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Correct choice is (d) Uvr D

For explanation I would SAY: Uvr D is the FACTOR responsible UNWINDING the DNA in PROKARYOTES and excising the sequence with the lesion. On the other hand, TFIIH is also a helicase for eukaryotes.

Correct ANSWER is (b) 5 Bromo uracil

For explanation I would say: 5 Bromo Uracil when incorporates in the DNA in a NORMAL state and then undergoes TRANSITION to a rare state in the next replication cycle leads to an AT->GC MUTATION. While when it incorporates against a G in excited state and then undergoes replication when in the normal state it gives GC->AT transition.

The correct choice is (d) Rec FOR

The explanation is: Rec FOR is necessary for STRAND migration which causes the REVERSE FORK migration during REPAIR of bulky lesions in DNA.

Correct OPTION is (c) It helps to generate 3’ overhang

Easy EXPLANATION: Rec BCD acts as an exonuclease that SELECTIVELY chews off the 5’ end of DNA to generate a 3’ overhang. This is necessary for strand INVASION by this 3’ overhang which brings about recombination REPAIR.

Right ANSWER is (c) RAD 23

For explanation: In eukaryotes, XPC WORKS in CONJUGATION with hHR23B which is simply the human homologue of RAD 23. RAd 51 is INVOLVED in DSB repair.

Correct choice is (b) Guanine METHYL TRANSFERASE

Explanation: While Alk B is ALSO a de-methylating agent, but in E. coli the UV INDUCIBLE de-alkylating agent is 8-oxo guanine methyl transferase.

Right ANSWER is (c) One larger circular chromosome or TWO smaller CHROMOSOMES

For explanation I would say: Resolution of Holliday intermediate will GIVE rise to two smaller chromosomes if RESOLVED in one way. And it can also give rise to one larger chromosome if it is resolved in the other way, but both of these have 50% chance and it doesn’t occur together.

Right answer is (C) ECORI

Easy explanation: ECoRI is a restriction endonuclease that can cut both the DNA strands generating an overhand. On the other hand Ruv C in recombination repair, Mut H in MMR and S1 endonuclease can only make single strand nicks.

Right choice is (C) XPG on 3’ end and XPF on 5’ end

To EXPLAIN I would SAY: While XPD and XPB bind to the 3’ and 5’ end RESPECTIVELY, they are helicases and not exonucleases. On the other hand XPF ALONG with ERCCI that binds to 5’ end and XPG are exonucleases.

The correct ANSWER is (a) 50 NT

Easiest explanation: A stretch of free 100 nt is needed by NER machinery to excise out and REPLACE the lesion. Thus, either the TRANSCRIBING RNA POLYMERASE then has to backtrack or it falls off.

The CORRECT answer is (d) XPB

The BEST I can explain: XPB and XPD are two subunits of the transcriptional factor TFIIH that ACTS as a helicase unwinding the region around the bulky lesion. It also marks the junctions of the SS and dsDNA where the cleavage would occur.

Correct answer is (B) False

Easiest explanation: DSB is in the endonuclear REGION. So the ends created lack telomere. DNA ends produced by synthesis in vitro have 5’ tri-phosphate and that in between strands has 5’ monophosphate, DSB can thus be distinguished from normal chromosomal ends.

The correct answer is (b) PCNA

To elaborate: PCNA or the beta clamp analogue in eukaryotes ASSISTS the FEN1 to cleave the long flap GENERATED by long patch BER. This can’t be TAKEN CARE of by the pol beta alone.

Correct ANSWER is (c) CPD

To explain: BER usually deals with single base damages, and NER takes care of the damages involving bulky lesions like CPD. Abasic site generation is the 1st STEP of BER PROCESS so it can carry out the REACTION easily.

Right answer is (c) G1

The explanation is: For homologous recombination to TAKE PLACE there MUST be two sister chromatids with same or similar composition. This is available only after DNA replication, thus it can’t take place in G1 phase before replication.

The correct option is (a) Nitrous acid acting on A

Explanation: All options other than OXIDATION of A by nitrous acid LEAD to GC->AT transition. Nitrous acid converts A to Hypoxanthine that in place of pairing with T pairs with C. THUS, it LEADS to AT->GC transition.

The CORRECT ANSWER is (d) C—C

Explanation: It has been OBSERVED that although MMR can effectively detect and repair most of the mismatches like T—T, A—C, G—T, A—Aetc, it can’t detect C—C, this may be due to c—C causing LESS distortion in the DNA.

Correct answer is (d) SENSE

The best I can explain: All the other MUTATION LEAD to wild phenotype, so unless a GENETIC analysis is performed, they are indistinguishable. But in case of sense a stop codon is changed to a coding codon which has a serious phenotypic effect on the protein.

The CORRECT ANSWER is (c) Mutation in sperm mother cells

The EXPLANATION is: Option a and d are somatic tissue, their mutation will not be propagated. A single gamete will only have a slight chance to fertilize and lead to mutation, but sperm mother mutation will PRODUCE many LETHAL gametes that will increase the probability.

Correct answer is (c) Another FRAME shift mutation

To explain: If OVERALL the two frame shift MUTATIONS LEAD to a change of 3 or its multiple in the genetic code, the codons will again be in the same frame. This is an internal compensation, so intragenic SUPPRESSION.

Correct choice is (b) Transposition

Explanation: Point MUTATION concerns a single base. Thus SUBSTITUTION, which INCLUDES transversion, and insertion or deletion, FALLS within point mutation. On the other hand, translocation INVOLVES a stretch of bases so it is not a point mutation.

The correct option is (c) Guanine

Explanation: This is a BASE ANALOGUE that is a PYRIMIDINE. Under normal CONDITIONS it behaves LIKE T and base pairs with A; but under abnormal conditions, it behaves as C and Base pair with G. As a pyrimidine can’t base pair with another pyrimidine, G is the only available option.

The correct answer is (b) Thalassemia

Explanation: In Thalassemia the alpha globin is 141 amino ACIDS long as the UAA codon is changed to CAA, i.e a stop codon is changed to a glutamic acid. While sickle CELL anemia is due to a mis sense MUTATION and c is due to a nonsense mutation and d due to trisomy.

The correct OPTION is (C) Silent

Easiest EXPLANATION: Silent mutation is the one where two codons CODE for the same gene. Thus, CONVERSION of one codon to another doesn’t effectively change the protein at all.

Right option is (B) Ruv B

The explanation: The Ruv B BINDS to the DNA in Holliday intermediate to two branch points. This FACTOR helps in branch migration.

The correct OPTION is (B) Gamma-P H2AX

For explanation: In DSB the canonical bases are replaced by the non-canonical bases, which it also phosphorylated by ATM kinase.H2AX is one such non-cannonical base, while there is no H3BX or H5G. H2AX is phosphorylated at gamma not alpha.

Correct choice is (B) Bond between base and sugar

To elaborate: The glycosidic bond between the sugar residue and the nitrogenous base is the 1st to be BROKEN by BER glycosylases GENERATING an ABASIC SITE.

Correct answer is (d) DNA POLYMERASE III

Explanation: Polymerase III is responsible for re-synthesizing the stretch of DNA in MMR between the nick made my Mut H and the mis PAIRED base.

Correct option is (b) Mis-sense mutation

Best explanation: In this disease, a GAG is CONVERTED to a GUG, i.e a glutamic ACID is converted to a valine. This is example of a mis-sense mutation where a CHANGE of base codes for a different amino acid from the ORIGINAL.