CSIR NET Life Science Exam  >  CSIR NET Life Science Tests  >  CSIR NET Life Sciences Mock Test - 7 - CSIR NET Life Science MCQ

CSIR NET Life Sciences Mock Test - 7 - CSIR NET Life Science MCQ


Test Description

30 Questions MCQ Test - CSIR NET Life Sciences Mock Test - 7

CSIR NET Life Sciences Mock Test - 7 for CSIR NET Life Science 2024 is part of CSIR NET Life Science preparation. The CSIR NET Life Sciences Mock Test - 7 questions and answers have been prepared according to the CSIR NET Life Science exam syllabus.The CSIR NET Life Sciences Mock Test - 7 MCQs are made for CSIR NET Life Science 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for CSIR NET Life Sciences Mock Test - 7 below.
Solutions of CSIR NET Life Sciences Mock Test - 7 questions in English are available as part of our course for CSIR NET Life Science & CSIR NET Life Sciences Mock Test - 7 solutions in Hindi for CSIR NET Life Science course. Download more important topics, notes, lectures and mock test series for CSIR NET Life Science Exam by signing up for free. Attempt CSIR NET Life Sciences Mock Test - 7 | 75 questions in 180 minutes | Mock test for CSIR NET Life Science preparation | Free important questions MCQ to study for CSIR NET Life Science Exam | Download free PDF with solutions
CSIR NET Life Sciences Mock Test - 7 - Question 1

Kanak ranks ninth from the top and from the bottom in a class. How many students are there in the class?

Detailed Solution for CSIR NET Life Sciences Mock Test - 7 - Question 1

Kanak position from the top

Kanak position from the bottom

Total number of students in the class (Kanak position from the top) (Kanak position from the bottom)

Therefore, there are  students in the class.

CSIR NET Life Sciences Mock Test - 7 - Question 2

If 1/4, 1/x and 1/10 are in HP, then find the value of x.

Detailed Solution for CSIR NET Life Sciences Mock Test - 7 - Question 2

Given,
1/4, 1/x and 1/10 are in HP.
4, x and 10 is in AP.
x - 4 = 10 - x
x = 7

1 Crore+ students have signed up on EduRev. Have you? Download the App
CSIR NET Life Sciences Mock Test - 7 - Question 3

In _______ communication, transmission occurs through extracellular fluid and primarily limited to local area ?

Detailed Solution for CSIR NET Life Sciences Mock Test - 7 - Question 3

In paracrine communication, transmission occurs through extracellular fluid and primarily limited to local area.

A system called “paracrine signaling” allows cells to communicate with each other by releasing signaling molecules that bind to and activate surrounding cells.

CSIR NET Life Sciences Mock Test - 7 - Question 4

Polythene chromosomes are found because of _____?

Detailed Solution for CSIR NET Life Sciences Mock Test - 7 - Question 4

Endomitosis is the replication or duplication of the chromosome in the absence of nuclear division or cell, resulting in numerous copies within each cell, which occurs in the Drosophila salivary glands. 

Polytene chromosomes are popularly called salivary chromosomes. They contain 1000- 16000 times DNA, as compared to ordinary somatic chromosomes and can reach a length of 2000 μm. During the interphase stage of the cell division in the nuclei of the salivary gland cells of the larvae of Drosophila melanogaster, polytene chromosomes are formed due to endoreduplication, duplication without separation and replication of DNA without cell division. These chromosomes undergo somatic pairing to form identical chromosomes which are joined along their length to one another.

CSIR NET Life Sciences Mock Test - 7 - Question 5

Which of the following form of lipids are also referred as neutral lipids?

Detailed Solution for CSIR NET Life Sciences Mock Test - 7 - Question 5

A triglyceride (TG, triacylglycerol, TAG, or triacylglyceride) is an ester derived from glycerol and three fatty acids (from tri- and glyceride). Triglycerides are the main constituents of body fat in humans and other vertebrates, as well as vegetable fat. They are also present in the blood to enable the bidirectional transference of adipose fat and blood glucose from the liver, and are a major component of human skin oils.

CSIR NET Life Sciences Mock Test - 7 - Question 6

Which of the following has a major role in the regulation of chloroplast movement?

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

Phototoropin acts as a blue light receptor, it is a flavoprotein which has a major role in regulation of chloroplast movement and photropism.
Chloroplasts accumulate in areas irradiated with weak light to increase photosynthetic efficiency.

CSIR NET Life Sciences Mock Test - 7 - Question 7

Which of the following is the best breeding method for animals which are below average in productivity?

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

Out-crossing is the practice of mating of animals within the same breed, but having no common ancestors on either side of their pedigree. The offspring of such a mating is known as an outcross. It is the best breeding method for animals that are below average in productivity in milk production, the growth rate in beef cattle, etc.

CSIR NET Life Sciences Mock Test - 7 - Question 8

Which of these functions is not regulated by intracellular hormone-receptor complexes?

Detailed Solution for CSIR NET Life Sciences Mock Test - 7 - Question 8

Intracellular hormone-receptor complexes transmit signals mainly to the nucleus for the regulation of gene expression, and regulation of chromosome function through indirect interaction with the genome.

CSIR NET Life Sciences Mock Test - 7 - Question 9

If a four nucleotide sequence codes for an amino acid instead of three theoretically how many unique amino acids could be coded by such a system:

Detailed Solution for CSIR NET Life Sciences Mock Test - 7 - Question 9

The central dogma of molecular biology states that the flow of genetic information takes place from DNA to RNA to proteins. The transcription of DNA to RNA is based on the complementarity of their strands. However, no such complementarity exists between the RNA and the proteins for translation to take place. Evidence shows that changes in nucleic acids also caused changes in the amino acids of the proteins. This suggested the existence of a Genetic Code that directs the synthesis of amino acids that form the proteins. It was George Gamow who suggested that if the 4 nitrogen bases of nucleic acids have to code for the 20 non-essential amino acids produced in the body, the code should constitute a combination of bases. The nitrogen bases include Adenine (A), Guanine (G), Cytosine (C), and Uracil (U) for an RNA. A permutation combination of 2 bases would give 42 = 16 amino acids only.

So, it has to be a combination of 3 bases giving: 43 = 64 codons.

From the above points, we can get the formula as (No. of N-bases available)No. of bases in per sequence code = No. of codons

Therefore, if we take 4 nucleotide sequence codes,

No. of codons = 44 = 256

  • Similarly, if we take a 5 sequence code, we will get 45 = 1024 codons.
  • Please note that these are only theoretical possibilities.

CSIR NET Life Sciences Mock Test - 7 - Question 10

If VV produces violet flowers and vv produces white flowers, what will be the phenotype and genotype of the F1 progeny?

Detailed Solution for CSIR NET Life Sciences Mock Test - 7 - Question 10

The VV and vv upon crossing will produce all progeny with genotype Vv. This is the heterozygous state. Violet being dominant over white, all the progeny will have violet flowers.

CSIR NET Life Sciences Mock Test - 7 - Question 11
Who is known as the father of Molecular biology?
Detailed Solution for CSIR NET Life Sciences Mock Test - 7 - Question 11

Dr. Max Perutz is known as the father of Molecular biology.

Dr. Max Perutz, whose success in elucidating the structure of the hemoglobin molecule helped give birth to the field of molecular biology and brought him the Nobel Prize in Chemistry in 1962, died on Wednesday at a hospital near his home in Cambridge, England. He was 87.

CSIR NET Life Sciences Mock Test - 7 - Question 12

Where are the parental genotypes mentioned in a Punnett square?

Detailed Solution for CSIR NET Life Sciences Mock Test - 7 - Question 12

In a Punnett square, the parental genotypes are represented in the left column and the top row of the square.

CSIR NET Life Sciences Mock Test - 7 - Question 13
Based on which of the following, the neurons are divided into three major types?
Detailed Solution for CSIR NET Life Sciences Mock Test - 7 - Question 13

Based on the number of axons and dendrites the neurons are divided into three major types. 

  • Multipolar neurons-with one axon and two or more dendrite
  • Bipolar-with one axon and one Dendrite
  • Unipolar-cell body with one axon only

CSIR NET Life Sciences Mock Test - 7 - Question 14
High concentration of glucose 6-phosphate is inhibitory to ___________
Detailed Solution for CSIR NET Life Sciences Mock Test - 7 - Question 14

High concentration of glucose 6-phosphate is inhibitory to Hexokinase.

Hexokinase is the initial enzyme of glycolysis, catalyzing the phosphorylation of glucose by ATP to glucose-6-P. It is one of the rate-limiting enzymes of glycolysis. Its activity declines rapidly as normal red cells age. Hexokinase catalyzes the reaction involving conversion of glucose to glucose 6-phosphate. 

CSIR NET Life Sciences Mock Test - 7 - Question 15
The three most common cell junctions in animals are:
Detailed Solution for CSIR NET Life Sciences Mock Test - 7 - Question 15

The three most common cell junctions in animals are: adhesive junctions, tight junctions and gap junctions. Tight junction made up of claudins and occludins make the closest contact between the adjacent cells to prevent free passage of molecules. Gap junctions are channels present between cell membranes that allow ions and small molecules to pass directly from one cell to another cell. These channels are primarily formed of membrane proteins called connexins. Adhesive junctions attach cells to other cell or ECM and helps in holding tissues together. Whereas plasmodesmata is present only in plants not in animal cells.

CSIR NET Life Sciences Mock Test - 7 - Question 16
Ethylene receptor does not
Detailed Solution for CSIR NET Life Sciences Mock Test - 7 - Question 16

Ethylene receptor is evolutionary related to bacteria's two-component histidine kinases. Histidine kinases are found in most bacteria, archaea, and lower eukaryotic species such as slime molds, fungi, and plants where they function in two-component signal transduction pathways. Two proteins are crucial for interacting ethylene with the receptors, namely constitutive triple response 1 (CTR1) and ethylene insensitive 2 (EIN2) and Ethylene receptor is active in the absence of ethylene, hence it is negative regulator that represses the hormone response in the absence of the hormone.

Ethylene receptor does not involves the phosphorylation relay system instead ethylene binding to the receptor disrupts the EIN2 phosphorylation.

CSIR NET Life Sciences Mock Test - 7 - Question 17

Coagulation of blood in blood vessels is prevented by:

Detailed Solution for CSIR NET Life Sciences Mock Test - 7 - Question 17

Heparin works in both in-vivo & in-vitro conditions, thus heparin is to be added to the blood sample drawn from the patient for analysis of blood corpuscles and plasma.

  • It is secreted by the basophil cells
  • Heparin is the most powerful anticoagulant. It activates the Antithrombin III present in the blood plasma
  • The antithrombin binds and deactivates the serum clotting factors, this prevents the clotting of blood.

CSIR NET Life Sciences Mock Test - 7 - Question 18

The β cells in the pancreas secrete insulin. What do α cells secrete?

Detailed Solution for CSIR NET Life Sciences Mock Test - 7 - Question 18

The pancreas is a compound (both exocrine and endocrine) elongated organ situated between the limbs of the ‘C’ shaped duodenum. 99% part of the pancreas is exocrine while only 1% part is endocrine. The exocrine portion (acini cells) secretes an alkaline pancreatic juice containing enzymes and the endocrine portion (islets of Langerhans) secretes hormones, insulin, and glucagon. islets of Langerhans have 2 types of cells α cells & β cells, the α cells surround the β cells. Insulin is secreted by the β cells, while α cells secrete glucagon.

Both insulin & glucagon are very important for sugar metabolism. Insulin is released into the bloodstream to lower down blood sugar levels and prevent hyperglycemia. Glucagon is released into the bloodstream to prevent blood sugar levels from dropping too low (hypoglycemia)

Oxytocin: It is an endocrine gland secretion. It is also known as the birth hormone. It stimulates uterine contraction during the time of childbirth.

Thyroxine: It is an endocrine secretion of the thyroid gland. The thyroid gland requires 120 micrograms of Iodine per day for the production of thyroxine It regulates the basal metabolic rate of the body.

Vasopressin: Also called Antidiuretic hormone secreted by the posterior pituitary gland. It mainly acts on the kidney and stimulates the reabsorption of water by the distal convoluted tubules and thereby reducing the loss of water through urine (Diuresis) so-called antidiuretic hormone.

CSIR NET Life Sciences Mock Test - 7 - Question 19

In which of the following viruses, the flow of information is opposite ie, RNA to DNA?

Detailed Solution for CSIR NET Life Sciences Mock Test - 7 - Question 19

In HIV viruses, the flow of information is opposite ie, RNA to DNA.

In the infected person's body, the HIV enters into macrophages where the RNA genome of the virus replicates to form viral DNA with the help of the enzyme reverse transcriptase. This viral DNA gets incorporated into the host cell’s DNA and directs the infected cells to produce virus particles. The macrophages continue to produce viruses and in this way acts as an HIV factory.

CSIR NET Life Sciences Mock Test - 7 - Question 20

Which of the following is the most reliable method of soil remediation?

Detailed Solution for CSIR NET Life Sciences Mock Test - 7 - Question 20

Bio-remediation is the most efficient method as it uses biological organisms which is also environment friendly, Chemical method requires comparatively higher temperature and pressure, lyophilization is a process of freeze drying, and Bio-magnification is the process by which toxic substances get deposited in the food chain.

CSIR NET Life Sciences Mock Test - 7 - Question 21

_____ removes urea In urecotelic animals.

Detailed Solution for CSIR NET Life Sciences Mock Test - 7 - Question 21

Ornithine cycle removes urea In urecotelic animals.

In ureotelic animals, urea is formed by Ornithine cycle. It helps in conversion of excess of amino acids into urea in the liver. It takes place in five major steps. The extra proteins in the body are degraded by the process of deamination where the  group is removed and it is converted into ammonium ions in the liver. The ammonium ions enter urea cycle and combines with carbon dioxide and get converted into urea. The urea formed in liver is transported to kidney through circulation and finally gets excreted. So, the correct answer is 'Ornithine cycle'.

CSIR NET Life Sciences Mock Test - 7 - Question 22
G-protein coupled receptors contain ________ transmembrane alpha helices.
Detailed Solution for CSIR NET Life Sciences Mock Test - 7 - Question 22

The G-protein coupled receptors (GPCRs) are a family of cell surface receptors that consist of 7 transmembrane alpha helices. These receptors translate the binding of extracellular messengers to the activation of intracellular GTP-binding proteins.

CSIR NET Life Sciences Mock Test - 7 - Question 23
The sequence of DNA from where replication starts is called _______.
Detailed Solution for CSIR NET Life Sciences Mock Test - 7 - Question 23

The sequence of DNA from where the replication starts is called the origin of replication. It is also written as ‘ori’. It is also responsible for controlling the copy number of target DNA. Bacteria usually have only one origin of replication.

CSIR NET Life Sciences Mock Test - 7 - Question 24

The net reproductive rate (R0) is 1.5 for a given population. If Nt, the population of females at generation t, is 500, then what will be the population of females after four generations (Nt+4)?

Detailed Solution for CSIR NET Life Sciences Mock Test - 7 - Question 24

net reproductive rate (given),

Now apply the formula-

Where, population at time

population at initial time

Put all the values in formula,

=2531.250

CSIR NET Life Sciences Mock Test - 7 - Question 25

The terms expressing some of the developmental events or specific body structures are given in column X and the names of animals that are associated with them in column Y:

The correct match of the terms in column X with the name of animals in column Y is:

Detailed Solution for CSIR NET Life Sciences Mock Test - 7 - Question 25

1. Torsion (apple snail) occurs in gastropods which is class including snails and slugs. 
2. Metagenesis (Obelia) is observed in cnidarians and Obelia belongs to phylum cnidaria. 
3. Apolysis (Taenia) is the separation of the cuticle from the epidermis observed in arthropods and related groups.
4. Pedicellaria (starfish) is a defensive organ like a minute pincer present in large numbers on an echinoderm.

CSIR NET Life Sciences Mock Test - 7 - Question 26

During wing development in chick, if Apical Ectodermal Ridge (AER) is removed, the limb development ceases, on the other hand placing leg mesenchyme directly beneath the wing AER, distal hindlimb structures develop at the end of the wing, and if limb mesenchyme is replaced by non-limb mesenchyme beneath AER, the AER regresses. This may demonstrate that:
(A) the limb mesenchyme cells induce and sustain AER.
(B) the mesenchyme cells specify the type: wing or limb.
(C) the AER is responsible for specifying the type: wing or limb.
(D) the AER is responsible for sustained outgrowth and development of the limb
(E) the AER does not specify the type of wing or limb

Which combination of the above statements is demonstrated by the experiment?

Detailed Solution for CSIR NET Life Sciences Mock Test - 7 - Question 26

The proximal-distal growth and differentiation of the limb bud are made possible by a series of interactions between the limb bud mesenchyme and the AER. Although the mesenchyme cells induce and sustain the AER and determine the type of limb to be formed, the AER is responsible for the sustained outgrowth and development of the limb. The AER keeps the mesenchyme cells directly beneath it in a state of mitotic proliferation and prevents them from forming cartilage. if they cut away a small portion of the AER in a region that would normally fall between the digits of the chick leg, an extra digit emerged at that place. Statement A, B D, and E are correct according to the explanation.

CSIR NET Life Sciences Mock Test - 7 - Question 27

What are the complex organic remains such as dead animal remains, dead plant remains, and fecal matter called?

Detailed Solution for CSIR NET Life Sciences Mock Test - 7 - Question 27

Detritus are the complex organic remains such as dead animal remains, dead plant remains, and fecal matter. 

The organic remains such as dead plants remain, dead animal remains and fecal matter is called detritus. It acts as the raw material for various organisms such as decomposers. In terrestrial ecosystems, it is called “soil organic matter” while in the aquatic system it is known as “marine snow”. 

CSIR NET Life Sciences Mock Test - 7 - Question 28

Which of these features cannot help us differentiate between C3 and C4 plants?

Detailed Solution for CSIR NET Life Sciences Mock Test - 7 - Question 28

C4 plants show Kranz anatomy in which bundle sheath cells surrounds the vascular bundles. They lack photorespiration and show presence of bundle sheaths. However, chloroplasts are present in both types of plants.

C3 plants are defined as plants that exhibit the C3 pathway. These plants use the Calvin cycle in the dark reaction of photosynthesis. The leaves of C3 plants do not show Kranz anatomy. Here the photosynthesis process takes place only when the stomata are open. Approximately 95% of the shrubs, trees, and plants are C3 plants.

On the other hand, C4 plants are defined as the plants that use the C4 pathway or Hatch-Slack pathway during the dark reaction. The leaves possess kranz anatomy, and the chloroplasts of these plants are dimorphic. About 5% of plants on earth are C4 plants.

CSIR NET Life Sciences Mock Test - 7 - Question 29

Which of the following is Edman's reagent?

Detailed Solution for CSIR NET Life Sciences Mock Test - 7 - Question 29

The Edman reagent, phenyl isothiocyanate reacts with the amine group of the N-terminal amino acid. Phenyl isothiocyanate is an isothiocyanate having a phenyl group attached to the nitrogen used for amino acid sequencing in the Edman degradation. It has a role as an allergen and a reagent.

CSIR NET Life Sciences Mock Test - 7 - Question 30

Which of the following options correctly represents the lung conditions in asthma and emphysema, respectively ?

Detailed Solution for CSIR NET Life Sciences Mock Test - 7 - Question 30

Inflammation of bronchioles; Decreased respiratory surface correctly represents the lung conditions in asthma and emphysema, respectively.

Asthma is a difficulty in breathing causing wheezing due to inflammation of bronchi and bronchioles. Emphysema is a chronic disorder in which alveolar walls are damaged due to which respiratory surface is decreased.

View more questions
Information about CSIR NET Life Sciences Mock Test - 7 Page
In this test you can find the Exam questions for CSIR NET Life Sciences Mock Test - 7 solved & explained in the simplest way possible. Besides giving Questions and answers for CSIR NET Life Sciences Mock Test - 7, EduRev gives you an ample number of Online tests for practice
Download as PDF