datavorous/entrance-exam-dataset huggingface dataset:
Match the following:

List-I (Ion)	List-II (Shapes)

(p)	Cassiterite	(1)	FeCO3

(q)	Rutile	(2)	2Fe2O3.3H2O

(r)	Cerussite	(3)	SnO2

(s)	Siderite	(4)	2CuCO3.Cu(OH)2

(t)	Limonite	(5)	PbCO3

(6)	TiO2

Options:
A p-6;q-3;r-5;s-4;t-2
B p-1;q-3;r-4;s-2;t-6
C p-3;q-6;r-5;s-1;t-2
D p-3;q-6;(r)-4;(s)-1;t-5

The correct answer is:
p-3;q-6;r-5;s-1;t-2

Solution:
Names of different ores and their molecular formula (Information based).

Topic: Chemistry, General Principles and Processes of Isolation of Metals, NEET
Match the following:

List-I	List-II

(i)	Statins	A	Propionibacterium sharmanii

(ii)	Swiss cheese	B	Streptococcus

(iii)	Cyclosporin A	C	Aspergillus niger

(iv)	Citric acid	D	Trichoderma polysporum

(v)	Clot buster	E	Monascus purpureus

Options:
A (i)-E, (ii)-A, (iii)-D, (iv)-C, (v)-B
B (i)-B, (ii)-A, (iii)-D, (iv)-E, (v)-C
C (i)-E, (ii)-A, (iii)-B, (iv)-C, (v)-D
D (i)-C, (ii)-E, (iii)-A, (iv)-D, (v)-B

The correct answer is:
(i)-E, (ii)-A, (iii)-D, (iv)-C, (v)-B

Solution:
The bioactive molecule, cyclosporin A, that is used as an immunosuppressive agent in organ-transplant patients is produced by the fungus Trichoderma polysporum. Statins produced by the yeast Monascus purpureus have been commercialized as blood-cholesterol lowering agents. Streptokinase produced by the bacterium Streptococcus and modified by genetic engineering is used as a âclot busterâ for removing clots from the blood vessels of patients who have undergone myocardial infarction leading to a heart attack. Aspergillus niger (a fungus) produces citric acid. the large holes in âSwiss cheeseâ are due to
production of a large amount of CO2 by a bacterium named Propionibacterium sharmanii.

Topic: Biology, Microbes in Human Welfare, NEET
Match the following organisms with the products they produce:

Column - I	Column - II

(a)	Clostridium butylicum	(i)	Lactic acid

(b)	Saccharomyces cerevisiae	(ii)	Butyric acid

(c)	Aspergillus niger	(iii)	Citric acid

(d)	Acetobacter aceti	(iv)	Alcohol

(v)	Acetic acid

Options:
A a-ii;Â b-iv;Â c-v;Â d-iii
B a-ii; b-iv; c-iii; d-v
C a-iii;Â b-iv;Â c-v;Â d-i
D a-ii;Â b-i;Â c-iii;Â d-v

The correct answer is:
a-ii; b-iv; c-iii; d-v

Solution:
Microbes	Products

Clostridium butylicum	Butyric acid

Saccharomyces cerevisiae	Alcohol

Aspergillus niger	Citric acid

Acetobacter aceti	Acetic acid

Lactobacillus acidophilus	Lactic acid

Topic: Biology, Microbes in Human Welfare, NEET, NEET 2019 (All India)
Match the following organisms with the products they produce:

Column - I	Column â II

(a)	Clostridium butylicum	(i)	Lactic acid

(b)	Saccharomyces cerevisiae	(ii)	Butyric acid

(c)	Aspergillus niger	(iii)	Citric acid

(d)	Acetobacter aceti	(iv)	Alcohol

(v)	Acetic acid

Options:
A a-ii;Â b-iv;Â c-v;Â d-iii
B a-ii; b-iv; c-iii; d-v
C a-iii;Â b-iv;Â c-v;Â d-i
D a-ii;Â b-i;Â c-iii;Â d-v

The correct answer is:
a-ii; b-iv; c-iii; d-v

Solution:
Microbes	Products

Clostridium butylicum	Butyric acid

Saccharomyces cerevisiae	Alcohol

Aspergillus niger	Citric acid

Acetobacter aceti	Acetic acid

Lactobacillus acidophilus	Lactic acid

Topic: Biology, Microbes in Human Welfare, NEET
Match the following organisms with their respective characteristics:

