Developmental Biology MCQ Quiz - Objective Question with Answer for Developmental Biology - Download Free PDF

Last updated on Jul 7, 2025

Latest Developmental Biology MCQ Objective Questions

Developmental Biology Question 1:

Which of the following statements in relation to cell movement during gastrulation in Sea urchin is/are correct?  

  1. Delamination leads to the formation of endoderm
  2. Ingression leads to the development of mesoderm 
  3. Involution leads to the development of ectoderm 
  4. Invagination leads to the development of endoderm 

Answer (Detailed Solution Below)

Option :

Developmental Biology Question 1 Detailed Solution

The correct answer is Option 2 and Option 4

Explanation:

  • Invagination: Refers to the infolding of a region of cells, much like poking a finger into a balloon. This is typically seen in the formation of the endoderm in sea urchins.
  • Involution: Describes the inward movement of an expanding outer layer so that it spreads over the internal surface of the remaining external cells. This is observed in the mesoderm formation in amphibians.
  • Ingression: Involves the migration of individual cells from the surface into the interior of the embryo. It is seen in the formation of mesoderm in sea urchins.
  • Delamination: Is the splitting of one cellular sheet into two parallel sheets. This is characteristic of the formation of the hypoblast in birds.
  • Epiboly: Refers to the movement of epithelial cells as a unit to enclose deeper layers of the embryo, often seen in the ectoderm of amphibians.

Developmental Biology Question 2:

During the development of a mammalian embryo, “yolk sac” is formed by which one of the following?

  1. Syncytiotrophoblast 
  2. Primitive endoderm (hypoblast)
  3. Amniotic ectoderm
  4. Embryonic epiblast

Answer (Detailed Solution Below)

Option 2 : Primitive endoderm (hypoblast)

Developmental Biology Question 2 Detailed Solution

The correct answer is Primitive endoderm (hypoblast)

Explanation:

  • During mammalian embryonic development, extraembryonic structures are formed to support the growth and nourishment of the embryo. One of these structures is the yolk sac, which plays a crucial role in early nutrition and hematopoiesis (formation of blood cells).
  • The yolk sac is an early extraembryonic structure that originates from the primitive endoderm, also known as the hypoblast.
  • It is essential for providing nutrients to the developing embryo before the establishment of placental circulation.
  • Primitive endoderm (hypoblast):
    • The yolk sac forms from the primitive endoderm (hypoblast), which is a layer of cells beneath the embryonic epiblast.
    • The hypoblast contributes to the formation of the outer layer of the yolk sac, which is involved in nutrient transfer and initial hematopoiesis.
    • Additionally, the yolk sac serves as a site for the migration of primordial germ cells to the developing gonads.
    • This structure is transient and diminishes in importance as the placenta develops to take over the role of nutrient and gas exchange.

qImage685d374a4d150da0547ecddb

Other Options (Incorrect):

  • Syncytiotrophoblast: The syncytiotrophoblast is a layer of the trophoblast that invades the uterine wall to facilitate implantation of the embryo. Its primary function is to form part of the placenta and help in nutrient and gas exchange, but it does not form the yolk sac.
  • Amniotic ectoderm: The amniotic ectoderm is derived from the embryonic epiblast and lines the amniotic cavity. Its role is to produce the amniotic fluid that cushions and protects the embryo, but it is not involved in the formation of the yolk sac.
  • Embryonic epiblast: The embryonic epiblast is the source of all embryonic germ layers (ectoderm, mesoderm, and endoderm) during gastrulation.

Developmental Biology Question 3:

Correctly match the following Arabidopsis genes (Group I) and the biological processes they primarily regulate (Group II).

