Category: CSIR NET Life Science Previous Year Questions and Solution on Morphogenesis and Organogenesis in Animals

Category: CSIR NET Life Science Previous Year Questions and Solution on Morphogenesis and Organogenesis in Animals

CSIR NET Life Science Previous Year Questions and Solution on Morphogenesis and Organogenesis in Animals

83. The mammalian genital ridge is bipotential. Which one of the following statements regarding determination of the fate of genital ridge is INCORRECT? (1) The activation of Sox9 gene promotes testis determining pathway. (2) The accumulation of β-catenin is critical for activating ovarian development. (3) R-spondin 1 (Rspo1) stimulates the Disheveled protein, thus promoting testis determining pathway. (4) Though Wnt4 is expressed in the bipotential gonads, it is an important factor in ovary determination.

Sex Determination in Mammalian Bipotential Genital Ridge: Key Genetic Pathways and Misconceptions

82. Following statements were made about sex determination in Drosophila melanogaster A. It is achieved by a balance of female determinants on the X-chromosome and male determinants on the autosomes. B. A Drosophila with 0.66 value of X:A ratio would develop intersex type. C. Due to non-involvement of Y chromosome in sex determination process, XO Drosophila develop as normal fertile male. D. The high value of X:A ratio facilitates activation of feminizing switch gene Sex - lethal (sxl). E. Sex specific expression of sxl causes selective activation of dosage compensation genes in female Drosophila. Select the option with combination of all correct statements. (1) A, B, E (2) A, B, D (3) B, C, E (4) B, C, D

Sex Determination in Drosophila melanogaster: Understanding the X:A Ratio and Gene Function

81. A mutant embryo of Drosophila in which one of the major sex determining gene, sex lethal, can only undergo default splicing, was allowed to develop. The following statements are towards explaining the determination of sex of the embryo: A. The embryo will develop into a male fly B. The embryo will develop into a female fly C. sex lethal gene product directly regulates sex specific alternate splicing of double sex RNA D. sex lethal gene product regulates sex specific splicing of transformer RNA which in turn regulates splicing of double sex RNA The correct combination of above statements to explain sex determination of the given embryo is: (1) A and C (2) A and D (3) B and D (4) B and C

Role of Sex-lethal (Sxl) Gene and Its Splicing Regulation in Drosophila Sex Determination

80. Sxl genes of Drosophila regulate expression at (1) Transcriptional level (2) Post transcriptional level (3) Translational level (4) post translational level

Sex-lethal (Sxl) Gene Regulation Mechanism in Drosophila

79. Individuals having X chromosome and short arm of Y chromosome are male while individuals having X chromosome and long arm of Y chromosome are female. This suggest that (1) Genes for maleness are located on short arm of Y chromosome (2) Genes for maleness are located on long arm of Y chromosome (3) Genes for maleness are located on X chromosome (4) Male determining genes are not located on Y chromosome

Location of Male-Determining Genes on the Y Chromosome: Short Arm vs Long Arm

78. In Drosophila XO are male and XXY are female while in humans XO are female and XXY are male. On the basis of given information which statement is NOT true (1) Y chromosome do not play any role in sex determination of drosophila (2) Y chromosome is sex determinant in humans (3) In humans sex determination is based on number of X chromosome to sets of autosomes. (4) In Drosophila sex determination is based on number of X chromosome to sets of autosomes.

Drosophila vs Humans: Understanding Roles of Y Chromosome and X:A Ratio in Sex Determination

77. It has been observed the transfer of certain X segment in XO male of Caenorhhabditis elegans, restores female character. It suggest that mode of sex determination is (1) X/A ratio (2) XX-XY type (3) Hormonal type (4) Ploidy difference

Sex Determination in Caenorhabditis elegans: The Role of X Chromosome Dosage and X/A Ratio

76. Drosophila with chromosome complement XXXXY/AA will be- (1) Meta female (2) meta male (3) normal male (4) Intersex

Sex Chromosome Complement XXXXY in Drosophila: Understanding Meta Female Phenotype

75. Sex determination in Drosophila is based on ratio of number of X chromosomes to- (1) Number of autosomes (2) Sets of autosomes (3) Number of Y chromosome (4) Sets of total chromosomes

Sex Determination in Drosophila: The Role of X Chromosome to Autosomes Ratio

74. The following assumptions were derived from the above experiment: A. Medium A contained FGF and PDGF. B. Medium B contained retinoic acid. C. Cells cultured in Medium B were determined to become functional neurons prior to addition of the medium. Which one of the following combinations represents correct statements? (1) A and B only (2) A, B and C (3) B and C only (4) A and C only

Understanding the Roles of FGF, PDGF, and Retinoic Acid in Neural Differentiation

Understanding What Is Not True About Lateral Inhibition in Cell Differentiation

71. Transforming the neural plate into a neural tube is an important event towards the formation of central nervous system, in which the following sub-events might take place: A. In primary neurulation, the cells surrounding the neural plate direct the neural plate cells to proliferate, invaginate and separate from the surface ectoderm to form a hollow tube. B. In secondary neurulation, the neural tube does not arise from the aggregation of mesenchyme cells into a solid cord. C. The morphogen, Sonic hedgehog, that is expressed in notochord, is required for induction of floor plate cells in the neural plate to form the medial hinge point. D. In mammals, secondary neurulation begins at the level of sacral vertebrae. E. In mammals, the primary neurulation forms brain regions while the secondary neurulation takes care of forming rest of the central nervous system from neck to tail. Which one of the following options represents the combination of all correct statements? (1) A, C and D (2) A, B and E (3) B, C and D (4) C, D and E

