datavorous/entrance-exam-dataset huggingface dataset:
The figure shows a hypothetical tetrapeptide portion with parts labelled a-d. Which is the correct option?

Options:
A $c$ is an aromatic amino acid tryptophan.
B $a$ is the C-terminal amino acid and $d$ is $\mathrm{N}$-terminal amino acid.
C $a$ is a sulphur containing amino acid methionine.
D $d$ is the acidic amino acid-glutamic acid.

The correct answer is:
$d$ is the acidic amino acid-glutamic acid.

Solution:
In the given figure a tetrapeptide is shown. Here, amino acids are indicated as (a), (b) (c) and (d). (a) indicates N-terminal amino acid and (d) indicates Cterminal residue. As it can be seen from the figure sulphur containing amino acid is $b$ and carboxyl group $(-\mathrm{COOH})$ containing amino acid is (d). Residue (d) indicates glutamic acid, an acidic residue due to presence of $(-\mathrm{COOH})$ group.
So, the correct answer is 'd is acidic amino acid - glutamic acid'.

Topic: Biology, Biomolecules (B), NEET, NEET 2013 (Karnataka)
The figure shows a logic circuit with two inputs $A$ and $B$ and the output $C$. The voltage wave forms across $A, B$ and $C$ are as given. The logic circuit gate is

Options:
A OR gate
B NOR gate
C AND gate
D NAND gate

The correct answer is:
OR gate

Solution:
From the given woveforms, the following truth table can be made
This truth table is required to 'OR' gate. So, logic circuit gate is OR gate.
Note The logic of the OR gate is if any of the input is one then output is one.

Topic: Physics, Semiconductors, NEET, NEET 2012 (Screening)
The figure shows a network in which the cell is ideal and it has an emf E. The potential difference across the resistance 2R is

Options:
A 2E
B 4E7
C E7
D 3E7

The correct answer is:
4E7

Solution:
In the given circuit, resistors 4R and 2R are connected in parallel while resistance R is connected in series to it.
Hence, equivalent resistance is, of parallel combination
1Râ²=14R+12R
â R'=86R =43R
â¹ R" =R+43R=7R3(series combination)
Given, emf is E volts, thereforeusing ohm's law,
i=ER''=3E7R
Potential difference across R is
V=ir=3E7RÃR=3E7
Potential difference across 2R is
Vâ²=E-3E7=4E7

Topic: Physics, Current Electricity, NEET
The figure shows a plot of photo current versus anode potential for a photo sensitive surface for three different radiations. Which one of the following is a correct statement?

Options:
A Curves $a$ and $b$ represent incident radiations of different frequencies and different intensities
B Curves $a$ and $b$ represent incident radiations of same frequency but of different intensities
C Curves $b$ and $c$ represent incident radiations of different frequencies and different intensities
D Curves $b$ and $c$ represent incident radiations of same frequency having same intensity

The correct answer is:
Curves $a$ and $b$ represent incident radiations of same frequency but of different intensities

Solution:
From the two graphs we can conclude that for the graph in question curves $a$ and $b$ represent incident radiations of same frequency but of different intensities.

Topic: Physics, Dual Nature of Matter, NEET, NEET 2009 (Screening)
The figure shows a region containing a uniform magnetic field B, which is increasing at the rate dBdt. The region is gravity free. A disc of mass m and radius R is kept in this magnetic field. Two charged particles having charge q1 and q2 and masses m and 2m are embedded to the non-conducting disc on the circumference at diametrically opposite points. The disc is free to translate as well as rotate.

If q1=-q and q2=q, then the initial acceleration of q1 will be

Options:
A qR3mdBdt
B 2qR3mdBdt
C 2qRmdBdt
D qRmdBdt

The correct answer is:
qR3mdBdt

Solution:
E=R2dBdt
a=2qE3m=qR3mdBdt

Topic: Physics, Electromagnetic Induction, NEET
The figure shows a smooth inclined plane of inclination Î¸ fixed in a car. A sphere is set in pure rolling on the incline. For what value of the acceleration of the car in the horizontal direction the sphere will continue pure rolling?

Options:
A g  cosθ
B gâ sinÎ¸
C g  cotθ
D g  tanθ

The correct answer is:
g  tanθ

Solution:
The sphere will continue pure rolling if
ma cosâ¡Î¸= mg sinâ¡Î¸
orâ a=g tanâ¡Î¸

Topic: Physics, Rotational Motion, NEET
The figure shows an axon terminal and synapse. Select the option giving correct identification of labels $a-d$.

