Category: CSIR NET Life Science Previous Year Questions and Solution on GENE MAPPING METHODS

Category: CSIR NET Life Science Previous Year Questions and Solution on GENE MAPPING METHODS

CSIR NET Life Science Previous Year Questions and Solution on GENE MAPPING METHODS

95. The following figure represents a physical map and a genetic map for 5 different genes (a to e) Which one of the following statements based on the above is correct? (1) The region between b and c is more recombinogenic than the other loci. (2) In comparison to the region between a and b, the region between d and e is more recombinogenic. (3) 1 CM is equal to 1 Kb (4) If more markers were mapped between d and e. the genetic distance between d and e is likely to decrease

Genetic Map vs Physical Map

94. The following figure depicts the relationship between a genetic map for four genes (A, B, C and D) and their corresponding physical map. The following statements are made to explain this relationship. A. More number of recombination events occur between A and B as compared to B and C. B. Lesser number of recombination events occur between C and D as compared to B and C. C. Although the physical distance between A and B is less than that between C and D, the region between A and B is more recombinogenic. D. The physical distance between A and B is less than that between C and D, and thus the region between A and B is less recombinogenic. E. Although the physical distance between C and D is more than that between B and C, the region between C and D is less recombinogenic. F. Although the physical distance between C and D is more than B and C. the region between C and D is more recombinogenic. Which statements are correct? (1) A and B (2) C and E (3) D and F (4) A, C and E

Genetic Map vs Physical Map

93. A series of ceil lines was created by fusing mouse and human somatic cells. In mouse-human somatic cell hybrids, human chromosomes tend to get lost before becoming a stable cell line. Some hybrid cell lines may carry human chromosome deletions. Each cell line was examined for the presence of chromosomes and for the production of an enzyme. The following results were obtained: Cell line Gene product Chromosomal segments 1p 1q 2p 2q 3p 3q 4p 4q 5p 5q A + + + - - + + - - + + B + - - + + - - + + + + C - + + - - + + - + - + D + - - + + + + - - + - E - + + + + - + - - - - Which segment of the chromosome has the gene encoding for the enzyme (1) 1p (2) 5p (3) 5q (4) 4p

Somatic Cell Hybridization

92. Three somatic hybrid ceil lines, designated as X, Y and Z, have been scored for the presence or absence of chromosomes 1 through 8, as well as for their ability to produce the hypothetical gene product A, B, C and D) as shown In the following table: Hybrid cell lines Human chromosome present Gene product expressed 1 2 3 4 5 6 7 8 A B C D X + + + + - + - + Y + + + + + - - + Z + + + + + + - + Which of the following option has most appropriately assigned chromosomes for each of the given genes? (1) Gene A on chromosome 5. Gene B on chromosome 3, Gene C on chromosome 8 and Gene D on chromosome 1 (2) Gene A on chromosome S and Gene B on chromosome 3 only (3) Gene D on chromosome 8, Gene C on chromosome 1, Gene B on chromosome S and Gene A on chromosome 4 (4) Gene A on chromosome 5. Gene B on chromosome 3 and Gene D on chromosome 1

Somatic Hybrid Cell Lines

91. A panel of six hybrid cell lines, each containing a different subset of human chromosome, was examined for the presence of the gene product as shown below: Cell Gene Human chromosome line product present present 1 2 3 4 5 6 7 8 9 10 A + + + ++ - - - - - - B + - - + + + + + - - - C - - + + - - - - + + + D - -+ -- - + +++ - E - -+ - -- +- - - - F + + + -+ ++- - - - The gene which codes for the given gene product is located on which chromosome? (1) Chromosomes 3, 4 or 5 (2) Chromosomes 3 (3) Chromosomes 3 or 4 (4) Chromosomes 4

Gene Mapping by Somatic Cell Hybridization

90. Somatic cell hybridization is used to assign a gene to aparticular chromosome. When two cell lines from twodifferent species are fused, they form a heterokaryonwhich tends to Jose chromosomes as they divide,preferentially from one species. A panel of cell lines wascreated from mouse-monkey somatic cell fusions. Eachline was examined for the presence of monkeychromosomes and for the production of a given enzyme.The following results were obtained: Cell line Presence of Enzyme Presence of Monkey Chromosomes 1 2 3 4 5 6 7 8 9 10 A + + + + + + - + - + + B + - + + - + + + - + - C - - + - - - - - + + + D + + + + - - + + + + - E - - - - + + - - - + + F + + + - + + + + - + - On the basis of these results, which chromosome has the gene that codes for the given enzyme? (1) Chromosome 10 (2) Chromosome 7 (3) Chromosome 1 (4) Chromosome 5

Somatic Cell Hybridization Gene Mapping

89 Interspecific somatic hybridization are often followed (1) Fusion of paternal chromosome (2) Fragmentation of Paternal chromosomes (3) Selective elimination of one of parental chromosome (4) Fusion of nuclei

Interspecific Somatic Hybridization Selective Chromosome Elimination

88. During fusion of Rat and Human cell it was found that chromosomes of one species was selectively eliminated, this had opened new doors to study (1) Transgenic study (2) Chromosomal incompatibility (3) Gene physical mapping (4) Dominance of genes

Rat Human Cell Fusion Chromosome Elimination Gene Mapping

87. use of inactivated Sendai virus and PEG for fusion of two cells has proved very beneficial in studies for- (1) Transgenic (2) Tissue culture (3) Physical mapping (4) Recombinant DNA technology

Sendai Virus PEG Cell Fusion Physical Mapping

86. The somatic cell hybridization of human and mouse cell can be effectively mediated by (1) CaCl2 treatment (2) Heat Shock (3) Inactivated Sendai virus (4) Enzymatic treatment

Somatic Cell Hybridization Human Mouse Cells

85. The following scheme represents deletions (1-4) in the rll locus of phage T4 from a common reference point: (The bars represent the extent of deletion in each case) Four point mutations (a to d) are tested against four deletions for their ability (+) or inability to give wild type (rll+) recombinants. The results are summarized below: a b c d 1 + + + + 2 + + + - 3 + - + - 4 - - + - Based on the above the predicted order of the point mutations is: (1) b-d-a-c (2) d-b-a-c (3) d-b-c-a (4) c-d-a-b

