Match the drug/chemicals listed in Column I with the developmental/physiological defects listed in Column II.Column IColumn IIP. Veratrum alkaloids(i) ObesityQ. Thalidomide(ii) Minamata syndromeR. Methylmercury(iii) CyclopiaS. Diethylstilbesterol(iv) Phocomelia

The correct match is P-(iii); Q-(iv); R-(ii); S-(i).Veratrum alkaloids such as cyclopamine and jervine, from Veratrum californicum, block Sonic hedgehog signalling; lambs of ewes that grazed the plant were born with cyclopia (iii).Thalidomide, taken in early pregnancy around 1957-1961, caused phocomelia, severely shortened limbs (iv).Methylmercury released into Minamata Bay, Japan, caused Minamata disease, a neurological syndrome that also affected children exposed before birth (ii).Diethylstilbestrol, a synthetic oestrogen, is studied as an obesogen: developmental exposure in mice leads to obesity in later life (i).Options B, C and D all pair the Veratrum alkaloids with something other than cyclopia; D also gives thalidomide cyclopia, and B and C give methylmercury cyclopia instead of Minamata syndrome.

Which one of the options correctly matches the human tissues/organs with their embryonic germ layers of origin?Tissues/organsEmbryonic germ layers(P) liver(I) ectoderm(Q) cerebellum(II) endoderm(R) femur(III) mesoderm

The correct match is P-II, Q-I, R-III.The liver grows from the embryonic gut tube, so it is endodermal.The cerebellum is part of the brain, which develops from the neural tube, so it is ectodermal.The femur, like other bones of the limbs, arises from mesoderm.The other options swap these, for example deriving the liver from mesoderm or the femur from ectoderm.

Animal Developmental Biology Questions and Answers

Q: Zygotic genes required for the formation of a group of adjacent segment in the developing Drosophila embryo is called

They are called Gap genes.Gap genes such as hunchback, Kruppel and knirps are the first zygotic segmentation genes expressed in the Drosophila embryo, in broad bands. A mutation in one of them deletes a block of adjacent segments, leaving a 'gap' in the larval pattern, which is how they were named.Maternal genes such as bicoid act from the mother, not the zygote. Pair-rule genes (even-skipped, fushi tarazu) affect alternate segments, and homeotic genes specify the identity of segments rather than their formation.

Q: To which class of Drosophila developmental genes does fushi tarazu (ftz) belong?

fushi tarazu is a pair rule gene.In the segmentation cascade of the Drosophila embryo, maternal effect genes set up the axes, gap genes divide the embryo into broad regions, pair rule genes are expressed in seven stripes that mark every other segment, and segment polarity genes then pattern each segment. ftz is expressed in seven stripes, and its mutants lose alternate segments, which is the defining pair rule phenotype (the name is Japanese for 'not enough segments').Gap genes include hunchback, Krüppel and knirps; segment polarity genes include engrailed, wingless and hedgehog; maternal effect genes include bicoid and nanos.

Q: Which of the following animals show “Bottle cells” during the gastrulation stage of development?

Bottle cells are seen in amphibians.In amphibian gastrulation, as in Xenopus, invagination begins at the dorsal lip of the blastopore when a group of cells constrict at their outer (apical) surface and elongate inward, taking a flask or bottle shape. These bottle cells start the inward movement of the endoderm and mesoderm.Gastrulation in birds and mammals proceeds by ingression of cells through the primitive streak, and in snails by invagination of a blastula without these characteristic flask-shaped cells, so bottle cells are the textbook hallmark of amphibian gastrulation.

Q: In an experiment, excess amount of bicod mRNA (more than wild-type expression level) was injected into the posterior pole of a wild-type Drosophila embryo at pre-blastodermal stage. Out of the following options, which one represents the best expected phenotype in the resulted developing embryo?