Group I Group II
(P) CLAVATA3 (1) Organ identity in flower
(Q) CONSTANS (2) Cell-type specification in root meristem
(R) SCARECROW (3) Meristem size in shoot
(S) AGAMOUS (4) Photoperiodic floral transition

  1. P-3; Q-4; R-1; S-2
  2. P-1; Q-3; R-2; S-4
  3. P-3; Q-4; R-2; S-1
  4. P-4; Q-1; R-3; S-2 

Answer (Detailed Solution Below)

Option 3 : P-3; Q-4; R-2; S-1

Developmental Biology Question 3 Detailed Solution

The correct answer is P-3; Q-4; R-2; S-1

Explanation:

  • Arabidopsis thaliana is a model organism in plant biology, and its genes play key roles in regulating various biological processes, such as meristem development, floral transition, organ identity, and cell specification.

Explanation:

  • (P) CLAVATA3 (3: Meristem size in shoot):
    • CLAVATA3 (CLV3) encodes a signaling peptide that regulates the size of the shoot apical meristem (SAM).
    • It works in conjunction with other genes like CLAVATA1 and WUSCHEL (WUS) to maintain the balance between stem cell proliferation and differentiation in the shoot meristem.
    • Mutations in CLV3 result in an enlarged meristem due to excessive stem cell proliferation.
  • (Q) CONSTANS (4: Photoperiodic floral transition):
    • CONSTANS (CO) regulates flowering time in response to photoperiod (day length).
    • It promotes the expression of the FLOWERING LOCUS T (FT) gene, which induces flowering under long-day conditions.
  • (R) SCARECROW (2: Cell-type specification in root meristem):
    • SCARECROW (SCR) is involved in cell-type specification and patterning in the root meristem.
    • It is required for the proper development of the endodermis and cortex layers in the root.
    • SCR works with SHORT-ROOT (SHR) to regulate asymmetric cell divisions and maintain the radial organization of the root.
  • (S) AGAMOUS (1: Organ identity in flower):
    • AGAMOUS (AG) is a homeotic gene that regulates organ identity in the flower, specifically the development of stamens and carpels (whorls 3 and 4).
    • It belongs to the MADS-box family of transcription factors and is essential for proper floral organ development and termination of floral meristem activity.

Developmental Biology Question 4:

Consider a diploid plant species where the cells in the epidermis (the outermost single cell layer) always divide in the anticlinal orientation. If one such cell within the central zone of the shoot apical meristem (SAM) spontaneously becomes tetraploid at the seedling stage, which one of the following cellular arrangements would be most likely observed if the meristem is examined at the adult stage?

  1. Only one tetraploid cell in the epidermis
  2. Many tetraploid cells in the epidermis
  3. All cells in the entire SAM tetraploid 
  4. All cells in the entire SAM diploid

Answer (Detailed Solution Below)

Option 2 : Many tetraploid cells in the epidermis

Developmental Biology Question 4 Detailed Solution

The correct answer is Many tetraploid cells in the epidermis

Explanation:

  • In plants, the shoot apical meristem (SAM) is responsible for the continuous growth of the plant in the vertical direction. It consists of layers of cells, with the outermost layer being the epidermis.
  • Cell division in the epidermis of the SAM typically occurs in the anticlinal orientation, which means the division plane is perpendicular to the surface, ensuring that the cell arrangement remains as a single-layered sheet.
  • A tetraploid cell arises when the chromosome number in a cell doubles, often due to errors in mitosis. In this case, a single epidermal cell in the SAM spontaneously becomes tetraploid at the seedling stage.
    • Only one tetraploid cell in the epidermis (Incorrect): This option assumes that the tetraploid cell does not undergo any further division or that its progeny cells do not remain in the epidermis. However, because epidermal cells divide in the anticlinal orientation, the tetraploid cell will divide and produce multiple progeny that remain in the epidermis.
    • Many tetraploid cells in the epidermis (Correct): Since the epidermal cells of the SAM divide anticlinally, the tetraploid cell will divide and its progeny will remain confined to the epidermis. Over time, as the plant grows and the SAM expands, the tetraploid lineage will give rise to many tetraploid cells in the epidermis.
    • All cells in the entire SAM tetraploid (Incorrect): The tetraploid cell is initially confined to the epidermal layer. Because SAM is organized into different layers, with cells dividing in specific orientations, the tetraploid cells will not invade other layers of the SAM. 
    • All cells in the entire SAM diploid (Incorrect): The tetraploid cell and its progeny will remain in the epidermis due to anticlinal divisions

Developmental Biology Question 5:

If an egg cell of a diploid plant species has 10 chromosomes, the expected number of chromosomes in a double trisomic somatic cell of this species would be _______________ (Answer in integer).

