datavorous/entrance-exam-dataset huggingface dataset:
Define the term inflorescence. Explain the basis for the different types of inflorescence in flowering plants.

Options:

None

Solution:
The arrangement and distribution of flowers on the floral axis is termed as inflorescence. Depending on whether the apex gets converted into a flower or continues to grow, two major types of inflorescences are defined - racemose and cymose. In racemose type of inflorescence, the main axis continues to grow, the flowers are borne laterally in an acropetal succession, i.e., older flower are at the base and younger flowers are at the top.
In cymose type of inflorescence, the main axis terminates in a flower, hence is limited in growth. The flowers are borne in a basipetal order, i.e., younger flowers are near the base and older flower are at the apex.

Topic: Biology, Morphology of Flowering Plants
Degree of the given differential equation $\left(\frac{d^2 y}{d x^2}\right)^3=\left(1+\frac{d y}{d x}\right)^{1 / 2}$ , is

Options:
A $2$
B $3$
C $\frac{1}{2}$
D $6$

The correct answer is:
$6$

Solution:
On squaring, we will have the result.

Topic: Mathematics, Differential Equations, JEE Main
Dehydration of alcohol to alkene by heating with conc. H2SO4, the initiation step is _______ followed with________ mechanism.

Options:
A Elimination of water, free radical
B Formation of an ester, free radical
C Protonation of alcohol, carbocation
D Protonation of alcohol, carbanion

The correct answer is:
Protonation of alcohol, carbocation

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Dehydration of the following in increasing order is
(I)
(II)
(III)
(IV)

Options:
A I < II < III < IV
B II < III < IV < I
C I < III < IV < II
D None of these

The correct answer is:
I < II < III < IV

Solution:
Dehydration of IV is most facile since, it gives an aromatic compound. Dehydration of III gives a conjugated diene which is stabilized by resonance.

Dehydration of II gives only cyclohexene which is not stabilized by resonance.

In contrast, phenol (I) does not undergo dehydration. Thus, the ease of dehydration is
IV > III > II > I.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Delicate hair-like feathers which remain sparsely distributed over the body are

Options:
A coverts
B filoplumes
C plumules
D remiges.

The correct answer is:
filoplumes

Solution:
No Solution Available

Topic: Biology, Animal Kingdom, AIIMS, AIIMS 2009
Delocalization of charge occurs in which of the following structures?
(a)
(b)
(c)
(d)

Options:
A (a), (b) and (e) but not (c) and (d)
B (c) and (e) but not (a), (b) and (d)
C (a), (b) and (d) but not (c) and (e)
D (a) and (d) but not (b), (c) and (e)

The correct answer is:
(a) and (d) but not (b), (c) and (e)

Solution:
Condition for delocalization of charge:

i) -ve charge will be delocalized if the lone pair of the atom bearing the -ve charge is in conjugation.

ii) +ve charge will be delocalized if the atom bearing the +ve charge has an empty orbital (like carbocation) and that orbital is in conjugation.

Bridgehead atoms cannot be planar due to the high bond angle constrain because of this:

a) Bridgehead carbon atom cannot be sp2 hybridised.

b) Lone pair or empty orbital of bridgehead carbon does not participate in conjugation.

Hence, the charge at bridgehead carbon will be localized.

In structure (a), the lone pair of carbanion is in conjugation with the pi-bond of O=C< and in structure (d), the empty orbital of carbocation is in conjugation with the lone pair of oxygen. Hence, charge is delocalized in (a) and (d).

Topic: Chemistry, General Organic Chemistry, JEE Main
\( \Delta_{1}=\left|\begin{array}{ccc}y^{5} z^{6}\left(z^{3}-y^{3}\right) & x^{4} z^{6}\left(x^{3}-z^{3}\right) & x^{4} y^{5}\left(y^{3}-x^{3}\right) \\ y^{2} z^{3}\left(y^{6}-z^{6}\right) & x z^{3}\left(z^{6}-y^{6}\right) & x y^{2}\left(x^{6}-y^{6}\right) \\ y^{2} z^{3}\left(z^{3}-y^{3}\right) & x z^{3}\left(z^{3}-y^{3}\right) & x y^{2}\left(y^{3}-x^{3}\right)\end{array}\right| \) and \( \Delta_{2}=\left|\begin{array}{ccc}x & y^{2} & z^{3} \\ x^{4} & y^{5} & z^{6} \\ x^{7} & y^{8} & z^{9}\end{array}\right| \), then \( \Delta_{1} \Delta_{2} \) is equal to