The expected phenotype is head structure at both anterior and posterior poles of the embryo.bicoid mRNA is normally localised at the anterior pole. Its protein forms a gradient that specifies head and thorax structures. Injecting excess bicoid mRNA at the posterior pole creates a second high point of Bicoid there, so the posterior end is also instructed to form a head, while the embryo's own anterior Bicoid still produces a head at the front. The result is a double-headed (bicephalic) embryo, with no normal tail structures.A normal embryo would ignore the injection. A head only at the posterior would require the anterior Bicoid to be lost, and a tail at the anterior would need posterior determinants there, neither of which the injection provides.

Q: In mammals, hematopoietic stem cells that give rise to different types of blood cells are known as

The Commission's answer is pluripotent stem cells.A single hematopoietic stem cell in the bone marrow can give rise to every blood cell type, red cells, platelets, granulocytes, monocytes and lymphocytes, and many textbooks therefore call it the pluripotent hematopoietic stem cell. (In stricter modern usage, cells limited to one tissue are called multipotent, but that term is not offered.)Totipotent cells, such as the zygote, can form the whole organism including extra-embryonic tissues. Myeloid and lymphoid progenitors are the more restricted daughters of the stem cell, each giving only one branch of the blood lineages.

Q: During the development of a mammalian embryo, "yolk sac" is formed by which one of the following?

The yolk sac is formed by the primitive endoderm (hypoblast).In the mammalian blastocyst the inner cell mass separates into the epiblast and the hypoblast. Hypoblast cells spread out over the inner surface of the blastocyst cavity as extra-embryonic endoderm and line it, forming the primary and then the secondary yolk sac, which is later covered by extra-embryonic mesoderm.The syncytiotrophoblast derives from the trophectoderm and invades the uterine wall to form the placenta. The amniotic ectoderm and the epiblast belong to the amnion and the embryo proper, not to the yolk sac.

Q: Five oogonia yield ten primary oocytes, then how many ova are produced by the completion of oogenesis ?

The number of ova is 10.In oogenesis each primary oocyte undergoes meiosis I to give one secondary oocyte and a small first polar body, and the secondary oocyte gives one ovum and a second polar body in meiosis II. So every primary oocyte yields exactly one ovum; the other products are polar bodies, which degenerate.Taking the stem's ten primary oocytes, there are ten ova. 40 is what spermatogenesis would give, where each primary spermatocyte yields four sperm; 20 and 5 do not follow from the one-ovum rule.

Q: Following are the functions of sertoli cells.1. Phagocytosis of excess sperm cytoplasm.2. Protection of developing spermatogenic cells.3. Nourishment of spermatocytes.4. Control of movements of spermatogenic cells.

All four are functions of Sertoli cells: 1, 2, 3 and 4.Sertoli cells line the seminiferous tubules and support sperm formation. They phagocytose the excess cytoplasm (residual bodies) shed by maturing spermatids (1), protect the developing germ cells by forming the blood-testis barrier (2), nourish the spermatocytes and spermatids (3), and guide the movement of the developing cells towards the lumen and their release (4).Options that leave out any of these list only some of the functions.

Zygotic genes required for the formation of a group of adjacent segment in the developing Drosophila embryo is called

A.Maternal gene
B.Pair rule gene
C.Homeotic gene
D.Gap gene

Exam Year(s)

GATE Life Sciences - 2012,

Answer: D

They are called Gap genes.

Gap genes such as hunchback, Kruppel and knirps are the first zygotic segmentation genes expressed in the Drosophila embryo, in broad bands. A mutation in one of them deletes a block of adjacent segments, leaving a 'gap' in the larval pattern, which is how they were named.

Maternal genes such as bicoid act from the mother, not the zygote. Pair-rule genes (even-skipped, fushi tarazu) affect alternate segments, and homeotic genes specify the identity of segments rather than their formation.

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To which class of Drosophila developmental genes does fushi tarazu (ftz) belong?

A.Gap genes
B.Segment polarity genes
C.Pair rule genes
D.Maternal effect genes

Exam Year(s)

GATE Life Sciences - 2022,

Answer: C

fushi tarazu is a pair rule gene.