Deletion Mapping of rII Locus in Phage T4

84. The location of six deletion (Shows as solid line underneath the chromosomes) has been mapped to the Drosophila chromosome as shown in the diagram given below: The recessive mutations a, b, c, d, e f and g are known to be located in the region of deletions, but order of mutations on the chromosome is not known. When flies homozygous for the recessive mutations are crossed with flies homozygous for the deletion, the following result were obtained where the letter 'm' represents a mutant phenotype and ‘+’ represents the wild type. The relative order of seven mutant genes onchromosome is: (1)bceafgd (2) b c d f g e a (3) bc deafg (4) c d e a g f b

Deletion Mapping in Drosophila

83. The locations of five overlapping deletions have been mapped to a Drosophila Chromosome as shown below (Horizontal lines in the above figure indicate the deleted regions) Recessive mutations a. b. c, d and e are known to be located within this region, but the order of mutations on the chromosome is not known. When the flies homozygous for the recessive mutations are crossed with flies homozygous for the deletions, the following results are obtained (letter "m" represents mutant phenotype and "+" represents the wild type). On the basis of the above data, the relative order of the five mutant genes on the chromosome is (1) bc de a (2) a b c d e (3) bc e a d (4) c d be a

Deletion mapping in Drosophila

82. The following is a schematic representation of region(showing six bands) of the polytene chromosome ofDrosophila, along With the extent of five deletions (Del1 to Del5): Recessive alleles a, b. c. d. e and f are known to correspond to each of the bands (1 to 6), but their order is not known. When the recessive alleles are placed against each of these deletions, the following results are obtained. The plus (+) in the table mdtcates wild type phenotype of the corresponding allele, while a minus indicates the phenotype governed by the corresponding mutant allele. a b c d e f Del 1 + - - - + + Del 2 + + - - + + Del 3 - + - - + + Del 4 - + + - - + Del 5 - + + + - - Which one of the following indicates the correct tocauonof the recessive alleles on the bands of the polytene chromosomes? (1) a-3; b-1; c-2; d-4; e-5; f-6 (2) a-2; b-1; c-3; d-4; e-5; f-6 (3) a-4; b-1; c-2; d-3; e-5; f-6 (4) a-6; b-2; c-3; d-4; e-1; f-5

Deletion Mapping on Drosophila Polytene Chromosomes

81. A novel enzyme was Identified in humans. The followingapproaches are available to identify the chromosome on which the gene encoding the enzyme is present: (A) Suppress the activity of enzyme by RNAi. (B) Identify polymorphism in the population and carry out pedigree analysis to study its inheritance. (C) Purify the enzyme, decipher Its amino acid sequence, predict its DNA sequence and search for its presence in the available human genome sequence. (D) Create chromosome addition lines by making somatic hybrids between human and mouse cells, identify lines showing the enzyme activity and the human chromosome present in it. Which of the above approaches can be used? (1) (A) or (B) (2) (B) or (C) (3) (C) or (D) (4) (A) or (C)

Methods to Identify the Chromosome Location of a Novel Human Gene

80 To detect mutation (GAGGTG) allele specific hybridization method is used. Four members of an affected family are investigated. DNA isolated from blood samples of parents and two offsprings are spotted on a membrane after appropriate processing and probed with either TGACTCCTGAGGAGAAGTC (first probe) or TGACTCCTGTGGAGAAGTC (second probe) after labelling. While probed with first oligonucleotide. signals are obtained for the positions where DNA are spotted from parents and offspring II. When probed with second oligonucleotide, signals are obtained at position where DNA from the parents and offspring I are spotted. Results are shown below: Father Mother Off-spring I Off-spring II First Probe + + - + Second Probe + + + - On the basis of the result, which of the following statements is correct? (1) Parents are affected (2) Offspring I is affected (3) Offspring II is carrier (2) Offspring II is affected

Allele-specific hybridization (GAG→GTG) dot-blot

79.interspecific somatic cell hybrids have proved very useful for (1) Gene mapping (2) Genetic manipulation (3) Gene structure (4) Gene function

Interspecific Somatic Cell Hybrids in Gene Mapping

78 'TaqMan' assay for detection of base substitutions (DNA variants).probes (oligonucleotides) with fluorescent dyes at the 5’-end and quencher at 3’-end are used. While the probe is intact, the proximity of the quencher reduces the fluorescence emitted by reporter dye. If the target sequences (wild type or variant) are present, the probe anneals to the target sequence, downstream to one of the primers used for amplifying DNA sequence flanking the position of variants. For an assay two flanking PCR primers. two probes corresponding to the Wild type and variant allele and labeled with two different reporter dyes and quencher were used. During extension the probe may be cleaved by the Taq-polymerase separating the reporter dye and the quencher. Three individuals were genotyped using the assay. Sample for individual I shows maximum fluorescence for the dye attached to the wild type probe, sample for individual II shows maximum fluorescence for the dye attached to variant probe and sample for individual Ill exhibits equal fluorescence for both the dyes, Which of the following is correct? (1) Individual is homozygous for the variant allele. (2) Individual II is homozygous for the variant allele. (3) Individual is homozygous for the Wild type allele. (4) Individual lit is homozygous for the Wild type allele.

TaqMan Assay Genotyping

77. The pedigree given below represents the genotype at four different loci for the children in generation III. Which one of the given genotypes is likely to represent the genotype of individual-II?

Genotype of Individual II-2 at Four Loci

76. In the following pedigree three STR loci A, B and C are linked on the long arm of the X-chromosome in the order centromere-A-B-C-telomere. Further in the table, the STR alleles present in each individual is indicated. Based on the above, X-chromosome(s) in which of the following individuals are recombinant? [Hint: X-chromosome in males will help phase of the alleles) (1) II-1, III-1 and III-2 (2) II-2, III-1 and III-2 (3) III-1 and III -3 (4) III-2 and III-4

X‑Linked STR Loci Pedigree Problem

75. Polymorphic DNA sequences are used for molecular identification. Short tandem repeats (STRS) and Single Nucleotide Polymorphism (SNPs) are used as polymorphic markers. The table below summarizes the status of autosomal. SNP, autosomal STR, mitochondrial SNP, Y- linked STR for four individuals related to each other, representing parents and their two children. Based on the above data, identify the individuals representing the two parents. (1) Individuals A and D (2) Individuals A and C (3) Individuals B and C (4) Individuals C and D