Options:
A \( \Delta_{2}^{3} \)
B \( \Delta_{2}^{2} \)
C \( \Delta_{2}^{4} \)
D None of these

The correct answer is:
\( \Delta_{2}^{3} \)

Solution:
Given Δ2=xy2z3x4y5z6x7y8z9

Calculating co-factors of Δ2:

Δ2 1,1=y5z6z3-y3

Δ2 1,2=x4z6x3-z3

Δ2 1,3=x4y5y3-x3

Δ2 2,1=y2z3y6-z6

Δ2 2,2=xz3z6-y6

Δ2 2,3=xy2x6-y6

Δ2 3,1=y2z3z3-y3

Δ2 3,2=xz3z3-y3

Δ2 3,3=xy2y3-x3

Here, we can see that the given determinant Δ1 is obtained by corresponding co-factors of determinant Δ2.

Hence, Δ1=Δ22

∴Δ1Δ2=Δ22Δ2=Δ23

Topic: Mathematics, Determinants, JEE Main
\(\Delta_{\text {vap }} \mathrm{H}^{\ominus}\) for water is \(+40.79 \mathrm{~kJ} \mathrm{~mol}^{-1}\) at 1 bar and \(100{ }^{\circ} \mathrm{C}\). Change in internal energy for this vapourisation under same condition is \(\qquad\) \(\mathrm{kJ} \mathrm{mol}^{-1}\). (Integer answer) (Given \(\mathrm{R}=8.3 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\))

Options:

The correct answer is:
38

Solution:
$\begin{aligned} & \mathrm{H}_2 \mathrm{O}(\ell) \rightleftharpoons \mathrm{H}_2 \mathrm{O}(\mathrm{g}) \quad \Delta \mathrm{H}_{\text {vap }}^0=40.79 \mathrm{~kJ} / \mathrm{mole} \\ & \Delta \mathrm{H}_{\text {vap }}^0=\Delta \mathrm{U}_{\text {vap }}^0+\Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT} \\ & 40.79=\Delta \mathrm{U}_{\text {vap }}^0+\frac{1 \times 8.3 \times 373.15}{1000} \\ & \Delta \mathrm{U}_{\text {vap }}^0=40.79-3.0971 \\ & =37.6929 \\ & \Delta \mathrm{U}_{\text {vap }}^0 \simeq 38\end{aligned}$

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2024 (08 Apr Shift 2)
Denaturation is a step in the procedure called Polymerase Chain Reaction. This step does not involve which of the following process?

Options:
A Breaking of hydrogen bonds of the original DNA double helix
B Heating of starting solution to 94°C
C Addition of nucleotides to the primer by DNA polymerase
D Creation of single-stranded DNA templates

The correct answer is:
Addition of nucleotides to the primer by DNA polymerase

Solution:
PCR stands for Polymerase Chain Reaction. In this reaction, gene (segment of DNA) of interest is synthesized in vitro and multiple copies are generated. The main steps of PCR- Denaturation - where the double-stranded template DNA is heated to separate it into two single strands. Annealing - where the temperature is lowered to enable the DNA primers to attach to the template DNA. Extension - where the temperature is raised and the new strand of DNA is made by the Taq polymerase enzyme.

Topic: Biology, Biotechnology - Principles and Processes, NEET
Denatured alcohol is

Options:
A ethanol + methanol
B rectified spirit + methanol + naphtha
C undistilled ethanol
D rectified spirit

The correct answer is:
ethanol + methanol

Solution:
Ethyl alcohol is made unfit for drinking purpose by adding some poisonous substance as methanol, naphtha, pyridine, rubber etc. It is called denatured alcohol.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Density of a $2.05 \mathrm{M}$ solution of acetic acid in water is $1.02 \mathrm{~g} / \mathrm{mL}$. The molality of the solution is

