datavorous/entrance-exam-dataset huggingface dataset:
The fossils of the Java man discovered in 1891 belonged to the species of ______________. They had a larger brain capacity than its ancestors of about _________ cc.

Options:
A Homo erectus, 900
B Homo habilis, 750
C Homo sapiens neanderthalensis, 1400
D Homo sapiens fossilis, 350

The correct answer is:
Homo erectus, 900

Solution:
Fossils discovered in Java in 1891, known as Java man, revealed the next stage of evolution as Homo erectus by Mayer. Homo erectus had a large brain around 900cc. They probably ate meat.

Topic: Biology, Evolution, NEET
The four sketches (A,B,C and D) given below, represent four different types of animal tissues. Which one these are correctly identified in the options given, along with its correct location and function?

Options:
A

Gland	Location	Function

Glandular epithelium	Intestine	Secretion

B

Gland	Location	Function

Collagen fibres	Cartilage	Attach skeletal muscles to bones

C

Gland	Location	Function

Smooth muscle tissue	Heart	Heart contraction

D

Gland	Location	Function

Columnar epithelium	Nephron	Secretion and absorption

The correct answer is:

Gland	Location	Function

Glandular epithelium	Intestine	Secretion

Solution:
The intestine is lined by glandular epithelium which is secretory in function. The glands found in the intestine are exocrine and may be unicellular or multicellular. When unicellular glands secrete mucus, they are called mucus cells or goblet cells and are common in the columnar epithelium of intestine. When unicellular glands secrete a clear watery fluid, they are called serous cells which are also present in intestinal glands. On the other hand, multicellular glands consist of a duct and secretory portion both formed of epithelial cells. They are further of two types: tubular and saccular. In tubular glands, the secretory portion is tube-like for example, Crypts of Lieberkuhn (a type of simple straight tubular glands found in the intestine) and Brunner's gland (a type of simple branched tubular glands found in the intestine).

Topic: Biology, Structural Organisation in Animals, NEET
The fraction of atoms of a radioactive element that decays in 6 days is 78. The fraction that decays in 10 days will be

Options:
A 7780
B 7180
C 3132
D 1516

The correct answer is:
3132

Solution:
78 fraction decays in 6 days
18 fraction is active after 6 days
By using, N=N012 t/T1/2
N=N02n
N08=N02n
n=3

6 days = half lives
t12=2 days
10 days=5 t1/2
N=N025=N032
So decay=N0-N032=3132N0

Topic: Physics, Nuclear Physics, NEET
The fraction of energy lost when an organism proceeds from one trophic level to the next trophic level is

Options:
A 10 percent
B 20 percent
C 30 percent
D 90 percent

The correct answer is:
90 percent

Solution:
As per the Ten percent energy transfer law introduced by Lindeman-'During the transfer of energy from organic food from one trophic level to the next, only about ten percent of the energy from organic matter is stored as flesh.' In a typical food chain, only ten percent of the energy is passed to the next consumer in the unidirectional manner and 90% is lost to the environment as heat. In a typical food chain, the apex predator receives the least amount of energy and hence, a single food chain typically does not have more than 4 levels.

Topic: Biology, Ecosystem, NEET
The fraction of total volume occupied by the atoms present in a simple cube is:

Options:
A $\frac{\pi}{3 \sqrt{2}}$
B $\frac{\pi}{4 \sqrt{2}}$
C $\frac{\pi}{4}$
D $\frac{\pi}{6}$

The correct answer is:
$\frac{\pi}{6}$

Solution:
Number of atoms per unit cell $=1$
Atoms touch each other along edges.
Hence $r=\frac{a}{2}$
$\begin{aligned}
& \text {Therefore % fraction }=\frac{\frac{4}{3} \pi r^3}{(2 r)^3}=\frac{\pi}{6} \\
& =0.52
\end{aligned}$

Topic: Chemistry, Solid State, NEET, NEET 2007
The fragmented DNA can be visualized by staining DNA with

Options:
A NaCl
B Ethidium bromide
C Ethylene bromide
D NaBr

The correct answer is:
Ethidium bromide

Solution:
In gel electrophoresis, the fragmented DNA can be visualized only after staining the DNA with a compound known as ethidium bromide followed by exposure to UV radiation, as pure DNA fragments cannot be seen in visible light without staining.