Identifying Parents Using Autosomal SNP, STR, mtDNA SNP, and Y‑STR Markers

74. The following represents sequences of different alleles of a gene found in family represented by mother, (allele1/ allele2), father (allele1/allele 2) and their two sons: Son1 (allele1/allele 2) and Son 2 (allele1/allele2). Further, a new mutation was observed in one of the alleles of the son, which is marked with a triangle. Mother allele 1 CAGCATAGTCATTCGTCCATGGACTAG Mother allele 2 CAGCATTCTCAUCGTCCATGGACTAG Father allele 1 CAGCATTGTCATTCGTCCATGGGCTAG Father allele 1 CAGCTTAGTCATTCGTCCATGGTACTAG Son 1 allel 1 CAGCATAGTC.ATTCGTCCATGGACTAG Son 1 allele2 CAGCTTACTCATTCCTCCATTGTACTAG Son 2 allele 1 CAGCATTGTCATTCGTCCATGGACTAG Son 2 allele 2 CAGCTTAGTCATTCGTCCATTGTACTAG The following statements were made about the mutation: A. The mutation arose in the germline of the father. B. The mutation arose in the son. C. The given DNA sequences are present on the X chromosome. D. There is a possibility to use RFLP for tracking this variation. Which one of the following options presents a combination of correct statements? (1) B only (2) A and D (3) A and C (4) B and D

Germline Mutation Analysis in a Family Pedigree Using DNA Sequence Variation and RFLP

73. Microsatellite are used as marker for identifying individuals via DNA fingerprinting os the alleles may differ in the number of repeals. From the Southern blot shown below identify the progeny (A, B, C and D) for the given parents (M = mother, F = father). (1) A, B, C and D (2) A, B and D (3) A and D only (4) B, C and D

Microsatellite DNA Fingerprinting

72. The pedigree given below follows the inheritance pattern of a late-onset (after age of 30 years) genetic disease that is 100% penetrant. Affected individuals are indicated by a solid circle (woman) or solid square (males). RFLP analysis of DNA from each individual is shown below in the pedigree. Which grandchildren (IIIb to IIId) will be affected by the disease after attaining the age of 30 years? (1) onlyIIIb (2) Both IIIb and IIIe (3) Both IIIc and IIId (4) Both IIIb and IIId

Late-onset autosomal dominant pedigree analysis with RFLP

71. An analysis of four microsatellite markers was carried out in a family showing a genetic disorder. The results are summarized below Based on the above, which ot the markers shows linkage to the disorder? (1) M1 (2) M2 (3) M3 (4) M4

Microsatellite Marker Linkage Analysis

70. Sickle cell anemia is a recessive genetic disease caused due to a point mutation in the 6th codon abolishing one of the MspIIendonuclease digestion site present in the ß-globin gene. MspII digested DNA from a normal person gives two bands, 115O bp and 200 bp. in ß-globin gene. A family with a pro-band (based on the disease phenotype) gave thc following MspII digestion pattern: The following conclusions were drawn: A. Son (I) is the proband and the given mutation is not present in Son (II). B. The daughter is a carrier for the given mutation. C. The gene is X-linked and thus Son becomes the proband. D. The father and daughter are affected E. A de novo mutation in same site on normal allele has allowed appearance of diseased phenotype in the proband Which of the following combination of conclusions wilt be the most appropriate for the figure given above? (1) A. Band E (2) A. B and C (3) B. C and E (4) C and D

Sickle Cell Anemia Restriction Digest Problem

69. In normal individuals, there are three Mstll restrictionsites. two flanking the ß-globin gene and one within thegene. In individuals affected by a disease, a singlenucleotide polymorphism in the ß-globin gene abolishesthe internal Mstll recognition site. The RFLP pattern forthis locus, obtained by hybridization using a probeinternal to the flanking Mstll sites, from three siblings ofa family is shown below: Based on the above profile, what is the nature of the genetic disorder? (1) X-linked Recessive (2) Autosomal Dominant (3) Autosomal Recessive (4) X-linked Dominant

RFLP Analysis of β‑Globin MstII Sites

68. The following picture represents a gel profile of a pair of DNA markers observed P1 and P2, their F1 progeny and F2 progeny. Four different profiles were observed in case of F2 progeny showing a given profile is indicated in brackets Based on the above observation, which one of the following statements is correct? (1) Co-dominant DNA markers were used for this study. (2) The polymorphic DNA bands represents two independent genes. (3) If the P1 parent was crossed to the Ft individual, the progeny will show all the four profiles as observed in the case of F2 progeny. (4) If an F2 progeny which does not show either of the DNA markers (last lane of the above gel) is crossed to a P1individual the obtained progeny will have two types of individual, one which shows a band and the other where no band is observed

DNA Marker Gel

67. The AFLP technique generates polymorphic DNA fragments that are generally scored as dominant markers. However, a pair of DNA fragments (say 'a'and 'b') generate by AFLP can be termed as co dominant, if on analysis of a large progeny of doubted haploids (DH) derived from an F1 (resulting from a cross between two parents one with fragment 'a' and the other with it is observed that: (1) 1.50% of the progeny has both ‘a' and ‘b’ fragments and the rest have none. (2) 50% of the progeny has fragment 'a' and the remaining have fragment 'b' (3) 25% of the progeny has fragment 'a', 50% both 'a' and 'b' and the rest fragment (4) 75% of the progeny has both the fragments. While 25% has either ‘a' or 'b'.

