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51.

During Strisinger’s model, what happens?(a) The Ribosome slips which causes duplication(b) Parent strand slippage polymerase(c) Daughter strand slippage polymerase(d) The polymerase back tracks and resynthesizesI had been asked this question during an interview.This is a very interesting question from Chromosomal Abnormalities : Duplication in division Chromosomal Basis of Inheritance of Cytogenetics

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Right answer is (C) DAUGHTER strand slippage polymerase

The best explanation: Daughter strand slippage will CAUSE looping which is stabilized by repetitive SEQUENCES. This leads to an increase in a number of repeats in the daughter.

52.

Choose the wrong one out for insertion sequence.(a) Insertion sequences are transposons(b) Insertion sequences range from 5kb to 30kb(c) Insertion sequences only carry genes for their transposition(d) Insertion sequences have terminal IRI have been asked this question in an internship interview.Question is taken from Chromosomal Abnormalities : Translocation in chapter Chromosomal Basis of Inheritance of Cytogenetics

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Correct option is (b) Insertion sequences range from 5kb to 30kb

Best EXPLANATION: Insertion sequences are very short transposons which range from 768 BP to 5 kb, they indeed have genes only NEEDED for TRANSPOSITION and there is an inverted repeat at two TERMINALS.

53.

______________ inversions reduce crossing over in ______________(a) Paracentric, Heterozygous(b) Pericentric, Heterozygous(c) Paracentric, homozygous(d) Pericentric homozygousThe question was posed to me by my college professor while I was bunking the class.The question is from Chromosomal Abnormality : Inversion topic in portion Chromosomal Basis of Inheritance of Cytogenetics

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Correct option is (b) PERICENTRIC, Heterozygous

Easiest explanation: In heterozygous female the PAIRING is DUE to looping and then recombination in pericentric region can produce abnormal chromosomes which don’t SEPARATE properly in anaphase. Thus, it prevents recombination in heterozygous females.

54.

Translocation in plants was discovered by_____________(a) Stern(b) Barbara McClintok(c) Sutton and Boveri(d) MorganI have been asked this question during an online interview.Origin of the question is Chromosomal Abnormalities : Translocation topic in division Chromosomal Basis of Inheritance of Cytogenetics

Answer» CORRECT choice is (b) Barbara McClintok

The explanation is: Barbara McClintok 1st OBSERVED translocation in maize. Sutton and Boveri gave the chromosomal THEORY of INHERITANCE.
55.

The chromosomal theory of inheritance was proposed by ____________(a) Mendel(b) Watson and Crick(c) Darwin(d) Sutton and BoveriThis question was posed to me in an online interview.My question is from Chromosomal Theory of Inheritance in chapter Chromosomal Basis of Inheritance of Cytogenetics

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The CORRECT answer is (d) Sutton and Boveri

To ELABORATE: The chromosomal theory of inheritance was unknown in Mendel’s and Darwin’s time. It was proposed by Sutton and Boveri while Watson and Crick gave the STRUCTURE of the DNA.

56.

At what maternal age the probability of having a child with Down’s syndrome is increased drastically?(a) 25(b) 30(c) 35(d) 45This question was addressed to me in unit test.I want to ask this question from Chromosomal Abnormalities : Aneuploidy topic in portion Chromosomal Basis of Inheritance of Cytogenetics

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The correct OPTION is (c) 35

Explanation: After the age of 35 the PROBABILITY of non-disjunction in GAMETE leading to Down’s syndrome INCREASES to 1/290. With increasing age it increases still further.

57.

A non-disjunction in the 1^st phase of meiosis will lead to ______________(a) Monosomy(b) Nullysomy(c) Disomy(d) TrtrasomyThis question was addressed to me at a job interview.I'd like to ask this question from Chromosomal Abnormalities : Aneuploidy topic in portion Chromosomal Basis of Inheritance of Cytogenetics

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The correct option is (b) Nullysomy

Best explanation: Nullysomy in chromosome is a LOSS of 2 CHROMOSOMES i.e. loss of a CHROMOSOMAL pair. This can RESULT from passing of both homologous chromosomes to the same POLE during meiosis 1- this pole will produce trisomic gametes while the counterpart will produce nullisomic gametes.