Options:
A
$1.14 \mathrm{~mol} \mathrm{~kg}^{-1}$

B
$2.28 \mathrm{~mol} \mathrm{~kg}^{-1}$

C
$3.28 \mathrm{~mol} \mathrm{~kg}^{-1}$

D
$0.44 \mathrm{~mol} \mathrm{~kg}^{-1}$

The correct answer is:
$3.28 \mathrm{~mol} \mathrm{~kg}^{-1}$

Solution:
No Solution Available

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2006
Density of ice is $\rho$ and that of water is $\sigma$. What will be the decrease in volume when a mass $M$ of ice melts

Options:
A $\frac{M}{\sigma-\rho}$
B $\frac{\sigma-\rho}{M}$
C $M\left[\frac{1}{\rho}-\frac{1}{\sigma}\right]$
D $\frac{1}{M}\left[\frac{1}{\rho}-\frac{1}{\sigma}\right]$

The correct answer is:
$M\left[\frac{1}{\rho}-\frac{1}{\sigma}\right]$

Solution:
The correct option is $\mathbf{3}$
$M\left[\frac{1}{\rho}-\frac{1}{\sigma}\right]$
Volume of ice $=\frac{M}{\rho}$, volume of water $=\frac{M}{\sigma}$
$\therefore$ Change in volume $=\frac{M}{\rho}-\frac{M}{\sigma}=M\left(\frac{1}{\rho}-\frac{1}{\sigma}\right)$

Topic: Physics, Mechanical Properties of Fluids, JEE Main
Density of substance at $0^{\circ} \mathrm{C}$ is $10 \mathrm{gm} / \mathrm{cc}$ and at $100^{\circ} \mathrm{C}$, its density is $9.7 \mathrm{gm} / \mathrm{cc}$. The coefficient of linear expansion of the substance will be

Options:
A $10^2$
B $10^{-2}$
C $10^{-3}$
D $10^{-4}$

The correct answer is:
$10^{-4}$

Solution:
Initial temperature \(\left(\mathrm{T}_1\right)=0^{\circ} \mathrm{C}\)
Initial density \(\left(\rho_1\right)=10 \mathrm{~g} / \mathrm{cc}\)
Final temperature \(\left(\mathrm{T}_2\right)=100^{\circ} \mathrm{C}\) and
Final density \(\left(\rho_2\right)=9.7 \mathrm{~g} / \mathrm{cc}\)
Coefficient of volumetric expansion
\(\begin{aligned}
\gamma & =\frac{\Delta \rho}{\rho_1 \times \Delta T}=\frac{\rho_1-\rho_2}{\rho_1\left(T_2-T_1\right)} \\
& =\frac{10-9.7}{10(100-0)}=\frac{0.3}{1000} \\
& =3 \times 10^{-4}
\end{aligned}\)
Therefore coefficient of linear expansion,
\(\alpha=\frac{\gamma}{3}=\frac{3 \times 10^{-4}}{3}=10^{-4}\)

Topic: Physics, Thermal Properties of Matter, JEE Main
Depolarization of axolemma during nerve conduction takes place because of

Options:
A an equal amount of Na+ and K+ move out across axolemma
B Na+ moves inside
C there is more Na+ outside than inside
D K+ moves inside

The correct answer is:
Na+ moves inside

Solution:
The axon is immersed in the extracellular fluid (ECF). In the axoplasm negatively charged protein molecules are present, which are neutralized due to the presence of large amount of K+ ions. In the ECF (outside the axon) the negatively charged CIâ ions are neutralized by the presence of +vely charged Na+Na+ ions.

To maintain resting potential, sodium potassium metabolic pump operates. This pump which is located on the axon membrane pump Na+ from axoplasm to ECF and K+ from ECF to axoplasm. It pumps more positive charges (3 Na+) from axoplasm to ECF than in the reverse direction (2K+), and is run by an enzyme called as sodium potassium-ATPase.

When a stimulus is applied to the membrane of the nerve fibre, its permeability changes and sodium potassium pump stop operating. Sodium ions rush inside and potassium ions rush outside. This results in the positive charge inside and negative charge outside. The nerve fibre is said to be in action potential or depolarized.