Answer (Detailed Solution Below) 22

Developmental Biology Question 5 Detailed Solution

The correct answer is 22

Concept:

  • Trisomy: A specific type of aneuploidy where a diploid cell has an extra copy of a particular chromosome. Instead of the normal two copies, there are three copies of that chromosome. This condition is represented as 2n + 1. For example, Down syndrome in humans is Trisomy 21 (2n=46, so 2n+1=47 chromosomes).
  • Double Trisomy refers to a condition where a diploid cell has an extra copy of two different chromosomes. For example, if a normal diploid cell has chromosomes 1 to N, a double trisomic cell might have three copies of chromosome 1 and three copies of chromosome 5.
  • The total chromosome number in such a cell would be the normal diploid number plus two extra chromosomes, represented as 2n + 1 + 1, or simply 2n + 2.

Explanation:

  • An egg cell of the plant species has 10 chromosomes.
  • An egg cell is a gamete and is haploid. Therefore, the haploid chromosome number (n) for this plant species is 10.
  • So, n = 10 chromosomes.
  • A normal somatic cell (body cell) of a diploid species contains 2n chromosomes.
  • Normal somatic chromosome number = 20 chromosomes.

"Double trisomic" means that there are two extra chromosomes, each belonging to a different homologous pair.

  • The general formula for a double trisomic cell is 2n + 1 + 1, which simplifies to 2n + 2.
  • Chromosome number = 2n + 2 = 20 + 2 = 22 chromosomes.

Top Developmental Biology MCQ Objective Questions

Which of the following is a part of apical meristem found in roots.

  1. Protoderm
  2. Axillary bud
  3. Differentiating vascular tissue
  4. Leaf primordium

Answer (Detailed Solution Below)

Option 1 : Protoderm

Developmental Biology Question 6 Detailed Solution

Download Solution PDF

The correct answer is Protoderm.

Key Points

  • Protoderm is the outermost primary meristem in plants.
  • In roots, it differentiates to form the epidermis.
  • Produces epidermal cells, including root hairs.
  • Root hairs play a crucial role in water and nutrient absorption.
  • Acts as a protective barrier.
  • Facilitates root-soil interactions.

Additional Information

  • Axillary buds, located at leaf-stem junctions, can grow into branches or flowers, influenced by hormonal signals and environmental factors. 
  • A leaf primordium is the initial embryonic tissue from which a leaf develops, found at the shoot apex or growing tip.
  • Vascular tissue in plants consists of xylem and phloem.
  • Xylem transports water and minerals from roots, providing structural support with tracheids, vessels, and fibers. 
  • Phloem transports sugars and nutrients throughout the plant.   

In Caenorhabditis elegans, blastomere identity occurs both through conditional and autonomous modes of cell specification. Which one of the following options is a correct statement in this regard?

  1. If the AB and P1 blastomeres are experimentally separated, the AB cell will generate all cells it would normally make.
  2. When AB divides to form daughter cells, ABp becomes different from ABa through its interaction with the P2 cell.
  3. The specification of AB cell is determined by the presence of cytoplasmic determinants.
  4. The P2 cell produces a morphogen for the determination of the ABp cell.

Answer (Detailed Solution Below)

Option 2 : When AB divides to form daughter cells, ABp becomes different from ABa through its interaction with the P2 cell.

Developmental Biology Question 7 Detailed Solution

Download Solution PDF

The correct answer is When AB divides to form daughter cells, ABp becomes different from ABa through its interaction with the P2 cell.

Concept: 

Autonomous Specification:

  • Autonomous specification refers to the inherent ability of a cell or group of cells to differentiate into specific cell types based on their intrinsic factors or developmental program.
  • Cells have the inherent potential to follow a predetermined developmental pathway and differentiate into specific cell types without relying heavily on external signals or interactions with neighboring cells.
  • Autonomous specification is often associated with the early stages of embryonic development, where cells possess predetermined information or cues that guide their differentiation.
  • One classic example of autonomous specification involves the early development of the nematode Caenorhabditis elegans.