AFLP codominant markers in doubled haploid progeny

66. The following represents selected AFLP bands (l to V) observed in parents (P1 and P2), Fl progeny and 20 doubled haploid (DH) progeny developed from the F1. DH are created through chromosome doubling of pollengrains in anther culture. The following statements were made about the above AFLP bands: A Bands I and IV are allelic. B. Bands II and V assort independently. C. Band III is uninformative. Which one of the following options represents a combination of all correct statements? (1) A only (2) C only (3) A and B only (4) A, B and C

AFLP bands, doubled haploids, and allelism

64. Doubled haploids (DH) are plants derived from single immature pollen and doubled artificially to form diploids. A DH population was created from progeny derived from a cross between two parents (P1 and P2), one resistant (R) and the other sensitive (S) to white rust. The parents, Fi and DH population were profited with a set of co-dominant markers, which is represented below. The following table summarizes the proposed percentage of the 4 different types (1 to 4) of DH progeny, assuming that the DNA marker is (i) unlinked to the gene governing the trait and (n) linked at a distance of 10 cM. Which one of the following options correctly represents the expected ratio for unlinked and linked, respectively? (1) A, ii (2) A, i (3) B, i (4) A, iii

Doubled Haploids, Marker–Trait Linkage and Recombination

65. The figure below represents a profile of DNA markers in two parents (Pl and P 2), progeny (F1) from a cross between P1 and P2 and that of gametes produced from F1. Eight different patterns (DH1 to DH8) were observed in case of gametes. The numbers below, DHI to DHS indicate the number of individuals observed in each case Based on the above observations, the following statements were made: A. Markers 'b' and 'f' are likely to be allelic in nature B. Markers 'c' and 'd' are linked in trans with a map distance of 24 CM C. Marker 'b' assorts independently from marker Which one of the following have a combination of all correct statements? (1) A, B and C (2) A and B (3) A only (4) C only

DNA Marker Mapping

63. In 1990, Bhattacharya et al identified that the wrinkled seed character of pea as described by Mendel is caused by a transposon — like insertion in a gene encoding Starch Branching Enzyme (encoded by the R allele). This leads to an RFLP pattern, when genomic DNA of round and wrinkled seed is digested With EcoRI and probed with the cDNA of the R gene product. The following is a representation of the hybridization pattern. Choose the INCORRECT statement? (1) Lane 3 represents genomic DNA from plant withwrinkled seeds (2) These DNA markers are codominant in nature (3) Lane 1 represents genomic DNA from plant with round seeds (4) If genomic DNA from F2 progeny as obtained in Mendel's work was analyzed by RFLP, the ratio of progeny with patterns in late 1, 2 and 3 will be 2:1:1

RFLP analysis of wrinkled vs round pea seeds

62. Alleles A, a, B and b can be distinguished on the basis of their mobility on an agarose gel. These genes are present on the same chromosome. In the gel imagebelow, the band pattern reflects the alleles in parents and their progeny (number reflects the progeny counted). Which one of the following statements correctly explains the band pattern? (1) In the heterozygous parent, the alleles are in coupling configuration. (2) In the heterozygous parent, the alleles are in repulsion configuration. (3) The alleles A and B are in different linkage groups. (4) The information is insufficient for any conclusion.

Coupling vs Repulsion: Inferring Linkage Phase from Gel-Band Progeny Counts

61. The RFLP pattern observed for two pure parental lines (P1 and P2) and their F1 progeny is represented below. Further, the P1 plant had red flowers while the P2 had white flowers. The F 1progeny was backcrossed to P(2)The result Obtained, showing the number of progeny with red and white flowers and their RFLP patterns is also represented below. Which one of the following conclusions made is correct? (1) The DNA marker and the gene for the flower colour are 10 cM apart. (2) The marker and the gene for the flower colour are 5 CM apart. (3) The marker and the phenotype are independently assorting. (4) The marker and the gene for the colour segregate from one another.

RFLP Testcross Mapping

60. A pair of alleles govern seed size in a crop plant. 'B' allele responsible for bold seed is dominant over 'b' allele controlling small seed. An experiment was carried out to test if an identified dominant DNA marker (5kb band) is linked to alleles controlling seed size. A plant heterozygous for the marker and the alleles was crossed to a small seeded plant lacking the 5kb band. 100 progeny obtained from the cross wereanalysed for the presence and absence of the DNA marker. The result are tabulated below: Phenotype Plant with bold seed Plant with small seed No. of progeny showing presence or absence of DNA marker Prsent Absent 22 23 Present Absent 27 28 Based on the above observations which one of the following conclusions is correct? (1)The DNA marker assorts independently of the phenotype (2) The 5kb band is linked to the allele 'B' (3) The 5kb band is linked to the allele 'b' (4) The DNA marker assorts independently with bold seed but is linked to the small seed trait

Seed Size Gene and 5 kb DNA Marker

59. A new mutant called early ripening (EL) is identified in a plant. The wild type phenotype is late ripening (LR). Further DNA marker(s) is/are observed to be polymorphic between EL and LR plants. A cross was made between pure lines of LR and ER. The F1 progeny was test crossed and its progeny was analyzed. The parental, F1 and progeny of test cross were also analyzed for the DNA marker. The table below summarizes thephenotype of the progeny and the pattern of DNA marker observed in each case: Based on the above information, the following statements were made: A. LR is dominant to ER B. The DNA marker used is a dominant marker C. The DNA marker is linked to the phenotype Which of the above statements are correct? (1) A only (2) A and B only (3) A and C only (4) A, B and C only

Early vs Late Ripening in Plants

58. Two homozygous individuals (Pl and P2), were genotyped using dominant DNA markers A and B, as shown below. The F1 progeny obtained was test crossed and frequency of progeny with which different genotypes appear, is given below: The following conclusions were made: A. In the Fl markers A and B are linked and in coupling phase (cis) B. In the Fl markers A and B are linked and in repulsion phase (trans) C. The distance between A and B is 10 CM D. The distance between A and a is 5 CM Which of the above conclusions are correct? (1) A and C (2) A and D (3) B and C (4) B and D

Test Cross Mapping with Dominant Markers

57. In recent decades.the use of genetic markers has allowed the rapid introgression and selection of desired breeding stocks in advance generations. In this regard, following statements are given: A. AFLP markers can discriminate between homozygote and heterozygote genotypes. B. Microsatellites (eg. SSR) are capable of detecting higher level of polymorphism than RFLP. C. SNPs are more prevalent in the coding regions of the genome. D. SNP markers are the most suitable markers for bothforeground and background selection. Which one of the following combination of the above statements is correct? (1) A, B and C (2) A, B and D (3) B and C only (4) B and D only

Genetic Markers in Introgression Breeding

56. The following are statements above molecular markers in the context of plant breeding A. Molecular markers can be used for elimination of undesirable traits B. Molecular markers cannot be used for elimination of genetic contribution of each individual parent in a segregating population C. Molecular markers are used for mapping of QTLs, which is also possible by conventional techniques. D. Molecular markers can be used for selection of individuals from a population that are homozygous for the recurrent parent genotype at loci flanking the target locus. Which of the above statements are TRUE? (1) A and B (2) A and C (3) A and D (4) B and C

