Molecular Biology MCQ Set 1
1. Pick the odd one out.
c) Jumping genes
d) Insertion sequences
Answer: a [Reason:] Introns are intervening sequences found within the gene itself which divides the gene into several exons and are spliced off from the hnRNA to produce the intact mRNA. The rest, cassettes, jumping genes and insertion sequences are the different names for transposable elements and are short sequences of DNA that have the ability to move from one location to the other within the genome.
2. Transposable elements are often found once in a gene and have a unique sequence for its own.
Answer: b [Reason:] Transposable elements have the ability to move from one location to the other within the genome and often make duplicate copies of them in this process. The process by which they are copied and inserted into a new site in the genome is called transposition.
3. Junk DNA are also a type of transposable element that can jump from one location to another within the genome.
Answer: b [Reason:] Transposable elements are considered as selfish DNA which replicates but is of no use to the host cell. Defective selfish DNA which is no longer capable of either moving or expressing its gene is termed as junk DNA.
4. On the mechanism of the movement of transposable elements they are of ___________ types.
Answer: b [Reason:] On the mechanism of the movement of transposable elements they are of two types. They are transposons or type II elements and retrotransposons or type I elements.
5. The type of mutation that is imposed by transposons is ___________
a) Silent mutation
b) Reverse mutation
c) Polar mutation
d) Frame shift mutation
Answer: c [Reason:] Transposable elements may contain transcription and/or translation termination signals. This blocks the expression of other genes downstream of the insertion site. This one-way mutational effect is known as polar mutation.
6. Transposons were first discovered in ___________
Answer: d [Reason:] The first transposons were discovered in maize in 1948. For this discovery Barbara McClintock was awarded a Nobel Prize in 1983.
7. Transposons are able to move by ___________ mechanism(s).
Answer: b [Reason:] Transposons can move by conservative or replicative process. In conservative process the copy number of the transposons remains same but in replicative process the copy number increases.
8. The transposons are more abundant than retrotransposons in higher eukaryotes.
Answer: b [Reason:] Transposable elements make upto 10% of higher eukaryotic genomes. Although retrotransposons are much more abundant in higher eukaryotes such as LINE, SINE, etc.
9. Silencing of transposons cannot be achieved by which of the following mechanisms?
b) DNA methylation
c) DNA phosphorylation
d) Chromatin remodeling
Answer: c [Reason:] As the movement of transposons can be destructive it is important to make the transposons silent to maintain the stability of the genome. They are kept inactive by epigenetic mechanisms such as RNA interference, DNA methylation and chromatin remodeling.
10. Pick the odd one out.
a) T4 phage
b) Mu phage
c) Coli phage
d) T2 phage
Answer: b [Reason:] The Mu is an unusual E. coli bacteriophage. It is both a virus and a transposon. It is a temperate phage having both lytic and lysogenic growth cycles. It integrates into the bacterial genome by transpositional mechanism rather than site-specific recombination.
Molecular Biology MCQ Set 2
1. What is the other name of DSB repair pathway?
a) RecBAD pathway
b) RecBCD pathway
c) RecABD pathway
d) RecDCB pathway
Answer: b [Reason:] The major DSB repair pathway is also known as the RecBCD pathway. Homologous recombination in eukaryotic cells and the proteins are involved in this pathway.
2. The RecBCD pathway involves the use of ligases.
Answer: a [Reason:] Many recombination proteins are involved in the RecBCD pathway. Beside these proteins topoisomerases and ligases are also recruited to play a critical role in the process of gene exchange.
3. What is branch migration?
a) Break and reformation of identical base pairs
b) Formation of lesion
c) Formation of heteroduplex DNA
d) Dissolution occurs
Answer: a [Reason:] Each time holiday junction moves, base pairs are broken in the parental DNA molecules and identical base pairs are created in the recombinants. This process of sliding of the holiday junction forming identical base pairs in the heteroduplex DNA is known as branch migration.
4. What is resolution?
a) Cleavage of holiday junction
b) Regeneration of duplex DNA molecule
c) Exchange of DNA fragments
d) Heterochromatin structure formation
Answer: b [Reason:] In the process of homologous recombination, DNA molecules are regenerated by the process of genetic exchange. This is known as the process of resolution.
5. How many processes are present to achieve resolution?
Answer: b [Reason:] Resolution can be achieved by two different processes. They are, cleavage of the holiday junction or by the process of dissolution.