Topic: Biology, Biotechnology - Principles and Processes, NEET
The framework system of classification in which various taxonomic categories are arranged in order of logical sequence is called

Options:
A Systematics
B Hierarchy
C Taxon
D Classification

The correct answer is:
Systematics

Solution:
Systematics (L-system: order). The comparative study of organisms based on morphology, anatomy, ecology, physiology, biochemistry, etc. is known as systematics. This comprises three fields such as
1. Nomenclature
2. Classification and
3. Taxonomy
Systematics is also the science of diversity of organisms from an evolutionary context.

Topic: Biology, The Living World, NEET
The free energy of formation of NO is 78 kJ mol-1 at the temperature of an automobile engine (1000 K). What is the equilibrium constant for this reaction at 1000 K?
1 2 N 2 g + 1 2 O 2 g →NO g
(Anti log 0.9263 = 8.459)

Options:
A 8.4×10-5
B 7.1×10-9
C 4.2×10-10
D 1.7×10-19

The correct answer is:
8.4×10-5

Solution:
Keq=antilog-ÎGo2.303RT
ÎGo=78 kJ/mol=78000 J/mol
R=8.314 JK-1mol-1,T=1000 K
Putting these values in eqn. (i) we get,
Keq=antilog-780002.303Ã8.314Ã1000
Keq=antilog(-4.0737) or Keq=8.4Ã10-5

Topic: Chemistry, Electrochemistry, NEET
The free-living, anaerobic nitrogen-fixer is

Options:
A Beijerinckia
B Rhodospirillum
C Rhizobium
D Azotobacter

The correct answer is:
Rhodospirillum

Solution:
Rhodospirillum is a free-living, anaerobic, nitrogen fixer. Both Beijerinckia and Azotobacter are free-living, nitrogen-fixing, aerobic microbes. Rhizobium is a symbiotic, nitrogen-fixer.

Topic: Biology, Mineral Nutrition, NEET, NEET 2010 (Screening)
The free-living fungus Trichoderma can be used for

Options:
A Killing insects
B Biological control of plant diseases
C Controlling butterfly caterpillars
D Producing antibiotics

The correct answer is:
Biological control of plant diseases

Solution:
The free-living fungus Trichoderma can be used for biological control of plant diseases.

Topic: Biology, Microbes in Human Welfare, NEET
The freezing point depression constant for water is $-1.86^{\circ} \mathrm{C} \mathrm{m}^{-1}$. If $5.00 \mathrm{~g} \mathrm{Na} \mathrm{SO}_4$ is dissolved in $45.0 \mathrm{~g} \mathrm{H}_2 \mathrm{O}$, the freezing point is changed by $-3.82^{\circ} \mathrm{C}$. Calculate the van't Hoff factor for $\mathrm{Na}_2 \mathrm{SO}_4$.

Options:
A 2.63
B 3.11
C 0.381
D 2.05

The correct answer is:
2.63

Solution:
$$
\begin{aligned}
\Delta T_f & =i \times K_f \cdot m \\
\text { or } i & =\frac{\Delta T_f \times W_{\text {solvent }}}{K_f \times n_{\text {solute }} \times 1000} \\
& =\frac{3.82 \times 45}{1.86 \times\left(\frac{5}{142}\right) \times 1000} \\
& =2.63
\end{aligned}
$$

Topic: Chemistry, Solutions, NEET, NEET 2011 (Screening)
The freezing point depression constant Kf of benzene is 5.12 K kg mol–1. The freezing point depression for the solution of molality 0.078 m containing a non-electrolyte solute in benzene is (rounded off upto two decimal places):

Options:
A 0.80 K
B 0.40 K
C 0.60 K
D 0.20 K

The correct answer is:
0.40 K

Solution:
ΔTf=Kf×m

=5.12×0.078

ΔTf=0.40K

Topic: Chemistry, Solutions, NEET, NEET 2020 (Phase 1)
The freezing point of a 3% aqueous solution of A is equal to the freezing point of 9% aqueous solution of B. If the molecular weight of A is 60, then the molecular weight of B will be

Options:
A 180
B 45
C 90
D 20

The correct answer is:
180

Solution:
ΔTf=Kf×m
As Kf is common and freezing points of K and b are equal so
ΔT f  of solution A ΔT f  of solution B = Concentration of A Molar mass of A( M A ) Concentration of B Molar mass of  B(M B )
1=3/MA9/MB
So, MB=93×60
=180

Topic: Chemistry, Solutions, NEET
The freezing point of equimolal aqueous solution will be highest for

Options:
A C6H5N+H3Cl- (aniline hydrochloride)
B Mg(NO3)2
C La(NO3)3
D NH2CONH2 (Urea)

The correct answer is:
NH2CONH2 (Urea)

Solution:
ΔTf∝m×i
m = morality, i = Vant Hoff factor
Except urea, all other are electrolytes undergoing dissociation so (i>1) . Urea is non electrolyte, i=1 . so it has least value of ΔTf hence its freezing point is maximum.