58.

Patau syndrome is a result of which of the following?(a) Non-disjunction of sex chromosome in female(b) Non- disjunction of sex chromosome in male(c) Non-disjunction of chromosome 21(d) Non-disjunction of chromosome 13This question was posed to me in examination.The question is from Chromosomal Abnormalities : Aneuploidy topic in chapter Chromosomal Basis of Inheritance of Cytogenetics

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Correct OPTION is (d) Non-disjunction of chromosome 13

To explain I would SAY: Non-disjunction of chromosome 13 will RESULT in a trisomy and monosomy if the GAMETE is fertilized. This results in Pataeu syndrome. It is an autosomal aneuploidy.

59.

Normal wheat Triticum aestivum is ___________(a) Monoploid(b) Tetraploid(c) Pentaploid(d) HexaploidI got this question during an interview for a job.Question is taken from Chromosomal Abnormalities : Euploidy in chapter Chromosomal Basis of Inheritance of Cytogenetics

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Correct choice is (d) Hexaploid

Best EXPLANATION: TRITICUM AESTIVUM is hexaploid i.e. there are 6 set of CHROMOSOMES in it. This fertile as the SEGREGATION in divalent is possible.

60.

Duplication of Vermillion eye colour gene leads to the expression of the mutant phenotype. Which of the following can lead to normal phenotype with this duplication present?(a) Reverse tandem duplication(b) Duplication of normal eye colour gene(c) Deletion of both the Vermillion gene(d) Translocation of Vermillion gene to another chromosomeI have been asked this question at a job interview.This intriguing question originated from Chromosomal Abnormalities : Duplication topic in chapter Chromosomal Basis of Inheritance of Cytogenetics

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The correct OPTION is (b) Duplication of normal eye colour gene

The best explanation: Duplication in normal eye colour gene would balance the mutant gene product and lead to EXPRESSION of wild type phenotype. Reverse tandem duplication will only create more copies of mutant making phenotype worse, ALSO DELETION of both genes will cause imbalance. Translocation makes no difference in expression UNLESS it’s beside a heterochromatin.

61.

Non-autonomous transposons can’t transpose. State whether the statement is true or false.(a) True(b) FalseI have been asked this question by my college professor while I was bunking the class.Enquiry is from Chromosomal Abnormalities : Translocation in section Chromosomal Basis of Inheritance of Cytogenetics

Answer» CORRECT ANSWER is (b) False

To explain I would say: Non-autonomous transposons LACK the RECOMBINASE that aids in transposition. However, they can transpose with the help of the recombinase from another trasposon.
62.

Which of the following is a result of reciprocal translocation?(a) Burkitt’s lymphpma(b) Trychothiodystrophy(c) Thalassemia(d) Cockyne’s syndromeI have been asked this question in an international level competition.My query is from Chromosomal Abnormalities : Translocation in section Chromosomal Basis of Inheritance of Cytogenetics

Answer» CORRECT answer is (a) Burkitt’s lymphpma

Best explanation: Burkitt’s lymphoma is a result of reciprocal translocation between chromosome 8 and 14 of HUMAN GENOME. Trichothiodystrophy is a defect in NER.
63.

Which of the following can result from non-disjunction?(a) XY and XO(b) YY and XY(c) X and Y(d) XY and XXYI have been asked this question in exam.My question is based upon Chromosomal Theory of Inheritance in division Chromosomal Basis of Inheritance of Cytogenetics

Answer» CORRECT choice is (a) XY and XO

To explain: Non-disjunction in the male will produce XY and XO RESPECTIVELY. The other options are not suitable results of non-disjunction. In this case some part having XY will develop a male character with the parts with XO will have FEMALE characters.
64.

You hybridize two cells one of which carries a deletion mutation on chromosome 1. What will you expect to observe?(a) Buckling of chromatin of 1^st cell(b) Buckling of chromatin of 2^nd cell(c) Twisting and loop formation between the two chromatins(d) No observable differenceThe question was asked in class test.Question is from Chromosomal Abnormalities : Deletion in division Chromosomal Basis of Inheritance of Cytogenetics

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Right OPTION is (b) Buckling of chromatin of 2^nd cell

Easiest explanation: As chromosome 1 of 1^st cell carries the deletion it will have less genes than that of the counterpart on the 2^nd cell. There will be an ERROR in a pairing where the EXTRA part of 2^nd cell chromatin CORRESPONDING to the deleted gene will buckle.