6. Which of the following promotes strand exchange?
a) DBS formation
b) Heteroduplex formation
c) Strand invasion protein
d) Branch migration
Answer: c [Reason:] Strand invasion initiates the exchange of DNA strands between the two parental DNA. The enzyme that catalyses strand invasion are called strand exchange proteins.
7. Which type of DNA is connected by a holiday junction?
a) Homologous DNA duplex
b) Heteroduplex DNA
c) Mutated DNA
d) Asymmetric DNA
Answer: a [Reason:] Finishing of recombination usually requires resolution of the holiday junction by cutting the DNA strand. Two homologous DNA duplex is connected by a single holiday junction.
8. From what the splice recombination product is formed?
a) Mutated DNA
b) Cross over DNA
c) Spliced DNA
d) Recombined DNA
Answer: c [Reason:] If the two DNA strands are cut and then covalently joined the resulting DNA molecules are referred to as the splice recombination product. Thus the splice recombination product is formed by two spliced DNA parents.
9. What is a crossover product?
a) Mutated DNA
b) Reassorted DNA
c) Crossover DNA
d) Spliced DNA
Answer: b [Reason:] The splice products result in the reassortment of genes that flank the site of recombination. This type of recombinant product is known as the crossover product.
10. What is a patch product?
a) Regions of hybrid DNA
b) Crossover DNA
c) Mutated DNA
d) Recombinant DNA
Answer: a [Reason:] After resolution and covalent joining of the strands the resulting DNA contains a region of hybrid DNA. This is known as a patch thus the DNA produced is known as the patch product.
11. Patch products are non-crossover products.
Answer: a [Reason:] In patch product formation recombination does not result in the reassortment of the genes flanking the site of initial cleavage. These molecules are therefore known as the non-crossover products.
12. Single strand break repair pathway involves genetic exchange.
Answer: b [Reason:] Homologous recombination is involved in the pathway to repair double strand breaks. This pathway is often initiated by the presence of double strand breaks in the DNA.
13. The double strand breaks, the DNA cleaving enzyme generates ____________
a) Double stranded DNA
c) Single stranded DNA
d) Single stranded over hangs
Answer: d [Reason:] DNA cleaving enzyme sequentially degrades the broken DNA on one strand. This produces a stretch of single stranded DNA known as overhanging DNA molecule.
14. In the DSB pathway the 5’-terminal of the broken strand are degraded.
Answer: a [Reason:] In some cases both strands at the double strand break region are processed. Although only the 5’-terminal ends undergo degradation causing the formation of a 3’-overhanging region.
15. What is gene conversion?
a) Change in the DNA sequence
b) Non-reciprocal step in the DSB repair pathway
c) Break of DNA strands
d) Branch migration
Answer: b [Reason:] If the two original DNA duplex were not identical in sequence near the site of break it leads to gene conversion. In the recombination event if the sequence information is lost from the DNA molecule the resultant DNA sequence is replaced to maintain the integrity of the DNA. This non-reciprocal step in DSB repair system is known as gene conversion.
Molecular Biology MCQ Set 3
1. The transposable elements of bacteria are generally retrotransposons.
Answer: b [Reason:] Most transposable elements in bacteria transpose directly as DNA. In contrast, most mobile elements in the eukaryotes are retrotransposons.
2. The IS elements can be identified by the presence of __________
a) Antibiotic resistance gene
b) Endonuclease cleavage site
c) 50 bp inverted repeat
d) Integrase site
Answer: c [Reason:] The IS elements are the most common type of transposons present in the bacterial genome. An inverted repeat of usually 50 bp is the hallmark of such elements present at each end of an insertion sequence.
3. Direct repeats in the IS element are present __________
a) Within the transposon
b) Upstream the inverted repeat
c) Within the inverted repeat
d) Downstream the inverted repeat
Answer: c [Reason:] A set of two direct repeats flank the IS element on either side. They are not transposed with the insertion element and are found within the inverted repeat sequence on either sides.
4. The direct repeat within the IS element has a length of __________
a) 20 bp
b) 11-15 bp
c) 5-11 bp
d) 3-7 bp
Answer: c [Reason:] The direct repeat on the either side of the insertion element is a hallmark of IS element. They are 5 to 11 bp long sequences immediately adjacent to the both ends of the inserted element.