Column – I	Column – II

(a)	Pila	(i)	Flame cells

(b)	Bombyx	(ii)	Comb plates

(c)	Pleurobrachia	(iii)	Radula

(d)	Taenia	(iv)	Malpighian tubules

Select the correct option from the following:

Options:
A a-iii; b-ii; c-i; d-iv
B a-iii; b-iv; c-ii; d-i
C a-ii; b-iv; c-iii; d-i
D a-iii; b-ii; c-iv; d-i

The correct answer is:
a-iii; b-iv; c-ii; d-i

Solution:
The radula is an anatomical structure for feeding (sometimes also compared to a tongue), It is a characteristic feature of the mollusks and is found in every class of mollusk except the bivalves. Eg. Pila

Malpighian tubules or MTs are very highly specific organs of arthropods for excretion and osmoregulation. Eg. Bombyx

Comb-plate is a locomotory organ made up of a row of very strong cilia fused together. Eg. Pleurobrachia

Flame cells are specialized for excretion present in some freshwater invertebrates like flatworms, rotifers, and nemerteans. Eg. Taenia

Topic: Biology, Animal Kingdom, NEET, NEET 2019 (All India)
Match the following :

Oxide	Nature

(a) CO	(i) Basic

(b) BaO	(ii) Neutral

(c) Al2O3	(iii) Acidic

(d) Cl2O7	(iv) Amphoteric

Which of the following is correct option?

Options:
A (a)-(ii); (b)-(i); (c)-(iv); (d)-(iii)
B (a)-(iii); (b)-(iv); (c)-(i); (d)-(ii)
C (a)-(iv); (b)-(iii); (c)-(ii); (d)-(i)
D (a)-(i); (b)-(ii); (c)-(iii); (d)-(iv)

The correct answer is:
(a)-(ii); (b)-(i); (c)-(iv); (d)-(iii)

Solution:
(a)	CO	(ii)	Neutral

(b)	BaO	(i)	Basic

(c)	Al2O3	(iv)	Amphoteric

(d)	Cl2O7	(iii)	Acidic

Topic: Chemistry, Classification of Elements and Periodicity in Properties, NEET, NEET 2020 (Phase 1)
Match the following parts of a nephron with their function.
Select the correct option from the following

Options:
A 1.(i) 2.(iii) 3.(ii) 4.(iv)
B 1.(ii) 2.(iv) 3.(i) 4.(iii)
C 1.(i) 2.(iv) 3.(ii) 4.(iii)
D 1.(iv) 2.(i) 3.(iii) 4.(ii)

The correct answer is:
1.(ii) 2.(iv) 3.(i) 4.(iii)

Solution:
The correct matches are

Topic: Biology, Excretory Products and their Elimination, NEET, NEET 2019 (Odisha)
Match the following :

Polymer	Monomer

i.	Nylon-6	a.	Ethene

ii.	Nylon-6,6	b.	Caprolactum

iii.	Polyethene	c.	Hexamethylene diamine and adipic acid

Options:
A (i)-(a),(ii)-(b),(iii)-(c)
B (i)-(a),(ii)-(c),(iii)-(b)
C (i)-(c),(ii)-(b),(iii)-(a)
D (i)-(b),(ii)-(c),(iii)-(a)

The correct answer is:
(i)-(b),(ii)-(c),(iii)-(a)

Solution:
(i) Nylon-6

(ii) Nylon-6, 6

n  H 2  N  ( CH 2 ) 6   NH 2 + n HOOC  ( CH 2 ) 4  COOH→— [NH  ( CH 2 ) 6 Nylon 6, 6  NHCO  ( NH 2 ) 4  CO ] n —+ nH 2 O

(iii) Polythene

nâC H 2 =C H 2 â Polymerisation ââââ [ ââC H 2 âC H 2 ] n â EtheneâââââââââââââââââââââââââââââââââPolythene

Topic: Chemistry, Polymers, NEET
Match the following prefixes with their multiples:

