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# IIT JAM Biotechnology Mock Test 6

## 60 Questions MCQ Test Mock Test Series of IIT JAM Biotechnology | IIT JAM Biotechnology Mock Test 6

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This mock test of IIT JAM Biotechnology Mock Test 6 for IIT JAM helps you for every IIT JAM entrance exam. This contains 60 Multiple Choice Questions for IIT JAM IIT JAM Biotechnology Mock Test 6 (mcq) to study with solutions a complete question bank. The solved questions answers in this IIT JAM Biotechnology Mock Test 6 quiz give you a good mix of easy questions and tough questions. IIT JAM students definitely take this IIT JAM Biotechnology Mock Test 6 exercise for a better result in the exam. You can find other IIT JAM Biotechnology Mock Test 6 extra questions, long questions & short questions for IIT JAM on EduRev as well by searching above.
QUESTION: 1

Solution:

QUESTION: 2

### To convert a galvanometer into an ammeter what is true about the placement and magnitude of the resistor with the galvanometer?

Solution:

A ammeter is a device which should have a small resistance. Therefore, a small resistance should be connected in parallel with the galvanometer.

QUESTION: 3

### The S.I unit of

Solution:

We know that

QUESTION: 4

Consider the matrix  A= then which of the following is true (|A| deontes determinents of matrix A)

Solution:

Determinant of all diagonal upper triangular and lower triangular matrices are the product of its diagonal elements.
|A| =abc,  |B| =abc,= |C|=abc
|A| = |B| = |C|

QUESTION: 5

In a leap year, the probability of having 53 Sundays or 53 Mondays is

Solution:

366= 7 x 52 + 2
A leap year contains 366 clay that is. 52 weaks and 2 days.
Rest two days are either (S. M). (M. T). (W. Th). (Th. F). (FL St). (St. S)
The probability of 53 Sunday or 53 Monday = 3/7

QUESTION: 6

Hormone whose receptor is neither membrane bound nor located in nucleus

Solution:

Nitric oxide is a gaseous hormone & has a cytosolic receptor.
Retinoid hormones have nuclear receptors & eicosanoids & insulin have membrane bound receptors.

QUESTION: 7

Enzyme needed for synthesis of eicosanoids from arachidonate is

Solution:

Cyclooxygenase or Cox is the enzyme that is needed for synthesis of thromboxanes & prostaglandins from arachidonate.

QUESTION: 8

Which one of the following relation holds true for the specific growth(µ) of a microorganism in the death phase?

Solution:

As we know that μ = ug-Kd where
μg= Specific growth rate at instant time
Kd = Death rate constant
μ = Net specific growth rate
As in case of death phase death rate constant is higher than that of growth rate then m that case net specific value must be less than 0. So the relation μ < 0 holds true in case of death phase.

QUESTION: 9

The northern blot is used for the probing of:

Solution:

A northern blot is a laboratory method used to detect specific RNA molecules among a mixture of RNA. It can be used to analyze a sample of RNA from a particular tissue or cell type in order to measure the RNA expression of particular genes. Northern blot analysis reveals information about RNA identity, size and abundance, allowing a deeper understanding of gene expression levels.

QUESTION: 10

The part of seed which goes on to form the shoot is

Solution:

Plumule is the outgrowth which goes on to form shoot.

QUESTION: 11

In the given configuration, a resistor ‘R’ is moving across a region of perpendicular magnetic field ‘B’. What is the current I flowing across the resistor when its velocity is v?

Solution:

I is clockwise in the given configuration.

QUESTION: 12

A plane electromagnetic wave travelling in certain medium is described by an electric field.

Choose the correct option

Solution:

The part ol seed which goes on to torm the shoot is
(a) Radicle (b) Plumule (c) Coleoptie
Plumule is the outgrowth which goes on to tonn shoot.

QUESTION: 13

Two capacitors have capacitances in the ratio 2 : 3. They are connected in parallel in a circuit. The energy stored in the capacitors are in the ratio?

Solution:

Parallel connection means V1 = V2

QUESTION: 14

Variation of electric potential of a charged hollow sphere is can be shown as

Solution:

Electric field inside a hollow charged sphere is zero. Therefore, potential should be constant from r= 0 to

QUESTION: 15

Consider the statements
I. The equation x +1 = 0 has no real root.
II. The equation x2 +1 = 0 has no real root.
III. The equation x3 +1 = 0 has no real root.
Than which of the following is false

Solution:

x + l = 0 = > . x = - 1
x2 +1 = 0 => no real root exists as x2 is non-negative
x3 +1 = 0 =» x3 = -1 => x = -1
statement I and III are false.