Topic: Biology, Neural Control and Coordination, NEET
Depressants are

Options:
A Medicine that provide energy to the body
B Medicine that reduce brain function
C Medicine that cause mind confusion
D Medicine that cause drowsiness

The correct answer is:
Medicine that reduce brain function

Solution:
Medicine that reduce brain function

Topic: Chemistry, Chemistry in Everyday Life, JEE Main
Depression in freezing point of 0.01 molal aqueous solution of HA is 0.01924âK. The pKa value of HA is
[Given = molarity = molality, K f H 2 O â=â1 .85 Kkg mol â1 ,âlogâ2â=â0.3

Options:

The correct answer is:
4.80

Solution:
ΔTf=i×m×Kf
0.01924 = i × 0.01 × 1.85
i = 1.04
i = 1 + α(n−1)
1.04 = 1 + α(2−1)
α=0.04
K a  =  α 2 c =  0.04 2  × 0.01 = 16 ×  10 −6
pK a  = 4.8

Topic: Chemistry, Solutions, JEE Main
Depression in freezing point of 0.01 molal aqueous HCOOH solution is 0.02046°C. 1 molal aqueous urea solution freezes at –1.86°C. Assuming molality equal to molarity, calculate the pH of HCOOH solution.

Options:

The correct answer is:
3.00

Solution:
1.86 = Kf(1)
0.02046 = (1 + α) × 1.86 × 0.01
⇒1 + α = 0 . 0 2 0 4 6 0 . 0 1 8 6 = 1 . 1
⇒ α = 0.1
[H+] = C α = 0.01 × 0.1 = 10–3
so, pH = 3.

Topic: Chemistry, Solutions, JEE Main
Derivative of nitrogen (III) act as

Options:
A Oxidizing agent only
B reducing agent only
C both oxidizing and reducing agent
D nitrating agent

The correct answer is:
both oxidizing and reducing agent

Solution:
Derivatives of nitrogen (III) has both oxidizing and reducing properties as it is having intermediate oxidation state
2NaNO2+3Oxidant+2KI+2H2SO4→I2+2NO+2+K2SO4+Na2SO4+2H2O
2KMnO4+5NaNO2+3(Reductant )+3H2SO4→2MnSO4+5NaNO+53+K2SO4+3H2O

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Describe a method for the identification of primary, secondary and tertiary amines. Also write chemical equations of the reactions involved.

Options:

None

Solution:
The three type of amines can be distinguished by Hinsberg test. In this test, the amine is shaken with benzenesulphonyl chloride \(\left(\mathrm{C}_6 \mathrm{H}_5 \mathrm{SO}_2 \mathrm{Cl}\right)\) in the presence of excess of aqueous \(\mathrm{NaOH}\) or \(\mathrm{KOH}\). A primary amine reacts to give a clear solution, which on acidification yields an insoluble compound.
A secondary amine forms an insoluble compound, which remains insoluble even on acidification.
A tertiary amine does not react with the reagent, but dissolves is acid.

Topic: Chemistry, Amines
Describe briefly the followings:
(a) Origin of replication
(b) Bioreactors
(c) Downstream processing

Options:

None

Solution:
(a) Origin of Replication: This is a sequence from where replication starts and any piece of DNA when linked to this sequence can be made to replicate within the host cells. This sequence is also responsible for controlling the copy number of the linked DNA. So, if one wants to recover many copies of the target DNA it should be cloned in a vector whose origin support high copy number.
(b) Bioreactor : Bioreactors can be thought of as vessels in which raw materials are biologically converted into specific products by microbes, plant and animal cell and/or their enzymes. The bioreactor provides optimum growth conditions and facilitates achieving the desired products. A bioreactor has an agitator system, an oxygen delivery system, a foam control system, a temperature control system, pH control system and sampling ports.
(c) Downstream Processing: The product obtained is subjected to a series of processes collectively called downstream processing before it is made into a finished product ready for marketing. The two main processes are separation and purification. The product is then formulated with suitable preservatives. Such formulations have to undergo clinical trials, in case of drugs.

Topic: Biology, Biotechnology - Principles and Processes
Describe briefly the four major groups of protozoa.