65.

If the genes are located in a chromosome as A—B—C—D—E—O—-T. Which of the gene pairs will have least probability of being inherited together?(a) C and D(b) A and T(c) A and B(d) O and TThis question was posed to me during an interview for a job.This interesting question is from Chromosomal Theory of Inheritance in portion Chromosomal Basis of Inheritance of Cytogenetics

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The correct answer is (b) A and T

Easiest EXPLANATION: Longer is the distance, LESSER will be the probability of the two genes being inherited together as then there will be a greater CHANCE of RECOMBINATION.

66.

How will you recognize a terminal deletion from breakage and loss at the terminal end?(a) Indistinguishable(b) Terminal break will lead to shorter chromosome than that due to chunk deletion(c) Terminal break will be sticky(d) Deletion will be recognized by trans factorsThe question was asked during an interview for a job.My doubt stems from Chromosomal Abnormalities : Deletion in chapter Chromosomal Basis of Inheritance of Cytogenetics

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The CORRECT option is (c) TERMINAL break will be STICKY

The explanation is: In terminal deletions the telomere is present; it is sort of removing the BASES before telomere not including it. Terminal breaks will HOWEVER lack telomerase and will be sticky. Shortening in both cases could be similar.

67.

In Drosophila Crossing over in paracentric region is lethal and leads to cross-over suppression. This leads to reduction in fertility.(a) True(b) FalseThis question was addressed to me in an interview for job.My enquiry is from Chromosomal Abnormality : Inversion in chapter Chromosomal Basis of Inheritance of Cytogenetics

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The correct ANSWER is (b) False

To EXPLAIN: In Drosophila only females undergo recombination. In the case of EGG formation out of the 4 meiotic products, only one forms the egg and the 3 other forms the polar body. Thus, the cross over suppressed gamete simply forms the polar body so doesn’t affect fertility.

68.

Which of the following could result in homologous chromosomes moving to the same pole after anaphase 1?(a) Doubling(b) Inversion(c) Breaking(d) TranslocationI have been asked this question during a job interview.Enquiry is from Chromosomal Abnormalities : Translocation topic in chapter Chromosomal Basis of Inheritance of Cytogenetics

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Correct option is (d) Translocation

The EXPLANATION is: In case of translocation there will be 4 homologous chromosomes INVOLVED in the pairing in MEIOSIS 1, in that case segregation where homologous PAIRS move to same pole is possible.

69.

Aneuploidy is usually deleterious as _________________(a) Chromosomal pairing is hampered(b) Gene balance is disrupted(c) Size of individual may vary(d) Chromosomal disintegration is increasedI got this question in an internship interview.My question comes from Chromosomal Abnormalities : Aneuploidy in section Chromosomal Basis of Inheritance of Cytogenetics

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The correct choice is (b) Gene BALANCE is disrupted

The BEST I can EXPLAIN: In case of aneuploidy there is loss or gain in single CHROMOSOME which results in dis-balance of the genes and their products. This is a reason that aneuploids are usually lethal.

70.

A male ant is _______________(a) Polyploid(b) Triploid(c) Diploid(d) MonoploidI have been asked this question at a job interview.The above asked question is from Chromosomal Abnormalities : Euploidy topic in chapter Chromosomal Basis of Inheritance of Cytogenetics

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Correct choice is (d) Monoploid

For explanation: Male ants, bees and wasps are produced by PARTHENOGENESIS, THUS they have only ONE set of genes that was present in their MOTHER- being monoploid.

71.

Long pericentric inversions generally don’t act as cross over suppressors. Why?(a) Long stretches of DNA recombination not recognized(b) Mechanism is different for short and long inversions(c) Two events of crossing over take place(d) Cross over product in this is viableI had been asked this question in my homework.I'd like to ask this question from Chromosomal Abnormality : Inversion topic in division Chromosomal Basis of Inheritance of Cytogenetics

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The correct option is (c) Two events of crossing over take place

The explanation: Longer stretches of inversion favour formation of Double Crossover, then the DNA are resolved normally between all 4 meiotic products and no dicentric CHROMOSOME is formed. Thus, Long pericentric inversion GENERALLY don’t act as cross over suppressors.