Options:
A a-Action potential, c - Neurotransmitter
B b- Neurotransmitter, d - Receptor capsules
C c - Receptor, d - Synaptic vesicles
D a - Axon terminal, b - Serotonin complex

The correct answer is:
c - Receptor, d - Synaptic vesicles

Solution:
Related Theory
A nerve impulse is transmitted from one neuron to another through junctions called synapses. A synapse is formed by the membranes of a pre-synaptic neuron and a post-synaptic neuron, which may or may not be separated by a gap called synaptic clefi. At electrical synapses, the membranes of pre-and posi-synaptic neurons are in very close proximity. Electrical current can flow directly from one neuron into the other across these synapses. Impulse transmission across an electrical synapse is always faster than that across a chemical synapse.

Topic: Biology, Neural Control and Coordination, NEET, NEET 2013 (Karnataka)
The figure shows blood circulation in humans with labels a to b. Select the option which gives correct identification of label and functions of the part.

Options:
A b - Capillary-Thin without muscle layer and wall two cell layers thick
B c - Vein-Thin walled and blood flows in jerks/spurts
C d - Pulmonary vein-Takes oxygenated blood to heart $\mathrm{pO}_2=95$ $\mathrm{mm} \mathrm{Hg}$
D a - Artery-Thick walled and blood flows evenly

The correct answer is:
d - Pulmonary vein-Takes oxygenated blood to heart $\mathrm{pO}_2=95$ $\mathrm{mm} \mathrm{Hg}$

Solution:
a- Artery- Carry blood from heart to body parts. It is thick walled and elastic, under high blood pressure.
b- Capillary- thin walled, with only one layer, facilitates the diffusion of nutrients, gases, etc., into tissues
c- Vein - thin walled, brings blood from body parts to heart, act as low resistance conduct for blood flow.
d- Pulmonary vein-take oxygenated blood to heart, $p \mathrm{O}_2=95 \mathrm{~mm} \mathrm{Hg}$.
(Ref. to Fig. 14.2)
Caution
Pulmonary artery takes deoxygenated blood $\quad$ carries oxygenated blood from lungs to from heart to lungs, while pulmonary vein heart.
$$
\text { Related Theory }
$$
(Ref. to Fig. 14.3)

Topic: Biology, Body Fluids and Circulation, NEET, NEET 2013 (Karnataka)
The figure shows elliptical orbit of a planet $m$ about the sun $S$. The shaded area $S C D$ is twice the shaded area $S A B$. If $\mathrm{t}_1$ is the time for the planet to move from $C$ to $D$ and $\mathrm{t}_2$ is the time to move from $A$ to $B$, then

Options:
A $t_1>t_2$
B $t_1=4 t_2$
C $\mathrm{t}_1=2 \mathrm{t}_2$
D $\mathrm{t}_1=\mathrm{t}_2$

The correct answer is:
$\mathrm{t}_1=2 \mathrm{t}_2$

Solution:
Key Idea Apply Kepler's law of area of planetary motion.
The line joining the sun to the planet sweeps out equal areas in equal time interval $i e$, areal velocity is constant.
$$
\frac{\mathrm{dA}}{\mathrm{dt}}=\text { constant }
$$
or
$$
\frac{\mathrm{A}_1}{\mathrm{t}_1}=\frac{\mathrm{A}_2}{\mathrm{t}_2}
$$
$$
\begin{array}{llrl}
\Rightarrow & \mathrm{t}_1 & =\frac{\mathrm{A}_1}{\mathrm{~A}_2} \mathrm{t}_2 \\
\text { Given } & \mathrm{A}_1 & =2 \mathrm{~A}_2 \\
\therefore & \mathrm{t}_1 & =2 \mathrm{t}_2
\end{array}
$$

Topic: Chemistry, Solid State, NEET, NEET 2009 (Screening)
The figure shows the acceleration-time graph of a particle. Which of the following represents the corresponding velocity-time graph? ( consider initial velocity zero )

Options:
A
B
C
D

The correct answer is:

Solution:
Since acceleration is constant, therefore there is uniform increase in velocity. So, the v-t graph is a straight line slopping upward to the right. When acceleration becomes zero, velocity is constant. So v-t graph is a straight line parallel to the time-axis.

