This set of Molecular Biology Multiple Choice Questions & Answers (MCQs) focuses on “Palindromes, Inverted Repeats and Stem and Loop Structures”.
1. Double – stranded structure of nucleic acid is the basic requirement for palindromic sequences.
Explanation: Palindromic sequence is a nucleic acid sequence present in double – stranded nucleic acid. This is because 5’ to 3’ reading frame on one strand matches the 5’ to 3’ reading frame of its complementary strand, which is the requisite for any sequence to be called a palindrome.
2. Which of the following words represents a palindrome?
Explanation: A palindrome is a word which can be read as the same from both forward and backward. Thus, MADAM is the same when spelt in either way.
3. Which of the following will form a palindromic sequence?
Explanation: The complementary sequence of ATTGCAAT is TAACGTTA. Thus, when the first is read from left to right and the later read from right to left the sequence of the bases is exactly the same. This is the criteria for a sequence to be palindromic.
4. Which of the following is affected by palindromic sequences?
a) Acetylation site
b) Phosphorylation site
c) Methylation site
d) Promoter site
Explanation: Palindromic sites are frequently methylated in many organisms. These are the sites recognized by methylases where a methyl group is attached to inactivate a gene or mark a restriction site for an endonuclease.
5. Palindromic sequences play a very important role in gene manipulation.
Explanation: Many restriction endonucleases recognize and cut the palindromic sequence, for example, EcoR1. This is important for the genetic manipulation to meet the purpose of insertion and excision of gene or genome fragment from certain genetic material (eg. plasmid).
6. Which of the following palindromes is not a restriction site?
Explanation: GAATTC serves as the restriction site for the EcoR1 endonuclease. CCTAGG serves as the restriction site for the endonuclease BamH1. Again AGCT acts as the restriction site for the endonuclease Alu1. Only TACGTA does not serve as a restriction site or else it is not yet known.
7. Molecular cutters do not recognize palindromic sequence.
Explanation: Molecular cutters or endonucleases always recognize a palindromic sequence. Type I endonucleases generally cuts the DNA 1000 bp away from the 5’ end of its recognition sequence. Similarly Type II cuts within the sequence and Type III cuts 25-27 bp away.
8. Which of the following is an example of an inverted repeat?
Explanation: Inverted repeat is a single stranded sequence of nucleotides followed downstream by its reverse complement separated by a few or more nucleotides. When the intervening length iin between the repeats is zero, the composite sequence is a palindromic sequence.
9. Inverted repeat have a number of biological functions. Which of the following is a biological function of an inverted repeat?
b) Central dogma
c) Cellular metabolism
d) Genetic stability
Explanation: The Inverted repeat defines the region of self complementation thus can be identified by many proteins, enzymes and transposons. They also serve as sites for many mutations. Thus all of these factors may lead to alteration in the basic genome sequence. If any such alteration occurs in any vital gene it could cause genetic disorders and diseases. Example of certain disease is osteogenesis imperfecta.
10. Most commonly known hairpin structures are found in _____________
Explanation: Hairpin structure is a type of stem – loop structure involving intra molecular base pairing. This is generally found in single – stranded DNA and RNA. Most commonly known hairpin structures are observed in the cloverleaf model of tRNA namely the anticodon loop, D loop and the ΨU loop.
11. Which of the following does not promote stability in stem – loop structure?
a) Length of the stem
b) Size of the loop
c) A : U base pairing
d) π orbital of aromatic ring
Explanation: A : U base pairing has only two hydrogen bonds to hold them together whereas G : C has three hydrogen bonds. Thus a G : C rich stem promotes stem – loop structure more by providing more stability to the stem.
12. What is the minimum number of bases required for loop stability?
Explanation: Loops less than 3 bases long are sterically impossible to form. The optimal loop length is 4 – 8 bases which perfectly stabilize the loop of the stem – loop structure.
13. Which of the following does not contain a stem – loop structure?
Explanation: tRNA contains three stem loops to meet the cloverleaf model. The pseudoknot is the structure with two nested loops. Again the ribozyme also contains a stem – loop structure, best example of which is the hammerhead ribozyme containing three stem – loops.
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