Key Processes in Neural Tube Formation: Role of Primary and Secondary Neurulation, Sonic Hedgehog, and Mammalian Development

70. Given below are some of the statements in connection with neural tube formation in vertebrates: A. In primary neurulation the cells surrounding the neural plate direct the neural plate cells to proliferate, invaginate and separate from the surface ectoderm to form an underlying hollow tube. B. In secondary neurulation the neural tube arises from the aggregation of mesenchyme cells into a solid cord that subsequently forms cavities to create a hollow tube C. In birds primary neurulation generates the neural tube from anterior up to the hind limb developing region. D. In mammals, secondary neurulation begins at the level of sacral vertebrae E. Anencephaly results when a failure to close the neural tube occurs, resulting in the forebrain remaining in contact with amniotic fluid. Which one of the following options gives all correct statements? (1) A, B, C, D and E (2) A only (3) B and E only (4) C, D and E only

Comprehensive Overview of Primary and Secondary Neurulation in Vertebrate Neural Tube Formation

69. In two major ways the neural plate is converted into a neural tube, primary neurulation and secondary neurulation. A. In primary neurulation, a hollow neural tube is formed. B. In secondary neurulation, the neural tube arises from the coalescence of mesenchymal cells into a solid cord. C. The solid cord formed in secondary neurulation subsequently gets hollowed. Which of the above processes is/are involved in brain development in vertebrates? (1) A and B (2) B and C (3) only A (4) only B

Primary and Secondary Neurulation in Vertebrate Brain Development

67. A type of regeneration in which the differentiated cells divide, maintaining their differentiated function without dedifferentiation and production of undifferentiated mass, is known as (1) Epimorphosis (2) Morphallaxis (3) Compensatory regeneration (4) Stem cell mediated regeneration

Compensatory Regeneration: Differentiated Cell Division Without Dedifferentiation

66. Given below are some of the statements regarding regeneration: A. The type of regeneration characteristic of mammalian liver is considered as compensatory regeneration. B. Regrowth of hair shaft from follicular cells exemplifies stem cell mediated regeneration. C. Regeneration occurring through the re-patterning of existing tissues with little new growth is known as morphallaxis. D. Adult structures undergoing dedifferentiation forming a blastema, that then re-differentiates to form the lost structure is called epimorphosis. Choose the most appropriate combination of correct statements: (1) D only (2) C and D only (3) A, Band C only (4) A, B, C and D

Understanding Types of Regeneration: Compensatory, Stem Cell-Mediated, Morphallaxis, and Epimorphosis

65. Following statements were made regarding regeneration in different organisms: A. The regenerating blastema cells in amphibians retain their specification even when they dedifferentiate. B. A transgenic Hydra when made to misexpress β- catenin will show numerous ectopic tentacles. C. In Planaria, if the Wnt pathway is activated, then the posterior blastemal would regenerate a head. D. A regenerating blastema is formed in the mammalian liver. Which one of the following options represents all correct statement(s)? (1) A only (2) C only (3) B and C (4) C and D

Correct Statements About Regeneration in Various Organisms

64. Clonogenic neoblasts are involved in planarian (flatworm) regeneration. This is an example of: (1) epimorphosis (2) morphallaxis (3) stem cell-mediated regeneration (4) compensatory regeneration

Clonogenic Neoblasts and Planarian Regeneration: An Example of Stem Cell-Mediated Epimorphosis

63. In case of Hydra, the major head inducer of the hypostome organizer is a set of Wnt proteins acting through the canonical β- catenin pathway. What would be the result, if a transgenic Hydra is made to globally mis-express the downstream Wnt effector β- catenin? (1) Ectopic buds will be formed all along the body axis and even on the top of the newly formed buds. (2) Ectopic tentacles form at all levels. (3) Both ectopic tentacles and buds would be formed along the body axis. (4) There would be no change observe

Effects of Global β-Catenin Overexpression in Hydra: Induction of Ectopic Buds and Axes

62. Hydra shows morphallactic regeneration and involves which one of the following signal transduction pathway in its axis formation? (1) Wnt/β-catenin pathway (2) Retinoic acid pathway (3) FGF pathway (4) Delta-Notch pathway

Role of Wnt/β-catenin Pathway in Axis Formation During Hydra Morphallactic Regeneration

61. In case of morphallactic regeneration: (1) there is re-patterning of the existing tissues with little new growth (2) there is re-patteming of the existing tissues after the stem cell division has taken place (3) there is cell division of the differentiated cells which maintain their differentiated state to finally form a complete organism. (4) there is dedifferentiation of the cells at the cut surface which become undifferentiated. These undifferentiated cells then divide to re- differentiate to form the complete structure

Morphallaxis Explained: Tissue Repatterning with Minimal New Growth in Regeneration

60. If hydra is fragmented into various parts, separate group of cells re-pattern themselves into various small hydra. Such an mode of development is termed as (1) Regeneration (2) Morphallaxis (3) Epimorphogenesis (4) Teratogenesis

Hydra Regeneration: Morphallaxis as the Mode of Development Through Tissue Repatterning

58. In hydra if any part is lost remaining portion re-pattern itself and give rise to complete organism. Such a pattern of regeneration is termed as (1) Epimorphosis (2) Morphallaxis (3) Regeneration (4) Healing

Morphallaxis in Hydra: Repatterning of Existing Tissue for Complete Regeneration

57. Which one of the following statements regarding regeneration in Hydra is correct? (1) It follows only stem cell-mediated regeneration. (2) It follows only stem cell-mediated regeneration and morphallaxis. (3) It follows stem cell-mediated regeneration, morphallaxis and epimorphosis. (4) It follows only morphallaxis.