Topic: Physics, Motion In One Dimension, NEET
The figure shows the electric lines of force emerging from a charged body. If the electric field at A and B are EA and EB respectively. If the distance between A and B is r, then

Options:
A EA>EB
B EA<EB
C EA=EBr
D EA=EBr2

The correct answer is:
EA>EB

Solution:
Figure indicates the presence of some positive charge to the left of A.
â´ EA>EB âµ rA<rB

Topic: Physics, Electrostatics, NEET
The figure shows the elliptical orbit of a planet m about the sun S. The shaded area SCD is twice the shaded area SAB. If t1 is the time for the planet to move from C to D and t2 is the time to move A to B then

Options:
A t1=4t2​

B t1=2t2​

C t1=t2​

D t1>t2​

The correct answer is:
t1=2t2​

Solution:
According to Kepler's second law
Equal areas are swept in equal time.

t1, the time taken to go from C to D = 2t2 where t2 is the time taken to go from A to B. As it is given that area SCD=2SAB.​

Topic: Physics, Gravitation, NEET
The figure shows the experiment of T.H. Morgan on the linkage. If in cross-I, genes are tightly linked and in cross-II, genes are loosely linked then what will be the percentage of recombination produced in cross-I and cross-II respectively?

Options:
A 98.7% and 62.8%
B 1.3% and 37.2%
C 37.2% and 1.3%
D 62.8% and 98.7%

The correct answer is:
1.3% and 37.2%

Solution:
Morgan and his students have found that linked genes show varied recombinations, some being more tightly linked than others, (i) In Drosophila, the crossing of yellow bodied (y) and white-eyed (w) female with brown bodied (y+) red-eyed (w+) male-produced F1 to be brown bodied and red-eyed females, while males were yellow bodied and white eyed. On intercrossing of F1 progeny, Morgan observed that the two genes did not segregate independently of each other and, therefore, the F2 ratio deviated significantly from expected 9: 3: 3: 1 ratio. He found that 98.7% to be of the parental type and only 1.3% recombinant type.

(ii) In a second cross of Drosophila between white-eyed and miniature winged (wwmm) female with wild type or red-eyed normal winged (w+m+) males, all the F1 were found to be of wild type, i.e., red-eyed and normal winged in females, while it was white-eyed and miniature wings in males. An F1 female fly was then test crossed with white-eyed and miniature winged male. 62.8% of the progeny was of parental types while 37.2% were recombinant types.

Topic: Biology, Principles of Inheritance and Variation, NEET
The figure shows the location of a source and detector at time t=0. The source and detector are moving with velocities vS=5i^ m s-1 and vD=10j^ m s-1 respectively. The frequency of signals received by the detector at the moment when the source crosses the origin is (the frequency of the source is 100 Hz. velocity of sound 330 m s-1 )

Options:
A 97 Hz

B 47 Hz

C 90 Hz

D 60 Hz

The correct answer is:
97 Hz

Solution:
Time taken by 'S' to travel from S to 0 = 20 s
In this time detector goes from D1 to D2
The signal received by the detector at t=20 s is not that produced by the source at 'O' but produced from a position S1 prior to 'O'
rv=dvscos⁡θ=dr=vsv=5330=166
sinθ= 1− ( 1 66 ) 2 ≈1
⇒fD=fv-v0sin⁡θv-vscos⁡θ
=100330-10330-566=97 Hz

Topic: Physics, Waves and Sound, NEET
The figure shows the total acceleration a=32 m s-2 of a moving particle moving clockwise in a circle of radius R=1 m. What are the centripetal acceleration a and speed of the particle at the given instant?

Options:
A 163 m s-2, 43 m s-1
B 163 m s-2, 4 m s-1
C 16 m s-2, 16 m s-1
D 16 m s-2, 4 m s-1

The correct answer is:
16 m s-2, 4 m s-1

Solution:
Given,
Total acceleration, a=32 m s-2
∴ Centripetal acceleration, ac=32cos⁡60°
=32×12
=16 m s-2
Also, we know that, ac=v2R=16 m s-2
⇒ v21=16 ∵ R=1
⇒v2=16 m2 s-2
∴ Speed of the particle v=4 m s-1