5. The enzyme that catalyzes the transposition of an IS element is called __________
Answer: a [Reason:] The enzyme that catalyzes the transposition of an IS element is called a transposase. The transposase recognizes the inverted repeats and moves the segment of DNA bound by them from one site to another.
6. Transposase restriction mechanism of IS element restricts the transposon and the target DNA in a combination of which of the following?
a) Blunt end cut for transposon and sticky end cut for target DNA
b) Blunt end cut for both transposon and target DNA
c) Sticky end cut for transposon and blunt end cut for target DNA
d) Sticky end cut for both transposon and target DNA
Answer: a [Reason:] The transposase is encoded by the IS element itself. The transposase on production makes blunt-ended cut in the transposable element and staggered/sticky-ended cut in the target DNA.
7. Which of the following functions is not performed by transposase?
a) Restriction of the IS element
b) Integration of the transposon
c) Formation of the RNA intermediate
d) Restriction of the host genome
Answer: c [Reason:] The transposase performs three functions in the transposition of the IS element. First, it precisely excises the IS element from the transposon. Second, it makes staggered cut in the target DNA segment. Last, it ligates the IS element within the target site.
8. The central block of the composite transposable element consists a gene for __________
b) Antibiotic resistance
Answer: b [Reason:] Composite transposons contain one or more protein-coding genes in addition to those required for transposition. Many of the well known composite transposons carry genes for antibiotic resistance. Two of the best known elements are Tn9 (chloramphenicol resistance) and Tn10 (tetracycline resistance).
9. Which of the following is a non-composite transposon?
Answer: c [Reason:] Tn5, Tn10 and Tn9 are examples of composite transposons. Tn3 and Tn21 are examples of non-composite transposons.
10. Which of the following about the P-element is false?
a) 25 bp inverted repeats
b) 2907 bp long element
c) Encodes for transposase
d) DNA transposon
Answer: a [Reason:] The P-element in Drosophila is one of the best examples of transposable elements in eukaryotes. This element is 2907 bp long and features a 31 bp inverted repeat at each end. This element encodes a transposase. P-element does not utilize an RNA intermediate during transposition and are able to insert at many different positions.
11. Hybrid dysgenesis is caused by which of the following transposable element?
a) Non-composite transposon
Answer: c [Reason:] Interbreeding of two incompatible strains of Drosophila displays dysgenic traits. These include a series of defects including mutations, chromosomal aberrations, distorted segregation at meiosis and sterility. This genetic phenomenon is called hybrid dysgenesis.
Molecular Biology MCQ Set 4
1. Dysgenesis occurs in every cell of the Drosophila.
Answer: b [Reason:] Hybrid dysgenesis is principally a phenomenon of the germ cells. In crosses involving the P-M system, the F1 generation hybrid flies have normal somatic tissues. Their gonads, however, do not develop.
2. P-strains of the Drosophila have active P-elements.
Answer: b [Reason:] P-elements reside passively in the P-strains because they express a repressor of transposition that keeps the transposons silent. The repressor is also inherited maternally through the egg cytoplasm thus the offspring thus produced will be normal.
3. Hybrid dysgenesis occurs in which of the following cases?
a) Crossing between P-female and P-male
b) Crossing between M-female and M-male
c) Crossing between P-female and M-male
d) Crossing between M-female and P-male
Answer: d [Reason:] Hybrid dysgenesis is observed when crossing P-males with M-females. The eggs of M-strain do not contain the repressor protein that prevents the transposition. This leads to transposition of the P-elements from the sperm of the P-strain leading to hybrid dysgenesis in the progeny.
4. Retrotransposons transposes through the RNA intermediates.
Answer: a [Reason:] The elements that transpose through an RNA intermediate utilizing a reverse transcriptase are called the retrotransposon. They are primarily found in the eukaryotic organisms.
5. Retrotransposons are of how many types?
Answer: b [Reason:] Retrotransposons can be divided into two types on the basis of presence of long terminal repeats. If the LTR is present on either side of the transposon they are known as LTR-transposons and if they are absent then it is known as non LTR-Transposons.
6. LTR-retrotransposons are abundantly found in __________
Answer: c [Reason:] LTR-retrotransposons are also known as viral retrotransposons. They are most abundantly found in lower eukaryotes such as yeast and Drosophila.