Regeneration in Hydra: Integration of Stem Cell-Mediated Regeneration, Morphallaxis, and Epimorphosis

56. Which of poikilotherm have endogenous cardiac stem cells regenerative heart? (1) Snake (2) Lizard (3) Zebrafish (4) Toad

Endogenous Cardiac Stem Cell-Mediated Heart Regeneration in Poikilotherms: Zebrafish as a Model

55. Axolotal Larvae is due to- (1) Absence of thyroid gland in organism (2) Lack of iodine in habitat (3) Lack of pituitary hormone (4) Unfavorable environment

Why Axolotls Remain Larval: The Role of Iodine Deficiency in Failed Metamorphosis

54. Human polysyndactyly (joining of extra digits) syndrome results from a homozygous mutation at (1) antennapedia complex locus (2) one of the genes of Hox D (3) one of the genes of Hox C (4) β-catenin locus

Genetic Basis of Human Polysyndactyly: The Role of HOXD Gene Mutations

53. The hedgehog pathway is extremely important in vertebrate limb development, neural differentiation and facial morphogenesis. In accordance with the above statement, what would happen in mice that are homozygous for a mutant allele of sonic hedgehog (shh)? (1) Limbs would form normally but the mice would have facial abnormalities. (2) Midline of the face would be reduced and a single eye would form in the center of the forehead. (3) Eyes would form normally but digits would be malformed. (4) Mutations of shh may activate tumor formation if Patched protein can inhibit the Smoothened protein.

Effects of Sonic Hedgehog (Shh) Mutation in Mice: Midline Facial Defects and Limb Malformations

52. Species to expression of some of the pattern-forming genes in vertebrate limb bud: A. Lmx1b gene is expressed in dorsal mesenchyme B. Shh is expressed in the posterior region. C. Wnt7a gene is expressed in dorsal ectoderm. D. Hoxa13 and Hoxd13 are expressed in the distal region. E. Tbx5 and FGF10 are expressed in the lateral plate mesoderm involved in formation of limb bud. The Nail-Patella syndrome in human and the syndrome in mouse exhibiting footpads on both dorsal and ventral surfaces of limb are associated with which of the above mentioned gene functions? (1) B, D and E (2) B and D only (3) A, B and D (4) A and C only

Gene Expression in Vertebrate Limb Buds and Their Association with Nail-Patella Syndrome and Limb Patterning

51. Which one of the following statements is NOT correct regarding the tetrapod limb development? (1) As the limb grows outward, the stylopod forms first, then the zeugopod and the autopod is formed last. Each phase is characterized by a specific pattern of Hox gene expression. (2) The zone of polarizing activity (ZPA) is maintained by the interaction of the FGFs from the AER and Shh expressed from the mesenchyme. (3) Although cell death in the limb is necessary for the formation of digits and joints, it is never mediated by the BMPs, which is only responsible for differentiating mesenchyme cells into cartilage. (4) The dorsal-ventral axis is formed in part by the expression of Wnt7a in the dorsal portion of the limb ectoderm, which maintains expression level of Shh in the ZPA and Fgf4 in the posterior AER.

Understanding Incorrect Statements in Tetrapod Limb Development: The Role of BMP in Cell Death and Hox Gene Patterns

50. Homeobox transcription factor (Hox proteins), play important roles in specifying whether a particular mesenchymal cell will become stylopod, zeugopod or autopod. Based on the expression patterns of these genes, a model was proposed wherein these HOX genes specify the identity of a limb region. What would be the observed phenotype for human homozygous for a HOXD13 mutation? (1) No zeugopod formation (2) Abnormalities of the hands and feet wherein the digits fuse (3) Deformities in stylopods (4) No femur of patella formation.

HOXD13 Mutation and Its Impact on Limb Development: Understanding Syndactyly and Digital Malformations

49. When a wrist blastema from a recently cut Axolotl forelimb is placed on a host hindlimb cut at the mid thigh level, it will generate only a wrist. The host (whose own hind limb was removed) will fill the gap and regenerate the limb upto the wrist. However if the donor blastema is treated with retinoic acid on grafting, the wrist blastema will regenerate a complete limb and will not allow the host to fill the gap. This happens because retinoic acid (1) helps in proximalization of the blastema and activates the Hox genes differentially across the blastema. (2) helps in the distalization of the blastema and activates the Hox genes differentially across the blastema. (3) helps block the receptors of FGF essential for limb development (4) helps vigorous proliferation of the cells at the cut surface.

