datavorous/entrance-exam-dataset huggingface dataset:
Match List-I with List-II relating to microbes and their products:
$\begin{array}{|l|l|r|l|} \hline & \begin{array}{l} \text{List-I} \\ \text{(Microbes)} \end{array} & & \begin{array}{l} \text{List-II} \\ \text{(Products)} \end{array} \\ \hline A. & \text{Streptococcus} & I. & \text{Citric acid} \\ \hline B. & \text{Trichoderma polysporum} & II. & \text{Clot buster} \\ \hline C. & \text{Monascus purpureus} & III. & \text{Cyclosporin A} \\ \hline D. & \text{Aspergillus niger} & IV. & \text{Statins} \\ \hline \end{array}$
Choose the correct answer from the options given below:

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

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

Solution:
Streptokinase is produced by Streptococcus and is used as a clot buster for removing clots from the blood vessels of patients who have undergone myocardial infarction leading to heart attack.
- Cyclosporin A, that is used as an immuno-suppressive agent in organ-transplant patients, it is produced by the fungus Trichoderma polysporum
- Statins produced by Monascus purpureus have been commercialized as blood-cholesterol lowering agent.
- Citric acid is produced by Aspergillus niger

Topic: Biology, Microbes in Human Welfare, NEET, NEET 2024 (Re-NEET)
Match List-I with List-II with respect to convergent evolution:
\begin{array}{|l|l|l|l|}\hline & List-I & & List-II \\\hline A. & Lemur & I. & Flying phalanger \\\hline B. & Bobcat & II. & Numbat \\\hline C. & Anteater & III. & Spotted cuscus \\\hline D. & Flying squirrels & IV. & Tasmanian tiger cat \\\hline\end{array}
Choose the correct answer from the options given below:

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

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

Solution:
The correct answer is option (1), because
- Lemur is a placental mammal, shows convergent evolution with spotted cuscus which is an Australian marsupial.
- Bobcat is a placental mammal, shows convergent evolution with Tasmanian tiger cat which is an Australian marsupial.
- Anteater is a placental mammal, shows convergent evolution with Numbat which is an Australian marsupial.
- Flying squirrel is a placental mammal, shows convergent evolution flying Phalanger which is Australian marsupial

Topic: Biology, Evolution, NEET, NEET 2024 (Re-NEET)
Match the abnormal conditions given in Column A with their explanations given in Column B and Choose the correct option

Column A

Column B

A. Glycosuria

i. Accumulation of uric acid in joints

B. Renal calculi

ii. Inflammation in glomeruli

C. Glomerular nephritis

iii. Mass of crystallised salts within the kidney

D. Gout

iv. presence of glucose in the urine

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

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

Solution:
Glycosuria is the presence of glucose in the urine. Renal calculi is the accumulation of mass of crystallised salts within the kidney. Glomerular nephritis is the inflammation in glomeruli. Gout is the accumulation of uric acid in joints.

Topic: Biology, Excretory Products and their Elimination, NEET
Match the abnormal conditions given in Column A with their explanations given in Column B and choose the correct option.

Column A	Column B

(A)	Glycosuria	(i)	Accumulation of uric acid in joints

(B)	Renal calculi	(ii)	Inflammation in glomeruli

(C)	Glomerular nephritis	(iii)	Mass of crystallised salts within the kidney

(D)	Gout	(iv)	Presence of glucose in urine

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

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

Solution:
A. Glycosuria is a disorder characterized by excretion of glucose in urine.
B. Renal calculi is a mass of crystallised salts which gets formed in the kidney. It is also known as kidney stone.
C. Glomerular nephritis is the inflammation of the glomeruli. It can be caused due to immunological or infectious reasons.
D. Gout is a condition which develops due to accumulation of uric acid in the joints. It is a very painful condition. The great toe is most likely to be affected by this condition.

Topic: Biology, Excretory Products and their Elimination, NEET
Match the amino acid with its correct R group.