Conditional Specification:

  • Conditional specification implies that the fate or differentiation of cells is influenced by external conditions, signals, or interactions with neighboring cells.
  • Cells may require specific signals, cues, or environmental factors to trigger their differentiation into particular cell types. In the absence of these external influences, the fate of the cells may be different.
  • Conditional specification is often observed in later stages of development, where cell fate decisions are influenced by the surrounding microenvironment or interactions with adjacent cells.

Explanation:

In Caenorhabditis elegans, the specification of blastomeres occurs through a combination of autonomous and conditional modes of cell fate determination:

  • ABp and ABa are the daughter cells formed when the AB cell divides. Initially, these cells are similar, but ABp becomes different from ABa due to its interaction with the neighboring P2 cell. This interaction is an example of conditional specification, where cell fate is influenced by external signals from neighboring cells.
  • Autonomous specification occurs in cells like the P1 blastomere, where cell fate is determined by internal cytoplasmic determinants.

Option 1: If the AB and P1 blastomeres are experimentally separated, AB will not generate all the cells it would normally make because some cell fates (like ABp) are determined conditionally through interaction with P2.
Option 3: The specification of the AB cell is not determined by the presence of cytoplasmic determinants. This applies more to P1 cells, which rely on autonomous specification.
Option 4: The P2 cell does not produce a general morphogen for the determination of ABp. Instead, it signals specifically to ABp via cell-cell interactions (such as the signaling molecule Notch/Delta).

MOM

Fig:Cell-cell signaling in the 4-cell embryo of C. elegans. The P2 cell produces two signals: (1) the juxtacrineprotein APX-1 (a Delta homologue), which is bound by GLP-1 (Notch) on the ABp cell, and (2) the paracrine protein MOM-2 (Wnt), which is bound by the MOM-5 (Frizzled) protein on the EMS cell. 

Conclusion:

Thus, Option 2 is the correct statement, as the ABp cell becomes different from ABa due to its interaction with the P2 cell, which is a form of conditional specification

In which of the following stages of Arabidopsis embryogenesis do the visible distinctions between the adaxial and abaxial tissues of the cotyledons become initially apparent?

  1. Globular stage
  2. Zygotic stage
  3. Torpedo stage
  4. Mature stage

Answer (Detailed Solution Below)

Option 3 : Torpedo stage

Developmental Biology Question 8 Detailed Solution

Download Solution PDF

The correct answer is Option 3 i.e. Torpedo stage.

Key Points

Arabidopsis Embryogenesis - 

  1.  The zygotic stage - haploid egg and sperm fuse to form a single-celled zygote, which is the initial stage of diploid life. Small apical and extended basal cells are produced during this stage of division.
  2. The globular stage embryo - An eight-cell (octant) globular embryo is produced by the apical cell following the first zygotic division, which occurs 30 hours after fertilization. Protoderm, which later develops into the epidermis, is created through cell division.
  3. The heart stage embryo - Two areas on either side of the eventual shoot apex undergo fast cell divisions to generate this stage. These two areas generate growths that subsequently give rise to the cotyledons, creating bilateral symmetry in the embryo.
  4. The torpedo stage embryo -  Caused by cell elongation along the embryonic axis and additional cotyledon development. The difference between the cotyledons' adaxial and abaxial tissue may be seen.
  5.  The maturation stage embryo - The embryo and seed lose water during the end of development and enter dormancy, becoming metabolically inactive. At advanced phases, the cell begins to collect storage chemicals. 

F3 Vinanti Teaching 05.07.23 D8
Explanation:

Option 1 - INCORRECT

  • It involves the first division, which is the second stage of fertilization Hence no adaxial or abaxial axis is seen.

Option 2 - INCORRECT

  • It involves no division and is just the beginning of fertilization. Therefore, no axis is visible.

Option 3 - CORRECT

  • This stage involves rapid cell division at the embryo axis which leads to cell elongation such that the adaxial and abaxial axis are seen for the first time.

Option 4 - INCORRECT

  • This stage is the last stage of embryogenesis, which is the quiescent stage where the seeds enter dormancy.

Which one of the following statements about the cortical reaction in sea urchins is correct?

  1. The entry of Ca2+ ions into the egg initiates development.
  2. The exocytosed cortical granules during egg maturation contain the components of the zona pellucida.
  3. The depolarization of the plasma membrane after sperm entry helps to block polyspermy.
  4. The release of the cortical granules after sperm entry converts the vitelline membrane into the fertilization membrane which blocks polyspermy.

Answer (Detailed Solution Below)

Option 4 : The release of the cortical granules after sperm entry converts the vitelline membrane into the fertilization membrane which blocks polyspermy.

Developmental Biology Question 9 Detailed Solution

Download Solution PDF

The correct answer is The release of the cortical granules after sperm entry converts the vitelline membrane into the fertilization membrane which blocks polyspermy.

Explanation:

The cortical reaction is a crucial event in the process of fertilization, particularly in sea urchins, and it plays a key role in preventing polyspermy (the entry of more than one sperm into the egg).

  • After sperm entry, cortical granules located just beneath the plasma membrane of the egg are released (exocytosis) into the space between the plasma membrane and the vitelline membrane.
  • The contents of these cortical granules modify the vitelline membrane, converting it into the fertilization membrane, which serves as a physical barrier that prevents additional sperm from entering the egg, thus blocking polyspermy.

Other Options:

  • The entry of Ca²⁺ ions into the egg initiates development: While calcium is crucial for initiating the cortical reaction, development is not initiated solely by calcium entry.
  • The exocytosed cortical granules during egg maturation contain components of the zona pellucida: Sea urchins do not have a zona pellucida; this structure is present in mammals. The correct term here is vitelline membrane in sea urchins.
  • The depolarization of the plasma membrane after sperm entry helps to block polyspermy: This is related to the fast block to polyspermy, which occurs immediately after sperm entry, but it is not related to the cortical reaction (which is the slow block).

Thus, Option 4 correctly describes the role of the cortical reaction in preventing polyspermy by forming the fertilization membrane.

In animal cells, typically which organelle is only provided by the sperm to the oocyte following fertilization?

  1. Nucleolus
  2. Peroxisomes
  3. Mitochondria
  4. Centrioles

Answer (Detailed Solution Below)

Option 4 : Centrioles

Developmental Biology Question 10 Detailed Solution

Download Solution PDF

The correct answer is Centrioles.

Explanation:

During fertilization, the sperm contributes specific cellular components to the oocyte (egg), and the contribution of different organelles can vary:

  • Nucleolus: The nucleolus is not specifically contributed by the sperm; it is a structure found in the nucleus of the oocyte.
  • Peroxisomes: Peroxisomes are organelles involved in metabolic processes and are not uniquely provided by sperm during fertilization.
  • Mitochondria: While sperm do contribute mitochondria, they are typically degraded after fertilization. In most animal species, the mitochondria in the zygote are predominantly derived from the oocyte, not the sperm.
  • Centrioles: Centrioles are cylindrical structures involved in cell division. In many species, sperm contribute centrioles to the fertilized egg, which are important for organizing the mitotic spindle during the first cell division. The oocyte typically lacks centrioles, making the sperm's contribution crucial for proper cell division.

In a transplantation experiment, the area of presumptive ectoderm from an early frog gastrula was transplanted to a region of the newt gastrula destined to become parts of the mouth. The resulting salamander larvae had frog like mouth parts (frog tadpole suckers) instead of balancers as observed during development of wild type newt embryo. This is an example of

  1. Determination
  2. Genetic specificity of interaction
  3. Regional specificity of interaction
  4. Autonomous specification

Answer (Detailed Solution Below)

Option 2 : Genetic specificity of interaction

Developmental Biology Question 11 Detailed Solution

Download Solution PDF

The correct answer is Option 2 i.e. Genetic specificity of interaction.