QUESTION: 16

In case of matrix of order n × n which of the following is true (I is identity matrix of order n)

Solution:

Matrix multiplication is always associative => (AB)C = A(BC)

QUESTION: 17

The angle between vactor

Solution:

QUESTION: 18

The reflection of a point ( α, β, γ ) in XY - plane is

Solution:

Two different side otAT-plaiie r-co-ordinate is of opposite sign.

The reflection of a point in XY-plane changes the sign of z -co-ordinate but x and y-co-ordinate remains same.
The reflection of ( α, β, γ ) ( α, β, –γ )

QUESTION: 19

The region of eye that is the site of highest visual activity is called as

Solution:

Fovea centralis or macula lutea is the spot that is rich only in cone cells. Due to cone cells it is point of highest visual acuity and has sharpest colour vision.

QUESTION: 20

Match the coloumns

Solution:

QUESTION: 21

Match the coloumns

Solution:

QUESTION: 22

Match the coluumns

Solution:

QUESTION: 23

What is the purpose of bacterial endopores ?

Solution:

An endospore is a dormant, tough and non-reproductive structure produced by certain bacteria in order to protect themselves from adverse conditions.

QUESTION: 24

Which of the following groups of bacteria is considered as a connecting link between bacteria and virus?

Solution:

Mycoplasmas is a group of bacteria that lack a cell wall around their cell membrane. It is considered as a connecting link between bacteria and virus.

QUESTION: 25

The ability of bacteria to change their morphological form is termed as

Solution:

Pleomorphism is the ability of micro-organisms to alter their shape and size in response to environmental conditions.

QUESTION: 26

Pseudopeptidoglycan is composed of:

Solution:

Pseudopeptidoglycan (also known as pseudomurein  is a major cell wall component of some Archaea that differs from bacterial peptidoglycan in chemical structure, but resembles bacterial peptidoglycan in function and physical structure. The basic components are N-acetylglucosamine and N-acetyltalosaminuronic acid (peptidoglycan has N-acetylmuramic acid instead), which are linked by â-1,3-glycosidic bonds.

QUESTION: 27

Why dideoxynucleoside triphosphates (ddNTPs) are used in DNA sequencing?

Solution:

Dideoxy derivatives of nucleoside triphosphates (ddNTPs) are used in DNA sequencing, because if one of these molecules is added to a growing DNA chain, synthesis stops. This is because DNA polymerase adds a nucleotide to the 3’-OH of a growing DNA chain and ddNTPs lack the 3’-OH group. Hence, although they can be added to a chain via their 5’-phosphate, the chain is then terminated. In sequencing reaction the fraction of newly synthesized DNA chains that are terminated will depend on the relative proportions of dNTP and ddNTP.

QUESTION: 28

The separation techniques of proteins are often based on all of the following properties except,

Solution:

A protein must be purified before its structure and the mechanism of its action can be studied. However, because proteins vary in size, charge and water solubility, no single method can be used to isolate all proteins.

QUESTION: 29

In polyacrylamide gel electrophoresis, SDS is used for the mixture of proteins for their efficient separation on the gel. Why SDS, in this experiment is used?

Solution:

Most of the proteins bind SDS with constant-weight ratio (one SDS molecule per two amino acids) leading to identical charge densities. However, unless proteins are completely unfolded, SDS cannot bind uniformly to protein amino acid residues as some amino acid residues may not be accessible due to folding of protein. Thus reducing agent further denatures the proteins by reducing disulfide linkages and enables SDS to binds the proteins uniformily.

QUESTION: 30

At which end does the pollen tube enter the ovary of the flower ?

Solution:

Pollen tube enters the ovary through the micropylar end where synergids are present.

*Multiple options can be correct
QUESTION: 31

Bipolar neurons are found in

Solution:

Bipolar neurons are found in retina and olfactory epithelium.

*Multiple options can be correct
QUESTION: 32

Solution:

*Multiple options can be correct
QUESTION: 33

Dopamine is

Solution:

Dopamine is
-a catecholamine & they act as ligand for adrenergic receptors.
-a neurotransmitter involved in behaviour.
-a hypothalamic hormone.

*Multiple options can be correct
QUESTION: 34

Hormones that are produced in male testis (humans) is/are-

Solution:

Leydig cell-Testosterone
FSH - Anterior pituitary.

*Multiple options can be correct
QUESTION: 35

The characteristics of acid dyes include:

Solution:

Acid dyes have a negative charge and hence bind to positively charged cell structures. Its examples include, eosin, rose bengal, and acid fuchsin.