How Retinoic Acid Proximalizes Blastema and Activates Hox Genes to Induce Complete Limb Regeneration in Axolotls

47. The programmed cell death that separates the digits during a tetrapod limb development is dependent on which one of the following signaling pathways? (1) BMP (2) FGF (3) Wnt (4) Shh

Role of BMP Signaling in Programmed Cell Death for Digit Separation During Tetrapod Limb Development

43. Which of the following is true about amphibian limb regeneration? (A) It requires a minimum number of functional nerves. (B) The blastema of an amputated limb, if transplanted in the trunk region between two existing limbs in host, will still give rise to a limb. (C) The size of the regenerated limb is often grossly different from the original limb. (1) (A) (2) (B) and (C) (3) (A) and (C) (4) (A), (B) and (C)

Key Aspects of Nerve Dependence and Pattern Regulation in Amphibian Limb Regeneration

42. Dose-dependence of retinoic acid treatment supports the notion that a gradient of retinoic acid can act as a morphogen along the proximo-distal axis in a developing limb. Following are certain facts related to the above notion. A. Treatment with high level of retinoic acid causes a proximal blastema to be re-specified as a distal blastema and only distal structures are regenerated. B. Treatment with high level of retinoic acid causes a distal blastema to be re-specified as a proximal blastema and regeneration of a full limb may be initiated. C Treatment with retinoic acid affects only distal blastemas and causes them to form only proximal structures. D. Treatment with high level of retinoic acid causes any blastema to form only distal structures. Which one of the following is correct? (1) B and D (2) Only C (3) A and C (4) Only B

Dose-Dependent Effects of Retinoic Acid on Positional Identity During Limb Regeneration

41. What would be the effect of retinoic acid (RA) treatment on the 'positional information' of blastema cells present on the amputated newt limb? (1) RA will have no effect (2) The cells will become re-specified to more proximal position (3) The cells will become re-specified to more distal position (4) Cells will lose their positional information and remain as dedifferentiated cells

Effect of Retinoic Acid on Positional Information in Blastema Cells During Newt Limb Regeneration

40. In tadpoles, if the tail is amputated it can regenerate. However, if the tail is amputated and then exposed to retinoic acid, it develops limbs instead of regenerating the tail. This could be due to the following reasons: A. Retinoic acid is a morphogen and induces genes responsible for limb formation. B. Retinoic acid raises the positional values in that region for limb development to take place. C. This is a random phenomenon and is not well understood. D. Retinoic acid possibly acts as a mutagen and the phenotype observed is a result of several mutations. Which combination of the above statements is true? (1) A and B (2) C and D (3) B and D (4) B and C

Retinoic Acid-Induced Limb Formation in Tadpole Tail Regeneration: Mechanisms and Implications

37. Sonic hedgehog (Shh) specifies the anterior-posterior axis during limb development. Which one of the following statements regarding it is correct? (1) Shh secreting cells undergo apoptosis after performing its function. (2) Descendants of Shh cells become the bone and muscle of the anterior limb (3) When the genes for Shh and G/i3 are conditionally knocked out in the mouse limb, the resulting limbs do not form any digit. (4) Specification of the digit is primarily dependent on the amount of time the Shh gene is expressed and to a small extent on the concentration of the Shh protein that other cells receive.

How Sonic Hedgehog (Shh) Specifies the Anterior-Posterior Axis During Tetrapod Limb Development

36. Following figure shows the early interactions between the Apical Ectodermal Ridge (AER) and the limb bud mesenchyme. The red lines with block head indicate repression while the black lines indicate activation. The following statements were made regarding the development of a tetrapod limb: A. When the limb bud grows Shh creates a new signalling centre that induces the posterior- anterior polarity. B. When the concentration of FGFs rises, it can inhibit Gremlin thus allowing BMPs to begin repressing the AER-FGFS. C. FGFs 4, 9 and 17 from the AER inhibit Shh to stabilize the ZPA. D. Repression of Gremlin synthesis helps maintain the AER. Which one of the following options represents the combination of all correct statements? (1) A and B (2) A and C (3) B and D (4) C and D

Molecular Interactions Between the Apical Ectodermal Ridge and Limb Bud Mesenchyme During Tetrapod Limb Development

35. The mechanisms specifying the three axes of a tetrapod limb are interrelated and coordinated. In case of chick limb development the following statements were made: A. In the positive feedback loop, the fibroblast growth factor (FGF) from the apical ectodermal ridge (AER) activates sonic hedgehog (Shh) transcription thereby stabilizing the zone of polarizing activity (ZPA). B. In the inhibitory loop, Shh from the ZPA inhibits Gremlin which blocks the bone morphogenetic proteins (BMPs) thus preventing the BMP mediated inactivation of the FGFs in the AER. C. The dorsal-ventral axis is formed by the expression of Wnt7α in the ventral portion of the limb ectoderm. D. The BMPs are involved both in inducing apoptosis and in differentiating the mesenchymal cells into cartilage. Which of the above combination of statements is true? (1) A and B (2) A and D (3) B and C (4) C and D

Coordination of Axes Specification in Chick Limb Development: Roles of FGF, Shh, Wnt7a, and BMP

34. During wing development in chick, if Apical Ectodermal Ridge (AER) is removed, the limb development ceases, on the other hand placing leg mesenchyme directly beneath the wing AER, distal hindlimb structures develops at the end of the wing, and if replaced by non-limb mesnchyme sheath AER, the AER regresses. This may demonstrate that: A. the limb mesenchyme cells induce and sustain AER. B. the mesenchyme cells specify the type: wing or limb. C. the AER is responsible for specifying the type: wing or limb. D. the AER is responsible for sustained outgrowth and development of the limb E. the AER does not specify the type: wing or limb. Which combination of above statements is demonstrated by the experiment? (1) A, B, C and D only (2) A, B, D and E only (3) C, D and E only (4) A and E only