Concept:

Determination:

  • This is a term in developmental biology that refers to the point in the developmental process at which a cell or group of cells becomes committed to a particular fate.
  • Once a cell's role is determined, the next steps are differentiation and morphogenesis which lead to the formation of a specific type of tissue or organ.
  • However, this determination is not always irreversible.

Genetic specificity of interaction:

  • This term refers to the concept that the fate of a cell or group of cells is determined by the specific genes they carry.
  • Genetic specificity implies that every cell or piece of tissue carries a specific set of genetic information that governs its development and dictates what form and function it will have, regardless of its environment.
  • When transplanted to a different location, these cells will still follow their inherent genetic program.

Regional specificity of interaction:

  • This concept suggests that a cell's or tissue's developmental behavior and fate might depend on its location within the embryo.
  • In other words, cells or tissues will interact with their immediate environment and adjust their developmental path based on these interactions.
  • Here, the environment and cell-to-cell interactions can influence the developmental outcome and modify a cell's fate from what was genetically programmed.

Autonomous specification:

  • This describes a mode of development in which the fate of a cell is established early on and is independent of interactions with neighboring cells.
  • Essentially, certain cells in the early embryo are programmed to follow a specific developmental path, regardless of their surroundings.
  • The genetic information in these cells directs them to form specific parts of the organism, even if they are isolated from the rest of the embryo or transplanted into a different region.
  • This is similar to the concept of genetic specificity but places more emphasis on the independence of the developmental path from neighboring cell influence.

Explanation

  • The transplanted frog tissue interacts with the newt embryo following the laws of its own species-specific developmental program.
  • Despite being placed into a different region and species, the transplanted tissue commits to developing into the structure it was originally destined to become, as per its inherent genetic program.
  • This shows that the developmental fate is attributed more to the genetic composition of the transplanted cells rather than the influence of their new location.
  • The interaction between the gene expression in the transplanted cells and the new host's cells results in the development of frog-like mouth parts in the newt larvae, illustrating the concept of genetic specificity of the interaction.
  • This is distinct from the concept of regional specificity of interaction which suggests that the fate of a cell or tissue is affected by its surrounding cells or tissues and will change its developmental path to match its new environment.
  • In this case, the cells contributed to the structure they were originally "programmed" to create via genetic specification, not the structure typical of the region to which they were transplanted.

Hence the correct answer is Option 2

During normal development of sea urchin, β‐catenin accumulates predominantly in the micromeres, which are fated to become endoderm and mesoderm. If GSK‐ 3 is blocked in the developing embryo:

  1. β‐catenin accumulation in the nuclei of large micromeres will be inhibited leading to formation of ectodermal ball.
  2. β‐catenin will accumulate in the nuclei of all blastula cells leading to an ectodermal ball.
  3. β‐catenin will accumulate in the nuclei of all blastula cells leading to animal cells getting specified as endoderm and mesoderm.
  4. β‐catenin which accumulate in the nuclei of large micromeres will be inhibited leading to animal cells getting specified as endoderm and mesoderm.

Answer (Detailed Solution Below)

Option 3 : β‐catenin will accumulate in the nuclei of all blastula cells leading to animal cells getting specified as endoderm and mesoderm.

Developmental Biology Question 12 Detailed Solution

Download Solution PDF

The correct answer is Option 3 i.e.β‐catenin will accumulate in the nuclei of all blastula cells leading to animal cells getting specified as endoderm and mesoderm.

Concept:-

  • Beta-catenin acts as a transcription factor in wnt signaling.
  • Beta-catenin is found throughout the embryo.
  • Beta-catenin is Active on the dorsal side.
  • GSK-3 inhibits Beta-catenin at the ventral side.
  •  During fertilization, Dsh and Wnt 11 protein from the vegetal pole translocated to the dorsal side of the egg.
  • Disheveled (Dsh) inhibits Gsk-3, thereby activating Beta-catenin.

