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CSIR NET Life Sciences Mock Test - 6 - UGC NET MCQ


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30 Questions MCQ Test CSIR NET Exam Mock Test Series 2025 - CSIR NET Life Sciences Mock Test - 6

CSIR NET Life Sciences Mock Test - 6 for UGC NET 2025 is part of CSIR NET Exam Mock Test Series 2025 preparation. The CSIR NET Life Sciences Mock Test - 6 questions and answers have been prepared according to the UGC NET exam syllabus.The CSIR NET Life Sciences Mock Test - 6 MCQs are made for UGC NET 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for CSIR NET Life Sciences Mock Test - 6 below.
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CSIR NET Life Sciences Mock Test - 6 - Question 1

There is a digit at the 10th place of 6n, in which n is a natural number that is greater than 1, it can be:

Detailed Solution for CSIR NET Life Sciences Mock Test - 6 - Question 1

Given:

There is a digit at the 10th place of 6n.

Where n is a natural number that is greater than 1.

Calculation:

Let be n= 2, ,3, 4, 5,...

⇒ 6 × 2 = 12 = 1

⇒ 6 × 3 = 18 = 1

⇒ 6 × 4 = 24 = 2

⇒ 6 × 5 = 30 = 3

⇒ 6 × 6 = 36 = 3

⇒ 6 × 7 = 42 = 4

⇒ 6 × 8 = 48 = 4

⇒ 6 × 9 = 54 = 5

∴ It can be 1, 2, 3, 4, 5.

CSIR NET Life Sciences Mock Test - 6 - Question 2

5 coins are tossed simultaneously. Find the probability of exactly 3 heads?

Detailed Solution for CSIR NET Life Sciences Mock Test - 6 - Question 2

Given:
The number of coins tossed is 5 coins.

Formula used:


The number of ways for obtaining three heads from five trials
Total number of outcomes from the trial of five coins  = 25 = 32
The required probability = 10/32 = 5/16

∴ The probability of exactly 3 heads is 5/16.

CSIR NET Life Sciences Mock Test - 6 - Question 3

Find the area of a square, one of whose diagonals is 3.8 m long.

Detailed Solution for CSIR NET Life Sciences Mock Test - 6 - Question 3

Given,

Length of diagonal  = 3.8 m

Area of the square  
= 7.22 m2

CSIR NET Life Sciences Mock Test - 6 - Question 4

Which of the following is known as the “window of the brain”?

Detailed Solution for CSIR NET Life Sciences Mock Test - 6 - Question 4

Sensory organs are considered as the “window of the brain”. 

When external stimuli contacted with sensory receptors, it sends signals to the brain in the form of nerve impulse which results in the sensation.

CSIR NET Life Sciences Mock Test - 6 - Question 5

C-value in genome represents___________.

Detailed Solution for CSIR NET Life Sciences Mock Test - 6 - Question 5

The C-value is the amount of DNA in the haploid genome of an organism. It varies over a very wide range, with a general increase in C-value with complexity of organism from prokaryotes to invertebrates, vertebrates, plants.

CSIR NET Life Sciences Mock Test - 6 - Question 6

KCI  (100 mM ) was entrapped inside large unilamellar vesicles. A diffusion potential across the bilayer can be generated by diluting with buffer containing :

Detailed Solution for CSIR NET Life Sciences Mock Test - 6 - Question 6

innism that has peroxidase and superoxide dismutase but lacks catalase is most likely anenergrhed Ause- loumm Mall and a k especific ia'nophore. It has been used exten Tensively to Obtain measure of al in Biological system, from a differene in the permeability of soga Be Livets to Mat and kt Inldich creáte a diffusion potential ir ke leaks out faster than Na boat in that they should be compact and akt specifie l'onoplose.

CSIR NET Life Sciences Mock Test - 6 - Question 7

Which of these is the primary carbon dioxide acceptor in the Hatch and Slack pathway?

Detailed Solution for CSIR NET Life Sciences Mock Test - 6 - Question 7

The primary carbon dioxide acceptor in the Hatch and Slack pathway is PEP or phosphoenol pyruvate. PEP is present in the mesophyll cells. Phosphoenol pyruvate is a 3-carbon molecule. The carboxylation takes place with the help of an enzyme called PEP carboxylase. They yield a 4-C molecule called oxaloacetic acid (OAA).

CSIR NET Life Sciences Mock Test - 6 - Question 8

Which of the following is a signaling molecule for bacteria?

