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


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30 Questions MCQ Test CSIR NET Exam Mock Test Series 2025 - CSIR NET Life Sciences Mock Test - 7

CSIR NET Life Sciences Mock Test - 7 for UGC NET 2025 is part of CSIR NET Exam Mock Test Series 2025 preparation. The CSIR NET Life Sciences Mock Test - 7 questions and answers have been prepared according to the UGC NET exam syllabus.The CSIR NET Life Sciences Mock Test - 7 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 - 7 below.
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CSIR NET Life Sciences Mock Test - 7 - Question 1

Determine the ratio of volumes of cylinders c1 and c2 if their heights and radii are in the ratio 7 : 9 and 3 : 4 respectively.

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

Volume of Cylinder =    (r → radius and h → height)

⇒ Volume of cylinder c1 = V1 =

⇒ Volume of cylinder c2 = V2 =

⇒ V1 : V2 = =

=

= ( × () = () × () =

∴ V1 : V2 = 7 : 16

CSIR NET Life Sciences Mock Test - 7 - Question 2

A student takes an 18-question multiple-choice exam, with four choices per question. Suppose one of the choices is obviously incorrect, and the student makes an 'educated' guess among the remaining three choices for each question. What is the expected number of correct answers?

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

The exam has 18 questions. For each question, the probability of a correct answer is 1/3 after eliminating one obviously incorrect option. Therefore, the expected number of correct answers is 18 × 1/3 = 6.

CSIR NET Life Sciences Mock Test - 7 - Question 3

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 3

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 4

Varshika purchases a dress for Rs. 3200 but the shop had another paying option of Rs. 1000 as a down payment and two equal monthly installments of Rs. 1200 each. Find the rate of interest in the installment scheme.

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

Given:
Cost = Rs. 3200
Down payment = Rs. 1000
Installments = Rs. 1200 each
Calculation:
Let the rate percent be R % per annum,
Total payment due to installment = 2400 + 1000
= 3400
Interest amount = 3400 − 3200 = 200
So, Principal for the 1 st month = Rs. (3200 − 1000)
= 2200
So, Simple interest for 1 st month 
Now, Principal for the 2nd month = Rs. (2200 − 1200)
= Rs. 1000
So, Simple interest for 2 nd month 
So, According to the question,

CSIR NET Life Sciences Mock Test - 7 - Question 5
Find the mean of the given data {a, b, a, a, b, a, b, c, a, b, a, c, a, b, a}, where a is less than b and b is less than c.
Detailed Solution for CSIR NET Life Sciences Mock Test - 7 - Question 5

Concept:

Mean

Calculations:

Sum of the given data

 

Number of values

Mean

CSIR NET Life Sciences Mock Test - 7 - Question 6

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 6

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

CSIR NET Life Sciences Mock Test - 7 - Question 7

Polythene chromosomes are found because of _____?

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

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 8

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

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

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 9

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 9

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 10

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 10

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 11

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

Detailed Solution for CSIR NET Life Sciences Mock Test - 7 - Question 11

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 12

Name the movement of the plant in response to a magnetic field.

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

Tactic movement is dependent on the external stimulus. On the basis of different stimuli, it is of many types, i.e., phototaxis, magnetotaxis, geotaxis, etc.

CSIR NET Life Sciences Mock Test - 7 - Question 13

Cell cycle is controlled by

P. certain cyclins

Q. certain cyclin-dependent kinases

R. certain inhibitory proteins

S. certain phosphatases 

Detailed Solution for CSIR NET Life Sciences Mock Test - 7 - Question 13

Cell cycle is controlled by certain cyclins, cyclin-dependent protein kinases, inhibitory proteins and phosphatases.

Cell cycle control is carried by cyclin-dependent kinases-Cdks (protien kinases). The raise or fall of these cyclins plays a key role in cell cycle. An increase activity of Cdks causes phosphorylation of intracellular protiens that regulate the major events of cell cycle such as DNA replication, mitosis and cytokinesis.

CSIR NET Life Sciences Mock Test - 7 - Question 14

Mendel could not find recombination and crossing over as:

Detailed Solution for CSIR NET Life Sciences Mock Test - 7 - Question 14

Mendel was not able to say anything about recombination and crossing over because traits he choose, were not linked and were present on different chromosomes or were far apart.

