Cytogenetics Multiple Choice Questions on “Restriction Endonuclease”.
1. Type I restriction endonuclease cuts________________
A. Within the recognition sequence
B. On either side of the recognition sequence
C. 25 to 30 bp away from the recognition sequence
D. 1000bp away from the recognition sequence
Answer: D
Clarification: Type I restriction endonucleases are of very less practical utility, this is because although they recognize a particular sequence, they cut the DNA up to 1000bp away not providing much specificity.
2. Which of the following is not true for type 1 restriction endonuclease?
A. It is a single enzyme with 3 subunits
B. Using it is an energy requiring process
C. Methylation and cleavage occurs in the same sequence
D. The cleavage is up to 1000bp away
Answer: C
Clarification: Type 1 restriction endonuclease recognizes and cleaves at different points. While it methylates at the recognition sequence, the cleavage could be as far as 1000bp. Also both these processes require ATP.
3. Which of the following is not true about type II restriction endonuclease?
A. Here Clevage and Methylation are by different enzymes
B. In this case the recognition sequence is asymmetric
C. It cleaves within the recognition sequence
D. It can’t cleave at methylated sequences
Answer: B
Clarification: In case of recognition sequence type IIs the recognition site in asymmetric. While in case of type II the recognition site is symmetric.
4. Restriction endonucleases can recognize___________
A. Tandem repeats
B. Pallindromic sequences
C. GATC
D. No sequence specificity
Answer: B
Clarification: Restriction endonuclease will recognize the pallindromic sequences. GATC is one of such pallindromic sequence but not the only one, and this is very specific.
5. What does this sequence GC/GGCCGC recognized and cleaves by Not1 Doesn’t agree to?
A. The cleavage is after the 2nd C from 5’ end
B. The cleavage generates a 5’ overhang
C. This sequence has a probability of occurrence in every 1024 bases
D. This cleavage is asymmetric
Answer: C
Clarification: The probability of occurrence of an 8 bp recognition sequence can be determined as 48, this is because there are 4 bases and 8 specific bases in position in this sequence; which is equal to 65536. However the rest of the options are true.
6. Which of the sequence will Dcm methylate?
A. GATC
B. GAATC
C. CCGTT
D. CCAGG
Answer: D
Clarification: DCM methylates at the C base while DAM methylates at the A base. DCM methylates ta the internal C of the sequence CCAGG. The sequence CCGTT is not a palindrome and GATC is recognized and methylated by DAM.
7. In an experiment you introduce your gene of interest in a wild type E. cloli genome using an episome. However, you fail to see the product protein being formed. What could be the possible reason for this?
A. The gene was silenced
B. The gene product was produced in negligible quantity
C. The bacteria was in stress so it was not producing much protein
D. The gene was excised out
Answer: D
Clarification: While all the reasons stand a possibility of being true, under the experimental condition it might be due to the restriction enzyme cleaving out GOI considering it a virus. Thus, the E. coli strains need to be restriction deficient but methylation proficient.
8. Which of the following enzyme air is a neoisoschizomer?
A. SphI and BbuI
B. Not1 and HindIII
C. SmaI and XmaI
D. Pst1 and Sma1
Answer: C
Clarification: Neo-isoschizomers are the enzymes that recognize the same sequence and cleave at different points. Here, both Sma1 and Xma1 recognize GGGCCC; the 1st cleaving after the 3rd G and 2nd cleaving after the 1st.
9. Which of these restriction enzymes generate 3’ overhang?
A. Sph1
B. EcoR1
C. Sma1
D. Xma1
Answer: A
Clarification: Sph1 recognized the sequence AGTAC/G, as it cleaves near the 3’ end of the sequence it generates an asymmetric cut with 3’ overhang. While Sma1 produces a blunt end and Xma1 and EcoR1 produces a 5’ overhang.
10. Restriction enzymes cleave the bond by which type of reaction?
A. SN1
B. SN2
C. SE1
D. SE2
Answer: B
Clarification: The reaction is by hydrolysis of the phosphodiester bond between two sugar residues by a nucleophilic attack. Inverted product configuration shows that the substrate was inverted only once. As only bimolecular nucleophilic reaction causes only one time inversion so that is the correct option.
11. The Mg2+ ion in restriction endonuclease is not bound to ________
A. Water molecule
B. Carboxylates of the enzymes aspertate group
C. Oxygen
D. Histidine
Answer: D
Clarification: In restriction endonucleases we don’t have His as a ligand to the central metal ion Mg2+. While it’s binding to the other groups assists in the nucleophilic attack.
12. In EcoRV the enzyme is a ____________
A. Homo dimer
B. Hetero dimer
C. Trimer
D. Hetero tetramer
Answer: A
Clarification: As the enzyme recognizes the palindrome sequences, the homodimer would be able to recognize the same sequence from the other end. Thus, restriction endonucleases are homodimers.
13. When the DNA enzyme interacts with the cognate DNA, there is _______________
A. Increase in free energy
B. Reduction in free energy
C. Breakdown of the complex
D. Repulsion
Answer: B
Clarification: On restriction endonuclease binding to cognate DNA there is a reduction in free energy to a greater extent than that in case of non-cognate DNA. This additional energy is used to distort the sequence and produce a kink.
14. Methylated DNA doesn’t form kink as________________
A. As they don’t fit in the active site
B. The enzyme recognized methylated A as different base than non-methylated A
C. They are less flexible so no kink is formed
D. They don’t interact with the side chains in an active site
Answer: D
Clarification: The kink in the DNA is formed not by just the recognition of the base sequence in the restriction site, but also by the interactions of the side chains. Asp 185 is unable to interact with methylated A, so no kink is formed.
15. A methylated recognition sequence can act as a competitive inhibitor for the restriction endonuclease.
A. True
B. False
Answer: A
Clarification: The methylated bases can bind to the enzyme but it will not be cut by the enzyme. Thus it can effectively as a competitive inhibitor as it competes with the non-methylated DNA for the same binding site.
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