*Multiple options can be correct
QUESTION: 36

In a balanced growth phase of bacterial cell, which of the following statements are correct ?

Solution:

In a balanced growth phase, the growth rate decreases with decrease in the concentration of the substrate. The specific growth rate is determined by the cell population. Only the increase in number of  cells takes place at a constant rate.

*Multiple options can be correct
QUESTION: 37

Which statement is not true for both prokaryotic and eukaryotic cells?

Solution:

Prokaryotic as well as eukaryotic cells vary in their size and shape. Both the cells essentially contain ribosomes. And any living cell (be it prokaryotic or eukaryotic) must be surrounded by plasma membrane. All prokaryotic and eukaryotic cells contain DNA as their primary genetic material.

*Multiple options can be correct
QUESTION: 38

The DNA migration rate through the gel is determined by,

Solution:

The electrophoretic migration rate of DNA through agarose gel is dependent upon the molecular size of the DNA as well as agarose gel concentration and composition. Molecules of linear duplex DNA travel through agarose gels at a rate which is inversely proportional to the log of their molecular weight. There is an inverse linear relationship between the logarithm of the electrophoretic mobility and gel concentration.

*Multiple options can be correct
QUESTION: 39

To find out the sequence of a hypothetical protein X following techniques can be used.

Solution:

Sequence of a protein can be found out using following techniques
(a) de-novo sequencing using tandem mass spectrometry or MS-MS
(b) Sanger sequencing for proteins
(d) cDNA sequencing and using translation software

*Multiple options can be correct
QUESTION: 40

If a matrix has 12 elements then possible order of matrix is/are

Solution:

12 = 1 × 12 = 12 × 1 = 6 × 2 = 2 × 6 = 4 × 3 = 3 × 4 all.

*Answer can only contain numeric values
QUESTION: 41

A parallel plate capacitor is maintained at a potential difference of 10V. An electron starts from rest from the negative end. The final momentum (in units of 10–24 kg m/s) it gains when it reaches the positive plate would be ___________  (Upto 2 decimal places)

Solution:

Kinetic energy, K= qv

*Answer can only contain numeric values
QUESTION: 42

A particle is travelling with a velocity of 10 m/s making an angle of 30º with a magnetic field of magnitude 0.5T. If the charge to mass ratio of the particle is 1.75 × 1011 C/kg, the time period (in pico seconds) of revolution of the particle due to the magnetic field would be ________ (upto 2 decimal places).

Solution:

(it does not depend on velocity and the angle it makes with the magnetic field)

*Answer can only contain numeric values
QUESTION: 43

The reading of the voltmeter (in mV) in the given circuit (assume the voltmeter and the batteries to be ideal) in mV would be _________(upto 2 decimal places).

Solution:

By Kirchhoff's voltage rule.
2V - I1 x Ω - I x 1Ω- 1V -I x 2Ω =0
⇒1-I1-3I=0
⇒3I+I1=1     ......(i)
Again, for the other circuit,
2V-2Ωx(I-I1)-Ix1Ω-1V-2ΩxI=0
⇒1-5I+2I1=0
⇒5I-2I1=1    ......(ii)

Multiplying (i) by 5 and (ii) by 3 and subtracting, we have
11I1 = 2 ⇒ I1 =2/11A
Therefore, voltage across the 1Ω resistor
=1Ωx2/11A=181.81mV
Since, the voltmeter is parallel to 1Ω resistor, the reading of the voltmeter is 181.81mV.

*Answer can only contain numeric values
QUESTION: 44

The work required to move a 2C charge slowly from the point in the given configuration is__________________

Solution:

The yz plane m the given configuration is an equipotential surface (V= 0) because each point in the yz plane is equidistant from -q and q. Hence, no work is required to be done to move the particle from .

*Answer can only contain numeric values
QUESTION: 45

If α and β are two solution of quadratic equation x2+x-1, then α2 + β2 is equal to___________.

Solution:

*Answer can only contain numeric values
QUESTION: 46

The sum of direction cosines of the vector

Solution:

*Answer can only contain numeric values
QUESTION: 47

The number of permultation of the letter of the word 'BOOK' is ____________.

Solution:

B,O,O,K
Total number permultation

*Answer can only contain numeric values
QUESTION: 48

If the mid-point of the line segment joining the points A(3, 4) and B (k, 6) is P(x, y) and x + y -10 = 0 then value of k is__________.