Amino acid	R- group

P	Cysteine	I	-CH2OH

Q	Serine	II	-CH2-CH2-CH2-COOH

R	Glutamic acid	III

S	Tyrosine	IV	-CH2-SH

Options:
A P-I, Q-IV, R-II, S-III
B P-IV, Q-I, R-II, S-III
C P-III, Q-IV, R-II, S-I
D P-II, Q-I, R-IV, S-III

The correct answer is:
P-IV, Q-I, R-II, S-III

Solution:
Amino acids are monomers of proteins. These are substituted methanes and has a variable group R attached to the carbon. Due to presence of R group various types of amino acids are formed. Cysteine and Methionine are Sulphur containing amino acids. Tyrosine are amino acids with benzene ring. Glutamic acid is acidic and that means have acidic group. Serine has alcoholic group.

Topic: Chemistry, Biomolecules, NEET
Match the animals given in column A with their location in column B. Choose the correct match from the following

Column I

Column II

A. Dodo

i. Africa

B. Quagga

ii. Russia

C. Thylacine

iii. Mauritius

D. Steller's sea cow

iv. Australia

Options:
A i—A, ii—C, iii—B, iv—D
B i—D, ii—C, iii—A, iv—B
C i—C, ii—A, iii—B, iv—D
D i—C, ii—A, iii—D, iv—B

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

Solution:
Correct Answer is : (D)
i—C, ii—A, iii—D, iv—B

Topic: Biology, Biodiversity and its Conservation, NEET
Match the catalyst with the process
Which of the following is the correct option?

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

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

Solution:
(1) $\mathrm{V}_2 \mathrm{O}_5$-(iii) Oxidation of $\mathrm{SO}_2$ in the manufacture of $\mathrm{H}_2 \mathrm{SO}_4$ by contact process.
(2) $\mathrm{TiCl}_4+\mathrm{Al}\left(\mathrm{CH}_3\right)_3-$ (Ziegler-Natta catalyst)- (iv) Polymerisation of ethylene.
(3) $\mathrm{PdCl}_2-$ (i) Oxidation of ethyne to ethanal.
(4) Nickel complexes-(ii) Polymerisation of alkynes. Thus, the correct match is
(1) $\rightarrow$ (iii), (2) $\rightarrow$ (iv), (3) $\rightarrow$ (i), (4) $\rightarrow$ (ii)

Topic: Chemistry, d and f Block Elements, NEET, NEET 2019 (Odisha)
Match the catalysts (Column I) with products (Column II)

Column I	Column II

(1) V2O5	(i) Polyethylene

(2) TiCl4/Al(Me)3	(ii) ethanal

(3) PdCl2	(iii) H2SO4

(4) Iron Oxide	(iv) NH3

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

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

Solution:
V2O5 is used as a catalyst in contact process for H2SO4 .
TiCl4/Al(Me)3 i.e., Ziegler Natta salt is used as catalyst for polymerisation of ethene.
PdCl2 is used as catalyst for ethanal (Wacker process).
Iron oxide is used as catalyst in Haber's synthesis.

Topic: Chemistry, d and f Block Elements, NEET
Match the chromosomes with their properties:

Chromosome	Property

1	Metacentric chromosome	i	Terminal centromere

2	Acrocentric chromosome	ii	Equal arms

3	Submetacentric chromosome	iii	One slightly shorter arm, one slightly longer arm

4	Telocentric chromosome	iv	One extremely short arm and one very long arm

Options:
A 1-iv,Â 2-iii,Â 3-ii,Â 4-i
B 1-i,Â 2-ii,Â 3-iii,Â 4-iv
C 1-ii, 2-iv, 3-iii, 4-i
D 1-iv,Â 2-i,Â 3-iii,Â 4-ii