Molecular Markers in Plant Breeding

55. Marker-assisted selection (MAS) defined as selection based on molecular markers should have some important criterion for plant breeding activities. Some statements about these criteria are mentioned below: A. Marker should co-segregate with the desired trait of interest. B. Marker should not co-segregate with the desired trait of interest. C. Marker should be un-linked with the desired trait of interest. D. Marker is used for indirect selection of a genetic determinant or determinants of a trait of interest. Which one of above combinations is correct? (1) A and B (2) B and C (3) C and D (4) A and D

Essential Criteria for Effective Marker-Assisted Selection (MAS) in Plant Breeding

54. Which one of the following DNA markers can be used to distinguish between a homozygote and heterozygote? (1) RAPD (2) AFLP (3) RFLP (4) ISSR

DNA Markers to Distinguish Homozygote and Heterozygote

53. In scanning Simple Sequence Repeats (SSR) primers are used against (1) Random sequence (2) Repetitive sequence (3) Flanking region of repetitive sequence (4) Conserved region of exon of gene

Understanding SSR Primers and Their Target Regions in Simple Sequence Repeat Scanning

52. Centromere positions can be mapped in linear tetrads in some fungi. A cross was made between two strains a b and a+ b+ and 100 linear tetrads were analysed. The genes a and b are located on two arms of the chromosome. The tetrad were divided into 5 classes as shown below. Class 1 2 3 4 5 a b a+ b a b+ a+ b+ a b a b+ a+ b a+ b+ a b a b a+b+ a+b+ a b a+b+ a b a+b+ a b+ a+b a+ b a b+ Linear tetrad 15 29 52 2 2 Based on the above observation the following conclusion were drawn: (A) Class 1 is a result of cross over between 'a' and the centromere. (B) Class 2 is result of double crossover involving 3 strands between 'a' and the centromere. (C) Class 5 is a result of a double crossover between 'a' centromere and centromere. (D) Class 4 is a result of a double cross over involving all the 4 strands. Which one of the following options represents all correct statements? (1)A and B (2) A and C (3) B and D (4) C and D

Centromere Mapping in Linear Tetrads

51. Fertilization between two mating types (Pl and P2) of Neurospora, led to a diploid ascus cell, which gave rise to ascus containing 8 haploid ascospores. A set of DNA markers representing two linked loci was analyzed in P1, P2 and the octads labeled 01 to 08 arranged from the tip to the base of the ascus. The observed profile is represented below: Which one of the following is a correct conclusion of the above observation? (1) Bands labeled (a) and (c) are allelic and segregation occurred during meiosis II (2) Bands labeled (b) and (d) are allelic and segregation occurred during meiosis II (b) (3) Bands labeled (a) and (d) are allelic and segregation occurred during meiosis II (4) Bands labeled (c) and (d) are allelic and segregation occurred during meiosis I

Neurospora linked loci mapping from ordered octads

50. The following table summarizes the result of a cross between two strains of Neurospora having the alleles D and d, respectively. The table shows the different patterns of octad arrangement and the number of ascus observed of each type. Octads D d D d D d D d D d D d D d d D d D D d d D d D d D D d d D d D D d d D d D d D D d d D d D D d 115 125 14 16 17 13 Number of ascus observed Based on the above, fill in the blanks from the options given below. "The first two columns are from meiosis with nocrossover between locus D and __[A]__. The patternfor these two columns represent __[B]__ segregation pattern. The distance between the locus D and thecentromere is __[C]__ map units". A B C (1) d allele first division 10 (2) Centromere first division 10 (3) d allele second division 20 (4) Centromere second division 10

Neurospora Octad Analysis

49. In Neurospora a cross between the genotypes 'A' and 'a' results in an ascus with ascospores of genotypes as shown below: Statements A to D areevents that could have occurred during meiosis. A) Crossing over between the centromere and the gene. B) Segregation of alleles 'A' and 'a' in meiosis I. C) Segregation of alleles 'A' and 'a' in meiosis II. D) Assortment of alleles 'A' and 'a'. Which of the above events could correctly explain the observation shown in the figure? (1) A followed by C (2) C alone (3) A followed b (4) D alone

Neurospora tetrad analysis

48. Segregation of alleles can occur either at anaphase I or anaphase II of meiosis. Which one of the following is an ideal model system for identifying the stage at which allelic segregation occurred? (1) Arabidopsis thaliana (2) Drosophila melanogaster (3) Neurosporacrosso (4) Saccharomyces cerevisiae

Ideal Model System to Study Allelic Segregation in Meiosis

47. A pedigree shown below depicts that the individual l-I is heterozygous for a dominant disease allele D and for molecular markers MI/M2. The paternal molecular markers present in the progeny individuals are indicated in the pedigree. The following statements may be drawn from the above pedigree: A. The two loci D/d and MI/M2 appears to be linked B. The recombination frequency between the two loci is 20% C. If LOD score comes out to be 3, then it ensures that the two loci are independently assorting D. A LOD score

Pedigree LOD Score

46. The table given below shows the Lod score values of three different pairs of genes studied for assessing if they are linked pairs: Gene Pair ‘1’ Gene Pair ‘2’ Gene Pair ‘3’ Lod Score 1 2 3 The following conclusions were made from the data given above: A. For gene pair 1 the probability of the genes being linked is 10 times more likely than them assorting independently. B. For gene pair 2, theLod score 2 indicates that the probability of the genes being linked is twice more likely than assorting independently. C. The genes of pair1 and 2, can both be considered as linked, while the genes of pair 3 exhibits independent assortment. D. The genes of pair 3 can be considered as linked. Which one of the following options represents statement(s) that is/are correct? (1) A and C (2) B and D (3) A and D (4) A and B

LOD Score Interpretation

45. DNA markers have enormous potential to improve theefficiency and precision of conventional animal breedingvia marker-assisted selection (MAS). Among thefollowing which would be most suitable marker forselection of animals with agronomic traits? (1) RFLP (2) RAPD (3) EST-SSR(4) Minisatellite

Choosing the Best DNA Marker for Animal Breeding

44. Molecular marker cannot be utilized for (1) Mapping of genes (2) Identifying the clones (3) Identifying the locus of gene on chromosome (4) Identifying the expressed product