Solution:

*Answer can only contain numeric values
QUESTION: 49

A rod shaped bacteria has a diameter of 0.6µm and length of 3µm . The average cell density of the bacteria is 1.5 × 103g/L. The mass of a single cell will be _________ ×10–12 grams. (Round off to two decimal places)

Solution:

Volume of rod shaped cell = πr2h
r = Radius = 0.3μm or 0.3 x 10-6 m
h = Length = 3μm or 3 x 10-6 m
Volume of cell =πr2h
= 3.14 x (0.3 x l0-6)2 x 3 x 10-6 m3
= 0.84 x l0-18 m3
Given density = 1.5 x 103 g L
Since Im3 = 1000 litres
Density = 1.5 x 106g 1000L or 1.5 x 106g/m3
Mass = Volume x Density
= 0.84 x I0-18x 1.5 x 106
= 1.26 x 10-12 g

*Answer can only contain numeric values
QUESTION: 50

Doubling time of a microbe is 1 min. and it takes 30 mins. to fill the cup. The time required to fill half the cup is ______ minutes.

Solution:

Doubling time 1 min means microbial population get doubled every minute. After 29 minutes the cup will half filled and during 30th minute, the population get doubled to completely fill the cup.

*Answer can only contain numeric values
QUESTION: 51

A neuron is stimulated with an electric pulse. The action potential is detected at a point 3.40 cm down the axon 0.0052s later. When action potential is detected 7.20 cm from the point of stimulation, the time required is 0.0063s. The speed of electric pulse along the neuron is  ______ m/s [in nearest integer].

Solution:

Speed is change in position per unit time

*Answer can only contain numeric values
QUESTION: 52

An axon is 10 cm long and has a radius of l0μm. If typical membrane thickness is 10-8 m and a dielectric
constant tor cellular fluid is 3. the capacitance of the axon would be_________x 10-8F. (upto one decimal
point).

Solution:

Cross-sectional area of axon A =2πrl

*Answer can only contain numeric values
QUESTION: 53

A rich medium is inoculated with a bacterium that divides every 30 minutes. The number of bacteria at end of 50 hours is _______ x 2100.

Solution:

*Answer can only contain numeric values
QUESTION: 54

A bacterial culture was diluted to 10000 fold and 0.1 ml of this diluted sample was spread on plate on nutrient agar. In a triplicate run the number of colonies found is 219, 195, 186. The number of colonies in the a signal culture is ..............× 107 cells/ml. [answer in nearest integer]

Solution:

Average cell after triplicate run

*Answer can only contain numeric values
QUESTION: 55

A growing bacterial culture with a doubling the of 20 minutes reaches cell density of 3 x 108 cell ml in 4 hours. The time it would take to reach 2 x 108 cells would be____________minutes, [answer innearest integer]

Solution:

As g = 20 minutes
t = 4 horns = 240 minutes.

N = N0 x 2n
3 x 108 = N x 221 => 3 x 108 = N0 x 4096

From here we get the initial number of cells/ml now
N = N0 x 2n

n=11
Generation time = 20 minutes
Total time taken to reach such 2 x 108 cells = 20 x 11 = 220 minutes

*Answer can only contain numeric values
QUESTION: 56

A sphere contains charges as shown in the figure The flux of the electric field through the surface of the sphere (in units of V /m)? ___________ x 1011 V m. (∈0 = 8.85 * 10-12 Farad/m)

Solution:

*Answer can only contain numeric values
QUESTION: 57

A ring has charge 2C uniformly spread over it. It rotates with a constant angular velocity about an axis passing through the centre and perpendicular to the ring. The magnetic field at the centre (in mT) is __________ (upto 2 decimal places)

Solution:

*Answer can only contain numeric values
QUESTION: 58

An silver wire having a cross-sectional area o f 5.25 x 10-6 m2 carries a current of 10A. The average distance travelled by an electron in 2 min i s _______________ (Assume each silver atom contributes 1 electron to the current).
Density of silver = 10.5 g/cm3 Molar mass of silver = 107.87 g. Avogadro number = 6.02 x 1023

Solution:

*Answer can only contain numeric values
QUESTION: 59

The variance of the number of heads in three tosses of a coin is _____ [answer upto two decimal places]

Solution:

Number of tosses = n = 3
S = (HHH. HHT. HTH. HTT. THH. THT. TTH.TTT)
Let X = number of heads.
Possible values of X are 0.1.2. 3
p = Probability ot getting head in one toss = 1/2
q = Probability oi getting tail in one toss =1/2

*Answer can only contain numeric values
QUESTION: 60

If the 7th term from beginning and 7th term from end in the expansion of  are equal the value of n is _______ [Answer ni nearest integer]

Solution: