datavorous/entrance-exam-dataset huggingface dataset:
In genetic engineering, the antibiotics are used

Options:
A as selectable markers
B to select healthy vectors
C as sequences from where replication starts
D to keep the cultures free of infection

The correct answer is:
as selectable markers

Solution:
The integration of selectable marker gene along with a gene of interest is helpful for identifying transformed cells. However, the presence of slectable marker gene for antibiotic resistance in a genetically modified organism that is released into enviornment is not desirable.

Topic: Biology, Biotechnology - Principles and Processes, NEET, NEET 2012 (Mains)
In germinating seeds fatty acids are degraded exclusively in the

Options:
A Mitochondria
B Proplastids
C Glyoxysomes
D Peroxisomes

The correct answer is:
Glyoxysomes

Solution:
It is regulated by glyoxylate cycle.

Topic: Biology, Cell - The Unit of Life, NEET, NEET 2008 (Mains)
In given figure, a wire loop has been bent so that it has three segments ab (a quarter circle), bc (a square corner) & ca (straight line). Here are three choices for a magnetic field through the loop -

(1) Bâ1=3i^ + 7j^ â 5tk^

(2) Bâ2=5ti^ â 4j^ â 15k^

(3) Bâ3=2i^ â5tj^ â 12k^

Where B is in milli tesla and t is in second. If the induced current in the loop due to B1â,B2â,B3â are i1, i2, i3 respectively then -

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

The correct answer is:
i2> i1> i3

Solution:
i1=dϕ1dt 1R=dB1×AreaRdt=ddt 5t×πa24×1R=5πa24R
i2=dϕ2dt×1R=d5a2tdt×1R=5a2R
i3=dϕ3dt×1R=d0.5a2dt× 1R=a22R
∴ i2 > i1 > i3

Topic: Physics, Electromagnetic Induction, JEE Main
In gobar gas, the maximum amount is that of

Options:
A butane
B methane
C propane
D carbon dioxide

The correct answer is:
methane

Solution:
Biogas is a methane rich fuel gas produced by anaerobic breakdown of biomass with the help of methanogenic bacteria. It is composed of methane (50-70\%), $\mathrm{CO}_2$ (30-40\%) and traces of $\mathrm{H}_2, \mathrm{~N}_2$ and $\mathrm{H}_2 \mathrm{~S}$.

Topic: Biology, Microbes in Human Welfare, NEET, NEET 2012 (Mains)
In Goldschmidt aluminothermic process which of the following reducing agents is used :

Options:
A
calcium

B
coke

C
Al-powder

D
sodium

The correct answer is:
Al-powder

Solution:
Reduction by powdered aluminium is known as Gold-Schmidt aluminothermic process. This process is employed in cases where metals have very high m.p. and are to be extracted from their oxides.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2013 (22 Apr Online)
In graph of atomic volume versus atomic weight, the elements corresponding to peaks in the curve belong to

Options:
A Group 1
B Group 18
C Group 4
D Group 14

The correct answer is:
Group 1

Solution:
If we see the atomic mass versus atomic volume curve, we will notice the following features:

Alkali metals such as Na, K, Rb which have similar properties occur as peaks of the curve.

Halogen elements like F, Cl, Br, which have similar properties occur at the rising or the ascending part of the curve.

Noble gases such as Ne, Ar, Kr which have similar properties occur just before the alkali elements.

Hydrogen and Helium seem to be exception to the rule.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
In graphite and diamond, the percentage of $p$ characters of the hybrid orbitals in hybridization are respectively:

Options:
A
$33$ and $25$

B
$67$ and $75$

C
$50$ and $75$

D
$33$ and $75$

The correct answer is:
$67$ and $75$

Solution:
$\%$ of $p$ - in graphite $\left(s p^2\right)=\frac{2}{3} \times 100=67 \%$
$\%$ of $p$ - in diamond $\left(s p^3\right)=\frac{3}{4} \times 100=75 \%$

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
In graphite and diamond, the percentage of p-characters of the hybrid orbitals in hybridization respectively, are