F3 Vinanti Teaching 05.07.23 D9

Explanation:-

Option 1:- β‐catenin accumulation in the nuclei of large micromeres will be inhibited leading to formation of ectodermal ball.

  • GSK-3 inhibits beta-catenin. if GSK-3 is blocked, beta-catenin will be free, which will enter each nucleus of the vegetal pole and will form endoderm and mesoderm.
  • ​  Hence, this statement is incorrect.   

Option 2:-  β‐catenin will accumulate in the nuclei of all blastula cells leading to an ectodermal ball.

  • If beta-catenin will accumulate, it will form an endoderm and mesoderm, not the ectodermal ball.
  •  Hence, this option is incorrect.

Option 3:- β‐catenin will accumulate in the nuclei of all blastula cells leading to animal cells getting specified as endoderm and mesoderm.

  • Beta-catenin will get accumulated in the nuclei of all blastula cells which was earlier inhibited by GSK-3. Therefore, beta-catenin will form endoderm and mesoderm.
  • Hence, this option is correct.

Option 4:- β‐catenin which accumulates in the nuclei of large micromeres will be inhibited leading to animal cells getting specified as endoderm and mesoderm.

  • If GSK-3 is blocked, beta-catenin will enter each nucleus of all dorsal cells, it will not get inhibited.
  • Hence, this option is incorrect.

What is the pattern of cleavage observed in mammals?

  1. Radial
  2. Spiral
  3. Rotational
  4. Bilateral

Answer (Detailed Solution Below)

Option 3 : Rotational

Developmental Biology Question 13 Detailed Solution

Download Solution PDF

The correct answer is Option 3 i.e.Rotational

Concept: 

  • Cleavage or cellulation is a series of mitotic cell divisions wherein the enormous volume of egg cytoplasm is divided into numerous smaller nucleated cells. 
  • The pattern of embryonic cleavage in a species is determined by two major factors:
    • The distribution and amount of the yolk in the cytoplasm.
    • Factors (mRNA, proteins) present in the cytoplasm of the egg cell. 
  • Types of cleavage - Holoblastic cleavage and meroblastic cleavage.

Holoblastic cleavage - 

  • In this type of cleavage, the yolk is present in a small amount and evenly distributed or the yolk is present in a moderate amount in a gradient.
  • In holoblastic cleavage, the first cleavage always occurs along the vegetal-animal axis of the egg, while the second cleavage is perpendicular to the first cleavage
  • From here the cleavage pattern varies in different species due to differences in the plane of the cleavage pattern. 
  • Types of holoblastic cleavage are as follows:
  1. Bilateral holoblastic cleavage- 
  • The first cleavage divides the zygote into left and right halves.
  • Cleavage planes of the following cleavages are centered on the axis and result in two halves that are the mirror image of one another.
  1. Radial holoblastic cleavage-
  • It is typically found in deuterostomes. 
  • The division planes are at 90-degree angles relative to one another.
  • The daughter cells (blastomeres) resulting from this division are aligned directly over or to the side of one another. 
  1. Rotational cleavage - 
  • It involves normal first division along the meridional axis, giving rise to two daughter cells. 
  • During the second cleavage division, one cell divides meridionally while the other divides equatorially.
  1. Spiral cleavage - 
  • It typically occurs in protostomes.
  • In this type of cleavage, the daughter cells (blastomeres) are not exactly aligned on top of one another, instead they are located at a slight angle
  • The first two cell divisions result in the four macromeres (A, B, C, D), each macromere representing one quadrant of the embryo. 

F2 Madhuri Teaching 23.08.2022 D2

Explanation: 

  • Cleavages in the mammalian egg are the slowest in the animal kingdom, about 12-24 hours apart. 
  • The first cleavage in mammals is along the meridional plane.
  • While in the second cleavage, one blastomere divides meridionally and another blastomere divides equatorially. 
  • This type of cleavage is rotational holoblastic cleavage

Hence the correct answer is option 3. 

The following are sketches of wild-type and mutant (I - III) embryos of Drosophila melanogaster.

