b) Protein contains phosphorus, DNA contains sulfur
c) Both DNA and protein contains phosphorus and not sulphur
d) Both DNA and protein contains sulfur and not phosphorus
Answer: a [Reason:] DNA contains phosphorus in phosphodiester linkage and do not contain any sulphur. DNA is primarily composed of phosphate, nucleotide bases and deoxyribose sugar. Whereas proteins are composed of sulfur containing amino acids.
9. What combination of radiolabelling is correct in case of Hershey and Chase’s demonstration of DNA as genetic material in T2 bacteriophage?
a) 31P, 35S
b) 31P, 32S
c) 31P, 14C
d) 31P, 12C
Answer: a [Reason:] The two factors considered by Hershey and Chase were the simple composition of T2 bacteriophage, that is, DNA and protein. Thus radiolabelling of only one element was required. 32P and 35S being the two radioactive isotopes, therefore, the combination of either 32P, 32S or 31P, 35S was used.
10. Recently, scientists have developed a procedure in which protoplasts of E. coli could be directly infected with the phage DNA. This process is termed as_________
Answer: c [Reason:] Transfection is the process of introduction of exogenous, naked and purified nucleic acid in any organism. In case of transformation the introduction of nucleic acid is via a plasmid or any such vector. Transduction is the process of introduction of nucleic acid via viral vector. Lastly mutation is the alteration of nucleic acid at its composition level, is has nothing to do with DNA transfer.
11. According to Mendel’s experiments gene carry genetic information from parents to offspring’s. On which plant did he perform his experiments?
a) Oryza sativa
b) Pisum sativam
c) Allium cepa
d) Vitis vinifera
Answer: b [Reason:] As we know, Gregor Mendel performed his experiments on garden pea plants whose scientific name is Pisum sativam. The other three scientific names Oryza sativa, Allium cepa and Vitis vinifera belongs to rice, onion and grapes respectively.
12. Plasmids are ____________
a) Double stranded DNA
b) Circular double stranded DNA
c) Linear double stranded DNA
d) Supercoiled circular double stranded DNA
Answer: b [Reason:] Found mostly in prokaryotic cells, plasmids are extra chromosomalauto replicating small circular double stranded DNA which generally carries certain genes for the advantage of bacteria, such as, antibiotic resistance.
13. What stores the genetic information in DNA?
c) Nitrogenous base
Answer: c [Reason:] Genetic information is stored in the sequence of nitrogenous base as they are of four types and they are A, T, G and C. The order of their occurrence decides the mRNA sequence which in turn codes amino acids and leading to synthesis of proteins. Sugars and phosphates form the backbone and are common to all, whereas polymerase helps in the replication of DNA template.
14. DNA always carries the genetic information for all organisms.
Answer: False [Reason:] There are certain viruses in which RNA carries the genetic information. They are known as RNA viruses, for example- retroviruses, influenza virus, etc.
15. What is the major type of RNA produced in the cell required for translation?
Answer: c [Reason:] Generally, rRNA is of 4 types 28S, 5.8S, 18S and 5S. These subunits comprises the 80% of the total amount of RNA produced in a cell and are majorly required for translation as these rRNAs along with proteins form ribosomes which reads the codes for translation.
Molecular Biology MCQ Set 2
1. For the translation to be initiated which of the following does not occur?
a) Ribosome recruitment to the mRNA
b) Positioning of ribosome on ‘GUG’
c) Addition of charged tRNA to the A site
d) Binding of the large and small subunits of the ribosome
Answer: c [Reason:] For translation to be successfully initiated, three events must occur. First, the ribosome must be recruited to the mRNA. Second, a charged tRNA must be placed into the P site of the ribosome. Third, the ribosome must be precisely positioned over the start codon (AUG or GUG).
2. Changing in the positioning of the ribosome leads to the synthesis of different proteins.
Answer: a [Reason:] The correct positioning of the ribosome over the start codon is critical, because this establishes the reading frame for the translation of the mRNA. Even a one-base shift in the location of the ribosome would result in the synthesis of a completely unrelated polypeptide.
3. What kind of interactions results in the binding of mRNA and the ribosome?
a) Hydrophobic interaction
b) Hydrophilic interaction
c) Ionic interaction
d) Hydrogen bonding
Answer: d [Reason:] The assembly of the ribosome on an mRNA occurs one subunit at a time. First, the small subunit associates with the mRNA. For prokaryotes the association of the small subunit with the mRNA is mediated by base pairing interactions, that is hydrogen bonding, between the ribosome binding site and the 16S rRNA.