Options:
A (i)-(2) , (ii)-(5) , (iii)-(4) , (iv)-(3) , (v)-(1)
B (i)-(1) , (ii)-(2) , (iii)-(4) , (iv)-(5) , (v)-(3)
C (i)-(3) , (ii)-(5) , (iii)-(1) , (iv)-(2) , (v)-(4)
D (i)-(4) , (ii)-(2) , (iii)-(3) , (iv)-(5) , (v)-(1)

The correct answer is:
(i)-(3) , (ii)-(5) , (iii)-(1) , (iv)-(2) , (v)-(4)

Solution:
Correct Option is : (C)
(i)-(3) , (ii)-(5) , (iii)-(1) , (iv)-(2) , (v)-(4)

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
Match the following respiratory volumes with their value and select the correct option:

Column-I	Column-II

(A)	Tidal Volume	(I)	2500 ml to 3000 ml

(B)	Residual Volume	(II)	1100 ml to 1200 ml

(C)	Expiratory Reserve Volume	(III)	1000 ml to 1100 ml

(D)	Inspiratory Reserve Volume	(IV)	500 ml

Options:
A A - IV, B - II, C - III, D - I
B A - IV, B - III, C - II, D - I
C A - I, B - II, C - III, D - IV
D A - IV, B - I, C - II, D - III

The correct answer is:
A - IV, B - II, C - III, D - I

Solution:
Tidal Volume is the amount of air inspired or expired during normal respiration. It is about 500 ml.
Inspiratory reserve volume is the additional volume of air that a person can inspire by a forcible inspiration. It is about 2500 ml to 3000 ml. Expiratory reserve volume is the additional volume of air that a person can expire by a forcible expiration. It is about 1000 ml to 1100 ml.
Residual volume is the amount of air which is left in the lungs after fully exhaling. It is about 1100 to 1200 ml.

Topic: Biology, Breathing and Exchange of Gases, NEET
Match the following RNA polymerases with their transcribed products
Select the correct option from the following

Options:
A 1.(i) 2.(iii) 3.(ii)
B 1.(i) 2.(ii) 3.(iii)
C 1.(ii) 2.(iii) 3.(i)
D 1.(iii) 2.(ii) 3.(i)

The correct answer is:
1.(ii) 2.(iii) 3.(i)

Solution:
The correct matches are

Topic: Biology, Molecular Basis of Inheritance, NEET, NEET 2019 (Odisha)
Match the following :
Select the correct option :

Options:
A $(\mathrm{A})-(\mathrm{IV}),(\mathrm{B})-(\mathrm{I}),(\mathrm{C})-(\mathrm{III}),(\mathrm{D})-(\mathrm{II})$
B $(\mathrm{A})-(\mathrm{III}),(\mathrm{B})-(\mathrm{I}),(\mathrm{C})-(\mathrm{II}),(\mathrm{D})-(\mathrm{IV})$
C $(\mathrm{A})-(\mathrm{I}),(\mathrm{B})-(\mathrm{II}),(\mathrm{C})-(\mathrm{IV}),(\mathrm{D})-(\mathrm{III})$
D $(\mathrm{A})-(\mathrm{I}),(\mathrm{B})-(\mathrm{II}),(\mathrm{C})-(\mathrm{III}),(\mathrm{D})-(\mathrm{IV})$

The correct answer is:
$(\mathrm{A})-(\mathrm{III}),(\mathrm{B})-(\mathrm{I}),(\mathrm{C})-(\mathrm{II}),(\mathrm{D})-(\mathrm{IV})$

Solution:
Perigynous flowers are found in Plum.
Zygomorphic flowers are found in Cassia.
Hypogynous flowers are found in Mustard.
Epigynous flowers are found in Cucumber.
Option (2) is correct.

Topic: Biology, Morphology of Flowering Plants, NEET, NEET 2023 (Manipur)
Match the following:

Set âI	Set â II

a	Rancher's rule	1	Body size

b	Bergman's rule	2	Wings

c	Allen's rule	3	Vertebra number

c	Jordan's rule	4	Organ size

Options:
A a-1, b-2, c-4, d-3
B a-3, b-4, c-2, d-1
C a-2, b-1, c-3, d-4
D a-2, b-1, c-4, d-3

The correct answer is:
a-2, b-1, c-4, d-3

Solution:
(i) Bergman's rule: Warm-blooded animals (birds and mammals) have larger body size in cold climate than in hotter areas.
(ii) Allen's rule: Extremities (legs, ears, tail, mouth) of warm blooded animals become smaller in colder areas as compared to animals of warmer areas.
(iii) Rancher's rule: In a colder climate, birds possess narrow and acuminate wings as compared to broader wings of birds found in warmer areas.
(iv) Jordan's Rule: As the temperature is lowered, some fishes possess a larger size with a larger number of vertebrae.