Experimental Demonstration of Apical Ectodermal Ridge (AER) and Limb Mesenchyme Interactions in Chick Wing Development

33. In the context of the proximal-distal growth and differentiation of a tetrapod limb following experiments were visualized (A) If the apical ectodermal ridge (AER) is removed at any time during the limb development, further development of distal limb skeletal elements ceases. (B) If leg mesenchyme is placed directly beneath the wing AER, proximal hind limb structures develop at the end of the limb (C) If an extra AER is grafted onto an existing limb bud, supernumerary structures are formed usually at the distal end of the limb. (D) If leg mesenchyme is placed directly beneath the wing AER, proximal hind limb structures develop at the end of the limb Which of the above experiments would show the possible interactions between the AER and the limb mesenchyme directly beneath it during limb development? (1) (A) and (B) only (2) (B) and (C) only (3) (C) and (D) only (4) (A) and (C) only

Interactions Between Apical Ectodermal Ridge and Limb Mesenchyme During Proximodistal Limb Growth

32. A set of experiments that were carried out to demonstrate the effect of Apical Ectodermal Ridge (AER) of the chick limb bud on the underlying mesenchyme are enlisted below, along with their expected outcomes: A. Removal of the AER of forelimb leads to cessation of limb development. B. If an extra AER is placed in the forelimb bud, duplication of the distal region of the wing takes place. C. If an extra AER is placed in the forelimb bud, a leg develops instead of a wing. D. If AER of forelimb bud is replaced with beads soaked in FGF2, a normal wing develops. E. If a non limb mesenchyme is placed below an AER, the AER directs the mesenchyme to form a normal wing. Which of the above statements are correct? (1) A, C and E (2) C, D and E (3) B, D and E (4) A, B and D

Apical Ectodermal Ridge (AER) Function and Limb Bud Mesenchyme Interactions in Chick Limb Development

31. The proximal distal growth and differentiation of the tetrapod limb bud are made possible by a series of interactions between the apical ectodermal ridge (AER) and limb bud mesenchyme directly beneath it. Some of the interactions performed in chick demonstrated the following results: A. When the AER was removed at any time of development, further development of distal limb skeletal elements ceased. B. When leg mesenchyme was placed directly beneath the wing AER, distal hindlimb structures developed at the end of the wing. C. When limb mesenchyme was replaced by a non- limb mesenchyme beneath the AER, the limb still developed D. When an extra AER was grafted onto an existing limb bud, the development of the limb ceased. Which of the above combinations is correct? (1) A and B (2) A and C (3) B and D (4) B and C

Apical Ectodermal Ridge and Limb Mesenchyme Interactions Governing Proximodistal Limb Development in Tetrapods

30. During vertebrate limb development, a specialized ectodermal structure, called Apical Ectodermal Ridge (AER), forms at the dorso-ventral ectodermal boundary at the distal tip of the developing limb bud. The following experimental facts about the AER is available: (A) FGF-2, 4, and 8 are expressed in the AER (B) Removal of the AER causes cessation of limb growth (C) Removal of AER along with implantation of beads soaked in FGF 8 or (D) FGF 4 or FGF 2 protein rescues the AER removal phenotype and gives rise to normal limb Which of the following statements cannot be made based on the above facts? (1) FGF 2, 4, and 8 are secreted proteins. (2) FGF 2, 4, and 8 are necessary and sufficient for AER function (3) FGF 2, 4, and 8 are sufficient for AER function (4) FGF 2, 4, and 8 have largely redundant functions

Role of FGFs in Apical Ectodermal Ridge Function During Vertebrate Limb Development

29. What would happen as a result of a transplantation experiment in a chick embryo where the leg mesenchyme is placed directly beneath the wing apical ectodermal ridge (AER)? (1) Distal hind limb structures develop at the end of the limb. (2) A complete hindlimb will form in the region where the forelimb should be. (3) The forelimb would form normally. (4) Neither a forelimb nor a hindlimb would form,since the cells are already determined.

Chick Embryo Limb Development: Effects of Transplanting Leg Mesenchyme Beneath Wing Apical Ectodermal Ridge

28. Apical ectodermal ridge induction is essential tetrapod limb development. Which of the following is NOT essential for the formation of a functional limb? (1) Tbx genes and Wnt (2) Androsterone (3) Fibroblast growth factor (4) Apoptotic gene

Essential and Non-Essential Factors for Apical Ectodermal Ridge Formation in Tetrapod Limb Development

27. Over-expression of a dominant negative FGF receptor during amphibian development would prevent formation of (1) trunk and tail (2) head and trunks. (3) trunk and fore limbs. (4) head and forelimbs.