In the segmentation cascade of the Drosophila embryo, maternal effect genes set up the axes, gap genes divide the embryo into broad regions, pair rule genes are expressed in seven stripes that mark every other segment, and segment polarity genes then pattern each segment. ftz is expressed in seven stripes, and its mutants lose alternate segments, which is the defining pair rule phenotype (the name is Japanese for 'not enough segments').

Gap genes include hunchback, Krüppel and knirps; segment polarity genes include engrailed, wingless and hedgehog; maternal effect genes include bicoid and nanos.

Discuss in Forum

Match the drug/chemicals listed in Column I with the developmental/physiological defects listed in Column II.

Column I	Column II
P. Veratrum alkaloids	(i) Obesity
Q. Thalidomide	(ii) Minamata syndrome
R. Methylmercury	(iii) Cyclopia
S. Diethylstilbesterol	(iv) Phocomelia

A.P-(iii); Q-(iv); R-(ii); S-(i)
B.P-(i); Q-(iv), R-(iii), S-(ii)
C.P-(ii), Q-(iv), R-(iii), S-(i)
D.P-(ii), Q-(iii), R-(iv), S-(i)

Exam Year(s)

GATE Life Sciences - 2022,

Answer: A

The correct match is P-(iii); Q-(iv); R-(ii); S-(i).

Veratrum alkaloids such as cyclopamine and jervine, from Veratrum californicum, block Sonic hedgehog signalling; lambs of ewes that grazed the plant were born with cyclopia (iii).
Thalidomide, taken in early pregnancy around 1957-1961, caused phocomelia, severely shortened limbs (iv).
Methylmercury released into Minamata Bay, Japan, caused Minamata disease, a neurological syndrome that also affected children exposed before birth (ii).
Diethylstilbestrol, a synthetic oestrogen, is studied as an obesogen: developmental exposure in mice leads to obesity in later life (i).

Options B, C and D all pair the Veratrum alkaloids with something other than cyclopia; D also gives thalidomide cyclopia, and B and C give methylmercury cyclopia instead of Minamata syndrome.

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Which of the following animals show “Bottle cells” during the gastrulation stage of development?

A.Snails
B.Amphibians
C.Birds
D.Mammals

Exam Year(s)

GATE Life Sciences - 2023,

Answer: B

Bottle cells are seen in amphibians.

In amphibian gastrulation, as in Xenopus, invagination begins at the dorsal lip of the blastopore when a group of cells constrict at their outer (apical) surface and elongate inward, taking a flask or bottle shape. These bottle cells start the inward movement of the endoderm and mesoderm.

Gastrulation in birds and mammals proceeds by ingression of cells through the primitive streak, and in snails by invagination of a blastula without these characteristic flask-shaped cells, so bottle cells are the textbook hallmark of amphibian gastrulation.

Discuss in Forum

In an experiment, excess amount of bicod mRNA (more than wild-type expression level) was injected into the posterior pole of a wild-type Drosophila embryo at pre-blastodermal stage. Out of the following options, which one represents the best expected phenotype in the resulted developing embryo?

A.Normal embryo with head structure at anterior and tail structure at posterior pole
B.Head structure only at posterior pole of the embryo
C.Tail structure at anterior and head structure at posterior poles of the embryo
D.Head structure at both anterior and posterior poles of the embryo

Exam Year(s)

GATE Life Sciences - 2023,

Answer: D

The expected phenotype is head structure at both anterior and posterior poles of the embryo.

bicoid mRNA is normally localised at the anterior pole. Its protein forms a gradient that specifies head and thorax structures. Injecting excess bicoid mRNA at the posterior pole creates a second high point of Bicoid there, so the posterior end is also instructed to form a head, while the embryo's own anterior Bicoid still produces a head at the front. The result is a double-headed (bicephalic) embryo, with no normal tail structures.

A normal embryo would ignore the injection. A head only at the posterior would require the anterior Bicoid to be lost, and a tail at the anterior would need posterior determinants there, neither of which the injection provides.