The correct answer is:
1-ii, 2-iv, 3-iii, 4-i

Solution:
A chromosome is supercoiled DNA which is visible during the divisional phase of the cell cycle. These chromosomes contain chromatids and show primary constriction sites called the centromere. Depending on the location of the centromere the chromosome is classified as -
Metacentric - centromere is at the middle
Telocentric - it has terminal centromere
Acrocentric - centromere is located close to end making one arm very long and the other one very short.
Sub-metacentric - centromere is close to the centre but not at the centre, thus making unequal sized arm (one short and one long)

Topic: Biology, Cell - The Unit of Life, NEET
Match the Column A with Column B -

Column A	Column B

1.	Caecum	A.	Villi

2.	Stomach	B.	Papillae

3.	Tongue	C.	Symbiotic micro-organisms

4.	Small intestine	D.	Pyloric sphincter

Options:
A 1âÂ C,Â 2âÂ D,Â 3âÂ A,Â 4âÂ B
B 1â C, 2â D, 3â B, 4â A
C 1Â âÂ D,Â 2Â âÂ C,Â 3Â âÂ B,Â 4Â âÂ A
D 1Â âÂ D,Â 2Â âÂ A,Â 3Â âÂ B,Â 4Â âÂ C

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

Solution:
Caecum is a small blind sac which is harbouring some symbiotic micro-organisms e.g. E.coli. The caecum opens into the colon.
The innermost layer lining the lumen of the alimentary canal is the mucosa. This layer forms irregular folds (rugae) in the stomach and small finger-like foldings called villi in the small intestine. The cells lining the villi produce numerous microscopic projections called microvilli giving a brush border appearance.
The tongue is a freely movable muscular organ attached to the floor of the oral cavity by the frenulum. The upper surface of the tongue has small projections called papillae, some of which bear taste buds.
The opening of the stomach into the duodenum is guarded by the pyloric sphincter.

Topic: Biology, Digestion and Absorption, NEET
Match the column and select the correct option.

Column âI	Column â II

A	UV rays	(i)	Biomagnification

B	Biodegradable organic matter	(ii)	Eutrophication

C	DDT	(iii)	Snow blindness

D	Phosphate	(iv)	BOD

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

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

Solution:
In human eyes, cornea absorbs UV B radiation, and causes inflammation of cornea, called snow-blindness. Biodegradble organic matter increases BOD of water, and non biodegradable substance like DDT gets accumulated in successive trophic levels of food chain leading to biomagnifications. Phosphate is plant nutrient which results in eutrophication.

Topic: Biology, Environmental Issues, NEET
Match the column and select the correct option.

Column-I	Column-II

(1)	Leaf tendril	(i)	Cactus

(2)	Leaf Spine	(ii)	Pea

(3)	Stem tendril	(iii)	Bougainvillea

(4)	Stem thorn	(iv)	Gourds

(v)	Turmeric

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

The correct answer is:
1-ii, 2-i, 3-iv, 4-iii

Solution:
Stem tendrils which develop from axillary buds, are slender and spirally coiled and help plants to climb such as in gourds (cucumber, pumpkins, watermelon) and grapevines. Axillary buds of stems may also get modified into woody, straight and pointed thorns. Thorns are found in many plants such as Citrus and Bougainvillea.
Leaves are often modified to perform functions other than photosynthesis. They are converted into tendrils for climbing as in peas or into spines for defence as in cacti

Topic: Biology, Morphology of Flowering Plants, NEET
Match the column and select the correct option.

Column-I	Column-II

(A)	Autogamy	(i)	Both male and female flower on one plant

(B)	Bisexual	(ii)	Male and female flower in different plants

(C)	Monoecious	(iii)	Both androecium and gynoecium on the same flower

(D)	Dioecious	(iv)	Self-pollination

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

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

Solution:
Autogamy is self-pollination in which pollen grain form stamen is transferred to stigma of same flower. Flower having both androecium and gynoecium is called as bisexual. Plant on which both male and female flowers are produced on same plant is called monoecious. When male and female flower are produced on different plant it is called as dioecious.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
Match the column and select the correct option from following.