4. In case of prokaryotes the first tRNA enters the ribosome in the ____________
a) A site
b) P site
c) E site
d) Already attached to the mRNA before ribosome association
Answer: b [Reason:] Typically charged tRNAs enter the ribosome in the A site and only reach the P site after a round of peptide bond synthesis. During initiation, however, a charged tRNA enters the P site directly.
5. The initiation of translation requires a special type of tRNA to bind to the ORF of the mRNA.
Answer: a [Reason:] When the polypeptide chain is elongated the aminoacyl tRNA enters the A site. In case of initiating the synthesis of polypeptide it can only be done by the initiator tRNA which contains the anticodon for the start codon on the mRNA.
6. The first amino acid to be incorporated in the prokaryotic polypeptide is ____________
c) N-formyl methionine
d) N-acyl valine
Answer: c [Reason:] The start codons being either AUG or GUG codes for methionine or valine respectively, but neither of them is attached to the initiator tRNA. Instead, it is charged with a modified form of methionine, N-formyl methionine that has a formyl group attached to its amino group.
7. Deformylase has a crucial role to play in the formation of mature polypeptide chains in prokaryotes.
Answer: a [Reason:] In case of prokaryotes the initiator amino acid is N-formyl methionine. This formyl group is removed from the polypeptide as we know that no polypeptide starts with a formyl group. This is done by the enzyme deformylase.
8. How many translation initiation factors work hand in hand to initiate prokaryotic translation?
Answer: c [Reason:] The initiation of prokaryotic translation commences with the small subunit and is catalyzed by the three initiation factor. They are of three types IF1, IF2 and IF3.
9. Which of the following is a GTPase?
Answer: b [Reason:] GTPase is a protein that binds and hydrolyses GTP. IF2 is a GTPase that binds to the three key components of translation, the small subunit, the charged initiator tRNA and the IF1 and facilitates their subsequent association.
10. Which of the following is not the function of IF3?
a) Binds to the small subunit
b) Helps dissociate the 70S ribosome after translation
c) Helps associate the large and small subunits during translation
d) Blocks the binding of the charged tRNA molecule to the small subunit
Answer: c [Reason:] IF3 binds to the small subunit and blocks it from reassociating with the large subunit. It also blocks the binding of small subunit to the charged tRNAs as it is critical for the small subunit to be free for initiating translation and binding with the correct initiator tRNA.
11. When the large and small subunit of ribosome associate which of the following is not found within the complex?
Answer: c [Reason:] When the start codon and fMet-tRNAifMet base-pair, the small subunit undergoes a change in conformation. This altered conformation results in the release of IF3. In the absence of IF3, the large subunit is free to bind to the small subunit with its cargo of IF1, IF2, mRNA and fMet-tRNAifMet.
12. Binding of the large subunit to the small subunit triggers a cascade of reactions. Arrange them in the order of their occurrence.
i. Release of IF1
ii. Hydrolysis of IF2-GTP
iii. Formation of vacant A site
iv. Reduced affinity of IF2-GDP
a) ii, iv, i, iii
b) iii, i, ii, iv
c) iv, ii, i, iii
d) i, iii, ii, iv
Answer: a [Reason:] The binding of the large subunit stimulates the GTPase activity of IF2-GTP, causing it to hydrolyze GTP. The resulting IF2-GDP has reduced affinity for the ribosome and the initiator tRNA leading to the release of IF2-GDP as well as IF1 from the ribosome. Thus, the net result of initiation is the formation of an intact (70S) ribosome assembled at the start site of the mRNA with fMet-tRNAifMet in the P site and an empty A site.
Molecular Biology MCQ Set 3
1. How many types of vector systems are used in eukaryotic plants?
Answer: c [Reason:] Three types of vector system have been used with varying degree of success:
i) Vector based on naturally occurring plasmid of Agrobacterium
ii) Direct gene transfer using various types of plasmid DNA
iii) Vectors based on plant viruses.
2. Ti plasmid is the only known plasmid of plant.
Answer: b [Reason:] There is no naturally occurring plasmid known in higher plants. The first known plasmid that is able to express itself in the plant is a bacterial plasmid known as Ti plasmid from the bacterium, Agrobacterium tumefaciens.
3. The Ti is referred to as _____________
a) Translocation inducing
b) Transfer inducing
c) Tumor inducing
d) Translation inducing
Answer: c [Reason:] The Ti of Ti plasmid refers to as tumor inducing plasmid. It is found in the bacterium Agrobacterium tumefaciens which when infects a plant causes tumor.
4. In which of the plant is the crown gall disease caused?
Answer: c [Reason:] The A. tumefaciens causes a crown gall disease in many dicotyledonous plant species. Crown gall occurs when a wound on the stem allow A. tumefaciens bacteria to cause tumor in the crown region of the stem.
