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Right CHOICE is (d) XIS from PHAGE and FIS from bacteria

Explanation: Excision of the genome of LAMDA phage from the bacterial genome requires IHF and XIS that is coded by the bacteriophage. It also requires the FIS coded by the bacterium.

Correct ANSWER is (d) The daughter strand without lesion

For explanation I would SAY: In this model to BYPASS the lesion the daughter strand synthesized from the other PARENT strand is used as a template. This codes for the correct bases against the lesion and is made possible by fork REGRESSION.

Correct answer is (B) Translocation

The EXPLANATION is: Translocation of a gene from ONE chromosome to another within the same GENOME will result in then being on same chromosome. Then they can show LINKAGE relationship.

Correct answer is (d) 6.2

Easy explanation: 6.2 monomers of Rec A are necessary for ONE turn AROUND the central CHROMOSOME. They form a RIGHT handed alpha HELIX around the chromosome.

Correct CHOICE is (b) Interference

For explanation: A single cross over in nearby REGION depreciates that probability of a DCO OCCURRING in nearby regions. This PHENOMENON is known as interference.

Right choice is (b) STERN

To explain I WOULD say: Barbara McClintok and Creigton observed CROSSING over in Plants while Stern showed cytological OBSERVATION of the same in ANIMALS.

The correct answer is (a) The 3’-OH end of the nick INVADES other COMPLETE strands

For explanation I would say: In single stranded nick the 5’ end of the nick forms the base pair and normal DNA strand with the available strand but the 3’ end has to invade the 5’ end to bypass the nick. In this bypass the nick is repaired and not left as it is.

Right CHOICE is (C) 0.022

Explanation: Expected double cross over FREQUENCY by probability rule is product of the two single CROSSOVER frequencies in three point MAPPING. Here 20.8X10.0/10000 results on 0.022.

The CORRECT answer is (C) It will be heterozygous

Easiest explanation: The exchange in TWO strand stage will lead to the initial strand having 2 COPIES of BAR becomes normal and the normal chromosome has two copies. But the resultant will still be heterozygous. Only exchange in 4 strand stage will give ultra-bar.

Correct option is (c) Decreases, Increases

The BEST EXPLANATION: As the distance between the two gens increase, the probability of recombination event OCCURRING between then is higher. Thus, linkage increases as the distance between two genes DECREASE and vice-versa.

The CORRECT option is (a) A single circular chromosome

Explanation: Site specific RECOMBINATION in BACTERIAL DNA and viral DNA results in a single circular chromosome. This is used by viruses to integrate into bacterial GENOME.

Right option is (d) Exponential

Easy EXPLANATION: The observed curve is seen to be exponential which reaches a SATURATION POINT when the recombination FREQUENCY is about 50%.

Right choice is (B) 50%

Explanation: Taking all types of a DOUBLE CROSS over in CONSIDERATION only 50% of the resultant will show recombination between the genes which are located on two sides of these recombination events.

Correct option is (b) Integrase and IHF

For EXPLANATION: Integration is brought ALONG by the integrase-which is the recombinase CODED by the lambda phage. It also requires IHF or Integration Host Factor coded by the host.

The correct option is (C) A HOMOLOGOUS stretch of 20-200bp

Best explanation: SITE specific recombination TAKES PLACE between regions of very short sequence homology about 20-200bp. It can occur between non-homologous chromosomes as well.

The correct ANSWER is (c) PACHYTENE

The BEST explanation: Pachytene stage in prophase 1 of MEIOSIS is when the recombination takes place. This is after the formation of a synaptonemal complex in zygotene when SYNAPSIS occurs.

Correct answer is (b) Two

To explain I would SAY: Due to the semiconservative nature of DNA replication in the 1^st ROUND the chromosome will have only one PARENTAL strand and one strand with BRDU. When two rounds are complete one chromosome had only BrdU strand and other has parental type. The BrdU strand is lighter so DISTINGUISHES the appearance.

Correct answer is (c) 8

Easiest explanation: In MEIOSIS 1 there is 1^st doubling of the chromosomes which are then DISTRIBUTED in two cells at its end. These are then further distributed in two cells leading to their number becoming HALF. THUS after 1^st meiosis the number should be 8.

The correct option is (a) True

Easiest explanation: It is stated in the chiasmatype THEORY, that strand breakage and reunion occurs PRIOR to chiasma formation. Both sides of a chiasma REPRESENT a reduction division of chromatids. These recombination SITES lead to the appearance of O shaped structures during anaphase.

Right OPTION is (d) Branch point

Explanation: The point where the two strands complementary to a SINGLE STRAND meet is known as a branch point. Chimera is a PROTEIN made by combining DOMAINS from two different proteins.