Impact of Dominant Negative FGF Receptor Overexpression on Amphibian Development: Consequences for Trunk and Tail Formation

26. In mice, the gene encoding Tbx5 is transcribed in limb fields of the forelimbs, while the genes encoding lslet1, Tbx4 and Pitx1 are expressed in presumptive hind limbs. Following statements are made about limb development in mouse: (A) Loss of Tbx5 gene results in complete failure of forelimb formation. (B) Hindlimb bud growth and initial patterning appears normal when Tbx4 is knocked out, although leg development is arrested prematurely. (C) Misexpression of Pitx1 in forelimb ceases development of muscles, bones and tendons. (D) When lslet1 is inactivated specifically in the lateral plate mesoderm, the hindlimbs still develop. Which one of the following options represents a combination of correct statements? (1) A and B (2) A and C (3) B and C (4) C and D

Key Genetic Factors in Mouse Limb Development: Functional Roles of Tbx5, Tbx4, Islet1, and Pitx1

25. Which one of the following statements regarding limb development in mice is true? (1) The gene encoding Tbx5 is transcribed in the limb fields of the hindlimbs. (2) The gene encoding Tbx4 is transcribed in the limb fields of the forelimbs. (3) Genes encoding lslet 1, Tbx4 and Pitx are expressed in the presumptive hindlimb. (4) Genes encoding lslet 1, Tbx4 and Pitx are expressed in the presumptive forelimb.

The Roles of Tbx4, Tbx5, Islet1, and Pitx1 in Mouse Limb Development: Distinguishing Forelimb and Hindlimb Gene Expression

24. Anterior-posterior limb axis and dorsal ventral nerve plate is determined by (1) Bicoid (2) Sonic hedgehog (3) Pax3 (4) Cactus

Sonic Hedgehog (Shh) as the Key Determinant of Anterior-Posterior Limb Axis Patterning and Dorsal-Ventral Nerve Plate Formation

23. The limb bud of a tetrapod is specified by (1) Hox genes only (2) retinoic acid only (3) both Hox genes and retinoic acid (4) fibroblast growth factor

Tetrapod Limb Bud Specification: The Integrated Role of Hox Genes and Retinoic Acid

22. In Amphibians, when due to some injury, the eye lens is damaged, the fully differentiated iris cells can regenerate the lens. It is achieved through the possible processes: A. Iris cells through some signaling undergo dedifferentiation and trans- differentiation into lens cells to regenerate the lens. B. Iris cells transform into lens cells spontaneously. C. Iris cells induce in a stepwise manner, specific genes responsible for their dedifferentiation and then conversion to lens cells. D. Stem cells present in iris tissue differentiate into lens cells. Which of the following is correct? (1) A and B (2) A and C (3) B and D (4) B and C

Mechanisms of Lens Regeneration in Amphibians: Dedifferentiation and Transdifferentiation of Iris Cells

21. Lens formation requires sequential events whereby the anterior neural plate signals the anterior ectoderm to promote secretion of Pax 6, which renders the anterior ectoderm more receptive to secretions from the optic vesicle. The above can be best explained by which of the following phenomenon? (1) Instructive interactions only (2) Epithelial- Mesenchymal interactions (3) Permissive interactions (4) Induction and competence

Induction and Competence in Lens Formation: The Sequential Role of Anterior Neural Plate and Optic Vesicle

20. Following are some statements given for vertebrate eye lens induction: A. Paired box6 (Pax6) is a transcription factor synthesized in specific region of head surface ectoderm. B. Pax6 is secreted by optic vesicle. C. Optic vesicle serves as an inducer for competent head surface ectoderm. D. Studies on amphibians suggest that the first inducers of lens may be the foregut endoderm and heart forming mesoderm. E. The genes for lens proteins get induced in the surface ectoderm of optic vesicle. Which combination of the above statements is correct towards vertebrate eye lens induction?

Understanding Pax6 and Inductive Signals in Vertebrate Eye Lens Formation

The functionality of the pax6 gene in the formation of optic and nasal structure may be attributed to the following (A) Pax6 makes the optic vesicle competent and allows lens formation. (B) The optic vesicle can induce any part of the head ectoderm to form the nasal and optic structures, due to presence of Pax6. (C) Pax6 renders the head ectoderm competent to receive signals from optic vesicle. (D) Apart from the optic vesicle, the head ectoderm may also be induced by BMP and FGF, so pax6 is not exclusive for lens formation. Which of the above attributes are true? (1) (A) and (D) (2) (C) and (D) (3) (B) and(C)           (4) (C) only  

The Role of Pax6 in Optic and Nasal Structure Formation: Competence and Induction in Eye Development

18. During lens formation in the Xenopus, the following statements have been proposed: A. Lens induction can be achieved in the absence of optic vesicle after priming of head ectoderm by the anterior neural plate. B. The optic vesicle can induce the presumptive trunk ectoderm to form the lens. C. Only the head ectoderm can respond to direct signals from the optic vesicle form the lens. D. The anterior neural plate primes the head ectoderm via BMP4 and Fgf8 prior to signals from the optic vesicle. Which of the above combinations is correct? (1) C and D (2) B and D (3) A and D (4) A and C

Mechanisms of Lens Formation in Xenopus: Signaling and Tissue Competence in Embryonic Development

17. In C. elegans the SKN-1 protein controls the fate of the EMS blastomere which generates the posterior pharynx. With reference to the above which one of the following statements is INCORRECT? (1) The MS blastomere is able to generate pharyngeal tissue even in isolation. (2) Embryos from skn-1 (skin excess) deficient mothers lack both pharyngeal mesoderm and endoderm derivatives of EMS (3) Embryos which are skn-1 null mutants will always make extra hypodermal (skin) and body wall tissue. (4) SKN-1 activates MED transcription factors whose level of activity controls the fate of EMS lineage