5. The size of the Ti plasmid is around ___________
a) Less than 100 kb
b) More than 200 kb
c) Less than 50 kb
d) More than 75 kb
Answer: b [Reason:] The Ti plasmid is a large plasmid that causes the crown gall disease in plants. The size of the Ti plasmid is more than 200 kb in size that contains the genes involved in the infective process.
6. The part of the bacterial plasmid incorporated in the plant genome includes ___________
a) Virulence region
b) T DNA, 25 base pair repeats
c) Entire plasmid
d) Only T-DNA
Answer: d [Reason:] The remarkable feature of Ti plasmid is that, after infection, a part of the molecule is incorporated into the plant genome. This segment of the Ti plasmid is called the T-DNA or transfer DNA.
7. The size of the T-DNA is around ____________
a) 5 – 10 kb
b) 10 – 20 kb
c) 15 – 30 kb
d) 25 – 30 kb
Answer: c [Reason:] The T-DNA contains the genes required for causing the crown gall disease. It is around 15 – 30 kb in length depending upon the strain of the bacterium.
8. The T-DNA contains how many genes for cancer in the plant?
Answer: d [Reason:] The T-DNA contains all the genes that are responsible for the cancerous property in the transformed cells. They are overall 8 in number which resides in the T-DNA that gets incorporated in the plant genome.
9. The sole purpose of the infection of plants by the Ti plasmid is formation of cancer.
Answer: b [Reason:] The formation of cancer is not the sole purpose of infection in the plants. These genes also direct the synthesis of unusual compound called opines, that bacteria use as nutrients.
10. How many process are used to form the chimera Ti plasmid?
Answer: b [Reason:] There are 2 processes that are used to form the chimera Ti plasmid. They are the binary vector strategy and the co-integration strategy.
Molecular Biology MCQ Set 4
1. The main problem in the formation of the chimera Ti plasmid is _____________
a) It is unstable
b) Large size
c) High virulence
d) No unique restriction site
Answer: d [Reason:] The main problem is that a unique restriction site is impossibility with a plasmid 200 kb in size. Novel strategies have to be developed for inserting new DNA into the plasmid.
2. The binary vector strategy uses two different vectors for the production of the single chimera molecule by recombination.
Answer: b [Reason:] The co-integration strategy uses an entirely new plasmid, based on an E. coli vector, but carrying a small portion of the T-DNA. The homology between the new molecule and the Ti plasmid means that if both are present in the same A. tumefaciens cell, recombination can integrate the E. coli plasmid into the T-DNA region.
3. Which of the following does not play any role in the infection of plant cell by the Ti plasmid of A. tumefaciens?
b) Virulence region
c) Host specificity region
d) 25 base pair repeats
Answer: a [Reason:] The host specificity region is required for the specificity of the cells to be infected. Virulence region is required for infection of the cells. The 25 base pair repeats are required for the transfer of the T-DNA from the bacterial cell to the plant cell. The T-DNA causes the cancer but plays no role in the process of infection.
4. Disarming of Ti plasmid is ____________
a) Removal of the Virulence region
b) Removal of the 25 base pair repeats
c) Removal of the T-DNA
d) Removal of the Host specificity region
Answer: c [Reason:] Disarming of Ti plasmid is done by the removal of the T-DNA. As T-DNA contains genes for cancer it is important to remove that part so that cancer does not occur in the transformed cell. And the gene of interest is placed within the 25 base pair repeats in place of the T-DNA to ensure correct transfer and integration of the DNA insert.
5. Insertion of the T-DNA into the host genome depends on the T-DNA itself.
Answer: b [Reason:] The T-DNA contains the genes to cause cancer and does not play any role in the incorporation of itself in the host genome. The 25 base pair repeats flanking on either sides of the T-DNA is responsible for the incorporation of the T-DNA into the host genome.
6. The Ti cloning vector pBIN19 contains which of the following genes?
a) Tetracycline resistant gene
b) Stress resistant gene
c) Kanamycin resistant gene
d) Ampicillin resistant gene
Answer: c [Reason:] The Ti cloning vector pBIN19 contains kanamycin resistant gene and a lac Z gene. These are incorporated within the 25 base pair repeats and contain a number of cloning sites. These are used for the transformant screening purposes.
7. Transformants of pBIN19 are selected by growing them in solid agar media with ampicillin.
Answer: b [Reason:] Transformants of pBIN19 are selected by growing them in solid agar media with kanamycin. This is because Ti cloning vector pBIN19 contains kanamycin resistant gene and a lac Z gene. If the cells are transformed they will be resistant to the kanamycin present in the culture media.