72.

Which of the following is not true about conservative transposition?(a) It creates direct repeat sequences on either side(b) It is Copy-Paste mechanism(c) Transesterification reaction takes place(d) Class II transposons have this mechanismI have been asked this question by my college professor while I was bunking the class.This interesting question is from Chromosomal Abnormalities : Translocation topic in division Chromosomal Basis of Inheritance of Cytogenetics

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Right choice is (b) It is Copy-Paste MECHANISM

Explanation: Conservative or non-replicative transposition acts through CUT and paste mechanism. The donor site loses the transposon which is INSERTED in the recipient GENERATING direct repeats. It is seen in class II transposons.

73.

Which of the following case of duplication involves more than one chromosome?(a) Tandem duplication(b) Reverse tandem duplication(c) Displaced(d) TransposedThis question was posed to me in an internship interview.This interesting question is from Chromosomal Abnormalities : Duplication in chapter Chromosomal Basis of Inheritance of Cytogenetics

Answer» RIGHT option is (d) Transposed

For EXPLANATION: In transposed the SEGMENT could be passed to another homologous chromosome or extra chromosome with a centromere. However, TANDEM, revere tandem and displaced occurs WITHIN the same chromosome.
74.

Which of the following is an example of inversion?(a) Chromosome 22 and 9(b) Chromosome 8(c) Chromosome 14(d) Chromosome 3This question was posed to me in a national level competition.This question is from Chromosomal Abnormality : Inversion in chapter Chromosomal Basis of Inheritance of Cytogenetics

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Correct ANSWER is (d) Chromosome 3

To explain: In chromosome 3 of human there is duplication-deletion which leads to inversion. The REST of the OPTIONS give EXAMPLES of genes participating in TRANSLOCATION.

75.

Which of the segregation of translocated chromosome will result in non- lethal gamete?(a) Homologous chromosomes moving to same pole(b) Alternate segregation(c) Adjacent with the homologous chromosomes moving to same pole(d) Adjacent with homologous chromosomes moving to different poleThe question was posed to me by my college director while I was bunking the class.My question is taken from Chromosomal Abnormalities : Translocation in section Chromosomal Basis of Inheritance of Cytogenetics

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The correct OPTION is (b) Alternate SEGREGATION

For explanation I would say: In CASE of alternate segregation all the genes pass to both the gametes ALTHOUGH in different chromosomes, this does not lead to lethality. But in other cases, the genetic balance is DISTURBED.

76.

If there is a non-disjunction in the X chromosome, the progeny being XXY, what will be the constitution of the chromosome(s) in the gamete?(a) X and Y(b) X and XY(c) XX and Y(d) XXY with no separationI had been asked this question in semester exam.Question is from Chromosomal Theory of Inheritance topic in division Chromosomal Basis of Inheritance of Cytogenetics

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Right choice is (b) X and XY

Best explanation: The non-disjunction results in the PRODUCTION of XXY. Then these ALLELES separate as X and XY respectively. The male progeny in this CASE will be XYY and the females XXY but the XXX dies DUE to no gene compensation.

77.

Transposition doesn’t require much sequence homology between two genetic molecules, based on this hey can help in _____________(a) Homologous DSB repair(b) Non-homologous end joining(c) Site-specific recombination(d) Homologous recombinationThe question was posed to me during an interview for a job.This intriguing question comes from Chromosomal Abnormalities : Translocation in portion Chromosomal Basis of Inheritance of Cytogenetics

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Correct answer is (c) Site-specific RECOMBINATION

To elaborate: In case of site specific recombination there WOULD be very little or no homology between the donor and recipient site, this can be IMPOSED by translocation. Translocation doesn’t play a part in homologous repair or NHEJ.

78.