Detailed Solution for CSIR NET Life Sciences Mock Test - 6 - Question 8

Homoserine lactones are the signaling molecules bacteria use in cell-cell interactions. These help in their growth and metabolism.

CSIR NET Life Sciences Mock Test - 6 - Question 9

Name the hormone which takes part in the release of FSH and LH from the anterior pituitary.

Detailed Solution for CSIR NET Life Sciences Mock Test - 6 - Question 9

GnRH is a Gonadotropin-Releasing Hormone, which is responsible for the release of follicle-stimulating hormone (FSH) and luteinizing hormone (LH).

CSIR NET Life Sciences Mock Test - 6 - Question 10

Three-dimensional images of the surface of the cells and tissues could be visualized through:

Detailed Solution for CSIR NET Life Sciences Mock Test - 6 - Question 10

Three-dimensional images of the surface of the cells and tissues could be visualized through scanning electron microscope. Transmission electron microscope produces two-dimensional image.

The scanning electron microscope (SEM) is a type of electron microscope which, by scanning with a high-energy electron beam, produces images of an object. The electron beam's behavior stimulates the emission from the surface of the specimen of high-energy back-scattered electrons and low-energy secondary electrons. A beam of electrons travels back and forth over the surface of a cell or tissue during scanning electron microscopy (SEM), producing a clear picture of the 3D surface.

CSIR NET Life Sciences Mock Test - 6 - Question 11

Bacteria that are resistant to penicillin and related antibodies produce an enzyme that breaks the__________ in antibiotics.

Detailed Solution for CSIR NET Life Sciences Mock Test - 6 - Question 11

Bacteria that are resistant to penicillin and related antibiotics produce an enzyme that breaks the Beta-lactam ring in antibiotics. β-lactam antibiotics (beta-lactam antibiotics) are antibiotics that contain a beta-lactam ring in their molecular structure. This includes penicillin derivatives (penams), cephalosporins (cephems), monobactams, carbapenems, and carbacephems. Most β-lactam antibiotics work by inhibiting cell wall biosynthesis in the bacterial organism and are the most widely used group of antibiotics.

CSIR NET Life Sciences Mock Test - 6 - Question 12

External genital organs are developed during:

Detailed Solution for CSIR NET Life Sciences Mock Test - 6 - Question 12

Limbs and digits, along with external genitalia, are formed during the first trimester. This is a period of 12 weeks after pregnancy.

CSIR NET Life Sciences Mock Test - 6 - Question 13

Name the red light-sensitive system.

Detailed Solution for CSIR NET Life Sciences Mock Test - 6 - Question 13

There are two light-sensitive systems involved in photomorphogenesis. One is the red light response and the other one is the blue light response. The red light-sensitive system is known as phytochrome while the blue light system is called cryptochrome.

CSIR NET Life Sciences Mock Test - 6 - Question 14

____ is the volume of air breathed in and out during normal breathing.

Detailed Solution for CSIR NET Life Sciences Mock Test - 6 - Question 14

Tidal volume is the volume of air breathed in and out during normal breathing.

Volume of air breathed in or out during normal breathing is called as tidal volume. Tidal volume is the lung volume representing the normal volume of air displaced between normal inhalation and exhalation when extra effort is not applied. In a healthy, young human adult, tidal volume is approximately 500 mL per inspiration or 7 mL/kg of body mass.

CSIR NET Life Sciences Mock Test - 6 - Question 15
Which of the following is DNA made up of?
Detailed Solution for CSIR NET Life Sciences Mock Test - 6 - Question 15

DNA is a linear molecule made up of four different types of nucleotide bases: adenine (A), cytosine (C), guanine (G), and thymine (T) (T). The DNA sequence is the order in which these bases appear. Human DNA consists of about 3 billion bases, and more than 99 percent of those bases are the same in all people.

CSIR NET Life Sciences Mock Test - 6 - Question 16
Which of the following does not affect the rate of diffusion?
Detailed Solution for CSIR NET Life Sciences Mock Test - 6 - Question 16

Diffusion is a passive process and it doesn’t require energy in the form of ATP. Temperature, Pressure and Concentration gradient do affect the rate of diffusion to varying degrees.

Diffusion is a result of the random movement of particles, hence there is a gross flow of matter from an area of higher concentration to an area of lower concentration. A good example of this is the dispersing of air particles by perfume is sprayed, permeating the still air of a room.