Topic: Biology, Organisms and Populations, NEET
Match the following sexually transmitted diseases (Column - I) with their causative agent (Column - II) and select the correct option.

Column–I	Column–II

(a) Gonorrhea	(i) HIV

(b) Syphilis	(ii) Neisseria

(c) Genital Warts	(iii) Treponema

(d) AIDS	(iv) Human Papilloma-Virus

Options:
A abciiiiiiv d(i)
B abciiiivi dii
C abciviiiii di
D abciviiiii di

The correct answer is:
abciiiiiiv d(i)

Solution:
GONORRHEA: It is a type of sexually transmitted disease caused by a gonococcus bacterium called Neisseria gonorrhoeae.

SYPHILIS: It is caused by a bacterium called Treponema pallidum.

GENITAL WARTS: It is caused by the Human Papilloma Virus (HPV).

AIDS: It is caused by a retrovirus called HIV.

Topic: Biology, Reproductive Health, NEET, NEET 2017
Match the following structures with their respective location in organs:

Column – I	Column – II

(a)	Crypts of Lieberkuhn	(i)	Pancreas

(b)	Glisson’s capsule	(ii)	Duodenum

(c)	Islets of Langerhans	(iii)	Small intestine

(d)	Brunner’s Glands	(iv)	Liver

Select the correct option.

Options:
A a-iii; b-i; c-ii; d-iv
B a-ii; b-iv; c-i; d-iii
C a-iii; b-iv; c-i; d-ii
D a-iii; b-ii; c-i; d-iv

The correct answer is:
a-iii; b-iv; c-i; d-ii

Solution:
An intestinal gland (also crypt of Lieberkühn and intestinal crypt) is a gland found in between villi in the intestinal epithelium lining of the small intestine and large intestine (colon). b) Glisson's capsule is the capsule of the liver. A layer of connective tissue surrounding the liver and ensheathing the hepatic artery, portal vein, and bile ducts within the liver. c) The pancreatic islets or islets of Langerhans are the regions of the pancreas that contain its endocrine (hormone-producing) cells, discovered in 1869 by German pathological anatomist Paul Langerhans. The pancreatic islets constitute 1 to 2% of the pancreas volume and receive 10-15% of its blood flow. d) Brunner's glands (or duodenal glands) are compound tubular submucosal glands found in that portion of the duodenum which is above the hepatopancreatic sphincter (a.k.a. sphincter of Oddi).

Topic: Biology, Digestion and Absorption, NEET, NEET 2019 (All India)
Match the following techniques or instruments with their usage:

(a)	Bioreactor	(i)	Separation of DNA fragments

(b)	Electrophoresis	(ii)	Production of large quantities of products

(c)	PCR	(iii)	Detection of pathogen, based on antigen-antibody reaction

(d)	ELISA	(iv)	Amplification of nucleic acids

Select the correct option from following:

Options:
A (a)-(iii), (b)-(ii), (c)-(iv), (d)-(i)
B (a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)
C (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)
D (a)-(ii), (b)-(i), (c)-(iii), (d)-(iv)

The correct answer is:
(a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)

Solution:
Techniques/Instruments	Usage

Bioreactor	It is used for the large scale production of desired products

Electrophoresis	Separation of DNA fragments based on molecular size. Large fragments are found near the cathode, while smaller fragments are found near the anode.

PCR	It is also called amplifier as it is responsible for Amplification of nucleic acids.

ELISA	Detection of the pathogen, based on antigen-antibody reaction. It is one of the confirmatory tests for detection of HIV.

Topic: Biology, Biotechnology - Principles and Processes, NEET, NEET 2020 (Phase 2)
Match the following types of leucocyte with their shape of nuclei.