Topic: Chemistry, Solutions, NEET
The freezing point of water is depressed by 0.37oC in a 0.01 molal NaCl solution. The freezing point of 0.02 molal solution of urea is depressed by

Options:
A 0.37oC
B 0oC
C 0.56oC
D 0.187oC

The correct answer is:
0.37oC

Solution:
The depression in freezing point is proportional to molal concentration of the solute i.e.,
ΔTf∝m
ΔTf=Kfmi
or Kf=ΔTfi×m=constant
So, ΔTƒNaCliNaCl×mNaCl=ΔTfi×m=constant
0.372×0.01=ΔTfUrea0.02×1
ΔTf=0.37×0.020.02=0.37 °C

Topic: Chemistry, Solutions, NEET
The frequencies of two sound sources are 256 Hz and 260 Hz. At t=0, the intensity of sound is maximum. Then the phase difference at the time t=116 s will be

Options:
A zero
B π
C π2
D π4

The correct answer is:
π2

Solution:
The time interval between two consecutive beats
T=1n1-n2=1260-256=14 s
So, t=116=T4 s
By using,
time difference =T2π× Phase difference
⇒ T4=T2π×ϕ
⇒ ϕ=π2

Topic: Physics, Waves and Sound, NEET
The frequencies of X-rays, γ-rays and ultraviolet rays are respectively a, b and c. Then

Options:
A a<b, b<c
B a<b, b>c
C a>b, b>c
D a>b, b<c

The correct answer is:
a<b, b>c

Solution:
γ rays has lowest wavelength and highest frequency among them while ultraviolet ray has highest wavelength and lowest frequency.

Topic: Physics, Electromagnetic Waves, NEET
The frequency f of vibration of mass m suspended from a spring of spring constant k is given by f=cmxky. Where c is a dimensionless constant. The values of x and y are respectively

Options:
A 12, -12
B -12, 12
C 12, 12
D -12, -12

The correct answer is:
-12, 12

Solution:
Frequency
M0L0T-1=MxML0T-2y
=Mx+y L0T-2y
Comparing the powers on M, L and T
-2y=-1
â y=12
And x+y=0
â´ x= -y=-12

Topic: Physics, Units and Dimensions, NEET
The frequency for a series limit of Balmer and Paschen series respectively are f1 and f3. If the frequency of the first line of Balmer series is f2 then the relation between f1, f2 and f3 is

Options:
A f1-f2=f3
B f1+f3=f2
C f1+f2=f3
D f2-f3=2f1

The correct answer is:
f1-f2=f3

Solution:
v=fÎ»,1Î»=fv
1Î»=R1P2-1n2
f=Rc1P2-1n2
f2=Rc122-132=Rc14-19
f1=Rc122=Rc4
f3=Rc132=Rc9
â´ f1-f3=Rc14-19
â´ f1-f3=f2
â´ f1-f2=f3

Topic: Physics, Atomic Physics, NEET
The frequency (Î½m) corresponding to which energy emitted by a black body is maximum, may vary with temperature T of the body as shown in the figure. The curve representing the correct variation is

Options:
A A
B B
C C
D D

The correct answer is:
A

Solution:
According to Wien's displacement law,
λm∝1T or cνm∝1T

νm ∝T
Hence, curve A represents the correct variation.

Topic: Physics, Thermal Properties of Matter, NEET
The frequency of a light wave in a material is $2 \times 10^{14} \mathrm{~Hz}$ and wavelength is $5000 Å$. The refractive index of material will be.