F1 Priya Teaching 21 10 2024  D2

Which one of the following options represents a correct match between the gene and its loss of function phenotype?

  1. dorsal : I
  2. torpedo : III
  3. gurken : II
  4. cactus : III

Answer (Detailed Solution Below)

Option 3 : gurken : II

Developmental Biology Question 14 Detailed Solution

Download Solution PDF

The correct answer is gurken : II

Concept:

  • Gurken (gene): This gene is essential for dorsal-ventral patterning in Drosophila. It encodes a protein involved in signaling from the oocyte to the surrounding follicle cells, influencing the patterning of the embryo. Loss of function of gurken leads to defective dorsal-ventral axis formation.
  • Dorsal (gene): Involved in specifying ventral structures. A loss of function leads to defects in ventral patterning, often resulting in an embryo that is fully dorsalized.
  • Torpedo (gene): This gene encodes a receptor that is part of the EGF signaling pathway, crucial for proper dorsal-ventral patterning. A loss of torpedo function also results in dorsal-ventral defects.
  • Cactus (gene): Acts as an inhibitor of Dorsal, regulating its nuclear localization. Loss of function in cactus leads to ventralization of the embryo.

Explanation:

  • Mutant I: This might show a phenotype corresponding to dorsalization, such as what is observed in dorsal mutants.
  • Mutant II: This indicates a dorsal-ventral patterning defect, characteristic of a gurken loss-of-function phenotype, where the embryo has lost proper dorsal-ventral polarity.
  • Mutant III: This might represent a cactus mutant, where the embryo becomes ventralized due to the uninhibited activity of the Dorsal protein.
  • Gurken loss-of-function: The embryo would lack the proper signaling for dorsal-ventral patterning, leading to a dorsalized phenotype, where the structures that would normally form along the dorsal side are missing. This is reflected in Mutant II, where the embryo lacks the proper dorsal-ventral axis.

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.

Answer (Detailed Solution Below)

Option 3 : It follows stem cell-mediated regeneration, morphallaxis and epimorphosis.

Developmental Biology Question 15 Detailed Solution

Download Solution PDF

The correct answer is It follows stem cell-mediated regeneration, morphallaxis and epimorphosis.

Concept:

Regeneration refers to the reactivation of the development in post-embryonic life so as to replace the missing tissues. Regeneration occurs in four major ways:

  1. Stem-cell mediated regeneration - In this type of regeneration, certain organs or tissues that are lost are regrow because of stem cells. For example, continuous replacement of blood cells from the bone marrow through hematopoietic stem cells.
  2. Epimorphosis - In this type of regeneration certain adult structure undergoes de-differentiation to from a relatively undifferentiated mass of cells that further undifferentiates to form a new structure. For example regeneration in flatworm and regeneration of amphibians limbs. 
  3. Morphallaxis - In this type of regeneration, certain existing tissues undergoes repatterning that results in new growth, even though only little new growth is seen. In this case, entire organism is regeneration from body pieces.
     For example, regeneration in hydra. 
  4. Compensatory regeneration - In this type of regeneration differentiated cells divide but they have maintain their differentiated function. In this case, new cells are produced but the does not comes from stem cells or from de-differentiated cells, rather each cells produces cells that is similar to itself. For example, mammalian liver shows compensatory regeneration. 

Explanation:

  • In hydra regeneration occurs through various mode of regeneration including stem cell-mediated regeneration, morphallaxis and epimorphosis.
  • In the case of hydra all cells of the body are replaced with new ones approximately every 20 days. This is possible due to present on stem cells population for all three cell-lineage (endodermal epithelial, ectodermal epithelial, and interstitial epithelial).
  • Hydra is know to developed missing body parts upon transverse or longitudinal amputation. Hence, hydra shows epimorphosis regeneration as well. 
  • Hydra shows morphallxis regeneration because pieces of body part can leads to generation of entire organism. When entire hydra is cut in several pieces then each piece will develop head at original apical end while it is develop foot at original basal end. 

Hence, the correct answer is Option 3. 

Get Free Access Now
Hot Links: teen patti joy official teen patti rules master teen patti