Leucocyte	The shape of the nucleus

A.	Neutrophil	I.	Rounded

B.	Monocytes	II.	Bilobed

C.	Lymphocytes	III.	Multilobed

D.	Eosinophil	IV.	Kidney shape

Options:
A C-I, D-II, A-III, B-IV
B C-I, B-II, D-III, A-IV
C D-I, C-II, A-III, B-IV
D C-I, D-II, B-III, A-IV

The correct answer is:
C-I, D-II, A-III, B-IV

Solution:
Lymphocyte has a large and rounded nucleus. Monocyte has a kidney shape or bean shape nucleus. Eosinophil has a bi-lobed or 2-lobed nucleus and neutrophils have a multilobed nucleus.

Topic: Biology, Body Fluids and Circulation, NEET
Match the following types of movements with their example and select the correct option.

Movements	Examples

A	Flagellar movement	I	macrophages and leucocytes

B	Amoeboid movement	II	muscular movement

C	Ciliary movement	III	non-muscular movement

D	Peristalsis movement	IV	passage of ova in the fallopian tube

Options:
A

A	B	C	D

III	I	IV	II

B

A	B	C	D

I	III	IV	II

C

A	B	C	D

I	II	IV	III

D

A	B	C	D

III	IV	II	I

The correct answer is:

A	B	C	D

III	I	IV	II

Solution:
The peristaltic movement is a type of muscular movement. The amoeboid movement is the phagocytic movement of macrophages and leucocytes by the formation of phagocytosis. Ciliary movement is present in our internal tubular organs which are lined by ciliated epithelium, such as the passage of ova in the oviduct. Flagellar movement is the movement of sperms with the help of the sperm tail.

Topic: Biology, Locomotion and Movement, NEET
Match the following with a CORRECT combination.

A. Carboxylation	1. Oxygen evolution

B. Phosphorylation	2. Photorespiration

C. Photolysis of water	3. Rubisco

D. Phosphoglycolate	4. Chemosynthesis

E. Nitrosomonas	5. ATP

Options:
A AÂ âÂ 1,Â BÂ âÂ 2,Â CÂ âÂ 3,Â DÂ âÂ 4,Â EÂ âÂ 5
B A â 3, B â 5, C â 1, D â 2, E â 4
C AÂ âÂ 3,Â BÂ âÂ 2,Â CÂ âÂ 5,Â DÂ âÂ 4,Â EÂ âÂ 1
D AÂ âÂ 5,Â BÂ âÂ 3,Â CÂ âÂ 4,Â DÂ âÂ 2,Â EÂ âÂ 1

The correct answer is:
A â 3, B â 5, C â 1, D â 2, E â 4

Solution:
Carboxylation is the first reaction in the Calvin cycle wherein the atmospheric carbon-di-oxide forms 3-PGA in presence of the enzyme RUBISCO. Phosphorylation takes place when the ADP gets converted to ATP due to the phosphate addition in the presence of light as the energy source. Photolysis of water occurs when the oxidized chlorophyll molecule takes an electron from water, which releases oxygen. 2-phosphoglycolate is formed when the enzyme RuBPoxygenase splits the molecule of Ribulose 1, 5-biphospate in presence of oxygen. The nitrosomonas bacteria synthesize organic compounds using the energy from reactions involving inorganic chemicals.

Topic: Biology, Photosynthesis in Higher Plants, NEET
Match the following with correct combination.

Column-I	Column- II

(A) Cascuta	(1) Saprophyte

(B) Eichhornia	(2) Pneumatophore

(C) Monotropa	(3) Insectivorous plant

(D) Rhizophora	(4) Parasite

(E) Uticularia	(5) Root pocket

Options:
A A B C D E 4 3 1 5 2
B A B C D E 4 5 1 2 3
C A B C D E 2 3 1 5 4
D A B C D E 3 1 5 4 2

The correct answer is:
A B C D E 4 5 1 2 3

Solution:
Column I	Column II

Cuscuta	Parasite

Eichhornia	Root pocket

Monotropa	Saprophyte

Rhizophora	Pneumatophore

Utricularia	Insectivorous

Topic: Biology, Mineral Nutrition, NEET
Match the following with reference to cockroach and choose the CORRECT option.