Correct choice is (b) LYSOGENY in lambda phage

The best I can explain: The phage particles attach to the BACTERIAL genome USING site-specific RECOMBINATION REACTION. This is via the means of att sites.

Correct answer is (b) It will be inverted

Easiest EXPLANATION: A flanking sequence between two oppositely oriented SYMMETRY ELEMENTS will be inverted on site specific recombination.

Right answer is (c) Cleaves at recombination site

For explanation: REC A HELPS to take up the homologous chromosome from the media and it aligns it in a manner helping in recombination. It ALSO helps in branch migration but it PLAYS no role in cleaving the DNA.

Correct OPTION is (B) Circular plasmid

Easy explanation: Circular chromosomes can appear as HOLLIDAY INTERMEDIATE as when they are TWISTED the twist points may appear as crossing over sites.

Correct option is (b) REPAIR

The explanation is: In prokaryotes, the repair is mainly the SOLE recombination as these organisms have only ONE copy of chromosomes. Two COPIES are available only during fission when this recombination takes place. Gene transfer processes and their INCORPORATION are very rare.

Correct CHOICE is (b) Three copies of bar are called ultra bar

Easiest explanation: Bar is present in single COPY in normal genes. Recombination results in two bar genes which in HOMOZYGOUS condition can produce a severe REDUCTION in NUMBER of eye facets. Recombination can also result in 3 copies of bar gene in a chromosome which reduces the number to 45, known as the ultra bar.

The CORRECT option is (c) No SYNAPSIS

Easy explanation: In Drosophila MALE these is no formation of synaptonemal COMPLEX, hence no synapsis is the pachytene phase of meiosis I. As there is no synapsis, there is no recombination.

The CORRECT choice is (b) Even number of DCO between two genes

Easy explanation: Even number of DCO between two genes LEADS to EXCHANGE of small PARTS between the two genes but the resultant is the same as the parental as the terminal genes remain unchanged.

Correct answer is (b) THREONINE

Explanation: The activated threonine RESIDUES of ONE pair OPPOSITE recombinase molecules first ATTACK creating the Holliday intermediate. Then the other opposite threonine pair resolves it.

Right OPTION is (d) 4

The best I can explain: DOUBLE cross over event between for STRANDS will result in all the strands exchanging their genes. Thus, all the resultant will be recombinant so recombination FREQUENCY is 100%.

The correct choice is (d) FOUR STAND stage

To explain I would say: The only possibilities in this case are RECOMBINATION in 2 strands and 4 strand stage. The study of recombinants show that recombination at 2 stand stage will produce effectively the same combination as in the ABSENCE of recombination. THUS it must be in the 4 strand stage.

Right option is (b) The gene under study will be deleted from SPECIFIC induced cell

Easy explanation: This method is used to selectively delete some genes from some tissues in vivo. In this case the CRE gene is ACTIVATED when an induction signal is provided and the Cre protein so formed ACTS as a trans factor on LoxP sequence to remove the flanking gene of interest.

Right CHOICE is (c) Exchange of middle portion between two strands

To EXPLAIN: Double CROSS over event occurring between two strands only result in the PRODUCTION of two strands that exchanged a portion between them in the middle. The other two strands remain unaffected.

The correct answer is (a) True

For explanation I would say: The transient DNA protein linkage in this RECOMBINATION process preserves the PHOSPHODIESTER BACKBONE. Thus no ATP or other high energy MOLECULE is necessary.

The correct option is (d) 30.8%

EASY explanation: RECOMBINATION frequency between terminal GENES will be as {(Single cross over frequency)+2(double cross over frequency)}/100. This will RESULT in 30.8%.

Right choice is (d) Recombinase

To explain I would SAY: Recombinase is a tetrameric PROTEIN that binds to the TWO chromosomes undergoing site-specific recombination reaction. It is a MUST for the reaction.

Correct choice is (b) At the branch point

Explanation: While Rub B binds on other side of the branch point AIDING in branch migration, RUV A binds at the CHI structure STABILIZING it.

The correct choice is (b) 50%

To elaborate: When TWO GENES are unlinked they assort independently. So there is an equal chance of production of parental and recombinant PHENOTYPE. THUS RECOMBINATION frequency is 50%.

Correct ANSWER is (b) p+/v+ p-/v- and p-/v- p-/v-

The BEST explanation: In MALE drosophila RECOMBINATION is ABSENT thus there can’t be formation of p+/v- and p-/v+ gametes at all. This can however be formed in case we use female F1 for the test cross.

Right answer is (B) Three point mapping

The explanation: When the DISTANCE between two genes is quite large there is increases possibility of double recombination event OCCURRING between them. This can be avoided by 3 point mapping.