Options:

None

Solution:
The four major group of protozoa are flagellated protozoan, amoeboid protozoan, sporozoan, ciliated protozoan. The main characters of these group are as follows:

Topic: Biology, Biological Classification
Describe how does the enthalpy of reaction remain unchanged when a catalyst is used in the reaction?

Options:

None

Solution:
A catalyst increases the speed of a reaction without itself undergoing any chemical change.
According to "intermediate complex formation theory" reactants first combine with the catalyst and forms an intermediate complex which is short-lived and decomposes to form the products and regenerate the catalyst.
This intermediate formed in presence of catalyst has much lower potential energy than the intermediate complex formed between the reactants in the absence of the catalyst. Therefore, the presence of catalyst lowers the potential energy barrier and the reaction follows a new alternate pathway which require less activation energy. Lower the activation energy, faster is the reaction because more reactant molecules can cross the energy barrier and change into products.
Enthalpy, $\Delta H$ is a state function. Enthalpy of reaction, i.e., difference in energy between reactants and product is constant, which is clear from potential energy diagram
given below.

Topic: Chemistry, Chemical Kinetics
Describe modifications of stem with suitable examples.

Options:

None

Solution:
Modifications of Stem
Food Storage: Stems are modified to perform different functions. Underground stems of potato, ginger, turmeric, Colocasia are modified to store food in them. They also act as organs of perennation to tide over conditions unfavourable for growth.
Tendrils 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.
Thorns: Axillary buds of stems may also get modified into woody, straight and pointed thorns. Thorns are found in many plants such as Citrus, Bougainvillea. They protect plants from grazing animals.
Some plants of arid regions modify their stems into flattened (Opuntia), or fleshy cylindrical (Euphorbia) structures. They contain chlorophyll and carry out photosynthesis. Vegetative Reproduction: Underground stems of some plants such as grass and strawberry, etc., spread to new niches and when older parts die, new plants are formed. In plants, like mint and jasmine a slender lateral branch arises from the base of the main axis and after growing aerially for some time arch downwards to touch the ground. A lateral branch with short internodes and each node bearing a rosette of leaves and a tuft of roots is found in aquatic plants like Pistia and Eichhornia. In banana, pineapple and Chrysanthemum, the lateral branches originate from horizontally beneath the soil and then come out obliquely upward giving rise to leafy shoots.

Topic: Biology, Morphology of Flowering Plants
Describe schematically the equipotential surfaces corresponding to
(a) a constant electric field in the Z-direction,
(b) a field that uniformly increases in magnitude but remains in a constant (say, $Z$ ) direction,
(c) a single positive charge at the origin, and
(d) a uniform grid consisting of long equally spaced parallel charged wires in a plane.

Options:

None

Solution:
(a) Planes parallel to $\mathrm{X}-\mathrm{Y}$ plane
Equipotential surfaces corresponding to a constant electric field in the Z-direction $(\mathrm{dV}=$ p.d between successive equipotential surfaces which is zero in this case)
(b) Same as in (a), except that planes differing by a fixed potential get closer as field increases.
Equipotential surfaces corresponding to a uniformly increasing electric field in the Z-direction
(c) Concentric spheres centered at the origin.

Topic: Physics, Electrostatic Potential and Capacitance
Describe the events taking place during interphase.

Options:

None

Solution:
Interphase is divided into three phases viz. $\mathrm{G}_1$ phase, $\mathrm{S}$ phase and $\mathrm{G}_2$ phase.
$G_1$ phase (Gap 1): $G_1$ phase corresponds to the interval between mitosis and initiation of DNA replication. During $\mathrm{G}_1$ phase the cell is metabolically active and continuously grows but does not replicate its DNA.
S phase (Synthesis): S or synthesis phase marks the period during which DNA synthesis or replication takes place. During this time the amount of DNA per cell doubles. If the initial amount of DNA is denoted as $2 \mathrm{C}$ then it increases to 4C. However, there is no increase in the chromosome number; if the cell had diploid or $2 \mathrm{n}$ number of chromosomes at $\mathrm{G}_1$, even after $\mathrm{S}$ phase the number of chromosomes remains the same i.e., $2 \mathrm{n}$.
$\mathrm{G}_2$ phase (Gap 2) : In animal cells, during the $\mathrm{S}$ phase, DNA replication begins in the nucleus, and the centriole duplicates in the cytoplasm. During the $\mathrm{G}_2$ phase, proteins are synthesised in preparation for mitosis while cell growth continues.

