This set of Molecular Biology Multiple Choice Questions & Answers (MCQs) focuses on “Types of Genetic Mutations”.
1. How many kinds of mutation are found in DNA which includes mutation of only one base?
Explanation: There are two kinds of mutation which is observed in the In the DNA that include only one base and are also known as point mutation. These mutations are transition where mutation occurs chancing a purine to purine and pyrimidine to pyrimidine, and transversion where there purine is converted to purine and vice versa.
2. What is the overall rate at which new mutations arise spontaneously at any given site on the chromosome per round of replication?
a) ≈ 10-8 – 10-12
b) ≈ 10-7 – 10-9
c) ≈ 10-6 – 10-11
d) ≈ 10-5 – 10-10
Explanation: The overall rate at which new mutations arise spontaneously at any given site on the chromosome ranges from ≈ 10-6 – 10-11 per round of DNA replication. With some sites on the chromosomes being hot spots, mutations arise at a high frequency in these sites.
3. The mutation occurs at a random basis within a genome.
Explanation: The mutation occurs at a particular mutation prone region known as the hot spots. The hot spots are rich in di- or tri- nucleotide repeat sequences known as microsatellites.
4. What is the dinucleotide sequence of microsatellites?
Explanation: Microsatellites involves the repeats of the dinucleotide sequence of CA. the CA repeat is found at many widely scattered sites in the genome of humans and other eukaryotes.
5. By which process miss-incorporated base can change into a permanent mutation?
Explanation: A potential mutation may be introduced by misincorporation in any round of replication. In the next round of replication if the mutation is not repaired it gets permanently incorporated in the DNA sequence.
6. Detection of mismatches and fidelity of replication is maintained by mutation repair system.
Explanation: Proofreading by the polymerase is not always perfect and some mismatches may escape the detection which can become a permanent mutation if not corrected. This fidelity check is done by the mutation repair system; more precisely mismatch repair system, of the cell itself ensuring that perfect matches occur in the complementary strands.
7. How many steps are required to attain mismatch repair?
Explanation: The mismatch repair system involves two steps. The first step involves the scanning of the genome for mismatches. The second step ensures the correction of mismatch that has occurred in the genome.
8. In E. coli mismatches are detected by which repair protein?
a) Mut H
b) Mut L
c) Mut S
d) Mut D
Explanation: Mut S scans the DNA and recognizes mismatches from the distortion formed by the unpaired bases. Mut S then embraces the mismatched region introducing a pronounced kink in the DNA and a conformational change in the enzyme itself.
9. Mut S recruits how many component(s) to the mismatched site?
Explanation: Mut S on binding with the mismatched region recruits two more enzymes to form a complex. They are Mut L and Mut H. Mut L activates Mut H, the enzyme that causes an incision or nick on the newly synthesized strand near the site of mismatch.
10. The nicking of DNA is followed by the adherence of a helicase known as __________
a) Uvr D
b) Uvr A
c) Uvr B
d) Uvr C
Explanation: Nicking of the DNA is followed by the addition of a specific helicase for the repair system known as Uvr D. This helicase unwinds the DNA from the site of mismatch to the mismatched site so that the endonuclease could excise the mismatch.
11. Mismatch repair system is ATP dependent.
Explanation: Mismatch repair system is an ATP dependent process. The activation of Mut S enzyme is ATP dependent and even the recruitment of Mut L and Mut H requires the assistance of ATP hydrolysis to bind the mismatched sequence.
12. By which enzyme does E. coli tags its parental DNA strand?
Explanation: The E. coli enzyme Dam methylase tags the parental DNA strand by methylation. It methylates adenine residues on both the strands of DNA in a specific sequence of 5’….GATC….3’ which is frequently distributed along the whole genome.
13. If the Mut H cuts the DNA at the 5’ side of the mismatch then which nuclease is activated?
a) Exonuclease VII
b) Exonuclease VIII
c) Exonuclease I
d) Exonuclease IX
Explanation: When the Mut H cut the DNA at 5’ side then exonuclease VII or Rec J is activated. Rce J degrades the DNA strand in 5’ – 3’ direction.
14. What are the eukaryotic components of Mut S and Mut L of E. coli?
a) MSH, MLH
b) MHS, MHL
c) MLS, PMS
d) PMS, MHL
Explanation: The homologous component of Mut S in eukaryotes is called MSH. The other homologous component of Mut L in eukaryotic cells is called MLH or PMS.
15. High level of mutation in the germ cells is acceptable but in somatic cells can be catastrophic.
Explanation: Mutations in both somatic and germ cells can be catastrophic. High rates of mutation in the germ line would destroy the species and high rate of mutation in the somatic cells would destroy the individual.
Sanfoundry Global Education & Learning Series – Molecular Biology.
To practice all areas of Molecular Biology, here is complete set of 1000+ Multiple Choice Questions and Answers.