The Role of SKN-1 Protein in Controlling EMS Blastomere Fate in Caenorhabditis elegans Development

16. The group of 6 cells (P3.p to P8.P) called vulval precursor cells (VPCs) of C. elegans form an equivalence group. The following statements were made as evidence that VPCs form an equivalence group: (A) If the anchor cell is destroyed the VPCs contribute to the formation of hypodermal tissues. (B) If the 3 central cells (P5.p to P7.p) are destroyed the remaining cells can generate vulval cells. (C) If expression of lin-3 is increased VPCs contributing to the secondary lineage can form cells of primary lineage. (D) Ectopic expression of let-23 in P5.p and P7.p VPCs converts them to primary cell lineage. Which one of the following options is a combination of all correct statements? (1) A and B only (2) B and C only (3) A, B and C (4) B, C and D

Evidence Supporting the Equivalence Group Model of Vulval Precursor Cells in C. elegans

15. In wild type C. elegans hermaphrodites, two adjacent cells, Z1.ppp and Z4.aaa, have the potential to become the anchor cell. They interact in a manner that causes one of them to become the anchor cell, while the other one becomes the precursor of the uterine tissue. The following statements are given to describe the interaction of the two cells: A. The cell secreting LAG-2 becomes the anchor cell. B. The cell secreting LIN-12 remains as the precursor of the uterine tissue. C. The LIN-12 secreting cell takes the fate of anchor cell while the LAG-2 secreting cell takes the fate of uterine precursor cell. D. The Hippo kinase signaling pathway brings lateral inhibition so that one cell is inhibited and the other cell is promoted to become the anchor cell. Which one of the following options represents the combination of all correct statements? (1) A and D (2) A and B (3) D only (4) C only

Cell Fate Decision Between Anchor Cell and Uterine Precursor in Caenorhabditis elegans: Role of Lateral Inhibition and Signaling Pathways

13. Following statements were made regarding vulval development in Caenorhabditis elegans: A. The six vulval precursor cells (VPCs) are influenced by the anchor cell to form an equivalence group. B. In the loss of function lin-12 mutants, both cells become uterine whereas in gain of function mutants, both become anchor cell. C. If the anchor cell is destroyed early in development, all the six VPCs divide once and contribute towards the formation of hypodermal cells. D. The anchor cell/ventral uterine precursor decision is due to Notch-Delta mediated mechanism of restricting adjacent cell fates. E. The paracrine factor secreted by the anchor cell directly activates the Notch-delta pathway. Which one of the following options represents a combination of correct statements? (1) A, C and D (2) A, B and D (3) C, D and E (4) B, D and E

Understanding Key Regulatory Mechanisms in Caenorhabditis elegans Vulval Development: Anchor Cell, VPCs, and Cell Fate Decisions

12. During vulva development in C. elegans, the anchor cell produces Lin-3 protein which interacts with the Let-23 protein present on the six vulval precursor cells (VPCs) that form an equivalence group. The central lineage cell (P6.p) adopts the primary fate, the adjacent VPCs (P5.p and P7.p) adopt the secondary fate and the rest VPCs adopt the tertiary fate. Few mutants (Column A) and phenotypes (Column B) are listed in the table given below. Match the correct mutant with the observed phenotype. (1) A-iv, B-ii, C-iii, D-i (2) A-iv, B-iii, C-i, D-ii (3) A-ii, B-iii, C-iv, D-i (4) A-iii, B-i, C-ii, D-iv

Matching Mutants with Phenotypes in C. elegans Vulval Development: A Comprehensive Guide

11. Development of vulva in C. elegans is initiated by the induction of a small number of cells by short range signals from a single inducing cell. With reference to this, following statements were put forward. A. When the anchor cell was ablated early in development no vulva formed. B. In a dominant negative mutant of let-23, a primary vulva formed but the secondary vulva formation did not take place. C. A cell adopting a primary fate inhibits adjacent cells from adopting the same fate by lateral inhibition involving LIN-39 and also induces the secondary fate in these cells. D, A constitutive signal from the hypodermis inhibits the development of both the primary and secondary fates but it is overruled by the initial signal from the anchor cell. Which of the above statements is true? (1) A and B (2) A and C (3) A and D (4) B and D

Key Insights into C. elegans Vulva Development: Anchor Cell Induction and Cell Fate Determination

Generally the number of cells present in gametophyte of angiosperm are- (1) 7 (2) 8 (3) 3 (4) 5 

Cellular Structure of Angiosperm Gametophytes: Understanding the 7-Cell Female Gametophyte

10. In C. elegans during embryogenesis, an anchor cell and 6 hypodermal vulval precursor cells (VPCs) get involved in forming the vulva. If 3 of the hypodermal VPCs are killed by a laser beam, a normal vulva is still formed. This could be due to the following possible reasons. A. Six hypodermal VPCs form equivalence group of cells, out of which only 3 participate in vulva formation and 3 cells remain as reserve cells. B. When 3 hypodermal VPCs are killed, the 3 neighboring hypodermal non- VPCs get freshly recruited. C. Anchor cell functions as an inducer which can induce epithelial cells of the gonad to get recruited to compensate for the loss. D. Anchor cell acts as an inducer which can spatially induce only 3 hypodermal cells to form the vulva. Which combination of the above statements is correct? (1) A and B (2) B and C (3) C and D (4) A and D