8. Introduction of new gene through Ti plasmid should be done in ____________
a) Fully grown plant
b) A sapling
c) Germ cells
Answer: d [Reason:] Introduction of new gene through the Ti plasmid should be done in a way that every cell of the plant is transformed. The simplest solution is to infect not the mature plant but the culture of plant cells, callus, or protoplasm in a liquid medium.
9. How the host specificity is achieved by the specificity gene of the Ti plasmid?
a) Opine released by a wounded plant
b) Acetosringone released by bacteria
c) Acetosringone released by wounded part of the plant
d) Opine released by bacteria
Answer: c [Reason:] Acetosringone is released by wounded part of the plant. This chemical is identified by the host specificity gene of the bacteria A. tumefaciens which then infects the plant through the wounded area.
Molecular Biology MCQ Set 5
1. Enhancers that can function over very long distances known as ________________
a) Cis-acting enhancers
b) Trans-acting enhancers
d) Allosteric enhancers
Answer: b [Reason:] Enhancers can function over very long distances, sometimes even from different chromosomes. This process, termed Transvection, has been best studied in Drosophila and involves trans-acting enhancers from one gene regulating the expression of its homolog on a separate chromosome.
2. Enhancers were first identified in __________
a) Rous sarcoma virus
c) SV40 virus
d) E. coli
Answer: c [Reason:] Enhancers were first identified in SV40 virus. In addition to a TATA box another two 72-base pair repeats located farther upstream are required for efficient transcription from this promoter known as the enhancer sequence.
3. Enhancers in which of the following positions is unable to enhance the process of transcription?
a) Just upstream the promoter
b) Several kilobases downstream promoter
c) Several kilobases upstream promoter
d) Within the promoter site
Answer: d [Reason:] Without an enhancer, the gene is transcribed at a low basal level. The enhancer is active not only when placed just upstream but also when inserted upto several kilobases upstream or downstream from the transcription start site.
4. Immunoglobulin enhancers are active in which of the following cell types?
a) B lymphocytes
b) T lymphocytes
Answer: a [Reason:] A well studied example is the enhancer that controls the transcription of immunoglobulin genes in B lymphocytes. Gene transfer experiments have established that the immunoglobulin enhancer is active in lymphocytes, but not in other types cells. Thus, this regulatory sequence is at least partly responsible dor tissue-specific expression of the immunoglobulin genes in the appropriate differentiated cell type.
5. Enhancers are a unit of multiple functional sequence elements that together regulates the expression of a gene.
Answer: a [Reason:] An important aspect of enhancers is that they usually contain multiple functional sequence elements that bind different transcriptional regulatory proteins. These proteins work together to regulate gene expression. The immunoglobulin heavy-chain enhancer, for example, spans approximately 200 base pairs and contains at least nine distinct sequence elements that serve as protein-binding sites.
6. The enhancers are gene specific regardless of the presence of insulators.
Answer: b [Reason:] Although DNA looping allows enhancers to act a considerable distance from promoters, the activity of any given enhancer is specific for the promoter of its appropriate target gene. This specificity is maintained by insulators, which divide chromosomes into independent domains and prevent enhancers from acting on promoters located in an adjacent domain.
7. Transcriptional activators consist of __________ domains.
Answer: b [Reason:] Transcriptional activators consist of two independent domains. One region of the protein specifically binds DNA; the region stimulates transcription by interacting with other proteins, including mediator or other components of the transcriptional machinery.
8. The activation domain of activators recruits transcriptional factors to proper sites on the DNA.
Answer: b [Reason:] The basic function of the DNA-binding domain is to anchor the transcription factor to the proper site on DNA. the activation domain then independently stimulates transcription through protein-protein interaction.
9. How many mechanisms are present to enhance transcription by the enhancer molecules?
Answer: a [Reason:] The activation domains of eukaryotic transcription factors stimulate transcription by two distinct mechanisms. First, they interact with mediator proteins and general transcription factor to recruit RNA polymerase and facilitate the assembly of a transcription complex on the promoter. Eukaryotic transcription factors interact with a variety of co-activators that stimulate transcription by modifying chromatin structure.
10. Which of the following steps for the production of RNA is not regulated by enhancers?
d) RNA processing
Answer: c [Reason:] It is also important to note that activators not only regulate the initiation of transcription. Elongation and RNA processing can also be regulated, both by direct modulation of the activity of RNA polymerase and by the effects on chromatin structure.
Synopsis and Project Report
You can buy synopsis and project from distpub.com. Just visit https://distpub.com/product-category/projects/ and buy your university/institute project from distpub.com