 Molecular Marker Uses and Limitations in Genetics

43. A Lod score of 3 represents a Recombination Frequency that is (1) 3 times as likely as the hypothesis of no linkage (2) 30 times as likely as the hypothesis of no linkage (3) 100 times as likely as the hypothesis of no linkage (4) 1000 times as likely as the hypothesis of no linkage

Understanding What a Lod Score of 3 Means in Genetic Linkage Analysis

42. A virgin Drosophila female was crossed with a wild type male. The Fi progeny obtained had four types of males as shown below. Phenotype White eyed Wild type Cross veinless White eyed and crossveinless Number 50 3 44 3 Assuming that white eye and crossveinless mutations are X-linked and recessive, the following statements were made: A. F1 females were also of four types as that of males. B. The white eyed crossveinless male flies appeared due to independent assortment. C. The map distance between the genes for white eye and crossveinless is estimated to be 12 cM. D. The map distance between white eye and crossveinless is estimated to be 6 cM. E. All Ft females are expected to be Wild type. F. The F1 wild type males appeared due to crossing over. The combination with correct statements is: (1) C, E, F (2) A, B, D (3) A, D, F (4) B, D, E

Drosophila White Eye and Crossveinless X-Linked Genetics

41. Which of the following molecular marker is non-PCRbased (1) RAPD (2) RFLP (3) SSR(4) SNP

Non-PCR Based Molecular Marker

40. Segregation of alleles can occur at Anaphase or at Anaphase n of meiosis. With reference to this statement, which one of the following organism is an ideal model system for identifying stage of allelic segregation at meiosis? (1) Neurosporacrossa (2) Saccharomyces cerevisiae (3) Pisumsativum (4) Drosophila melanogaster

Ideal Model Organism to Identify Stage of Allelic Segregation in Meiosis

39. E coli cells were simultaneously infected with two strains of phage λ, One strain of λ had a mutant host range, is temperature sensitive and known to produce clear plaques (genotype h st c); another strain of λCarried the wild type alleles (genotype h+st+ c+), progeny phage were collected from the lysed cells and were plated on bacteria. The following numbers of different progeny were obtained: Progeny phage genotype Number of plaques h+st+ c+ 350 h+ c st 86 h+ c+st 4 h c st 300 h+ c st+ 90 h c st+ 6 h c+st+ 114 h c+st 50 What will be the order of the three genes and the map distance between them? (1) h 36cM c 15cM st (2) c21cMh15cMst (3) h21cM st15cM c (4) h36cM c∞cM st

Lambda phage three‑point test cross

38. When F1 female Drosophila of he genotype a+a b+ b c+ c is test crossed, the following progenies were obtained: Progeny classes No. of progenies a+ b+ c+ 22 a+ b+ c 28 a b c+ 26 a b c 24 a+ b c+ 230 a+ b c 220 a b+ c+ 225 a b c 225 Total 1000 The progeny has been shown as classes derived from the female gamete. Statements A to F as given below are conclusions derived from the above result. A) Genes a and b are linked in cis. B) Genes a and b are linked in trans. C) Genes a and b are linked in cis while b and c are linked in trans. D) The genotype of the parents are a+ a+ b+ b+ and aabb E) The genotype of the parents are a+ a+ b b and aab+b+ F) Genes a and b are 10cM apart. Which of the above statements are correct? (1) C alone (2) A, E and F (3) B, E and F (4) A, D and F

Three-Point Test Cross in Drosophila

37. The genetic map of three genes in Drosophila melanogaster is given below: cu e se 20 cM12 cM A cross as given below individuals of the genotype: a+ b+ c a b c+ X a+ b+ c a b c+ The female F1 progeny are test-crossed and 1000 progeny are obtained. Assuming that there has been no double crossover, what is the expected number of progeny with the genotypes : a+b c+ a+ b+ c+ a+ b+c a b c a b c a b c Select the set which shows the correct number of expected progeny. SET (A) (B) (C) (1) 100 50 850 (2) 50 25 425 (3) 100 850 50 (4) 0 425 75

Expected Progeny Numbers Without Double Crossovers

36. Curled wing (cu), ebony body colour (e) and sepia eye (se) are three recessive mutations that occur in fruit flies. The loci for these mutations have been mapped and they are separated by the following hypothetical map distances: cue se 20 cM12 cM The interference between these genes is 0.4. A mutant cu e se fly was crossed with a homozygous wild type fly. The resulting Fl females were test crossed that produced 1800 progeny. What number of flies in each phenotype class is likely to be obtained in the progeny of the test cross? (1) Non recombinants will be 1250; single crossoverbetween cu and e 334; single cross over betweene and se 190; double cross over 26 (2) Non recombinants 1181; single crossover between cu and e 360; single cross over between e and se 216; double cross over 43 (3) Non recombinants 1198; single crossovers 576; double cross overs 26 (4) Non recombinants 1233; single crossover 524; double cross over 43

Three‑Point Test Cross in Drosophila

35. In Drosophila order of genes under investigation is as shown in figure. b+ cm+ w+ 5cM 10cM if drosophila with genotype b+ cm+ w+ b cm w is test crossed what would be observed frequency of progenies with genotype b+ cm w+ b cm w assuming zero interference ? (1) 0.005 (2) 0.02 (3) 0.0025 (4) 1.5

Drosophila Three‑Point Test Cross

34. In D. Melanogaster, traits 'b' and 'c' result from recessive alleles that are located on one of the autosomes. The resulting Fl females were test crossed and the following progeny (total of 1000) were obtained: a+b+c+ 350 a+bc 25 abc 380 ab+c+ 35 ab+c 100 abc+ 10 a+bc+ 85 a+b+c 15 The following conclusions were made from the above data: A. The order of genes isa b c and the distance between a and b is 8.5cM. B. The order of genes is a c b and the distance between a and c is 8.5 cM. C. The order of genes is b c a and the distance between b and c is 21 cM. D. The order of genes is c b a and the distance between b and c is 8.5 cM. Which one of the following options represent statement(s) that is/are correct? (1) A only (2) A and D (3) B and C (4) D only