7. LINE and SINE are examples of __________
a) LTR retrotransposon
b) Composite transposon
c) Non-LTR retrotransposon
Answer: c [Reason:] LINE and SINE are the most abundantly found non LTR-retrotransposons in the vertebrates. They lack the terminal LTR sequences and have a stretch of A/T at one end.
8. Which of the following about the L1 LINE family is correct?
a) Highly accumulated in autosomes
b) About 37% of the total human genome
c) Presence of short direct repeats on either end
d) Approximately 2 kb in length
Answer: c [Reason:] The corrected characteristics are:
i) Highly accumulated in sex chromosomes than autosomes
ii) Makes up about 17% of the total human genome
iii) Approximately 6 kb in length.
9. The consensus sequence of the L1 LINE family has three open reading frames.
Answer: b [Reason:] The consensus sequence co the L1 LINE family contains two long open reading frames. ORF1 encodes an RNA-binding protein and ORF2 encodes a protein with both endonuclease and reverse transcriptase activity.
10. SINES are transcribed by __________
a) Klenow fragment
b) Polymerase I
c) Polymerase II
d) Polymerase III
Answer: d [Reason:] SINEs are transcribed by RNA polymerase III. They require the machinery of the LINE elements as they are around 100-400 bp long non-autonomous sequences.
11. The SINE commonly contains the restriction site for the endonuclease __________
a) Pvu I
b) Spa I
c) Alu I
d) Xma I
Answer: c [Reason:] The most common SINE elements are Alu elements in the human genome. These are about 300 bp long sequences that are scattered throughout the human genome, constituting of almost 10% of the total DNA.
Molecular Biology MCQ Set 5
1. Which of the following is not true about nucleotides?
a) Monomeric units
b) Ubiquitous substances
c) Energy rich molecules
d) Non enzymatic molecules
Answer: d [Reason:] The nucleotides are known to be enzymatic molecules. For example, ribozymes are catalytic molecules. Also certain derivatives of nucleotides such as ATP, coenzyme A, nicotinamide adenine dinucleotide, etc., are important catalytic molecules that participate in various enzymatic processes.
2. The nitrogenous base is covalently linked to the _________ carbon of the pentose sugar.
Answer: a [Reason:] Nucleotides are phosphate esters of a five carbon sugar, either ribose of 2’-deoxyribose. The nitrogenous base is covalently linked to the C1 carbon of this pentose sugar to form the nucleotide.
3. In which carbon do the deoxyribonucleotides lack an –OH molecule?
Answer: b [Reason:] A deoxyribonucleotides lack an –OH molecule at the C2 position of the ribose sugar ring. This is the substrate for DNA synthesis and is known as the 2’-deoxyribonucleotide.
4. Which of the following is not a part of a nucleotide?
a) Ester linkage
b) Phosphate group
d) Hydrogen bond
Answer: d [Reason:] Nucleotides are phosphate esters of a five carbon sugar, either ribose of 2’-deoxyribose. The nitrogenous base is covalently linked to the C1 carbon of this pentose sugar to form the nucleotide. Hydrogen bond is made by the bases to hold the two strands of DNA together and is not a part of the nucleotide.
5. Which of the following is not a part of a nucleoside?
a) Deoxyribose sugar
b) Glycosidic linkage
Answer: c [Reason:] A nucleoside is the deoxyribose sugar linked to the base with a glycosidic linkage. Addition of a phosphate at the 5’-carbone leads to the formation of the nucleotide.
6. The linkage between the sugar and the base is a covalent linkage.
Answer: a [Reason:] A glycosidic bond is one which links a type of covalent linkage which forms between a sugar molecule and another molecule which may or may not be a sugar. In case of nucleotide/nucleoside the sugar and the base is a glycosidic linkage.
7. Which of the following is not a nucleotide?
Answer: b [Reason:] TMP is not a nucleotide. Thymine is not present in form of Thymidine monophosphate as it is not used in RNAs. Thymine is present in the form of dTMP deoxythymidine monophosphate as it is used in the synthesis of DNA only.
8. Purines and pyrimidines form glycosidic linkage with the purines and pyrimidines at the N6 position.
Answer: b [Reason:] The purines form glycosidic linkage to ribose via their N9 atoms. The pyrimidines form glycosidic linkage to ribose via their N1 atoms.
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