Column I	Column II

A	Completely condensed chromosome	i	Prophase

(B)	Division of centriole	ii	Metaphase

(C)	Nuclear membrane, Golgibody, ER disappear.	iii	S-phase

(D)	Splitting of centromere	iv	Anaphase II

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

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

Solution:
During cell cycle centriole duplicates during S Phase , condensation of chromosome starts during prophase and condensation is completed during metaphase. Nuclear membrane, golgi body and ER disappears during late prophase. Centraomere splits are chromatids are separated during anaphase of mitosis and anaphase II of meiosis.

Topic: Biology, Cell Cycle and Cell Division, NEET
Match the column:

Column I	Column II

A.	Acidic	1.	Valine

B.	Basic	2	Lysine

C.	Neural	3.	Glutamic acid

D.	Aromatic	4.	Phenylalanine

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

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

Solution:
Amino acids are classified as acidic, basic or neutral depending upon the relative number of amino and carboxyl groups in their molecule. Equal number of amino and carboxyl groups makes it neutral, e.g. valine, glycine, etc.; more number of amino than carboxyl groups makes it basic e.g. lysine, arginine and histidine; and more carboxyl groups as compared to amino groups makes it acidic e.g. aspartic acid and glutamic acid. If the side chain comprises of aromatic ring, it is called aromatic amino acid e.g. phenylalanine, tyrosine and tryptophan.

Topic: Chemistry, Biomolecules, NEET
Match the column.

Column I	Column II

(a)	Cristae	(i)	Flat membranous sacs in chloroplast

(b)	Cisternae	(ii)	Infoldings in mitochondria

(c)	Thylakoid	(iii)	Disc-shaped sacs in Golgi apparatus

Options:
A aÂ -Â ii,Â b-Â i,Â c-Â iii
B a - ii, b - iii, c âi
C aÂ -Â iii,Â bÂ -Â ii,Â cÂ âi
D aÂ -Â iii,Â bÂ -Â i,Â cÂ -Â ii

The correct answer is:
a - ii, b - iii, c âi

Solution:
The inner membrane of mitochondria is folded into finger-like projections called cristae, it increases the surface area for attachment of oxysomes. Golgi bodies are made up of flat disc-like structure piled one on top of other known as cisternae, which gets curved forming cis and trans face. Thylakoids are flat sac that is piled one on top of other forming bundles known as granum.

Topic: Biology, Cell - The Unit of Life, NEET
Match the column:

Column I	Column II

(A)	Marginal	(1)	Pea

(B)	Parietal	(2)	Mustard and Argemone

(C)	Axile	(3)	Dianthus and primrose

(D)	Free central	(4)	China rose, tomato, lemon

(E)	Basal	(5)	Sunflower and marigold

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

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

Solution:
Marginal placentation is observed in Fabaceae family, pea. Parietal placentation is present in Brassicaeae family, mustard and Argemone. Axile placentation is present in Solanaceae, Malvaceae and lemon family Free central palcentation is present in Dianthus and Primerose. Basal is most advance placentation present in Sunflower and marigold.

Topic: Biology, Morphology of Flowering Plants, NEET
Match the column:

Column I	Column II

A	NADH dehydrogenase	(1)	Complex IV

B	Cytochrome b c 1	(2)	Complex III

C	Cytochrome a, a 3	(3)	Complex I

D	ATP synthase	(4)	Complex V

Options:
A A–1, B–2, C–3, D–4
B A–3, B–2, C–1, D–4
C A–4, B–1, C–3, D–2
D A–1, B–4, C–2, D–3

The correct answer is:
A–3, B–2, C–1, D–4

Solution:
ETS is organized in the inner mitochondria membrane, organized in the form of IV complex and is associated with complex V. Complex I is NADH dehydrogenase, complex II is FADH2 dehydrogenase, Compex III is cytochrome b - c1 , associated with a mobile carrier cytochrome c, complex Iv refers to cytochrome c oxidase complex containing cytochromes a and a3 and two copper centers. Complex V is ATP synthase.