Options:
A 33 and 25
B 67 and 75
C 50 and 75
D 33 and 75

The correct answer is:
67 and 75

Solution:
Graphite	Diamond

sp2 hybridisation	sp3 hybridisation

% P=23×100=67%	% P=34×100=75%

For sp2: %P character
P−1P=cosθ
P−1P=cos120=−0.5
P−1=0.5P
1.5P=1
P=23
%P=23×100=66.67%	For sp3: %P character
P−1P=cos109'28
P−1P=0.33
P−1P=34
%P=75%

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2018 (15 Apr)
In guard cells, presence of postassium is essential for

Options:
A Maintaining osmotic pressure
B In controlling cell division
C In enzymatic reactions
D All the above

The correct answer is:
Maintaining osmotic pressure

Solution:
The guard cells of stomata are involved in the process of transpiration and the turgidity of guard cells is controlled by potassium, chloride, and malate. This turgidity helps in maintaining the opening and closing of stomata.

Topic: Biology, Mineral Nutrition, NEET
In gymnosperms, the ovule is naked because

Options:
A ovary wall is absent
B integuments are absent
C perianth is absent
D nucellus is absent

The correct answer is:
ovary wall is absent

Solution:
Gymnosperms are commonly known as 'naked seed plants' because the ovary wall is not present and hence, the ovules remain unprotected or naked.

Topic: Biology, Plant Kingdom, JIPMER, JIPMER 2012
In gymnosperms, where do the male and female gametophytes typically reside?

Options:
A independently in the environment
B within the sporangia on the sporophytes
C within the ovules on the megasporophylls
D within the archegonia on the ovules

The correct answer is:
within the sporangia on the sporophytes

Solution:
Unlike bryophytes and pteridophytes, in gymnosperms the male and the female gametophytes do not have an independent free-living existence. They remain within the sporangia retained on the sporophytes.

Topic: Biology, Plant Kingdom, NEET
In He⊕ ion sample e- is in ground state, if photon of energy 52.24eV is given to the sample all the atom goes to higher energy state. It again falls back up to ground state. If it is not emitting any lines in Balmer series then what is the maximum possible number of spectral lines observed

Options:

The correct answer is:
6

Solution:
13.6Ã221â1n2=52.24

â13.6â13.6n2=13.06

â13.6n2=0.54

ân=5

Total line =55â12=10

But it is not emitting balmer lines hence

âtotal line =10-3=7

But it will also bit enquit 2â1 line

Hence total line =7-1=6

Topic: Chemistry, Structure of Atom, JEE Main
In Haber's process for the manufacture of ammonia

Options:
A Finely divided iron is used as catalyst
B Finely divided molybdenum is used as catalyst
C Finely divided nickel is used as catalyst
D No catalyst is necessary

The correct answer is:
Finely divided iron is used as catalyst

Solution:
Finely divided iron is used as catalyst in the manufacture of ammonia by Haber's process

Topic: Chemistry, Surface Chemistry, JEE Main
In half wave rectification, if the input frequency is 60 Hz, then the output frequency would be

Options:
A 30 Hz
B 60 Hz
C 120 Hz
D zero

The correct answer is:
60 Hz

Solution:
As we can see from graph, frequency remains same that is 60 Hz.

Topic: Physics, Semiconductors, NEET, NEET 2022 (Phase 1)
In half-wave rectification, what is the output frequency if the input frequency is $50 \mathrm{~Hz}$ ? What is the output frequency of a full-wave rectifier for the same input frequency?

Options:

None

Solution:
In half wave rectification, output frequency remains same as input i.e., $50 \mathrm{~Hz}$. In full wave rectification, output shows two positive pulses for each input pulse hence frequency doubles i.e., $\mathbf{1 0 0} \mathrm{Hz}$.

Topic: Physics, Semiconductors
In Hall-Heroult process, the following is used for reducing Al2O3 :-

Options:
A Magnesium
B Na3AlF6
C Graphite
D CaF2

The correct answer is:
Graphite

Solution:
The process of extracting Al metal by electrolysis of a mixture of purified Al2O3 and Na3AlF6 is known as Hall and Heroult process.
Graphite is used as anode which is useful in the reduction of alumina oxide to the Al metal. The graphite electrode reacts with the oxygen released at cathode to form carbon dioxide and carbon monoxide.
Therefore, the role of graphite as an anode in the extraction of Al is to check the formation of oxygen. In other words, we can say its role is to prevent the oxidation of Al back to Al2O3 by oxygen.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2023 (08 Apr Shift 2)
In history of biology, human genome project led to the development of

Options:
A biotechnology
B biomonitoring
C bioinformatic
D biosystematics

The correct answer is:
bioinformatic

Solution:
Human Genome Project (HGP) was closely associated with the rapid development of a new area in biology called a Bioinformatics, the science of collecting and analysing complex biological data such as genetic codes.

Topic: Biology, Molecular Basis of Inheritance, NEET, NEET 2011 (Mains)
In how many minimum steps you will convert 4-nitrotoluene into 2-bromobenzoic acid?

Options:

The correct answer is:
5

Solution:
This conversion is possible in 5 steps as follows

Topic: Chemistry, Amines, JEE Main
In how many number of ways can 10 students be divided into three teams, one containing four students and the other three?

Options:
A 400
B 700
C 1050
D 2100

The correct answer is:
2100

Solution:
We know that the number of ways of dividing $(\mathrm{m}+\mathrm{n}+\mathrm{p})$ things into three groups containing $\mathrm{m}, \mathrm{n}$ and $\mathrm{p}$ things respectively
$$
=\frac{(\mathrm{m}+\mathrm{n}+\mathrm{p}) !}{\mathrm{m} ! \mathrm{n} ! \mathrm{p} !}
$$
Further if any two groups out of the three have same number of things then number of ways
$$
=\frac{(m+n+p) !}{m ! n ! p ! \times 2}
$$
Hence number of ways to divide 10 students into three teams one containing four students and each remaining two teams contain three
$=\frac{10 !}{4 ! 3 ! 3 ! \times 2}=\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5}{3 \times 2 \times 3 \times 2 \times 2}=2100$

Topic: Mathematics, Application of Derivatives, VITEEE, VITEEE 2008
In how many of the following all bond lengths are not equal?
CO3-2,O3,BF3,P4white,PCl,SF4,CIF3,XeF2,XeF4,ClF4+

Options:

The correct answer is:
4

Solution:
In PCl5,SF4,ClF3,CIF4+ all bond lengths are not equal.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
In how many of the following reactions aldehyde is formed as a product?

Options:

The correct answer is:
6

Solution:

Topic: Chemistry, Aldehydes and Ketones, JEE Main
In how many representation of the below-mentioned compounds, does not correctly represent the bonding capacities of the atoms involved?

Options:

The correct answer is:
1

Solution:
The star marked carbon has a valency of 5 and hence this formula is not correct.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
In how many ways 3 letters can be posted in 4 letter-boxes, if all the letters are not posted in the same letter-box

Options:
A $63$
B $60$
C $77$
D $81$

The correct answer is:
$60$

Solution:
Three letters can be posted in 4 letter boxes
in $4^{\wedge} 3=64$ ways but it consists the 4 ways
that all letters may be posted in same box.
Hence required ways $=60$.

Topic: Mathematics, Permutation Combination, JEE Main
In how many ways a committee of a 6 members can be formed out of 10 members but always consisting of a specified member ?

Options:
A \(\frac{1}{2}{ }^9 P_5\)
B \({ }^9 P_5\)
C \({ }^9 \mathrm{C}_5\)
D \(\frac{1}{2}{ }^9 C_5\)

The correct answer is:
\({ }^9 \mathrm{C}_5\)

Solution:
As the committee consists a specified member out of 6 members, then we need to select only 5 members out of 9 remaining members and it can be done is \({ }^9 C_5\) ways. Hence, option (c) is correct.

Topic: Mathematics, Permutation Combination, AP EAMCET, AP EAMCET 2020 (21 Sep Shift 2)
In how many ways a garland can be made from exactly 10 flowers

Options:
A $10 !$
B $9 !$
C $2(9$ !)
D $\frac{9!}{2}$

The correct answer is:
$\frac{9!}{2}$

Solution:
A garland can be made from 10 flowers in $\frac{1}{2}(9$ !) ways.
$\{\because n$ flowers' garland can be made in $\frac{1}{2}(n-1)!$ ways $\}$

Topic: Mathematics, Permutation Combination, JEE Main
In how many ways can 5 boys and 3 girls sit in a row so that no two girls are together

Options:
A $5!\times 3!$
B ${ }^4 P_3 \times 5!$
C ${ }^6 P_3 \times 5!$
D ${ }^5 P_3 \times 3!$

The correct answer is:
${ }^6 P_3 \times 5!$

Solution:
Since the 5 boys can sit in 5 ! ways. In this case there are 6 places are vacant in which the girls can sit in ${}^{6} P_3$ ways. Therefore required number of ways are ${}^{6} P_{3} * 5$ !.

Topic: Mathematics, Permutation Combination, JEE Main
In how many ways can 5 boys and 5 girls sit in a circle so that no two boys sit together

Options:
A $5!\times 5$ !
B $4!\times 5$ !
C $\frac{5!\times 5!}{2}$
D None of these

The correct answer is:
$4!\times 5$ !

Solution:
Since total number of ways in which boys can occupy any place is $(5-1)!=4!$ and the 5 girls can be sit accordingly in 5 ! ways. Hence required number of ways are 4 ! * 5 !

Topic: Mathematics, Permutation Combination, JEE Main
In how many ways can 5 boys and 5 girls stand in a row so that no two girls may be together 97]

Options:
A $(5!)^2$
B $5!\times 4$ !
C $5!\times 6$ !
D $6 \times 5$ !

The correct answer is:
$5!\times 6$ !

Solution:
5 boys can be stand in a row 5 ! ways
Now, two girls can't stand in a row together in ${ }^6 P_5$ ways.
Total no. of required arrangement
$=5!\times{ }^6 P_5=5!\times 6!$

Topic: Mathematics, Permutation Combination, JEE Main
In how many ways can 5 boys and 5 girls stand in a row so that no two girls may be together

Options:
A $(5!)^2$
B $5!\times 4$ !
C $5!\times 6$ !
D $6 \times 5$ !

The correct answer is:
$5!\times 6$ !

Solution:
5 boys can be stand in a row 5 ! ways
Now, two girls can't stand in a row together in ${ }^6 P_5$ ways.
Total no. of required arrangement
$=5!\times{ }^6 P_5=5!\times 6!$.

Topic: Mathematics, Permutation Combination, JEE Main
In how many ways can 5 keys be put in a ring

Options:
A $\frac{1}{2} 4!$
B $\frac{1}{2} 5!$
C $4 !$
D $5 !$

The correct answer is:
$\frac{1}{2} 4!$

Solution:
Required number of ways $=\frac{1}{2}(5-1)!=\frac{4!}{2}$.
\{Since clockwise and anticlockwise are same in case of ring\}.

Topic: Mathematics, Permutation Combination, JEE Main
In how many ways can 5 prizes be distributed among 4 boys when every boy can take one or more prizes?

Options:
A 1024
B 625
C 120
D 600

The correct answer is:
1024

Solution:
First prize may be given to any one of the 4 boys, hence first prize can be distributed in 4 ways. similarly every one of second, third, fourth and fifth prizes can also be given in 4 ways. the number of ways of their distribution $=4 \times 4 \times 4 \times 4 \times 4=4^{5}=1024$

Topic: Mathematics, Permutation Combination, JEE Main
In how many ways can 5 prizes be distributed among four students when every student can take one or more prizes

Options:
A $1024$
B $625$
C $120$
D $600$

The correct answer is:
$1024$

Solution:
Required number of ways are $4^{\wedge} 5$
$=1024$. \{Since each prize can be distributed in 4 ways\}

Topic: Mathematics, Permutation Combination, JEE Main
In how many ways can 5 red and 4 white balls be drawn from a bag containing 10 red and 8 white balls

Options:
A ${ }^8 C_5 \times{ }^{10} C_4$
B ${ }^{10} C_5 \times{ }^8 C_4$
C ${ }^{18} C_9$
D None of these

The correct answer is:
${ }^{10} C_5 \times{ }^8 C_4$

Solution:
Explanation for the correct option:
Step 1. Formula for Combination \(={ }^{\mathrm{n}} \boldsymbol{c}_{\mathbf{r}}=\frac{\mathrm{n}!}{\mathrm{r}!(\mathrm{n}-\mathbf{r})!}\)
Step 2. Number of ways can be drawn.
We have 10 red balls and 8 white balls. out which 5 red and 4 white balls can be drawn
\(\text {Number of ways }={ }^8 c_4 \times{ }^{10} c_5\)

Topic: Mathematics, Permutation Combination, JEE Main
In how many ways can 6 persons be selected from 4 officers and 8 constables, if at least one officer is to be included

Options:
A $224$
B $672$
C $896$
D None of these

The correct answer is:
$896$

Solution:
Required number of ways
$={ }^4 C_1 \times{ }^8 C_5+{ }^4 C_2 \times{ }^8 C_4+{ }^4 C_3 \times{ }^8 C_3+{ }^4 C_4 \times{ }^8 C_2$
$=4 \times 56+6 \times 70+4 \times 56+1 \times 28=896$.

Topic: Mathematics, Permutation Combination, JEE Main
In how many ways can 10 balls be divided between two boys, one receiving two and the other eight balls

Options:
A $45$
B $75$
C $90$
D None of these

The correct answer is:
$90$

Solution:
$A$ gets $2, B$ gets $8 ; \frac{10!}{2!8!}=45$
$A$ gets $8, B$ gets $2 ; \frac{10!}{8!2!}=45$
$\therefore 45+45=90$.

Topic: Mathematics, Permutation Combination, JEE Main
In how many ways can 10 true-false questions be replied

Options:
A $20$
B $100$
C $512$
D $1024$

The correct answer is:
$1024$

Solution:
Required number of ways are \(2^{10}=1024\), because every question may be answered in 2 ways.

Topic: Mathematics, Permutation Combination, JEE Main
In how many ways can 15 members of a council sit along a circular table, when the Secretary is to sit on one side of the Chairman and the Deputy Secretary on the other side

Options:
A $2 \times 12$ !
B $24$
C $2 \times 15$ !
D None of these

The correct answer is:
$2 \times 12$ !

Solution:
Since total members are 15 , but one is to left, because of circular condition, therefore remaining members are 14 but three special member constitute a member. Therefore required number of arrangements are 12 ! *2, because, chairman remains between the two specified persons and the person can sit in two ways.

Topic: Mathematics, Permutation Combination, JEE Main
In how many ways can a girl and a boy be selected from a group of 15 boys and 8 girls

Options:
A $15 \times 8$
B $15+8$
C ${ }^{23} P_2$
D ${ }^{23} \mathrm{C}_2$

The correct answer is:
$15 \times 8$

Solution:
$$
={ }^{15} C_1 \times{ }^8 C_1=15 \times 8
$$

Topic: Mathematics, Permutation Combination, JEE Main
In how many ways can a student choose a programme of 5 courses if 9 courses are available and 2 specific courses are compulsory for every student?

Options:

None

Solution:
Out of available nine courses, two are compulsory. Hence, the student is free to select 3 courses out of 7 remaining courses. If $\mathrm{P}$ is the number of ways of selecting 3 courses out of 7 courses, then
$\begin{aligned}
P &=\mathrm{C}(7,3)=\frac{7 !}{3 !(7-3) !}=\frac{7 !}{3 ! 4 !} \\
&=\frac{7 \times 6 \times 5 \times 4 !}{3 \times 2 \times 1 \times 4 !}=7 \times 5=35 \text { ways }
\end{aligned}$

Topic: Mathematics, Permutation Combination
In how many ways can the letters of the word "MULTIPLE" be arranged keeping the position of the vowels fixed?

Options:
A 60
B 360
C 600
D 300

The correct answer is:
360

Solution:
The word MULTIPLE consists of three vowels that are placed at the 2nd , 5th and 8th position.

The three vowels can be arranged at their respective places in 3! ways.

The remaining 5 consonants can be arranged at their respective places in 5!2!=60 ways

∴ Total number of arrangements =3!×5!2!=360

Topic: Mathematics, Permutation Combination, AP EAMCET, AP EAMCET 2022 (04 Jul Shift 2)
In how many ways can the letters of the word PERMUTATIONS be arranged if the
(i) words start with $P$ and end with $S$
(ii) vowels are all together,
(iii) there are always 4 letters between $P$ and $S$ ?

Options:

None

Solution:
(i) Letters between P and S are ERMUTATION. These 10 letters having $\mathrm{T}$ two times, these letters can be arranged in $\frac{10 !}{2 !}$ ways $=1814400$ ways
(ii) There are 12 letters in the word PERMUTATIONS. which have T two times.
Now the vowels a, e, i, o, u are together Let it be considered in one block. The letters of vowels can be arranged in 5 ! ways Thus, there are 7 letters and 1 block of vowel with $T$ two times
$\therefore$ Number of arrangements $=\frac{8 !}{2 !} \times 5 !=2419200$
(iii) There are 12 letters to be arranged in 12 place [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] there are 12 letters are to be filled in 12 place. $P$ may be filled up at place No. 1, 2, 3, 4, 5, 6, 7 and consequently S may be filled up at place no. $6,7,8,9,10,11,12$. leaving four places between. Now $\mathrm{P}$ and $\mathrm{S}$ may be filled up in 7 ways.
Similarly, $S$ and $P$ may be filled up in 7 ways
Thus, $P$ and $S$ or $S$ and $P$ can be filled up in $7+7$ $=14$ ways
Now, the remaining 10 places can be filled by $=\frac{10 !}{2 !}$ ways
$\therefore \quad$ No of ways in which 4 letters occur between $P$ and $S$
$=\frac{10 !}{2 !} \times 14=\frac{3628800}{2} \times 14=25401600$

Topic: Mathematics, Permutation Combination
In how many ways can you form a dancing couple from 3 boys and 3 girls so that no boy dances with his respective girlfriend?

Options:
A $5$
B $4$
C $2$
D $8$

The correct answer is:
$2$

Solution:
2 ways
This is clearly a case of derangement of 3 boys and 3 girls.
The value can be calculated as $D=3!(1-1 / 1!+1 / 2!-1 / 3!)=6(2 / 6)=2$ ways.

Topic: Mathematics, Permutation Combination, JEE Main
In how many ways $n$ books can be arranged in a row so that two specified books are not together

Options:
A $n!-(n-2)$ !
B $(n-1)!(n-2)$
C $n!-2(n-1)$
D $(n-2) n!$

The correct answer is:
$(n-1)!(n-2)$

Solution:
Total number of arrangements of $n$ books $=n!$.
If two specified books always together then number of ways $=(n-1)!\times 2$
Hence required number of ways
$=n!-(n-1)!\times 2$

Topic: Mathematics, Permutation Combination, JEE Main
In human adult females oxytocin

Options:
A Causes strong uterine contractions during parturition
B Is secreted by anterior pituitary
C Stimulates growth of mammary glands
D Stimulates pituitary to secrete vasopressin

The correct answer is:
Causes strong uterine contractions during parturition

Solution:
Oxytocin is called as Birth Hormone as it causes the contraction of smooth muscles of the uterus leading to the birth of the child.

Topic: Biology, Human Reproduction, NEET, NEET 2008 (Mains)
In human beings, the plasma membrane of the erythrocytes has approximately ____ protein and ____ lipids

Options:
A 40 percent; 52 percent
B 52 percent; 40 percent
C 80 percent; 20 percent
D 20 percent; 80 percent

The correct answer is:
52 percent; 40 percent

Solution:
The initial studies carried out about the cell membrane, before the advent of electron microscope, had showed that the plasma membrane in erythrocytes is composed of lipids arranged in a bilayer with the polar heads on the outer side and hydrophilic heads on the inner. Lipid component mainly consisted of phosphoglycerides. Later through biochemical investigations it was found that human erythrocytes have around 40 % lipids and 52% proteins in the plasma membrane. The proteins can be divided into two types - integral and peripheral, depending on how easily they can be extracted.

Topic: Biology, Cell - The Unit of Life, NEET
In human female, the blastocyst

Options:
A forms placenta even before implantation
B gets implanted into uterus three days after ovulation
C gets nutrition from uterine endometrial secretion only after implantation
D gets implanted in endometrium by the trophoblast cells

The correct answer is:
gets implanted in endometrium by the trophoblast cells

Solution:
Implantation is the attachment of blastocyst to the uterine wall (endometrium). It occurs after 6-7 days of fertilization. During implantation, the trophoectoderm (trophoblast) comes in contact with the endometrium of the uterus and sinks into a pit formed in the endometrium and get completely burried in the endometrium. Trophoblast cells form specialized structures called villi, which help in absorption of nutrient materials.

Topic: Biology, Human Reproduction, NEET, NEET 2010 (Mains)