Understanding Vulval Precursor Cell Recruitment in C. elegans: Explaining Developmental Flexibility

1. Apomixis is reproduction in which (1) Fusion of gametes but no fusion of central cell (2) Egg form complete plant (3) Pollen forms seeds (4) Seeds are formed without union of gametes

Apomixis in Plants: Asexual Seed Formation Without Fertilization

9. The following experiments related to vulva development in C. elegans were performed: • When the anchor cell was killed by laser beam at early embryonic development and embryo was allowed to develop, a vulva did not form. • When any three out of the SIX hypodermal cells were killed and embryo was allowed to develop, a normal vulva formed. • When any four out of six hypodermal cells were killed and embryo was allowed to develop a normal vulva did not form. • When anchor cell as well as anyone of the six hypodermal cells were killed, the vulva did not form. Based on the above observations the following conclusions were made: A. Out of six hypodermal cells, any three get involved in vulva formation. B. The fourth hypodermal cell plays a major role in vulva formation. C. Anchor cell as well as the middle three hypodermal cells only has the ability to participate in vulva formation. D. Anchor cell is essential for vulva formation. Which of the above statement or combination of statements is correct? (1) A and B (2) only C (3) A and D (4) A and C

Key Findings from Vulva Development Experiments in C. elegans: Which Hypodermal Cells and Signals Matter?

8. In C. elegans, an anchor cell and a few hypodermal cells take part in the formation of vulva. The experiment performed to understand the role of these cells in vulva formation and the results obtained are as follows: - If the anchor cell is killed by laser beam, hypodermal cells do not participate in vulva formation and no vulva develops. - If six hypodermal cells closely located with anchor cell (called vulval precursor cells) are killed, no vulva develops - If the three central vulval precursor are destroyed, the three outer cells, which normally form hypodermis, take the fate of vulval cells instead. Following are certain statements regarding vulva formation: A. Anchor cells acts as an inducer B. Six hypodermal cells with the potential to form vulva form an equivalence group. C. Three, out of six, hypodermal cells participate in vulva formation D. The central cell functions as the 10 cell and the two cells on both side act as the 20 cells E. The 10 cell secretes a short range juxtacrine signal Which combinations of the above statements have been derived from the above experimental results? (1) A, B and C (2) A, B and D (3) D and E (4) B, D and E

Dissecting Vulva Formation in C. elegans: Key Experimental Insights and Derived Statements

7. A group of six cells called 'equivalence group cells' divide to form the vulval structure in Caenorhabditis elegans. They are called so because (1) they have similar fates during development of vulva. (2) all the six cells are competent to form vulva and can replace each other under various experimental conditions. (3) they are all under the influence of the anchor cell, signals from which initiate vulval development. (4) they interact with each other to form the vulval structure.

Why Are C. elegans Vulval Equivalence Group Cells So Named?

6. The model organism to study cell lineage is (1) Xenopus (2) Yeast (3) Caenorhabditis elegans (4) Drosophila

Caenorhabditis elegans: The Prime Model Organism for Studying Cell Lineage

5 The first organ system to develop during organogenesis is the (1) Integumentary system (2) Respiratory system (3) Excretory system (4) Cardiovascular system

The First Organ System to Develop During Human Organogenesis: The Cardiovascular System

4. Fusion of precursor cells is an essential phenomenon in development of (1) Nerve (2) Skeletal muscle (3) Spleen (4) Liver

Fusion of Precursor Cells: The Building Block of Skeletal Muscle Development

3. cAMP signalling plays a very important role in the development of Dictyosteliumdiscoideum. Below are few statements related to it. A. Every amoeba at the time of aggregation has the potential to make, receive and relay cAMP. B. acb- mutants develop normally but the spores formed appear glassy and are unable to germinate. C. The spores formed by the acg- mutants germinate in the sorus itself. D. RegA is an extracellular phosphodiesterase. E. cAMP is continuously secreted in nanomolar amounts during aggregation. Which combination of the above statements is correct? (1) A and D (2) A and B (3) A and E (4) B and D

Decoding cAMP Signaling in Dictyostelium discoideum: Truths About Aggregation, Mutant Spores, and Key Enzymes

2. cAMP signalling plays a very important role in the development and differentiation of Dictyostelium discoideum. This morphogen' is synthesized by different adenyl cyclases expressed at different stages of its life cycle. The following statements (AD) refer to the effect of mutations in different adenyl cyclase genes: (A) aca deficient cells can be allowed to aggregate by exposing them to pulses of cAMP. (B) acb deficient cells would form normal fruiting bodies and the spores can germinate when exposed to favourable conditions. (C) acg deficient cells develop normally and the spores germinate in the spore head itself. (D) Spores formed from the acg deficient cells will germinate irrespective of the osmotic conditions. Which of the above statements are correct? (1) A and D (2) A only (3) A and B (4) C and D

Adenyl Cyclase Gene Mutations in Dictyostelium discoideum: Understanding cAMP’s Role in Development and Differentiation

1. The myxamoebae are in the vegetative stage where they feed on bacteria but enter the aggregation stage during nutrient starvation finally forming fruiting body with two types of differentiated cells. It is best example of evolution of (1) Multicellularity (2) Social behavior (3) Embryonic development (4) Pattern formation

Myxamoebae: Unlocking the Evolution of Multicellularity Through Aggregation and Fruiting Body Formation

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