Understanding Gene Order and Map Distances in Drosophila

33. Poplar is a dioecious plant. A wild plant with genes AABBCC was crossed with a triple recessive mutant aabbcc. The Fl male hybrid (AaBbCc) was then back crossed with the triple mutant and the phenotypes recorded are as follows: AaBbCc 300 aaBbCc 100 aaBbcc 16 AabbCc 14 AaBbcc 65 aabbCc 75 aabbcc 310 Aabbcc 120 The distance in map unit (mu) between A to B and B to C is (1) 25 and 17 mu, respectively (2) 33 and14 mu. Respectively (3) 25 and 14 mu, respectively (4) 33 and 11 mu. Respectively

Calculating Genetic Map Distances in Poplar Using Backcross Phenotypic Data

32. In Drosophila, Bar eye (B) is a dominant mutation whileminiature wing (m) and yellow body colour (y) arerecessive mutations. eterozygous females for thesemutations were crossed to normal eyed miniaturewinged and yellow body coloured males. Assume thefollowing progeny was obtained: Phenotypes Numbers B+ m+ y+ 30 B m y+ 25 B m+ y+ 165 B+ m y+ 120 B m y 20 B+ my 185 Bm+y 110 B+ m+ y 45 Based on the results obtained, the order of genes will be: A. B m y B. m B y The genetic distance between B and y will be: C. 40 cM D. 17.1 cM The correct combination of answer is (1) A and C (2) B and C (3) A and D (4) B and D

Point Test Cross in Drosophila

31. A three point test cross was carried out in Drosophila melanogaster involving three adjacent genes X, Y and Z. arranged in the same order. The distance between X to Y is 32.5 map unit (mu) and that between Y to Z is 20.5 map. The coefficient of coincidence = 0.886. What is the percentage of double recombinants in the progeny obtained from the testcross? (1) 6% (2) 8% (3) 12% (4) 16%

Calculating Percentage of Double Recombinants in a Three Point Test Cross in Drosophila

30. In Drosophila, balancer chromosomes are used to keep all the alleles on one chromosome together. A balancer contains multiple inversions; so that when it recombines with the corresponding wild type chromosome, no viable cross over products are formed. Balancers also carry an allele for a dominant phenotype. A Drosophila male with sepia eye color is crossed to a female carrying a third chromosome balancer (TM6B). The allele for sepia phenotype (se) is located on chromosome 3 and is recessive to the wild type eye color. The dominant marker for TM6B is a tubbyphenotype. Further, an individual homozygous for TM6B balancer does not survive. F1 progeny with tubby phenotype is sib-mated. The F2 progeny is expected to have: (1) only sepia eye color (2) sepia. tubby and wild type flies in a ratio of 1:2:1 (3) sepia and tubby flies in a ratio of 1:2 (4) sepia and wild type flies in ratio of 3:1

Genetics of Drosophila Balancer Chromosome TM6B and Sepia Eye Color

29. The additive nature of a genetic map as suggested by Alfred Sturtevant and T. H. Morgan is possible if there is: (1) no interference in crossovers. (2) complete interference in crossovers. (3) partial interference in crossovers. (4) variable interference in crossovers dependent on the genetic distances.

Additive Nature of Genetic Maps

28. Ratio of observed double crossover frequency to the expected frequency of double crossover gamete during three point test cross is termed as- (1) Coefficient of interference (2) Coefficient of Coincidence (3) Coefficient of variance (4) Coefficient of suppression

Understanding Coefficient of Coincidence in Three-Point Test Cross

27. The ratio of observed number of double crossover to expected number of cross over is termed as (1) Coefficient of coincidence (2) Coefficient of interference (3) Hitch hiking (4) Linkage suppression

Understanding the Ratio of Observed to Expected Double Crossovers in Genetics

26. In an experiment involving mapping of 3 genes (a, b andc) in Drosophila, a three point test cross is carried out.The parental cross was AAbbCCXaaBBcc. The genotypesof the double crossovers are: Aabbcc and aaBbCc. Basedon this, determine the order of the genes. (1) acb(2) cab (3) abc(4) bac

How to Determine Gene Order Using Three-Point Test Cross in Drosophila

25. In a haploid organism, the loci A/a and D/d are 8 map units apart. In a cross Ad X aD, what will be the proportion of each of the following progeny classes: (a) Ad (b) Recombinants. (1) 92%, 8% (2) 46%, 8% (3) 92%, 4% (4) 46%, 4%

Progeny Proportions in Haploid Organisms

24. A test cross was made with phenotypically wild type Drosophila files hiving genes for sepia eye and curled wing in heterozygous condition. The following results were obtained: Wild type 400 individuals Sepia eyed 150 individuals Curled winged 100 individuals Sepia eyed curled winged 350 individuals The result indicates (1) Independent assortment, as wild and the double mutant types are in almost equal proportion (2) Unequal segregation, as it is showing departure from 1:1:1:1 ratio (3) Complete linkage, as the two genes are very closely placed (4) Linkage, the two genes are separated by 25 CM distance

Understanding Linkage or Independent Assortment from Test Cross

23. Which statement is true for Drosophila melanogaster? (1) female are achiasmatic (2) recombination does not occur during male meiosis (3) recombination does not during female meiosis (4) males show high frequency of recombination

Understanding Recombination in Drosophila melanogaster

22. A cross was made between pure wild type males and brown eyed curled winged females of D. melanogaster. The F1 females were test crossed. The F2 progeny obtained was as follows: Wild type 200 Brown eyes, curled wings 150 Brown eyes, normal wings 30 Normal eyes, curled wings 20 Total400 The genetic distance (cM) between brown eye and curled wing loci is: (1) 12.5 (2) 50 (3) 150 (4) 25

Calculating Genetic Distance Between Brown Eye and Curled Wing Genes in Drosophila Melanogaster

21. In a linkage map, two genes A and B, are 70 CM apart. If individuals heterozygous for both the genes are test crossed number of progeny with parental phenotype will be: (1) equal to the number of progeny with recombinant phenotype (2) more than the number of progeny with recombinant phenotype (3) less than the number of progeny with recombinant phenotype (4) could be more or less than the number of progeny with recombinant phenotype depending on whether the genes are linked in cis or trans, respectively

