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Right answer is (c) POL BETA

Explanation: In eukaryotes, there is no REQUIREMENT of a phosphodiesterase to resolve the short flap generated by short patch BER. It is taken care of by pol beta.

Right option is (b) MMR

Explanation: MMR/mis-match repair system concerns the wrong bases incorporated by polymerase error while REPLICATION. It has to distinguish the PARENT and DAUGHTER strand for this repair. Since after replication COMPLETION and during division both the stands are indistinguishable, it is restricted to S phase.

The CORRECT option is (a) TTTTTTTTTTTTTTT

For explanation: Repetitive SEQUENCES have a greater CHANCE of frame shift mutation due to polymerase error. The polymerase tends to slide through these same sequences as even the base pairing is not much AFFECTED if it does so.

The CORRECT CHOICE is (c) Ligase IV

To EXPLAIN: Ligase IV has exceptional flexibility as it can even ligate dissimilar DNA ends. It can also ligate the overhand with a complex structure. This is NECESSARY for the immediate repair of the NHEJ breaks.

The correct answer is (d) Alpha helical

Best explanation: Rec A factors join TOGETHER to form right handed alpha helical structure AROUND the SINGLE strand of DNA. This has 6.2 Rec A per turn.

Correct CHOICE is (a) True

For explanation: While de-alkylating agents could lead to the DIRECT reversal of the damages, yet it is a suicide protease. One molecule is irreversibly inactivated in correcting single methylation damage. It is WASTEFUL, and eukaryotes have better ways of taking CARE of the methylation like BER.

Right ANSWER is (a) Deamination

To elaborate: Mutational hot SPOTS are REGIONS which are rich in 5MC and these on deamination give a T base. Under normal condition deamination of C gives a U which is spontaneously excised out, but as T is a normal DNA base the MUTATION is fixed.

Correct answer is (c) BRCA 1

The best EXPLANATION: Defects in the genes of BRCA 1 and BRCA 2 INCREASES the propensity of breast cancer to 80%.On the other hand OPTION a, B and d doesn’t directly play a ROLE in breast cancer.

The correct CHOICE is (C) GENERATING abasic site

Explanation: While APE1 has a 5’-> 3’ exonuclease activity, it is not a BIFUNCTIONAL glycosylase and can’t generate an abasic site.

Correct answer is (a) POL EPSILON

To elaborate: Major polymerase in eukaryotes is pol DELTA and pol epsilon. So, they are responsible of REPAIR synthesis of the excised out PATCH in NER.

The correct option is (c) Endonuclease

To EXPLAIN: Mut H can bind to hemi-methylated DNA and make a NICK in the strand with unmethylated GATC sequence. As it MAKES a nick within the DNA it is an endonuclease.

Right option is (b) MUT H

Explanation: While Mut S homologue (MSH) and Mut L homologue (MLH) has been DISCOVERED in eukaryotes, the Mut H homologue is not found. The nick in the lagging strand is expected to be provided by the unligated okazaki FRAGMENTS.

Right choice is (c) In both cases MMR is COMPROMISED

The best EXPLANATION: DAM knockdown leads to no METHYLATION, and then both strands are indistinguishable as both are not methylated. While in DAM over expression both are methylated at a shorter window giving lesser time for MMR to work. Thus, the MMR in EITHER case is compromised.

The correct answer is (C) PROFLAVIN

Explanation: Proflavin, like Ethidium bromide is an intercalating agent. This when inserted in DNA can LEAD to addition of a random base, and when lost can lead to DELETION.

Right answer is (b) KU-PKC

For explanation I would say: Ku PKC has the property of both kinase as WELL as nuclease. Thus, it can work in non-homologous DSB WITHOUT the help of additional nucleases.

Right option is (a) Pol I

For explanation I WOULD say: While pol III is RESPONSIBLE for the repair synthesis in MMR, pol I is however responsible for repair synthesis in BER in PROKARYOTES.

The CORRECT answer is (b) MUTATION frequency

Explanation: While mutation frequency is the number of times a particular mutation occurs in a POPULATION, mutation rate is the probability of a particular type of mutation per UNIT time.

Correct option is (b) 2

The BEST explanation: In case of DSB there are TWO BRANCH points produced. One where the 3’ overhang invades the HOMOLOGOUS STRAND and other when it hybridizes with the other 3’ end.

Correct ANSWER is (d) EXONUCLEASE X

To explain: Rec J is an exonuclease which is also known as exonuclease VII. It is responsible for CHEWING the DNA from 5’ to 3’ direction. While exonuclease I and X chew from 3’ to 5’ direction which will not be of much use in this case.

The correct answer is (B) Mut S

Best EXPLANATION: Mut S is responsible for recognizing the mismatch and then it complexed with Mut L which helps it to BIND to the nearest Mut H and carry out the repair process.

Right choice is (C) Liver; the mutagen is processed

Best explanation: Rat is a mammal similar to human being. In us the food undergoes several processing steps than it would in CASE of a simple prokaryote LIKE bacteria. Thus, using a rat liver extract ENSURES that the functional state of the substance is in the extract ensuring more efficiency. Liver processed all the toxins so its extract is used.

The CORRECT answer is (b) G

Explanation: Deamination of GUANINE will give Xanthine, which also BASE pairs with Cytosine. Thus the base PAIRING is not changed.

Right choice is (c) Replacement of LYS by Arg

For EXPLANATION I would say: Lys and Arg are both basic amino acid of similar SIZE, thus the mutation of one with another doesn’t have much effect on the final protein structure or function. On the other hand, option b is a nonsense mutation, and a and d are missense mutation.

The correct option is (a) LYNCH syndrome

To elaborate: Lynch syndrome, also known as Hereditary Non-Poyposis Colon Cancer is due to the slippage of replicative machinery in the microsatellite region of the CHROMOSOMES where it encounters repetitive SEQUENCES. Option b and d are errors in NER.

Right choice is (c) Flipping of bases

The explanation is: It has been SEEN that mis-incorporated bases have a tendency to APPEAR flipped out, these flipped out bases are recognized by the BER. This is because flipping REQUIRES less ENERGY than distortions like bulging. RNA pol STALLING can trigger NER mainly TC-NER.

The correct option is (C) RADIATION

Easiest explanation: 6,4 PYRIMIDINE photoproducts are formed when the DNA is exposed to strong radiation like UV or X-rays. These can be REPAIRED by BER.

Right option is (c) 50%

EXPLANATION: It has been experimentally SEEN that a mutation can effect at max 50% of the PROGENY cells. This is due to the FACT that there are two DNA strands and only one of them is carrying the mutation. So, only progenies receiving copies of that mutant strand will be mutant i.e. 50%.

Right option is (b) Rec F

To explain I would SAY: Rec B is a pard of Rec BCD complex that helps to generate the 3’ overhang. On the other HAND, Rec FOR complex is RESPONSIBLE for ASSEMBLING Rec A on the DNA stand and guides its INVASION.

The correct ANSWER is (b) FADH-

The best I can explain: When the chromophore MTHF poly Glu receives the UV light, it passes the energy on to FADH- which in turn passes the electron to CPD. The electron helps in rearrangement of the BONDS in CPD and RETURNS them to the form of two pyrimidine residues.

Right answer is (c) Insertion of 3 bases

For explanation: Frame shift mutations lead to lease effect when they lead to an insertion or deletion of 3 bases, as due to our CODON being triplet this doesn’t shift the reading frame. However, if the codon THUS lost or added is a very important ONE in the gene, there could be SERIOUS consequences.

The correct OPTION is (b) False

For EXPLANATION I would say: Suppressor mutations are visible in the back CROSS as they are DUE to compensation and not a reversion to wild codon LIKE reverse mutation. Thus, they can be distinguished by phenotypic study of back cross progeny.

Right answer is (d) 4 and 8

For explanation I would say: Transition is the replacement of a purine by another purine or a PYRIMIDINE by another pyrimidine. TRANSVERSION is the replacement of a purine by a pyrimidine, or a pyrimidine by a purine. So there are 4 possible transition A—G and C— T, and 8 possible transversions A—C, A—T, G—C, G—T.

The correct answer is (b) P21

For explanation: P53 is associated with DSB as ATM kinase phosphorylates 53BP1 that on activation codes for P53. On the other hand P21 is not associated with DSB. Ku and H2AX the non-canonical BASE is also associated with DSB REPAIR.

Right option is (b) Chi

Explanation: Initially the exonuclease Rec BCD can chew up both the 3’ and 5’ END of the DSB till it reaches the specific chi SEQUENCE in the 3’ strand. Then it STOPS chewing the 3’ end and increases its activity in the 5’ end.

The correct option is (c) XPC

Easy explanation: XPC in association with the hHR23B is responsible for RECOGNIZING the BULKY lesions and attaching the XPB and XPD.

The correct CHOICE is (b) False

To explain: Any change in genetic CONSTITUTION of the organism is termed as MUTATION, but only when the phenotypic expression is changed it is called a MUTANT.

Right answer is (a) 5’, 3’

The best explanation: The AP endonuclease can CLEAVE at the 5’ END of the Apurinic or Apyrimidine site, while the bifunctional AP LYASE cleaves at the 3’ end.

Correct choice is (c) Addition

For explanation: DNA microsatellites have several repetitive SEQUENCES within it, thus these regions are prone to polymerase slippage that can LEAD to INSERTION or DELETION of bases.

Correct choice is (c) Seed

The explanation: Delicious apple was PRODUCED first as a somatic mutation within a BRANCH and was since propagated vegetatively. As somatic mutations don’t AFFECT GERM cells, it can’t be propagated through seeds.

Right choice is (d) AMINO and Keto

Best explanation: The normal FORM of A and C is amino and that of G and T base is Keto. SHIFT to imino and enol form is due to tautomeric shift, it is a rare phenomenon.

Right ANSWER is (d) The 3’overhang CHEWED up to chi sequence invades to search for homology

The BEST I can explain: In NHEJ to REDUCE the LOSS of genetic information and for a quicker search of homology domains the 3’ over hand is only 4 nt long. This increases the probability of fining micro homology region as well as prevents loss.