Genetic Engineering Multiple Choice Questions on “Restriction Endonuclease & Phosphatases – 1”.
1. The term ‘endonuclease’ refers to cutting the DNA sequence from ______________
a) only within the polynucleotide chain, not at the ends
b) the ends of the chain
c) anywhere in the chain
d) exactly in the middle of the chain
Answer: a
Explanation: The cleavage is done within the polynucleotide chain and not at the ends. The enzyme which cuts the sequence at the ends is known as exonuclease.
2. The restriction endonuclease is having a defence mechanism in the bacterial system against foreign DNA such as viruses. But how it is able to protect its own DNA?
a) By methylation of bacterial DNA by restriction enzyme
b) By methylation of foreign DNA by restriction enzyme
c) By phosphorylation of bacterial DNA by restriction enzyme
d) By phosphorylation of foreign DNA by restriction enzyme
Answer: a
Explanation: The bacterial DNA is methylated by restriction enzyme and thus now it is not recognized by the restriction endonuclease. Thus methylation prevents the restriction endonuclease from cutting its own DNA.
3. Even after replication, how the modified DNA remains protected?
a) It remains protected because of conservative mode of replication
b) It remains protected because of semi-conservative mode of replication
c) The mode of replication has no role to play in the protection
d) It is again modified after replication
Answer: b
Explanation: Because of the semi-conservative mode of replication, one of the DNA strands remain methylated even after replication. One methylated strand is sufficient for providing protection against cleavage by a restriction endonuclease.
4. How many classes of restriction enzymes are there?
a) 2
b) 1
c) 3
d) 4
Answer: c
Explanation: Three classes of restriction enzymes are there, I, II and III. These classes are having different characteristics such as the site of cleavage on the basis of the recognition sequence.
5. Type II cuts the sequence in the following way __________
a) Within the recognition sequence
b) At 100-1000 nucleotides away from the recognition sequence
c) At 27-30 nucleotides away from the recognition sequence
d) It cuts randomly
Answer: a
Explanation: Recognition sequence is the set of nucleotides which are identified by the enzyme and then it cleaves. In class II, the cleavage is done within the recognition sequence.
6. After cleaving the sequence, the nature of the ends created by the type II endonuclease is __________
a) The ends created are always single stranded
b) The ends created are always double stranded
c) Either the ends are single stranded or they are double stranded
d) One end is single stranded and one end is double stranded
Answer: c
Explanation: Either the ends are both double stranded or are both single stranded. The double stranded ends are blunt ends whereas the single stranded ends are sticky ends.
7. A sequence is having two ends, 5’ and 3’. Which of the following statements is correct regarding the nature of the ends?
a) The 5’ end is having hydroxyl group
b) The 5’ end is having phosphate group
c) The 3’ end is having phosphate group
d) Any group can be present at any end
Answer: b
Explanation: The 5’ end is having a phosphate group. As in a DNA sequence, the 5’ end is characterized by phosphate group and the 3’ end is characterized by a hydroxyl group.
8. Blunt ends created by the restriction endonuclease can be joined.
a) True
b) False
Answer: a
Explanation: The blunt ends created can be joined by the enzyme responsible for ligation such as DNA ligase. It is not necessary to have sticky or single stranded ends.
9. The recognition sequence for BamHI is 5’ G|GATCC 3’. The ‘|’ represents the cutting site. What can be inferred about the ends from it?
a) The ends created are double stranded
b) The single stranded end is 5’ in nature
c) The single stranded end is 3’ in nature
d) To decide about the nature of the ends more information is needed
Answer: b
Explanation: The other strand is just complementary to it and can be written in the following way:
5’ G|GATCC 3’
3’ CCTAG|G 5’
After cleavage, the sequences are represented as:
5’ G 3’ 5’ GATCC 3’
3’ CCTAG 5’ 3’ G 5’
Thus, we can see that the ends generated are single stranded at 5’ end.
10. The recognition sequence of Sau3A is 5’ |GATC 3’ and that for DpnI is 5’ GA|TC 3’. Which of the statements is true?
a) The ends created by both the enzymes are compatible
b) The ends created by both the enzymes are not-compatible
c) The ends created by DpnI are single stranded
d) The ends created by Sau3A are single stranded
Answer: b
Explanation: Though the recognition sequence is the same, the ends are not compatible. It is so because both the enzymes leave double stranded ends and thus can’t be ligated.
11. The recognition sequence is at times palindromic in nature. Which of the following statements is correct with respect to it?
a) The molecules which are cut by the same enzyme, anneal only if the sequence is palindromic in nature
b) When the molecules are cleaved by the same enzyme and the recognition sequence is palindromic in nature, there is no effect on annealing
c) There are increased chances of annealing if the recognition sequence is palindromic in nature
d) The term ‘palindromic’ can be used whether the sequence is read from 5’ to 3’ or 3’ to 5’
Answer: c
Explanation: The term palindromic can be used only when a sequence is read along the same polarity ie either 5’ to 3’ or 3’ to 5’. When the recognition sequence is palindromic in nature, there are increased chances of annealing because now there are increased orientations. If the sequence is non-palindromic in nature, then also annealing would take place but in fewer orientations.
12. If all the nucleotides are present with equal frequencies and at random, what are the chances of having a particular four nucleotide long motif?
a) 1/256
b) 1/64
c) 1/16
d) 1/8
Answer: a
Explanation: There are four nucleotide bases present in a DNA sequence A, T, C and G. If the bases are present with equal frequency and at random the chances of having a particular 4 nucleotide long motif is 1/ (4*4)= 1/256.