Column I	Column II

P.	Anal style	(i)	10th tergite

Q.	Testis	(ii)	2nd - 6th segments

R.	Anal cerci	(iii)	9th stemite

S.	Ovary	(iv)	4th - 6th

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

The correct answer is:
P-(iii), Q-(iv), R-(i), S-(ii)

Solution:
In cockroaches, the male reproductive system consists of a pair of testes, vas deferns, ejaculatory duct, seminal vesicle, mushroom gland and external genitalia. Testes are present at 4th to 6th segments.
The female reproductive system consists of a pair of ovaries, oviduct, vagina, genital chamber. Ovaries are present at 2nd to 6th segments.
Male and female cockroaches show sexual dimorphism. Males show presence of anal cerci and anal styles while females have only anal cerci. Anal cerci arises from 10th tergite and anal style is from 9th sternite.

Topic: Biology, Structural Organisation in Animals, NEET
Match the following with respect to chromosomal mutation

Column I	Column II

a.	Tetrasomy	(i)	AABB

b.	Double trisomy	(ii)	2n â 1

c.	Hypoploidy	(iii)	2n + 1

d.	Allopolyploidy	(iv)	2n + 2

(v)	2n + 1 + 1

Options:
A a(iv), b(ii), c(i), d(v)
B a(iv), b(v), c(ii), d(i)
C a(iv), b(v), c(iii), d(ii)
D a(ii), b(iii), c(i), d(iv)

The correct answer is:
a(iv), b(v), c(ii), d(i)

Solution:
A tetrasomy is a form of aneuploidy with the presence of four copies, instead of the normal two ( 2n + 2) , Double aneuploidy leading to trisomy or monosomy of 2 different chromosomes arises because of 2 meiotic non-disjunctional events (2n + 1 + 1) , In Hypoploidy ,having a chromosome number lower by only a few chromosomes than the normal diploid number (2n - 1), Allopolyploids are polyploids with chromosomes derived from different species. Precisely it is the result of multiplying the chromosome number in an F1 hybrid.

Topic: Biology, Principles of Inheritance and Variation, NEET
Match the following (with respect to flowers and family).

I	II

(1)	Actinomorphic, trimerous	(a)	Solanaceae

(2)	Zygomorphic, pentamerous	(b)	Liliaceae

(3)	Actinomorphic, pentamerous	(c)	Cruciferae

(4)	Actinomorphic, tetramerous	( d )	Fabaceae

Options:
A 1-a,2-b,3-c,4-d
B 1-a,2-c,3-d,4- b
C 1-b,2-d,3-a,4-c
D 1-d,2-c,3-a,4-b

The correct answer is:
1-b,2-d,3-a,4-c

Solution:
Actinomorphic and trimerous flowers are found in the Liliaceae family. Actinomorphic and tetramerous flowers are found in the Brassicaceae family. Zygomorphic and pentamerous flowers are seen in the Fabaceae family.

Topic: Biology, Morphology of Flowering Plants, NEET
Match the following with respect to meiosis

(a)	Zygotene	(i)	Terminalization

(b)	Pachytene	(ii)	Chiasmata

(c)	Diplotene	(iii)	Crossing over

(d)	Diakinesis	(iv)	Synapsis

Select the correct option from the following

Options:
A (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)
B (a)-(i), (b)-(ii), (c)-(iv), (d)-(iii)
C (a)-(ii), (b)-(iv), (c)-(iii), (d)-(i)
D (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)

The correct answer is:
(a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)

Solution:
Prophase I:
Prophase in meiosis-I is typically longer and more complex than the prophase in meiosis II. Prophase I is subdivided into five phases, viz. Leptotene, Zygotene, Pachytene, Diplotene and Diakinesis.

Leptotene:
• During this stage, the chromosomes become gradually visible under the light microscope. The compaction of chromosomes continues throughout this phase.

Zygotene:
• Chromosomes start pairing together. This process is called synapsis. The paired chromosomes are called homologous chromosomes.
• Formation of synapsis is accompanied by the formation of the synaptonemal complex.
• The synaptonemal complex by a pair of homologous chromosomes is called a bivalent or a tetrad.