Options:
A 1.50
B 3.00
C 1.33
D 1.40

The correct answer is:
3.00

Solution:
Refractive index $\mu=\frac{C}{V}=\frac{C}{V \lambda}$
$=\frac{3 \times 10^8}{2 \times 10^{10} \times 5000 \times 10^{-10}}=3$

Topic: Physics, Ray Optics, NEET, NEET 2007
The frequency of a light wave in a material is 2 Ã 1014 Hz and wavelength is 5000 Aâ . The refractive index of material will be

Options:
A 1.40
B 1.50
C 3.00
D 1.33

The correct answer is:
3.00

Solution:
The velocity of light water in the material is
v=vÎ» â¦i
Refractive index of the material is
Î¼=cv â¦ii
Where c is the speed of light in vacuum or air
or Î¼=cvÎ» â¦iii
Given, v=2Ã1014 Hz,
Î»=5000 Ã=5000Ã10-10 m
c=3Ã108 m s-1
Hence, from Eq (iii), we get
Î¼=3Ã1082Ã1014Ã5000Ã10-10=3.00

Topic: Physics, Ray Optics, NEET
The frequency of a light wave in a material is 2 Ã 1014 Hz and wavelength is 5000 Aâ . The refractive index of material will be

Options:
A 1.40
B 1.50
C 3.00
D 1.33

The correct answer is:
3.00

Solution:
The velocity of light water in the material is
v=vλ …i
Refractive index of the material is
μ=cv …ii
Where c is the speed of light in vacuum or air
or μ=cvλ …iii
Given, v=2×1014 Hz,
λ=5000 Å=5000×10-10 m
c=3×108m s-1
Hence, from Eq (iii), we get
μ=3×1082×1014×5000×10-10=3.00

Topic: Physics, Ray Optics, NEET
The frequency of crossing over between any two linked genes is

Options:
A Higher if they are recessive
B Determined by their relative dominance
C The same as if they were not linked
D Directly proportional to the distance between them

The correct answer is:
Directly proportional to the distance between them

Solution:
Crossing over, a process in genetics by which the two chromosomes of a homologous pair exchange equal segments with each other. Crossing over occurs in the first division of meiosis. Crossing over results in recombination of genes found on the same chromosome, called linked genes, which would otherwise always be transmitted together. Because the frequency of crossing over between any two linked genes is proportional to the chromosomal distance between them, crossing over frequencies are used to construct genetic, or linkage, maps of genes on chromosomes.

Topic: Biology, Principles of Inheritance and Variation, NEET
The frequency of oscillation of current in the inductor is-

Options:
A 1 3 LC
B 1 6 Ï LC
C 1 LC
D 1 2 π LC

The correct answer is:
1 6 π LC

Solution:
Here L eq = L + 2 L = 3 L (in series)
C eq = C +2 C = 3 C (in parallel)
Ï = 1 L eq C eq
= 1 3 LC
frequency f= Ï 2 Ï = 1 6 Ï LC

Topic: Physics, Alternating Current, NEET
The frequency of recessive allele in a population is 0.2. The number of heterozygotes in a population of 500 is

Options:
A 150.
B 240.
C 160.
D 64.

The correct answer is:
160.

Solution:
According to Hardy-Weinberg principle :
Frequency of dominant allele (A) = p

Frequency of recessive allele (a) = q = 0.2

Sum total of all the allelic frequencies is 1, i.e. p+ q = 1
â p + 0.2 = 1 â p = 0.8

Frequency of homozygous dominant individuals = p Ã p = p2

Frequency of homozygous recessive individuals = q Ã q = q2

Frequency of hetrozygous dominant individuals = 2pq = 2 Ã 0.8 Ã 0.2 = 0.32

Total Population = 500
Number of heterozygotes in population = 0.32 x 500 = 160

Topic: Biology, Evolution, NEET
The frequency of recombination between gene pairs on the same chromosome as a measure of the distance gene was explained by:

Options:
A T.H. Morgan
B Gregor J. Mendel
C Alfred Sturtevant
D Sutton Boveri

The correct answer is:
Alfred Sturtevant

Solution:
Alfred Sturtevant (student of Morgan) used the frequency of recombination between gene pairs on the same chromosome as a measure of the distance between genes and ‘mapped’ their position on the chromosome. Two genes show higher frequency of crossing over if the distance between them is higher and lower frequency if the distance is small.

Topic: Biology, Principles of Inheritance and Variation, NEET, NEET 2019 (All India)
The frequency of the incident light falling on a photosensitive metal plate is doubled, the kinetic energy of the emitted photoelectron is

Options:
A Double the earlier value
B Unchanged
C More than doubled
D Less than doubled

The correct answer is:
More than doubled

Solution:
Let E1 and E2 be the KE of photoelectrons for incident light of frequency v and 2v respectively.
Then hv=E1+ϕ and h2v=E2+ϕ0
So, 2E1+ϕ0=E2+ϕ0 or E2=2E1+ϕ0
It means the KE of photoelectron becomes more than double

Topic: Physics, Dual Nature of Matter, NEET
The frequency of two alleles in a gene pool is 0.19 (A) and 0.81(a). Assume that the population is in Hardy-Weinberg equilibrium and
(a) Calculate the percentage of heterozygous individuals in the population.
(b) Calculate the percentage of homozygous recessives in the population.

Options:
A 60% and 40%
B 66% and 31%
C 40% and 60%
D 31% and 66%

The correct answer is:
31% and 66%

Solution:
According to the Hardy-Weinberg Equilibrium equation, the percentage of heterozygous individuals in the population is represented by the 2pq. Therefore, the number of heterozygous individuals (Aa) is equal to 2pq which equals 2 × 0.19 × 0.81 = 0.31 or 31%.
The percentage of homozygous recessives individuals in the population is represented by the q2, which equals to 0.81 × 0.81 = 0.66 or 66%.

Topic: Biology, Evolution, NEET
The frequency of X-rays, Gamma rays and visible light wave rays are a, b and c respectively, then

Options:
A a<b, b>c
B a<b, b<c
C a>b>c
D a>b, b<c

The correct answer is:
a<b, b>c

Solution:
The frequency of X-rays
a=3×1019 to 1×1016
The frequency of Gamma-rays
b=3×1022 to 3×1018
The frequency of visible light
c=8×1014 to 4×1014
So, b>a, b>c

Topic: Physics, Electromagnetic Waves, NEET
The frequency radiation emitted when the electron falls from $\mathrm{n}=4$ to $n=1$ in a hydrogen atom will be (Given ionization energy of $\mathrm{H}=2.18 \times 10^{-18} \mathrm{~J} \mathrm{atom}^{-1}$ and $h$ $\left.=6.625 \times 10^{-34} \mathrm{Js}\right)$ :

Options:
A $1.54 \times 10^{15} \mathrm{~s}^{-1}$
B $1.03 \times 10^{15} \mathrm{~s}^{-1}$
C $3.08 \times 10^{15} \mathrm{~s}^{-1}$
D $2.00 \times 10^{15} \mathrm{~s}^{-1}$

The correct answer is:
$3.08 \times 10^{15} \mathrm{~s}^{-1}$

Solution:
$\mathrm{E}=h \mathrm{v}$ or $\mathrm{v}=\frac{\mathrm{E}}{h}$
For $\mathrm{H}$ atom,
$$
\begin{aligned}
E & =\frac{-21.76 \times 10^{-19}}{n^2} \mathrm{~J} \mathrm{~atm}^{-1} \\
\Delta E & =-21.76 \times 10^{-19}\left(\frac{1}{4^2}-\frac{1}{1^2}\right)
\end{aligned}
$$
$$
\begin{aligned}
& =20.40 \times 10^{-19} \mathrm{~J} \mathrm{~atm}^{-1} \\
v & =\frac{20.40 \times 10^{-19}}{6.625 \times 10^{-34}} \\
& =3.079 \times 10^{15} \mathrm{~s}^{-1} .
\end{aligned}
$$

Topic: Chemistry, Structure of Atom, NEET, NEET 2004
The freshly prepared solution of sugar undergo change in optical rotation, with time, is known as

Options:
A Inversion
B Specific rotation
C Rotatory motion
D Mutarotation

The correct answer is:
Mutarotation

Solution:
α-D-Glucose[α]=+112°        (36%)   ⇌    Equilibrium  mixture  [α]D=+52°    (0.02%)  ⇌   β-D-Glucose[α]D=+19°       (64%)
Glucose has two forms α and β . When either of these two is dissolved in water and allowed to stand, it gets converted to an equilibrium mixture of α and β forms.

Topic: Chemistry, Biomolecules, NEET
The fringes produced in diffraction patterns _____.

Options:
A Are of equal width
B Are of unequal width
C Having equal intersity for bright fringes
D None of these.

The correct answer is:
Are of unequal width

Solution:
In single slit diffraction pattern, central maxima is double the fringe width.

Topic: Physics, Wave Optics, NEET
The fruit is chambered, developed from inferior ovary and has seeds with succulent testa in

Options:
A pomegranate
B orange
C guava
D cucumber

The correct answer is:
pomegranate

Solution:
Balausta is special type of false or pseudocarpic berry develops from multilocular, syncarpous inferior ovary. The whole fruit is enclosed by a hard rind made up of exocarp (epicarp fused with thalamus) and part of mesocarp. Plate-like infoldings are developed by mesocarp. The papery endocarp covers the individual group of seeds. The seeds possess bright red juicy testa that form edible part of fruit, $e g$, pomegranate.
The fruit of cucumber is pepo. In this the exocarp is not separable from mesocarp and the seeds from placentae.
The fruit of guava is berry. A berry is pulpy, indehiscent few to multiseeded fruit derived from multicarpellary syncarpous gynoecium.
The fruit of orange is hespiridium. It develops from multicarpellary, syncarpous, multilocular, superior ovary with axile placentation.