Topic: Physics, Laws of Motion, NEET
The figure shows the variation of force acting on a particle of mass 400 g executing simple harmonic motion. The frequency of oscillation of the particle is

Options:
A 4 s-1
B 5/2Ï s-1
C 1/8π s-1
D 1/2π s-1

The correct answer is:
5/2Ï s-1

Solution:
The slope of the curve is Fx=-0.55=-0.1Ncm=-10 N/m
But F=-mÏ2x or F/x=-mÏ2
so, -mÏ2=-10 or Ï2=10m
â´Ï2=104Ã10-1âÏ=102=5
â´f=Ï2Ï=52Ï s-1

Topic: Physics, Oscillations, NEET
The figure shows three circuits with identical batteries, inductors and resistances. Rank the circuits in decreasing order, according to the current through the battery just after the switch is closed.

Options:
A i2>i3>i1
B i2>i1>i3
C i1>i2>i3
D i1>i3>i2

The correct answer is:
i2>i3>i1

Solution:
Just after switching on inductor behaves as open circuit
⇒i1=E∞=0, i2=ER and i3=E3R
⇒i2>i3>i1

Topic: Physics, Alternating Current, NEET
The figure shows three circular arcs, each of radius R and total charge as indicated. The net electric potential at the centre of curvature is -

Options:
A Q2πε0R
B Q4ÏÎµ0R
C 2Qπε0R
D QÏÎµ0R

The correct answer is:
Q2πε0R

Solution:
Since electric potential is scalar quantity therefore :
V = V1+ V2+ V3
=14ÏâÎµ0.QR+14Ï Îµ0-2QR+14Ï Îµ03QR
=14ÏÎµ0.2QR

Topic: Physics, Electrostatics, NEET
The figure shows variation of photocurrent with anode potential for a photo-sensitive surface for three different radiations. Let Ia,Ib and Ic be the intensities and va,vb and vc be the frequencies for the curves a,b and c respectively. Then

Options:
A va=vb and Ia≠Ib
B va=vc and Ia=Ic
C va=vb and Ia=Ib
D vb=vc and Ib=Ic

The correct answer is:
va=vb and Ia≠Ib

Solution:
Saturation current is proportional to intensity while stopping potential increases with increase in frequency. Hence,
va=vb while Ia<Ib

Topic: Physics, Dual Nature of Matter, NEET
The final product 'C' in the above reaction is

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Aldehydes and Ketones, NEET
The final product C, obtained in this reaction

would be

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Amines, NEET
The final product ' $\mathrm{c}$ ', obtained in this reaction would be

Options:
A

B

C

D

The correct answer is:

Solution:
After bromination, bromoacetanilide is acid hydrolysed to yield the desired halogenated amine.

Topic: Chemistry, Amines, NEET, NEET 2003
The final product 'D' of the reaction,

Options:
A C6H5CH2OCH2CH3
B C6H5CH(OH)CH2CH2OH
C C6H5CH2CH2CH2OH
D C6H5CH2CH2OCH3

The correct answer is:
C6H5CH2CH2CH2OH

Solution:
Here the final product D is 3-phenyl propanol

Topic: Chemistry, Alcohols Phenols and Ethers, NEET
The final product (E) in the following reaction is

Options:
A
B
C
D

The correct answer is:

Solution:
Here Nylon-6 is formed as follows

Topic: Chemistry, Alcohols Phenols and Ethers, NEET
The final product formed when methylamine is treated with NaNO 2 and HCl followed by hydrolysis is:

Options:
A Nitromethane
B Methylcyanide
C Methyl alcohol
D Diazomethane

The correct answer is:
Methyl alcohol

Solution:
CH 3 NH 2 ⟶ HCl NaNO 2 CH 3 N 2 + Cl - ⟶ H 2 O CH 3 OH Methyl alcohol

Topic: Chemistry, Amines, NEET
The final product (III) obtained in the reaction

Options:
A

B
C

D

The correct answer is:

Solution:
Aldol condensation takes place as,

Topic: Chemistry, Aldehydes and Ketones, NEET
The final product in the following reaction sequence is

Options:
A p-chlorophenol
B p-chlorobenzamide
C p-chlorobenzylamine
D p-chlorobenzyl alcohol

The correct answer is:
p-chlorobenzylamine

Solution:
Here the sequence of reactions is as follows

Topic: Chemistry, Amines, NEET
The final product (X) in the following reaction is