Pachytene:
• Bivalent chromosomes clearly appear as tetrads, at this stage.
• Recombination nodules appear. These nodules are the sites at which crossing overtakes place between non-sister chromatids of the homologous chromosomes.
• Exchange of genetic materials between two homologous chromosomes takes place during crossing over. This leads to the recombination of genetic materials on the two chromosomes.

Diplotene:
• Synapotnemal complex is dissolved at this stage.
• The recombined homologous chromosomes of the bivalent separate from each other; except at the site of crossing over.
• The X-shaped structures; thus formed; are called chiasmata.

Diakinesis:
• Chiasmata is terminated at this stage.
• Meiotic spindles are formed to prepare the homologous chromosomes for separation.
• Nucleolus disappears and nuclear envelope breaks down by the end of diakinesis.

Topic: Biology, Cell Cycle and Cell Division, NEET, NEET 2020 (Phase 1)
Match the following with the correct combination:
A. Hyaluronidase 1. Acrosomal reaction
B. Corpus luteum 2. Morphogenetic movements
C. Gastrulation 3. Progesterone
D. Capacitation 4. Mammary gland
E Colostrum 5. Sperm activation

Options:
A A - 5, B - 2, C - 4, D - 1, E - 3
B A - 3, B - 2, C - 5, D - 4, E - 1
C A - 1, B - 2, C - 3, D - 4, E - 5
D A - 1, B - 3, C - 2, D - 5, E - 4

The correct answer is:
A - 1, B - 3, C - 2, D - 5, E - 4

Solution:
Hyaluronidase or sperm lysin is an enzyme released by acrosome of the sperm by breaking down of the membrane surrounding it.
Corpus luteum is a yellowish mass of tissue that forms from the granulosa cells in the cavity of a Graafian follicle in the ovary of a mammal after the release of the egg cell. It secretes the hormone progesterone.
Gastrulation is the phase of embryonic development during which the cells of blastula move in small masses or as a sheet of cells to attain the final locations. such movement is called morphogenetic movement.
Sperms in the female genital tract are made capable of fertilizing the egg by secretions of the female genital tract. These secretions of the female tract remove coating substances deposited on the surface of the sperms particularly those on the acrosome. It exposes the receptor sites on acrosome and leads to sperms activations. This phenomenon is called capacitation.
Colostrum is a liquid with high nitrogen content, antibodies and vitamins that is secreted from the mammary glands before and just after giving birth.

Topic: Biology, Human Reproduction, NEET
Match the followings and choose the correct option.

Group A	Group B

(A)	Aleurone layer	(i)	Without fertilization

(B)	Parthenocarpic fruit	(ii)	Nutrition

(C)	Ovule	(iii)	Double fertilization

(D)	Endosperm	(iv)	Seed

Options:
A A-ii, B-iv, C-iii, D-i
B A-ii, B-i, C-iv, D-iii
C A-i, B-ii, C-iv, D-iii
D A-ii,Â B-iv,Â C-i,Â D-iii

The correct answer is:
A-ii, B-i, C-iv, D-iii

Solution:
Aluerone layer is the outermost layer of endosperm in monocots which causes breakdown of stored food during germination of seed. Fruits formed without fertilization are called parthenocarpic fruits. Ovule after fertilization develops into seed.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
Match the followings and choose the correct option.

Group A	Group B

( A )	Pericarp	(i)	Without fertilization

( B )	Parthenocarpic fruit	(ii)	Ovary wall

( C )	Ovule	(iii)	Nutrition

( D )	Endosperm	(iv)	Seed

Options:
A (A)–(i), (B)–(ii), (C)–(iii), (D)–(iv)
B (A)–(ii), (B)–(i), (C)–(iv), (D)–(iii)
C (A)–(iv), (B)–(ii), (C)–(i), (D)–(iii)
D (A)–(ii), (B)–(iv), (C)–(i), (D)–(iii)

The correct answer is:
(A)–(ii), (B)–(i), (C)–(iv), (D)–(iii)

Solution:
After feritilisation ovary develops to form fruit and ovary wall develops to form fruit wall known as pericarp. Parthenocarpic fruits are formed without fertilization thus are seedless in nature. After fertilistaion ovule develops to form seed. Endosperm is the product of double fertilization and provides nutrition to developing embryo.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET