Given below are some statements with blank spaces indicated by '_________'. A. A plasmid cloning vector digested with an enzyme (with a single restriction site in the plasmid) that generates 3' overhangs can be made blunt-ended using _________. B. DNA with a nucleotide composition of 30% A, 35% G, 20% C and 15% T is most likely _________. C. Production of only truncated molecules of transgene-derived mRNA in transgenic plants generated using a transgene from a prokaryotic source is most likely due to _________. D. _________ is a method for identifying the positions where individual DNA-binding proteins attach to a genome. Which one of the following options has the correct sequence of terms that can be used to complete the above statements (from A to D) such that all statements become true?
Completing Statements on Molecular Biology Techniques and DNA Properties
Let's analyze each statement and determine the correct term to fill in the blank, based on the provided options.
A. Blunt-ending DNA Overhangs
The statement describes making a plasmid cloning vector, digested with an enzyme generating 3' overhangs, blunt-ended. Blunt-ending overhangs means removing the single-stranded extensions to create a flush end. Several enzymes can modify DNA ends.
- Mung bean nuclease is a single-strand specific nuclease. It is commonly used to remove single-stranded overhangs (both 5' and 3') from DNA molecules, thereby creating blunt ends.
- Taq polymerase and Pfu polymerase are DNA polymerases primarily used in PCR. While some polymerases have proofreading activity that can clean up ends, Mung bean nuclease is specifically used for removing overhangs to create blunt ends.
- Reverse transcriptase synthesizes DNA from RNA template.
Therefore, Mung bean nuclease is the correct enzyme for making 3' overhangs blunt-ended.
B. DNA Composition Analysis
The DNA sample has the following nucleotide composition: A = 30%, G = 35%, C = 20%, T = 15%.
In double-stranded DNA that follows Chargaff's rules, the amount of Adenine (A) is equal to the amount of Thymine (T), and the amount of Guanine (G) is equal to the amount of Cytosine (C). Let's check if this holds true for the given sample:
- A (30%) ≠ T (15%)
- G (35%) ≠ C (20%)
Since the base pairing rules (A=T and G=C) are not followed, this DNA is most likely not a typical double-stranded DNA molecule. This composition is characteristic of single-stranded DNA (ssDNA) or RNA (though options suggest DNA), where the base ratios are not necessarily paired.
Thus, the DNA is most likely single-stranded.
C. Truncated Transgene mRNA Production
The issue is the production of only truncated mRNA from a prokaryotic transgene in transgenic plants (a eukaryotic system). Prokaryotic gene expression signals (like transcription terminators) are different from eukaryotic ones (like poly-adenylation signals). Eukaryotic mRNAs typically undergo poly-adenylation at the 3' end, which is important for stability and translation. Poly-adenylation is directed by specific signals, often including the sequence AAUAAA, in the mRNA precursor.
If the prokaryotic transgene sequence contains sequences that resemble or function as premature eukaryotic poly-adenylation signals when transcribed in the plant cell, the transcription or processing machinery in the plant might recognize these internal signals. This would lead to cleavage and poly-adenylation occurring within the coding sequence or prematurely, resulting in a truncated mRNA molecule.
- Presence of potential poly-adenylation signals in the transgene sequence can cause premature termination or processing in a eukaryotic host.
- Codon usage variations primarily affect the efficiency or rate of translation, not mRNA truncation.
- Absence of a polyA signal would likely lead to unstable or non-polyadenylated mRNA, not necessarily truncated polyadenylated mRNA.
- Presence of mRNA instability sequences would lead to faster degradation of the mRNA, but not necessarily truncation during synthesis or processing.
Therefore, the presence of potential poly-adenylation signals in the prokaryotic transgene is the most likely cause for truncated mRNA in the eukaryotic plant system.
D. Identifying DNA-Binding Protein Positions
The statement asks for a method to identify where individual DNA-binding proteins attach to a genome.
- ChIP-seq (Chromatin Immunoprecipitation followed by sequencing) is a powerful molecular technique used to identify the locations of DNA-binding proteins on the genome. It involves crosslinking proteins to DNA, fragmenting the DNA, using antibodies to precipitate specific proteins along with the bound DNA, reversing the crosslinks, and then sequencing the DNA fragments to map the binding sites genome-wide.
- FISH (Fluorescence In Situ Hybridization) is used to detect and localize specific DNA sequences on chromosomes.
- PFGE (Pulsed-Field Gel Electrophoresis) is used to separate very large DNA molecules.
Thus, ChIP-seq is the appropriate method for identifying the positions where individual DNA-binding proteins attach to a genome.
Based on the analysis, the correct terms to complete the statements are:
- A - Mung bean nuclease
- B - single-stranded
- C - presence of potential poly-adenylation signals in the transgene sequence
- D - ChIP - seq
Important Questions from Molecular Biology Techniques
Which one of the following does NOT use RNA-sequencing?
Which one of the following statements related to molecular cloning procedures is INCORRECT?
Amongst the following, which one is the most appropriate strategy to sequence and assemble highly repeated regions of a genome?
Given below are statements related to various molecular techniques
A. During molecular cloning of DNA fragments, a vector and insert molecule digested with two different enzymes can never be ligated with each other.
B. Only 3' → 5' exonucleases and not 5' → 3' exonucleases can be used for digesting nucleic acids to generate blunt-ended fragments for cloning.
C. In Sanger's dideoxy sequencing method, each reaction consists of a mixture of three dNTPs and one ddNTP.
D. Self-ligation of a vector with compatible ends can be prevented by treatment with alkaline phosphatase.
Which one of the following options represents a combination of correct statements?
Given below are four DNA sequences and a set of forward and reverse primers for PCR amplification.
In the absence of any other factors such as (but not restricted to) Tm, length, percent GC, etc., which one of the above template-primers combinations would produce an amplified fragment?Sequence Primes A 5' ACAATCGT.....ACTAGTAC-3' FP: 5'-TGTTAG-3'
RP: 5'-TAGTAC-3'B 5' AGTCTTAG.....ATGCCAGT - 3' FP: 5'-AAGACT-3'
RP: 5'-ACTGGC-3'C 5' CTTGACTA.....GTACAGTCA-3' FP: 5'-CTTGAC-3'
RP: 5'-TGACTG-3'D D 5' GATCTAGC.....TCAAGCAGAC-3' FP: 5'-GATCTA-3'
RP: 5'-CAGACG-3'