Topic: Biology, Morphology of Flowering Plants, NEET, NEET 2008 (Screening)
The fruit juices available commercially are clearer than ones which are homemade. Which ingredient is added in commercially available fruit juices?

Options:
A Lipase and protease
B Pectinase and hydrolase
C Pectinase and protease
D Hydrolase and protease

The correct answer is:
Pectinase and protease

Solution:
Bottled juices brought from the market are more clear than home made juices as they are treated with enzyme pectinase and proteinase which dissolves the fibres present in the juice.

Topic: Biology, Microbes in Human Welfare, NEET
The fruit of Trifolium is a

Options:
A capsule.
B legume.
C berry.
D drupe.

The correct answer is:
legume.

Solution:
Trifolium is the member of family Fabaceae that was earlier called Papilionoideae, a subfamily of family Leguminosae. The fruit in this family is legume with one or multiple seed and are non-endospermic.

Topic: Biology, Morphology of Flowering Plants, NEET
The fruit that develops from monocarpellary superior ovaries and has one seed is/are

Options:
A mango.
B apple.
C coconut.
D More than one option is currect

The correct answer is:
More than one option is currect

Solution:
In mango and coconut, the fruit is known as a drupe. They develop from monocarpellary superior ovaries and are one-seeded. In mango, the pericarp is well-differentiated into a thin outer epicarp, a middle fleshy edible mesocarp and an inner stony hard endocarp. In coconut, which is also a drupe, the mesocarp is fibrous.

Topic: Biology, Morphology of Flowering Plants, NEET
The function of companion cells is to

Options:
A Provide energy to sieve elements for active transport
B Provide water to phloem
C Load sucrose into sieve elements by passive transport
D Load sucrose into sieve elements by active transport

The correct answer is:
Load sucrose into sieve elements by active transport

Solution:
Companion cells move sugar and amino acids into and out of the sieve elements. In "source" tissue such as the leaf, companion cells use transmembrane proteins to take up sugar and amino acids by active transport. Movement of sugars in the phloem begins at the source, where sugars are loaded (actively transported) into a sieve tube. Loading sets up a water potential gradient that facilitates the movement of sugar.

Topic: Biology, Anatomy of Flowering Plants, NEET
The function of copper ions in copper releasing IUD's is

Options:
A They suppress sperm motility and fertilizing capacity of sperms
B They inhibit gametogenesis
C They make uterus unsuitable for implantation
D They inhibit ovulation

The correct answer is:
They suppress sperm motility and fertilizing capacity of sperms

Solution:
Cu2+ interfere in the sperm movement, hence suppressing the sperm motility and fertilizing capacity of sperms. The copper T is a simple intra-uterine device (IUD) made of a flexible, "T" shaped piece of plastic wrapped with a thin copper-containing wire. This shape is chosen as it fits the area around the uterus, allowing for the copper T to sit in place for years without moving around.

Example: CuT, Multiload 375, Cu7, etc.

Topic: Biology, Reproductive Health, NEET, NEET 2017
The function of leghaemoglobin in the root nodules of legumes is

Options:
A oxygen removal
B nodule differentiation
C expression of nif gene
D inhibition of nitrogenase activity

The correct answer is:
oxygen removal

Solution:
Leghaemoglobin is an oxygen scavenger. It protects the nitrogen fixing enzyme nitrogenase.

Topic: Biology, Mineral Nutrition, NEET, NEET 2011 (Screening)
The function of nucleated cells in phloem is to _____

Options:
A provide energy to sieve elements for active transport.
B provide water to phloem.
C load sucrose into sieve elements by passive transport.
D load sucrose into sieve elements by active transport.

The correct answer is:
load sucrose into sieve elements by active transport.

Solution:
Companion cells move sugar and amino acids in and out of the sieve elements. In "source" tissue such as leaf, companion cells use transmembrane proteins to take up sugar and amino acids by active transport. Movement of sugars in the phloem begins at the source, where sugars are loaded (actively transported) into a sieve tube. Loading sets up a water potential gradient that facilitates the movement of sugar.

Topic: Biology, Transport in Plants, NEET