Understanding Parental and Recombinant Progeny in Linkage Maps

20. The maximum frequency of recombination that can occur between two loci is (1) 25% (2) 50% (3) 75% (4) 100%

 Maximum Frequency of Recombination Between Two Loci Explained

19. In an organism, allele governs grey body colour,while its mutant allele a gives yellow body colour.Further, presence of allele gives long and thin hairswhile b allele gives rise to short and thick hairs. Theallelesa+ and b+ are dominant over aand b,respectively. An individual with the genotype a+ b+ a b Has a patch of yellow cells with short and thick hairs. Which one of the following events is most likely to lead to the above? (1)Non disjunction of the homologous chromosomesduring mitosis (2) Somatic recombination involving a and b (3) Translocation occurring in a few somatic cells (4) Mutation of both and alleles in the somatic

Somatic recombination and mosaic yellow patches

18. Assume that the genes w+ and cv+ are located 20 CM apart on the X chromosome of Drosophila melanogaster. Mutations in w+and cv+ give rise to white eyes and crossveinless phenotypes, respectively, which are recessive to the wild-type phenotype. A homozygous wild-type female was crossed to a white- eyed, crossveinless male. The F1 progeny was sib-mated. What percentage of the progeny will be white-eyed and crossveinless? (1)20 (2) 40 (3) 10 (4) 5

Genetic Cross Analysis of White-Eyed and Crossveinless Traits in Drosophila melanogaster

17. Two mutations were isolated in bacteriophage, one causing clear plaque (c) and the other causing minute plaque (m). The genes responsible for these two mutations are 9 CM apart. The plaques with genotype c+m- and c-m+ were mixed to infect bacterial cells. The progeny plaques were collected, cultured and plated on bacteria. The expected number of the different types of plaques are shown below: A. c+m+455, c+m- 45, c-m+ 45, c-m- 455 B. c+m+ 455, c+m- 455, c-m+ 45, c-m- 45 C. c+m+45, c+m- 455, c-m+ 455, c-m- 45 D. c+m+65, c+m- 680, c-m+ 685, c-m- 70 Which one of the following options represents the combination of all correct statements? (1) A only (2) B only (3) C only (4) Cand D

Understanding Genetic Recombination in Bacteriophage Mutations Affecting Plaque Morphology

16. Two genes a and b are located at a distance of 10 cM. Individuals of the genotype AaBb are sib-mated. The two genes are linked in trans. What percentage of the progeny is expected to have the genotype aabb? (1) 0.25 (2) 0.01 (3) 6.25 (4) 25

Expected Percentage of aabb Progeny from Sib-Mating AaBb Individuals with Linked Genes in Trans at 10 cM

15. Genetic mapping reveals that distance between two genes 'A' and 'B' is 10 cM. What is chance of getting Aabb progeny if AaBb is test crossed? (1) 5 & (2) 10 % (3) 45 % (4) 90 %

Genetic Mapping and Probability of Aabb Progeny in AaBb Test Cross

14. A plant with genotype r+h+/ r- h- was test crossed. Out of total 280 progeny 260 are r+h+/ r-h- and r-h-/r-h-. The recombination frequency will be (1) 92.8 (2) 46.4 (3) 7.2 (4) 3.6

How to Calculate Recombination Frequency from a Test Cross in Genetics

13. The distance between gene A and B is 10 cM. If a genotype A b is selfed, the a B percentage of progeny with genotype aabb will be (1) 10% (2) 25% (3) 0.25% (4) 0.01%

Probability of aabb progeny after selfing Ab/aB

12. Distance between the two linked genes A and B is 20 cM. On test cross of A b a B with recessive parent how many offspring will have genotype A B a b (1) 10 (2) 20 (3) 40 (4) 80

Test Cross Problem

11. Distance between gene A and B is 10 cM. If the F1 genotype A B was test a b crossed then what is probability of obtaining A B genotype a b (1) 90 (2) 45 (3) 10 (4) 5

Probability of Obtaining AB/ab Genotype When Distance Between Gene A and B Is 10 cM

10. After meiosis the 20 % gametes are recombinant for two genes. The distance between two genes will be (1) 5 Cm (2) 10 cM (3) 20 cM (4) 40 CM

How to Calculate Distance Between Genes from Recombinant Gametes Percentage

9. Direct correlation between recombination frequency and distance between genes can be disturbed by presence of (1) Heterochromatin (2) Exons (3) Introns (4) Euchromatin

Impact of Heterochromatin on Recombination Frequency and Gene Distance Relationship

8. In an organism if number of linkage group is 12, then the number of haploid set of chromosome is (1) 12 (2) 6 (3) 24 (4) 4

Number of Haploid Chromosome Sets in an Organism with 12 Linkage Groups

7. In number of linkage groups present in human male is: (1) 21 (2) 22 (3) 23 (4) 24

Number of Linkage Groups in Human Male Explained

6.The gene map produced on based on recombination frequency data are termed as (1) Deletion map (2) Linkage map (3) Physical map (4) Cytogenetic

Understanding Gene Maps

5. Maximum recombination frequency between two genes during meiosis of diploid organism can be (1) 25% (2) 50 % (3) 75 % (4) 100 %

Maximum Recombination Frequency Between Two Genes During Meiosis Explained

4. Then typical Mendel test cross ratio in two point cross (1) 9:3:3:1 (2) 1:2:1 (3) 1:1 (4) 1:1:1:1

Understanding Mendel’s Test Cross Ratio in Two-Point Crosses

3. In genetic map, distance between two genes is measured in units of (1) cM (2) A0 (3) bp (4) nm

Understanding Genetic Map Distance

2. When a cross is made between dominant long and full endosperm (LLSS) and recessive short and shrunken endosperm (Ilss). The progeny obtained 98.3 % were long and full endosperm and short and shrunken, rest 1.7 % were long shrunken endosperm and short full endosperm. Such a ratio was observed due to (1) Epitasis (2) Linkage (3) Co-dominance (4) Over dominance

Understanding the Genetic Cross of Long and Full Endosperm with Short and Shrunken Endosperm in Maize

1. Frequency of crossing over is directly proportional to- (1) Length of chromosomes (2) Number of genes (3) Distance between genes (4) Number of chromosomes

Frequency of Crossing Over is Directly Proportional to Distance Between Genes

24. Of the following meiotic events (A) formation of synaptonemal complex (B) crossing over (C) arrangements of chromosomes at the equatorial plate at Metaphase I (D) separation of chromatids at Anaphase II Which one(s) lead to variation? (1) (B) only (2) (B) and (C) (3) (A) and (D) (4) (B) and (D)

Meiotic events that create genetic variation: crossing over and metaphase I alignment

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