Discuss in Forum

In mammals, hematopoietic stem cells that give rise to different types of blood cells are known as

A.totipotent stem cells.
B.pluripotent stem cells.
C.myeloid progenitor cells.
D.lymphoid progenitor cells.

Exam Year(s)

GATE Life Sciences - 2024,

Answer: B

The Commission's answer is pluripotent stem cells.

A single hematopoietic stem cell in the bone marrow can give rise to every blood cell type, red cells, platelets, granulocytes, monocytes and lymphocytes, and many textbooks therefore call it the pluripotent hematopoietic stem cell. (In stricter modern usage, cells limited to one tissue are called multipotent, but that term is not offered.)

Totipotent cells, such as the zygote, can form the whole organism including extra-embryonic tissues. Myeloid and lymphoid progenitors are the more restricted daughters of the stem cell, each giving only one branch of the blood lineages.

Discuss in Forum

Which one of the options correctly matches the human tissues/organs with their embryonic germ layers of origin?

Tissues/organs	Embryonic germ layers
(P) liver	(I) ectoderm
(Q) cerebellum	(II) endoderm
(R) femur	(III) mesoderm

A.P-II, Q-I, R-III
B.P-III, Q-I, R-II
C.P-I, Q-II, R-III
D.P-II, Q-III, R-I

Exam Year(s)

GATE Life Sciences - 2024,

Answer: A

The correct match is P-II, Q-I, R-III.

The liver grows from the embryonic gut tube, so it is endodermal.
The cerebellum is part of the brain, which develops from the neural tube, so it is ectodermal.
The femur, like other bones of the limbs, arises from mesoderm.

The other options swap these, for example deriving the liver from mesoderm or the femur from ectoderm.

Discuss in Forum

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

A.Syncytiotrophoblast
B.Primitive endoderm (hypoblast)
C.Amniotic ectoderm
D.Embryonic epiblast

Exam Year(s)

GATE Life Sciences - 2025,

Answer: B

The yolk sac is formed by the primitive endoderm (hypoblast).

In the mammalian blastocyst the inner cell mass separates into the epiblast and the hypoblast. Hypoblast cells spread out over the inner surface of the blastocyst cavity as extra-embryonic endoderm and line it, forming the primary and then the secondary yolk sac, which is later covered by extra-embryonic mesoderm.

The syncytiotrophoblast derives from the trophectoderm and invades the uterine wall to form the placenta. The amniotic ectoderm and the epiblast belong to the amnion and the embryo proper, not to the yolk sac.

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Five oogonia yield ten primary oocytes, then how many ova are produced by the completion of oogenesis ?

A.5
B.10
C.20
D.40

Exam Year(s)

MPPSC Forest Service - 2022,

Answer: B

The number of ova is 10.

In oogenesis each primary oocyte undergoes meiosis I to give one secondary oocyte and a small first polar body, and the secondary oocyte gives one ovum and a second polar body in meiosis II. So every primary oocyte yields exactly one ovum; the other products are polar bodies, which degenerate.

Taking the stem's ten primary oocytes, there are ten ova. 40 is what spermatogenesis would give, where each primary spermatocyte yields four sperm; 20 and 5 do not follow from the one-ovum rule.

Discuss in Forum

10.

Following are the functions of sertoli cells.

1. Phagocytosis of excess sperm cytoplasm.
2. Protection of developing spermatogenic cells.
3. Nourishment of spermatocytes.
4. Control of movements of spermatogenic cells.

A.1,2 and 4
B.1 and 2
C.1, 2 and 3
D.1,2,3 and 4

Exam Year(s)

MPPSC Forest Service - 2022,

Answer: D

All four are functions of Sertoli cells: 1, 2, 3 and 4.

Sertoli cells line the seminiferous tubules and support sperm formation. They phagocytose the excess cytoplasm (residual bodies) shed by maturing spermatids (1), protect the developing germ cells by forming the blood-testis barrier (2), nourish the spermatocytes and spermatids (3), and guide the movement of the developing cells towards the lumen and their release (4).

Options that leave out any of these list only some of the functions.

Discuss in Forum

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