Options:
A 2-Nitroaniline
B 3-Nitroaniline
C 4-Nitroaniline
D Sulphanilic acid

The correct answer is:
2-Nitroaniline

Solution:

Topic: Chemistry, Carboxylic Acid Derivatives, NEET
The final product (Z) is the following sequence of reactions
CH3CH2NH2→HNO2(X)→SOCl2(Y)→NH3(Z) is

Options:
A methanamine
B ethanamide
C ethanamine
D propan-1-amine

The correct answer is:
ethanamine

Solution:
Here ethyl amine is formed as the final product
CH3CH2NH2→HNO2CH3CH2OH(X)→SOCl2CH3CH2Cl(Y)→NH3CH3CH2NH2(Z)

Topic: Chemistry, Amines, NEET
The final proof for DNA as the genetic material came from the experiments of

Options:
A Griffith
B Hershey and Chase
C Avery, Mcleod and McCarty
D Hargobind Khorana

The correct answer is:
Hershey and Chase

Solution:
The final proof that DNA is the genetic material came from the experiments of Alfred Hershey and Martha Chase (1952). Griffith's experiment proved the existence of genetic material while Avery, MacLeod and McCarty worked to determine the biochemical nature of transforming principle.
Concept enhancer Hershey and Chase during their experiment grew viruses in two mediums, one containing 32P and other 35S, when these were allowed to infect bacteria, they observed that viruses containing 32P DNA were radioactive while those with 35S protein were not radioactive. Hence, DNA not protein coat entered bacterial cells from viruses.

Topic: Biology, Molecular Basis of Inheritance, NEET, NEET 2017
The Finches of Galapagos island provide an evidence in favour of:

Options:
A evolution due to mutation
B retrogressive evolution
C biogeographical evolution
D special creation.

The correct answer is:
biogeographical evolution

Solution:
The finches of Galapagos is lands providesan evidence in favour of biogeographical evolution.

Topic: Biology, Evolution, NEET, NEET 2007
The first 5-C dicarboxylic acid in Krebs' cycle which is used in nitrogen metabolism is ____.

Options:
A succinate
B oxalosuccinate
C α-ketoglutaric acid
D succinyl coenzyme A

The correct answer is:
α-ketoglutaric acid

Solution:
The TCA cycle or the Kreb's cycle is a form of aerobic respiration where the reaction starts with the acetyl group condensation by oxaloacetic acid and water, whose product is citric acid. The citrate is then isomerised to isocitrate followed by successive steps of decarboxylation. The first step of decarboxylation leads to the formation of a 5-C dicarboxylic acid, the α-ketoglutaric acid which is used in nitrogen metabolism as a precursor for many amino acids and purines.

Topic: Biology, Respiration in Plants, NEET
The first acceptor of electrons from an excited chlorophyll molecule of photosystem II is:

Options:
A iron-sulphur protein
B ferredoxin
C quinone
D cytochrome.

The correct answer is:
quinone

Solution:
Electrons from excited chlorophyll molecule of photosystem II are accepted first by Quinone. Photosystem II is a photosynthetic pigment system along with some electron carriers that is located in the appressed part of the grana thylakoids. Photosystem II has chlorophyll a, b and carotenoids. Other components of PS II are phaeophytin, plastoquinone (PQ), cytochrome complex and blue coloured copper containing plastocyanin.
Related Theory
Photosystem II is a photosynthetic pigment, along with some electron carriers, that is located in the appressed part of the grana thylakoids. It picks up electron released during photolysis of water and passes to pheophytin.
Caution
The first electron acceptor in the case of photosystem I is Cytochrome.

Topic: Biology, Photosynthesis in Higher Plants, NEET, NEET 2007
The first amino acid in the primary structure of the protein is placed in

Options:
A right end
B left end
C middle
D right end and left end

The correct answer is:
left end

Solution:
In the primary structure of the protein, the left end is represented by the first amino acid and the right end is represented by the last amino acid.

Topic: Chemistry, Biomolecules, NEET
The first and second dissociation constants of an acid H2A are 1.0Ã10-5 and 5.0Ã10-10 respectively. The overall dissociation constant of the acid will be

Options:
A 5.0×10-5
B 5.0×1015
C 5.0×10-15
D 0.2×105

The correct answer is:
5.0×10-15

Solution:
Given K 1 =1.0Ã 10 â5 , K 2 =1.0Ã 10 â5
so overall dissociation constant of the acid will be K=K1ÃK2
=1.8Ã10-5Ã5.0Ã10-10
=10Ã10-15

Topic: Chemistry, Ionic Equilibrium, NEET
The first and second ionisation enthalpies of a metal are 496 and 4560 kJ mol-1 , respectively. How many moles of HCl and H2SO4 , respectively, will be needed to react completely with 1 mole of the metal hydroxide?

Options:
A 1 and 1
B 1 and 2
C 2 and 0.5
D 1 and 0.5

The correct answer is:
1 and 0.5

Solution:
Metal: First ionization enthalpies = 496 kJ/mole
Second ionization enthalpies = 4560 kJ/mol
According to the given information of ionization enthalpies, metal belong to first group i.e. Monovalent cation
MOH1mole+HCl1mole→MCl+H2O
2MOH1mole+H2SO41/2mole→M2SO4+H2O
That is 1 mole HCl and 0.5 mole H2SO4 are needed respectively.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, NEET
The first cellular forms of life appeared on earth about

Options:
A 500 mya
B 2000 mya
C 2500 mya
D 350 mya

The correct answer is:
2000 mya

Solution:
About 2000 million years ago (mya), the first cellular forms of life appeared on earth. By the time of 500 mya, invertebrates were formed and jawless fish probably evolved around 350 mya. Seaweeds and few plants existed probably around 320 mya. We are told that the first organisms that invaded land were plants. They were widespread on land when animals invaded the land. Fish with stout and strong fins could move on land and go back to water. This was about 350 mya.

Topic: Biology, Evolution, NEET
The first clinical gene therapy was done for the treatment of:

Options:
A AIDS
B Cancer
C Cystic fibrosis
D SCID (Severe Combined Immuno Deficiency resulting from deficiency of ADA)

The correct answer is:
SCID (Severe Combined Immuno Deficiency resulting from deficiency of ADA)

Solution:
The first clinical gene therapy was done for the treatment of SCID (Severe Combined Immuno Deficiency resulting from deficiency of ADA).

Topic: Biology, Biotechnology and Its Applications, NEET
The first clinical gene therapy was given for treating

Options:
A diabetes mellitus
B chicken pox
C rheumatoid arthritis
D adenosine deaminase deficiency

The correct answer is:
adenosine deaminase deficiency

Solution:
The first clinical gene therapy was given for treating Adenosine deaminase deficiency. It was conducted at Maryland in 1990 on a four year old girl. She was suffering from ADA deficiency. This enzyme is important for formation of immune system in the body. Hence biotechnology principles were applied to treat this condition. A functional ADA gene was introduced ex vivo in the marrow cells through a vector which had functional ADA gene. The treated cells were reintroduced into the patient's marrow.

Topic: Biology, Biotechnology and Its Applications, NEET
The first clinical gene therapy was given for treating:

Options:
A Tetany
B Rheumatoid arthritis
C Cystic fibrosis
D Adenosine deaminase deficiency

The correct answer is:
Adenosine deaminase deficiency

Solution:
The first clinical gene therapy was given for treating adenosine deaminase deficiency. A four-year old girl became the first gene therapy patient on September 14, 1990, at the NH Clinical Center. Adenosine deaminase deficiency, also called ADA deficiency or ADA-SCID is an autosomal recessive metabolic disorder that causes immunodeficiency. ADA deficiency is due to a lack of the enzyme adenosine deaminase.

Topic: Biology, Biotechnology and Its Applications, NEET
The first clinical gene therapy was given for treating

Options:
A Diabetes mellitus
B Chicken pox
C Rheumatoid arthritis
D Adenosine deaminase deficiency.

The correct answer is:
Adenosine deaminase deficiency.

Solution:
Gene therapy is a collection of methods that allows correction of a gene defect that has been diagnosed in a child/embryo. Here genes are inserted into a person's cells and tissues to treat a disease. Correction of a genetic defect involves delivery of a normal gene into the individual or embryo to take over the function of and compensate for the nonfunctional gene. The first clinical gene therapy was given in 1990 to a 4 year old girl with adenosine deaminase (ADA) deficiency. This enzyme is very important for the immune system to function. SCID is caused due to defect in the gene for the enzyme adenosine deaminase.

Topic: Biology, Biotechnology and Its Applications, NEET