Topic: Biology, Cell Cycle and Cell Division
Describe the internal structure of a dorsiventral leaf with the help of labelled diagrams.

Options:

None

Solution:
Dorsiventral (dicotyledonous) leaf : The vertical section of a dorsiventral leaf through the lamina shows three main parts, namely, epidermis, mesophyll and vascular system. Epidermis : The epidermis which covers both the upper surface (adaxial epidermis) and lower surface (abaxial epidermis) of the leaf has a conspicuous cuticle. The abaxial epidermis generally bears more stomata than the adaxial epidermis. The latter may even lack stomata.
Mesophyll : The tissue between the upper and the lower epidermis is called the mesophyll. It possesses chloroplasts and carries out photosynthesis, is made up of parenchyma. It has two types of cells - the palisade parenchyma and the spongy parenchyma. The adaxially placed palisade parenchyma is made up of elongated cells, which are arranged vertically and parallel to each other. The oval or round and loosely arranged spongy parenchyma is situated below the palisade cells and extends to the lower epidermis.
There are numerous large spaces and air cavities between these cells.
Vascular system : This includes vascular bundles, which can be seen in the veins and the midrib. The size of the vascular bundles is dependent on the size of the veins. The veins vary in thickness in the reticulate venation of the dicot leaves. The vascular bundles are surrounded by a layer of thick walled bundle sheath cells.

Topic: Biology, Anatomy of Flowering Plants
Describe the motion of a charged particle in a cyclotron if the frequency of the radio frequency (rf) field were doubled.

Options:

None

Solution:
In cyclotron, frequency of charged particle is equal to the radio frequency. When the frequency of the radio frequency (rf) field were doubled, the time period of the radio frequency (rf) field become half, so, the duration in which particle completes half revolution inside the dees, radio frequency completes the cycle.
So, a charge particle will accelerate and decelerate alternatively and its moves in circular path between the dees during motion in Dees. Hence, the radius of path in the dees will remain same.

Topic: Physics, Magnetic Effects of Current
Describe the post-fertilization changes in a flower.

Options:

None

Solution:
In sexual reproduction, events that occur after the formation of zygote are called post-fertilization events. In flowering plants, the zygote is formed inside the ovule. After fertilization the sepals, petals and stamens of the flower wither and fall off. But the pistil remains attached to the plant. The zygote develops into the embryo and the ovules develop into the seed. The ovary develops into the fruit that develops a thick wall called pericarp which is protective in function. After dispersal, seeds germinate under favourable conditions to produce new plants.

Topic: Biology, Reproduction in Organisms
Describe the structure of the following with the help of labelled diagrams.
(i) Nucleus
(ii) Centrosome

Options:

None

Solution:
(i) Nucleus: The nucleus is a large organelle controlling all the activities of the eukaryotic cells. The nucleus is bounded by two membranes, which make the nuclear envelope. The outer and inner membranes are separated by a narrow space, perinuclear space. The outer membrane remains in continuation with endoplasmic reticulum (ER) and the inner one surrounds the nuclear contents. At some points, the nuclear evelope is interrupted by the presence of small structures called nuclear pores. These pores help in exchange of materials between nucleoplasm and cytoplasm.
The nucleoplasm contains chromatin and nucleolus. The nucleolus is a rounded structure. It is not separated from the rest of the nucleoplasm by membrane. Nucleolus is the "site for ribosomal RNA synthesis".
(ii) Centrosome : Under the electron microscope, each centriole is seen to be formed of nine sets of tubular structures arranged in circular fashion. Each of these sets is a triplet composed of three microtubules. Each microtubule has a diameter of about $250 Å$. The triplets are found in the matrix. Sometimes delicate strands appear to connect sets of the triplet to each other. Also can be seen radiating from the central core of the cylinder, delicate strands which connect sets of the triplets to each other giving a cartwheel appearance. Basal bodies are structures similar to the centrioles. They produce cilia and flagella.

Topic: Biology, Cell - The Unit of Life