Topic: Biology, Respiration in Plants, NEET
Match the column:

Column I	Column II

(i) Epiphyllous stamens	(p) China rose

(ii) Monoadelphous	(q) Citrus

(iii) Epipetalous stamens	(r) Lily

(iv) Polyadelphous	(s) Brinjal

Options:
A (i)-(r),Â (ii)-(s),Â (iii)-(p),Â (iv)-(q)
B (i)-(q),Â (ii)-(s),Â (iii)-(p),Â (iv)-(r)
C (i)-(q),Â (ii)-(p),Â (iii)-(s),Â (iv)-(r)
D (i)-(r), (ii)-(p), (iii)-(s), (iv)-(q)

The correct answer is:
(i)-(r), (ii)-(p), (iii)-(s), (iv)-(q)

Solution:
Stamens in the flower form the male reproductive part. Stamens have the filament stalk that holds the anther at tip region. Anther contains the pollen sac where pollen grains are formed and released during pollination process.
Stamens in flowers of brinjal are found attached to the petal and this condition is called as epipetalous. Epipyllous is the condition when stamens are attached to the perianth, as observed in lily.
When stamens are bundled together, it is called adelphous. If only one bundle is formed- monoadelphous- seen in china rose. And if many bundles then it is called polyaldelphous, seen in citrus plants.

Topic: Biology, Morphology of Flowering Plants, NEET
Match the column :

Column-(A)	Column-(B)

(i)	Use & disuse Theory	(A)	Hugo-de-Vries

(ii)	Survival of the fittest	(B)	Lamarck

(iii)	Spark Discharge apparatus	(C)	Darwin

(iv)	Mutation	(D)	Stanley Miller

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

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

Solution:
Lamarck incorporated two ideas into his theory of evolution, in his day considered to be generally true: Use and disuse – Individuals lose characteristics they do not require (or use) and develop characteristics that are useful. Darwin first used Spencer's new phrase "survival of the fittest" alongside "natural selection" in the fifth edition of On the Origin of Species, published in 1869, intending it to mean "better designed for an immediate, local environment". Stanley Miller, the chemist who was famous for his spark discharge experiment and the findings of the first-ever modern equipment and confirmed that the reaction created peptides. De Vries Hugo Marie a Dutch botanist who studied evolution by observing mutations rather than natural selection. He was an early proponent of the works of Gregor Mendel.

Topic: Biology, Evolution, NEET
Match the column:

Column-I	Column-II

A.	ÏÃ 174 DNA	(i)	3.3 Ã 109 bp

B.	Î» - phage DNA	(ii)	4.6 Ã l06 bp

C.	E. coli DNA	(iii)	48502 bp

D.	Haploid content of human DNA	(iv)	5386 bases

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

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

Solution:
The length of DNA in organism is measured in terms of the number of base pairs or the number of nucleotides. Each organism has a fixed quantity of DNA, ϕ×174 has single-stranded DNA thus its length is measured in the number of bases present and for other organisms having double-stranded DNA, the length is measured in the number of base pairs.

Topic: Biology, Molecular Basis of Inheritance, NEET
Match the column:

Column-I	Column-II

a.	Prophase	(i)	Nucleolus reappears.

b.	Metaphase	(ii)	Division of protoplast.

c.	Anaphase	(iii)	Chromosome completes its condensation.

d.	Telophase	(iv)	Spindle fibres start appearing near the nucleus.

e.	Cytokinesis	(v)	Chromosomes appear V, L, J and I shaped.