Which of the following doesn’t agree with the chromosomal theory of inheritance?(a) The genes are located on the chromosome(b) The genes on the same chromosome are always passed together(c) The genes are located linearly on the chromosomes(d) The distance between two genes can be mappedThe question was asked in examination.My doubt is from Chromosomal Theory of Inheritance in division Chromosomal Basis of Inheritance of Cytogenetics

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Correct choice is (b) The genes on the same chromosome are always passed TOGETHER

The best explanation: Genes from the same chromosome can show independent ASSORTMENT as well by recombination. However, very close gene TEND to be LINKED and are passed on together.

79.

If an organism has 14 chromosomes, the number of chromosome generated by nullisomy will be_____________________(a) 15(b) 7(c) 13(d) 12I have been asked this question in semester exam.This interesting question is from Chromosomal Abnormalities : Aneuploidy topic in section Chromosomal Basis of Inheritance of Cytogenetics

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Correct OPTION is (d) 12

Explanation: NULLISOMY RESULTS in (2n-2) set of chromosome, here n=7, so it will result in 12 chromosome over all.

80.

Colchicine is used to cause______________(a) Mitotic non-disjunction(b) Meiotic non-disjunction(c) Mitotic disjunction(d) Meiotic disjunctionI had been asked this question in an internship interview.The above asked question is from Chromosomal Abnormalities : Euploidy in section Chromosomal Basis of Inheritance of Cytogenetics

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The correct CHOICE is (a) Mitotic non-disjunction

Easiest explanation: Colchinine causes an arrest in anaphase which leads to mitotic non-disjunction HALTING the mitosis. This leads to DOUBLING of chromosome as the DUPLICATED chromatins fail to separate.

81.

Which of the following will be sterile?(a) Tetraploid(b) Triploid(c) Diploid(d) MonoploidI got this question in an online interview.My doubt stems from Chromosomal Abnormalities : Euploidy in chapter Chromosomal Basis of Inheritance of Cytogenetics

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82.

Euploidy is a chromosomal variation in ______________________(a) Size(b) Position of genes(c) Number(d) StructureI have been asked this question in examination.The doubt is from Chromosomal Abnormalities : Euploidy topic in section Chromosomal Basis of Inheritance of Cytogenetics

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Right option is (C) NUMBER

The BEST I can explain: EUPLOIDY is chromosomal VARIATION in number which shows increase or decrease in complete set of chromosome number.

83.

Which of the following will have the highest number of eye facete?(a) Ultra bar female Drosophila(b) Heterozygous Bar male Drosophila(c) Heterozygous Bar female Drosophila(d) Homozygous bar FemaleI had been asked this question by my school teacher while I was bunking the class.My query is from Chromosomal Abnormalities : Duplication in chapter Chromosomal Basis of Inheritance of Cytogenetics

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Correct answer is (C) Heterozygous BAR FEMALE Drosophila

For explanation I would say: The bar gene is present in the X chromosome, THUS, males can’t have two copies for being homozygous or heterozygous. Ultra bar condition is most severe, then is the homozygous bar and then follows bar.

84.

Positive effect Variegations can lead to___________(a) Translocation(b) Over expression(c) Under expression(d) SuppressionThis question was posed to me in an online quiz.This is a very interesting question from Chromosomal Abnormalities : Translocation topic in portion Chromosomal Basis of Inheritance of Cytogenetics

Answer» CORRECT choice is (d) Suppression

To explain I would say: DUE to translocation some part of the gene can get transcribed to a POSITION NEAR heterochromatin REGION. This will suppress the otherwise expressed gene.
85.

If you cross a white eyed female drosophila with a red eyed male drosophila, what will be the colour of eyes for their male and female offspring?(a) Both red eyed(b) Both white eyed(c) Red eyed daughter and white eyed son(d) Red eyed son and white eyed daughterThis question was posed to me at a job interview.I need to ask this question from Chromosomal Theory of Inheritance topic in chapter Chromosomal Basis of Inheritance of Cytogenetics

Answer»

Correct answer is (c) Red EYED daughter and white eyed son

Explanation: The EYE colour of Drosophila is an X linked gene. So as the female is homozygous recessive white and the male has on red ALLELE and one Y chromosome, sons will only get Y chromosomes from father. Sons will thus be white eyed while daughters will be red eyed.