CSIR NET Life Sciences Mock Test - 6 - Question 17
C-value in genome represents___________.
Detailed Solution for CSIR NET Life Sciences Mock Test - 6 - Question 17

The measure of DNA display in the genome of animal varieties is known as a C-value, which is characteristics of every species.

C-value is the amount, in picograms, of DNA contained within a haploid nucleus (e.g. a gamete) or one half the amount in a diploid somatic cell of a eukaryotic organism. 

CSIR NET Life Sciences Mock Test - 6 - Question 18

Which of the following contain ionic, co-ordinate and covalent bonds?

Detailed Solution for CSIR NET Life Sciences Mock Test - 6 - Question 18

NaCN contain ionic, co-ordinate and covalent bonds. 
Ionic Bond, also called the electrovalent bond, type of linkage formed from the electrostatic attraction between oppositely charged ions in a chemical compound.
Covalent Bond: In these types of bonding sharing of electrons between two atoms of the same or different elements. This type of bonding may be single, double or triple depending upon the number of sharing pairs of electrons.
Co-ordinate Bond: In this type one side sharing of one pair of electrons between two atoms happens. Here one atom which provides electron is known as a donor and another one is known which receive electron is known as acceptor.  

CSIR NET Life Sciences Mock Test - 6 - Question 19
Where are pattern recognition receptors PRRs located?
Detailed Solution for CSIR NET Life Sciences Mock Test - 6 - Question 19

Pattern Recognition Receptors (PRRs) are proteins capable of recognizing molecules frequently found in pathogens (the so-called pathogen-associated Molecular Patterns—PAMPs), or molecules released by damaged cells (the Damage-Associated Molecular Patterns—DAMPs). Endocytic pattern-recognition receptors are found on the surface of phagocytes and promote the attachment of microorganisms to phagocytes leading to their subsequent engulfment and destruction. They include mannose receptors, scavenger receptors, and opsonin receptors.

CSIR NET Life Sciences Mock Test - 6 - Question 20
Shikimic acid pathway is the biosynthetic pathway of which of the following amino acid?
Detailed Solution for CSIR NET Life Sciences Mock Test - 6 - Question 20

Shikimic acid pathway is related to secondary metabolite synthesis of phenolic compounds like coumarins,Flavanoids. eg: of phenolics are Aflatoxins B1,scopoletin.

CSIR NET Life Sciences Mock Test - 6 - Question 21

Why biodiversity is of great scientific value?

Detailed Solution for CSIR NET Life Sciences Mock Test - 6 - Question 21

During the usage of many species for research and in turn we get a lot of knowledge on plants, insects and animals from this we find better ways of making medicines, hybrid plants, and many other things that are helpful to humans.

CSIR NET Life Sciences Mock Test - 6 - Question 22

DnaB helicase is required for the unwinding of DNA at the replication fork in E. coli. Some statements on the fidelity of DnaB are given below:

  1. Unwinding involves ATP hydrolysis because energy is required to melt DNA.
  2. DnaB helicase takes a ring-shaped conformation and dsDNA moves through the hole.
  3. SSB protein inhibits DnaB helicase mediated unwinding if added after DnaB has bound the template.
  4. DnaB helicase binds to the leading-strand template DNA

Which of the above statement(s) is CORRECT?

Detailed Solution for CSIR NET Life Sciences Mock Test - 6 - Question 22

Unwinding is an energy-consuming process because hydrogen bonds are kinetic barriers to strand separation, thus, it requires ATP hydrolysis. DnaB helicase forms a ring-shaped structure around the DNA strand to ease the movement across the replication fork. The third statement is incorrect because SSB protein inhibits DnaB helicase mediated unwinding if added before DnaB has bound the template because SSB binds to ssDNA and prevents binding of helicase. And DnaB binds to lagging-strand template DNA, not the leading strand.

CSIR NET Life Sciences Mock Test - 6 - Question 23

Which of the following is wrongly paired?

Detailed Solution for CSIR NET Life Sciences Mock Test - 6 - Question 23

Nucleic acids, that is, DNA and RNA show phosphodiesterase linkage which is the major type of linkage. Without the phosphodiester bonds between the phosphate and adjacent 3’OH sugar molecule, the backbone will not be formed. Therefore, the nucleotides would not be able to attach and bond to form a nucleic acid.

CSIR NET Life Sciences Mock Test - 6 - Question 24

During respiration, which of the following processes occur only inside mitochondria and not cytoplasm?

Detailed Solution for CSIR NET Life Sciences Mock Test - 6 - Question 24

The pyruvate formed from glucose is oxidatively decarboxylated to yield acetyl CoA under aerobic circumstances. The citric acid cycle reactions occur inside mitochondria in eukaryotes. The first step in the Kreb cycle's cellular respiration is glycolysis, which takes place in the cytoplasm of a cell. When glycolysis takes place, glucose is broken down into pyruvic acid in the cytoplasm. It can be found in both the mitochondrial matrix and the cytoplasm of eukaryotic and prokaryotic cells. 

CSIR NET Life Sciences Mock Test - 6 - Question 25

Early detection of a disease is possible by:

Detailed Solution for CSIR NET Life Sciences Mock Test - 6 - Question 25

Early detection of a disease is possible by PCR and Gene therapy.

For effective treatment of a disease, early diagnosis and understanding its pathophysiology is very important. Using conventional methods of diagnosis (serum and urine analysis, etc.) early detection is not possible. Recombinant DNA technology, Polymerase Chain Reaction (PCR) and Enzyme Linked Immuno-sorbent Assay (ELISA) are some of the techniques that serve the purpose of early diagnosis.

CSIR NET Life Sciences Mock Test - 6 - Question 26

Which of the following is a type of RNA involved in protein synthesis?
1. snRNA
2. rRNA
3. yRNA
4. dsRNA

Detailed Solution for CSIR NET Life Sciences Mock Test - 6 - Question 26

Protein synthesis requires three different forms of RNA. Messenger RNA (mRNA), transfer RNA (tRNA), and ribosomal RNA are the three types of RNA (rRNA).
The synthesis of a protein is governed by the information in its DNA, with the help of messengers (mRNA) and translators (tRNA). In the nucleus, DNA is transcribed to RNA. The mRNA carries the message out of the nucleus to the ribosome in the cytoplasm where the tRNA helps translate the message to make a protein.

CSIR NET Life Sciences Mock Test - 6 - Question 27
Which messenger molecules are derived from arachidonic acid?
Detailed Solution for CSIR NET Life Sciences Mock Test - 6 - Question 27

Eicosanoids are non-polar, 20-carbon containing molecules derived from the fatty acid, arachidonic acid. Eicosanoids regulate processes like blood pressure, inflammation, blood clotting. Several over-the-counter drugs used to treat headaches inhibit eicosanoid synthesis.

CSIR NET Life Sciences Mock Test - 6 - Question 28

Phylogeny is defined as the study of evolutionary lineages of a species, or taxa. It is studied by drawing phylogenetic trees. Following are the statements about them-
P). A phylogenetic tree is also known as cladogram that represents evolutionary relationships among organisms or taxa.
Q). The pattern of branching in a phylogenetic tree reflects how species or other groups evolved from a series of common ancestors.
R). In trees, two species are less related if they have a more recent common ancestor and more related if they have a less recent common ancestor.
S). Phylogenetic trees are hypotheses, not definitive facts.

Which of the following are correct?

Detailed Solution for CSIR NET Life Sciences Mock Test - 6 - Question 28

A phylogenetic tree also known as cladogram that represents evolutionary relationships among organisms or taxa. They are just hypotheses, not definitive facts. The pattern of branching in a phylogenetic tree reflects how species or other groups evolved from a series of common ancestors. In trees, two species are more related if they have a more recent common ancestor and less related if they have a less recent common ancestor.

CSIR NET Life Sciences Mock Test - 6 - Question 29

With respect to the “tails” of the histone molecules which of the following is not true?

Detailed Solution for CSIR NET Life Sciences Mock Test - 6 - Question 29

The “tail” of the histone is not required for the association for the DNA with the histone octamer into a nucleosome. This is proved when the nucleosome is treated with the protease, trypsin. Trypsin is known to cleave proteins after positively charged amino acid thus, when the N – terminal tail is removed no structural variation is observable in the nucleosome.

CSIR NET Life Sciences Mock Test - 6 - Question 30

The inactive conformation of a G-protein coupled receptor is stabilized by ______________

Detailed Solution for CSIR NET Life Sciences Mock Test - 6 - Question 30

The inactive conformation of the G-protein coupled receptor is stabilized by non-covalent interactions between the residues in the transmembrane alpha helices. Rotation or movements in these loops cause changes in the conformation of cytoplasmic loops.

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