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

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

Solution:
Cell division is the phase in the cell cycle where the cell divides to form new daughter cells. Depending on the type of division - Mitosis - makes two new identical daughter cells and Meiosis - makes four new daughter cells with half the genetic content of the mother cell.
During the mitotic cell division, cell undergoes through stages of division ultimately dividing to form two new cells-
Prophase - It is the first stage of nuclear division and marked by condensation of chromosomal material. Centrosomes start moving towards to the opposite poles. Each of the centrosome radiate microtubules called spindle fibres that hold the chromosomes at centromere. The radiating centrosome looks like a start hence called as aster. Both centrosome and the spindle fibres are called as mitotic apparatus.
Metaphase - In this stage the nuclear envelope disintegrated completely and disappears. The condensation of the chromosome is complete. The condensed chromosomes start moving and arranging towards the equatorial plane with support from spindle fibres that bind to the kinetochore of centromere in each chromosome.
Anaphase - It is the shortest phase of the nuclear division. In this phase, each chromosome split and move away from equator towards to the poles. The sister chromatids separate and show various shapes of the chromatids like V, I, L and J shaped.
Telophase - The last phase of nuclear division where the separated chromosomes cluster at the poles. The nuclear envelope reappears with decrease in cytoplasmic viscosity. Nucleolus, Golgi complex along with endoplasmic reticulum are reformed.
Cytokinesis - The last stage of cell division process. Once the nuclear content has been divided and separated in the karyokinesis process, the whole cell then divides to form new daughter cells by separation of the protoplast.

Topic: Biology, Cell Cycle and Cell Division, NEET
Match the column I and II and identify the correct option.

Column I (Microbes)	Column-II (Use)

[A]	Streptococcus	(i)	Biogas

[B]	Methenobacterium	(ii)	Biofertilizer

[C]	Azospirullum	(iii)	Ethanol

[D]	Yeast	(iv)	Streptokinase

Options:
A [A]-(i);Â [B]-(ii);Â [C]-(iii);Â [D]-(iv)
B [A]-(iv); [B]-(i); [C]-(ii); [D]-(iii)
C [A]-(iv);Â [B]-(i);Â [C]-(iii);Â [D]-(ii)
D [A]-(i);Â [B]-(iv);Â [C]-(ii);Â [D]-(iii)

The correct answer is:
[A]-(iv); [B]-(i); [C]-(ii); [D]-(iii)

Solution:
Streptokinase is used as a clot-buster is obtained from Streptococcus. Methanobacterium produces biogas by the anaerobic breakdown of organic matter. Azospirullum form a loose symbiosis with maize and is used as a biofertilizer. Yeast is used for commercial production of ethanol.

Topic: Biology, Microbes in Human Welfare, NEET
Match the column I (Ecologist) with column II (Concept) and choose the correct option.

Column I (Ecologist)	Column II (Concept)

(a)	R. May	1.	Effect of species richness on productivity

(b)	A.V. Humboldt	2.	Effect of area exploration on species richness

(c)	D. Tilman	3.	Global species estimation

(d)	P. Ehrlich	4.	Effect of the decline of species on ecosystem

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

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

Solution:
Robert May told that the global species diversity of Earth is about 7 million.
A. V. Humboldt observed that within a region, species richness increased with increasing explored area but only up to a limit.
David Tilman concluded that plots with more species show less year to year variation in total biomass and increased diversity contributed to higher productivity.
Paul Ehrlich explained effect of species extinction by Rivet Popper Hypothesis.

Topic: Biology, Biodiversity and its Conservation, NEET
Match the column I with column II and choose the correct combination from the option given:

Column-I	Column-II

(1) Deuteromycetes	(A) Morels

(2) Ascomycetes	(B) Smut

(3) Basidiomycetes	(C) Bread mould

(4) Phycomycetes	(D) Imperfect fungi

Options:
A 1-C; 2-D; 3-A; 4-B
B 1-D; 2-A; 3-B; 4-C
C 1-B; 2-C; 3-D; 4-A
D 1-D; 2-C; 3-A; 4-B

The correct answer is:
1-B; 2-C; 3-D; 4-A

Solution:
Deuteromycetes are fungi imperfection where rust and smut causing fungi are members of Basidiomycetes. Phycomycetes have common bread mould and Morels belong to ascomycetes.

Topic: Biology, Biological Classification, NEET