Molecular Biology Techniques MCQ Quiz in मराठी - Objective Question with Answer for Molecular Biology Techniques - मोफत PDF डाउनलोड करा

The correct answer is (i) and (iii) 

Concept:-

Basic Principle of EMSA

• EMSA, also known as a gel shift assay, relies on the principle that DNA-protein complexes are generally larger and migrate more slowly during gel electrophoresis compared to unbound DNA fragments.
• By comparing the migration of a labeled DNA fragment in the presence and absence of various proteins, researchers can infer binding interactions.

Components of EMSA

• Labeled DNA: A specific DNA sequence is labeled, typically with a radioactive or fluorescent tag, to allow detection after electrophoresis.
• Proteins: These can be individual or mixtures of proteins that are suspected to bind to the DNA. 
• Gel Electrophoresis: After incubation of the DNA with the proteins, the mixture is run on a polyacrylamide gel. The gel allows separation based on size and charge.

Interpretation of EMSA Results

• DNA Binding: If a protein binds to the DNA, the mobility of the DNA will decrease. This results in a shifted band on the gel.
• Specificity of Binding: By using multiple combinations of proteins, the specificity of binding can be assessed. For instance, if a protein binds only in the presence of another protein, it might suggest a cooperative interaction or a requirement for a complex formation.
• Comparative Mobility: By comparing the mobility of the DNA in different lanes, researchers can infer whether proteins bind independently, cooperatively, or competitively.

Explanation:

EMSA Results Analysis

• Lane 1: (-A, -B, -C) – No shift in DNA mobility, baseline control.
• Lane 2: (-A, +B, -C) – No shift in DNA mobility, indicating that protein B does not bind to DNA by itself.
• Lane 3: (-A, -B, +C) – No shift in DNA mobility, indicating that protein C does not bind to DNA by itself.
• Lane 4: (+A, -B, -C) – There is a shift in DNA mobility, indicating that protein A binds to the DNA.
• Lane 5: (+A, +B, -C) – There is a shift in DNA mobility, and this shift appears similar to the shift seen with A alone, suggesting that the presence of B changes the DNA binding observed with A alone and Protein B binds to DNA - protein A complex
• Lane 6: (-A, +B, +C) –  No shift in DNA mobility indicating that Protein A possesses the DNA binding motif
• Lane 7: (+A, -B, +C) – There is a shift in DNA mobility, and this shift appears similar to the shift seen with A alone indicating that Protein C does not binds to DNA only when protein A is bound

Conclusion:

• Protein A binds to DNA directly.
• Protein B binds to DNA - protein A complex.
• Protein C does not alter the binding pattern significantly, whether A is present or not.

The correct answer is Option 1 i.e.Treatment of vector with Mung Bean Nuclease followed by treatment with Shrimp Alkaline Phosphatase

Explanation-

The insert is blunt ended. but the vector is digested by EcoRI that will produce sticky end DNA. Blunt end DNA cannot be ligated into sticked ended vector.
Both (insert and vector) have to either blunt ended or sticky ended. The overhangs in the vector after EcoRI digestion can be removed by nuclease and phosphatase will remove the 5'P to prevent the re-ligation of the vector ends.

Klenow DNA polymerase could also be used to convert sticky end into blunt end by filling the gap. But Klenow DNA polymerase has to work on the vector NOT on the insert as the insert is already blunt ended.
So answer will be option 1. Treatment with mung bean nuclease, followed by shrimp alkaliine phosphatase.

Key PointsFor ligating a blunt-ended insert fragment with a vector digested with EcoRI restriction enzyme (GAATTC), it is necessary to remove the 5' phosphate groups from the vector and insert to prevent self-ligation. Blunt-ended fragments lack cohesive (sticky) ends, so modifying the ends appropriately is crucial for successful ligation.

The correct approach involves the following steps:-

• Treatment of vector with Mung Bean Nuclease: Mung Bean Nuclease removes the 3' overhangs or "sticky ends" from the EcoRI-digested vector.
• Followed by treatment with Shrimp Alkaline Phosphatase (SAP): Shrimp Alkaline Phosphatase removes the 5' phosphate groups from both the vector and the insert.

Conclusion-Therefore, the correct option is:Treatment of vector with Mung Bean Nuclease followed by treatment with Shrimp Alkaline Phosphatase.

The  correct answer is Option 2 i.e.It has a mutation in RNAII (primer for replication initiation) and does not carry the rop gene

Concept:

• Plasmids are extra-chromosomal present in bacteria. 
• They are small double-stranded DNA molecule that is usually circular in shape.
• They can exist independently of the host chromosomes, meaning they can auto-replicated. 
• They are present majorly in bacteria but they are also found in yeast and some fungi.
• They have their own replication origin to initiate replication and hence are auto-replicable and can be inherited stably. 
• Plasmids only have a few genes as compared to the genomic DNA of the bacteria. 
• Also, genetic information in the plasmid is not essential for the survival of the host. Bacteria that lack plasmid usually function normally.

Explanation: 

• Replication of ColEl-type plasmids is regulated by two main factors - RNAI and RNAII.
• RNA I is plasmid-specific RNA (RNA I) and it binds to the transcript (RNA II) from the primer promoter resulting in the inhibition in the formation of the primer required for DNA replication.
• So, if RNAII is mutated then RNAI cannot bind to it and it connects and stops DNA replication initiation in the bacteria. 
• The repressor of primer (Rop) is a small dimeric protein that plays an important role in the mechanism of controlling the copy number of plasmid by increasing the affinity between two complementary RNAs.
• pUC18 does not have Rop protein. 
• Rop protein is a negative regulator of plasmid replication.
• So, if it is removed then it will 'loosen the control' on plasmid replication leading to increased copy number.
• So, pUC18 have a high copy number because both of the above condition is fulfilled i.e., mutation in RNAII and absence of Rop. 

Hence, the correct answer is Option 2 i.e. It has a mutation in RNAII (primer for replication initiation) and does not carry the rop gene . 

Key Points

• Molecular markers are DNA or RNA sequences that can be used to identify and track genetic variation in individuals, populations, or species.
• They are used to understand genetic diversity, inheritance patterns, and to identify genes and gene functions.
• Molecular markers can be classified into two main types:
• DNA-based markers:
  • These markers are based on differences in DNA sequences between individuals.
  • Examples of DNA-based markers include restriction fragment length polymorphisms (RFLPs), amplified fragment length polymorphisms (AFLPs), microsatellites, and single nucleotide polymorphisms (SNPs).
• RNA-based markers:
  • These markers are based on differences in RNA sequences between individuals.
  • Examples of RNA-based markers include differential display, RNA fingerprinting, and gene expression profiling.

Important Points

• In molecular marker analysis, markers are classified as either dominant or co-dominant based on their inheritance patterns and the information they provide about heterozygosity.
• P. SSR (Simple Sequence Repeat) and RFLP (Restriction Fragment Length Polymorphism) markers (CORRECT)
  • They are classified as co-dominant markers.
  • Co-dominant markers provide information about heterozygosity, meaning they can distinguish between two different alleles at a specific genomic locus.
  • SSR markers are based on variations in the number of repeats of short DNA sequences, while RFLP markers involve variations in DNA fragment lengths resulting from specific DNA cleavage sites recognized by restriction enzymes. In co-dominant markers, both alleles are detected and can be distinguished from each other.
  • The co-dominant nature of these markers allows for the identification of heterozygotes and homozygotes at a particular locus.
• Q. SSR and RAPD (Random Amplified Polymorphic DNA) markers are also co-dominant markers. (INCORRECT)
  • RAPD markers involve amplification of random DNA segments using short arbitrary primers.
  • Similar to SSR markers, RAPD markers can distinguish between different alleles at a specific genomic locus, making them co-dominant.
• R. RAPD and RFLP markers are classified as dominant markers. (INCORRECT)
  • Dominant markers do not provide information about heterozygosity and can only distinguish between the presence or absence of a specific allele.
  • In the case of RAPD markers, the presence or absence of amplified DNA fragments indicates the presence or absence of a specific allele.
  • RFLP markers, on the other hand, are based on DNA fragment length differences resulting from restriction enzyme digestion, where the presence or absence of specific fragments indicates the presence or absence of a particular allele.
• S. AFLP (Amplified Fragment Length Polymorphism) and RAPD markers are also classified as dominant markers. (CORRECT)
  • AFLP markers involve the selective amplification of specific DNA fragments using restriction enzymes and adapter ligation.
  • Similar to RAPD markers, AFLP markers can only distinguish between the presence or absence of a specific allele, making them dominant.

Key Points:

• Restriction fragment length polymorphism (abbreviated RFLP) refers to differences (or variations) among people in their DNA sequences at sites recognized by restriction enzymes. Such variation results in different sized (or length) DNA fragments produced by digesting the DNA with a restriction enzyme. RFLPs can be used as genetic markers, which are often used to follow the inheritance of DNA through families.To begin with, DNA is extracted from blood, saliva or other samples and purified.

Important Points:Figure 1: RFLP  protocol 

RFLP is protocol can be accomplished in four steps:

1. DNA Extraction
2. DNA Fragmentation
3. Gel Electrophoresis
4. Visualization of Bands

• DNA Extraction: 
  • ​ DNA is extracted from blood, saliva or other samples and purified.
• DNA Fragmentation: 
  • ​The purified DNA is digested using restriction endonucleases.
  • The recognition sites of these enzymes are generally 4 to 6 base pairs in length.
  • The shorter the sequence recognized, the greater the number of fragments generated from digestion.
  • For example, if there is a short sequence of GAGC that occurs repeatedly in a sample of DNA.
  • The restriction endonuclease that recognizes the GAGC sequence cuts the DNA at every repetition of the GAGC pattern.
  • If one sample repeats the GAGC sequence 4 times whilst another sample repeats it 2 times, the length of the fragments generated by the enzyme for the two samples will be different.
• Gel Electrophoresis:
  • The restriction fragments produced during DNA fragmentation are analyzed using gel electrophoresis.
  • The fragments are negatively charged and can be easily separated by electrophoresis, which separates molecules based on their size and charge.
  • The fragmented DNA samples are placed in the chamber containing the electrophoretic gel and two electrodes.
  • When an electric field is applied, the fragments migrate towards the positive electrode.
  • Smaller fragments move faster through the gel leaving the larger ones behind and thus the DNA samples are separated into distinct bands on the gel.
• Visualization of Bands:
  • The gel is treated with luminescent dyes in order to make the DNA bands visible. 

hence the correct answer is option 1

Key Points

• PCR involves using short synthetic DNA fragments called primers to select a segment of the genome to be amplified, and then multiple rounds of DNA synthesis to amplify that segment.

Steps in PCR reaction

1. Template Denaturation - 
   •  Denaturing – when the double-stranded template DNA is heated to separate it into two single strands in PCR the process is known as template denaturation.
2. Primer annealing - 
   • Primer annealing is a critical step in polymerase chain reaction or PCR.
   • In this step, the primers bind to flanking sequences of the target DNA for amplification.
   • The annealing temperature of this step should be determined from the melting temperature of the selected primers to help ensure specificity of primer binding and target amplification.
3. DNA polymerization - 
   • A collection of enzymes known as Taq DNA polymerases catalyze the synthesis of DNA during PCR amplification cycle.
   • When a PCR cycle is in  process, DNA polymerases' primary job is to replicate the DNA in-vitro.
   • They accomplish this by incorporating nucleotides into the developing DNA strand's 3'-OH group.

• ​These steps are repeated multiple times to obtain the amplified DNA.​

Hence the correct answer is option 4.

The correct answer is Annealing for A and Extension for B. 

Concept:

• Real-time PCR or quantitative PCR (QPCR) is a powerful technique that allows measurement of PCR product while the amplification reaction proceeds.
• It incorporates the fluorescent element into conventional PCR as the calculation standard to provide a quantitative result. In this sense, fluorescent chemistry is the key component in QPCR.
• Till now, two types of fluorescent chemistries have been adopted in the QPCR systems: one is nonspecific probe and the other is specific.
• As a brilliant invention by Kramer et al. in 1996, molecular beacon is naturally suited as the reporting element in real-time PCR and has been adapted for many molecular biology applications whereas TaqMan PCR uses a nucleic-acid probe complementary to an internal segment of the target DNA.

Explanation:

• A Molecular beacon is a single-stranded bi-labeled fluorescent probe held in a hairpin-loop conformation (around 20 to 25 nt) by complementary stem sequences (around 4 to 6 nt) at both ends of the probe.
• The 5’ and 3’ ends of the probe contain a reporter and a quencher molecule, respectively.
• The loop is a single-stranded DNA sequence complementary to the target sequence. The proximity of the reporter and quencher causes the quenching of the natural fluorescence emission of the reporter.
• Molecular beacons hybridize to their specific target sequence causing the hairpin-loop structure to open and separate the 5’ end reporter from the 3’ end quencher.
• As the quencher is no longer in proximity to the reporter, fluorescence emission takes place.
• The measured fluorescence signal is directly proportional to the amount of target DNA.​

​Fig 1.Molecular Beacons Work

TaqMan PCR is a type of real-time PCR.​

• TaqMan PCR uses a nucleic-acid probe complementary to an internal segment of the target DNA.
• The probe is labeled with two fluorescent moieties.
• The emission spectrum of one overlaps the excitation spectrum of the other, resulting in “quenching” of the first fluorophore by the second.
• The probe is present during the PCR and if product is made, the probe is degraded via the 5′-nuclease activity of Taq polymerase that is specific for DNA hybridized to template (=TaqMan activity).
• The degradation of the probe allows the two fluorophores to separate, which reduces quenching and increases intensity of the emitted light.
• Because this assay involves fluorescence measurements that can be performed without opening the PCR tube, the risk of contamination is greatly reduced. 

Fig 2: TaqMan PCR 

Hence the correct answer is Option 3

The correct answer is SUP4

Concept:

• Of all cloning systems, Yeast artificial chromosomes (YACs) have the highest insert capacity.
• This technique, created in 1987 by Burke and Olson, enables the replication of exogenous DNA segments that are hundreds of kilobases long.
• For the human, mouse, and even Arabidopsis genomes, YACs that represent continuous regions of genomic DNA (YAC contigs) have offered a physical map framework.
• In a YAC vector, SUP4 is a selectable marker that can be disrupted by the insertion of a DNA fragment at the cloning site. When SUP4 is interrupted by the insertion, the original suppressor phenotype is lost, allowing for the distinction between YACs that have re-ligated without an insert and those that have successfully incorporated a DNA fragment. The presence or absence of SUP4 function can be tested using appropriate selection media.

Explanation:​

• The SUP4 gene contains the foreign DNA cloning site for the YAC vector.
• This gene corrects a yeast host cell mutation that results in the buildup of red pigment.
• The host cells are typically red, and those that are exclusively altered with YAC will produce colonies that are colourless.
• When a foreign DNA fragment is cloned into the YAC, it results in insertional inactivation, which returns the red hue.
• Therefore, the colonies that contain the foreign DNA fragment are red.​

Key Points

Mapping transcription initiation sites - 

• RNA-sequencing (RNA-seq) can be used to map transcription initiation sites by capturing and sequencing the 5' end of RNA transcripts.
• This technique, known as cap analysis of gene expression (CAGE), allows researchers to identify the precise locations where transcription starts.

Long non-coding RNA profiling - 

• RNA-seq is widely used to profile and characterize long non-coding RNAs (lncRNAs).
• These are RNA molecules that are transcribed from the genome but do not encode proteins.
• RNA-seq enables the comprehensive identification and quantification of lncRNAs, providing insights into their expression patterns and potential functional roles.

Alternative polyadenylation profiling - 

• Alternative polyadenylation refers to the process by which different polyadenylation sites are used within the same gene, leading to the production of mRNA isoforms with variable 3' untranslated regions (UTRs).
• RNA-seq can be used to profile alternative polyadenylation events by capturing and sequencing the mRNA molecules and identifying the different polyadenylation sites used.

Mammalian epigenome sequencing - 

• Mammalian epigenome sequencing refers to the comprehensive profiling of epigenetic modifications, such as DNA methylation and histone modifications, across the genome.
• While RNA-seq can provide information on gene expression levels, it is not typically used to directly sequence or analyze epigenetic modifications.
• Specific techniques, such as bisulfite sequencing for DNA methylation or chromatin immunoprecipitation followed by sequencing (ChIP-seq) for histone modifications, are more commonly employed for studying the mammalian epigenome.

Hence, the correct answer is option 4.

Key Points

• Restriction enzymes, also known as restriction endonucleases, are enzymes that recognize specific DNA sequences and cleave the DNA at or near these recognition sites.
• They are commonly used in molecular biology and genetic engineering to manipulate DNA.
• Recognition Sequence:
  • Each restriction enzyme recognizes a specific DNA sequence, typically 4 to 8 base pairs in length. This sequence is usually palindromic, meaning it reads the same on both DNA strands when read in opposite directions.
• Cutting Types:
  • Restriction enzymes can cut DNA in two different ways: blunt ends and sticky ends.
• Blunt Ends:
  • Some restriction enzymes cut DNA straight across both DNA strands, resulting in blunt ends with no overhanging nucleotides.
• Sticky Ends:
  • Other restriction enzymes create staggered cuts, producing fragments with short single-stranded overhangs called sticky ends. These sticky ends can form base pairs with complementary sequences, allowing for the joining of DNA fragments from different sources.
• Nomenclature:
  • Restriction enzymes are named using a combination of letters and numbers. The letter(s) represent the genus or species from which the enzyme is derived, and the number indicates the order of discovery. For example, EcoRI is derived from E. coli and was the first enzyme discovered in the "R" series.
• Sources:
  • Restriction enzymes are naturally produced by bacteria and archaea as a defense mechanism against foreign DNA, such as bacteriophages. They recognize specific DNA sequences and cleave the DNA to prevent infection by foreign genetic material.
• Applications:
  • Restriction enzymes have numerous applications in molecular biology, including DNA cloning, gene expression analysis, DNA sequencing, and genetic engineering. They are used to cut DNA at specific sites, create recombinant DNA molecules, and analyze DNA fragments by gel electrophoresis.

Important Points

•  Based on the provided restriction profiles, we can infer the following:
  • E1 cuts the 4.8 kb plasmid into two fragments: a 500 bp fragment and a 4.3 kb fragment.
  • E2 cuts the 4.8 kb plasmid into two fragments: a 400 bp fragment and a 4.4 kb fragment.
  • E3 cuts the 4.8 kb plasmid into two fragments: a 2 kb fragment and a 2.8 kb fragment.
  • E4 cuts the 4.8 kb plasmid into two fragments: a 1.9 kb fragment and a 2.9 kb fragment.
• From these observations, we can determine theoverlapping regions of the cuts made by the enzymes:
  • E1 and E2 have a common 4.3 kb fragment.
  • E2 and E3 have a common 4.4 kb fragment.
  • E1 and E4 have a common 2.9 kb fragment.
• By combining this information, we can conclude that the enzyme E2 is the only one that cuts between the E1 and E3 sites.
• E2 leaves a 400 bp fragment when cutting the 4.8 kb plasmid, which is not observed with any other enzyme.
• Therefore, E2 must have recognition sites on both sides of the 4.4 kb fragment it produces.
• The resulting analysis shows the locations of the recognition sites for each enzyme and the sizes of the fragments they produce.