CSIR NET Life Sciences Mock Test - 7 - Question 15

In some flowering plants, the requirement of long day-length or low temperature treatment for flowering can be effectively substituted by:

Detailed Solution for CSIR NET Life Sciences Mock Test - 7 - Question 15

In some flowering plants, the requirement of long day-length or low temperature treatment for flowering can be effectively substituted by Gibberellins, while in short-day plants, they-inhibit flowering. Abscisic acid sometimes promotes flowering in short-day plants while inhibit in long-day plants (antagonistic to gibberellin). 2.4-D (2, 4 dichlorophenoxyacetic acid) are used as selective weed killers. They inhibit sprouting of potatoes. It also prevent premature fruit drop.

CSIR NET Life Sciences Mock Test - 7 - Question 16

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 16

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 17

The half-life of a radioisotope is:

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

The half-life of any radioisotope or radioactive element is the time taken for half of the atoms to undergo complete decay. The formula to find half-life is given by N(t) = N0e−λt.

CSIR NET Life Sciences Mock Test - 7 - Question 18

What is incorrect about the above figure representing DNA replication?


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

DNA polymerase can polymerize nucleotides only in direction on strand because it adds them at the end. Since, the two strands of DNA run in antiparallel directions, the two templates provide different ends for replication. Replication over the two templates thus proceeds in opposite directions. One strand with polarity forms its complementary strands continuously because end of the latter is open for elongation. It is called leading strand. Replication is discontinuous on the other template with polarity because only a short segment of DNA strand can be built in direction due to exposure of a small stretch of template at one time. Short segments of replicated DNA are called Okazaki fragments.

CSIR NET Life Sciences Mock Test - 7 - Question 19
Who is known as the father of Molecular biology?
Detailed Solution for CSIR NET Life Sciences Mock Test - 7 - Question 19

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 20
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 20

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 21

Run off transcription assays were performed to establish the specificity of three novel sigma factors for their promoters. Results of the experiments are shown below:

Following inferences were made from these results:
A) sA initiates transcription from P2 and sB from P1.
B) sC can initiate transcription from both promoters.
C) sB prevents initiation of transcription from P2.
D) sA initiates transcription from P1.
Choose the option that correctly interprets the results.

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

Sigma A initiate’s transcription from P2 and Sigma B from P1. Sigma C can initiate transcription from both promoters, Sigma B prevents initiation of transcription from P2

CSIR NET Life Sciences Mock Test - 7 - Question 22
Which one of the following biogeographical zones has the world’s richest goat and sheep community in the world?
Detailed Solution for CSIR NET Life Sciences Mock Test - 7 - Question 22

Biogeographic regions are those regions that are defined depending on the species found in them. India has 10 biogeographical regions.

Trans-Himalayan region: This region lies north of the great Himalayan range and is known for housing the world’s richest sheep and goat community.  Other animals found in this region are black-necked cranes, wolf, marmots, marbled cat, ibex, kiang, snow leopard, black bears and brown bears. Alpine steppe is the only flora found in the region.

Himalayan region: Alpine and deciduous forests are abundant in this region. Animals found in this region include: Hangul, musk deer, wild sheep, mountain goats, ibex, wolf, weasel, red panda.

Indian desert: Great Indian Bustard, one of the endangered species is found in this region. Large grasslands are characteristic of this region.

Semi-arid region: Several lakes and marshlands are found in this region. Asiatic lion, an endangered species is found in this region.

Western Ghats region: Extensive evergreen forests are found in this region. Nilgiri Langur and Lion-tailed Macaque are the endangered species found in this region. The Malabar grey hornbill is also found in this region.

Deccan peninsula region: This is the largest zone and it supports a lot of deer and antelope species. Some Asian elephants and water buffalos are also found in this region.

Gangetic plain region: One-horned Rhinoceros, Asian elephant, water buffalo, swamp deer, hog deer, hispid hare are found in this region. Sal and dry, deciduous forests are the predominant vegetation in this region.

Coasts: Mangrove forests are abundant in this region. Dungdong, Hump-back dolphin, turtles are found in this region.

North east region: rhinoceros, buffalo, elephant, swamp deer, hog deer, pygmy hog and hispid hare are the animals found here in this region. Alluvial grasslands are abundant in this zone.

Islands: Tropical evergreen forests are found in this region. Rodents are abundant.

CSIR NET Life Sciences Mock Test - 7 - Question 23
Ethylene receptor does not
Detailed Solution for CSIR NET Life Sciences Mock Test - 7 - Question 23

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 24

What are the possible phenotypes that can be observed after self-crossing violet flowered pea plants?

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

Violet is dominant over white. Self-crossing of violet flowering plants will produce 25% recessive plants, which will have white flowers.

CSIR NET Life Sciences Mock Test - 7 - Question 25

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

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

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 26

_____ removes urea In urecotelic animals.

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

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 27

Which of the following statement is not correct?
1. Six unusual bases are observed in a tRNA molecule.
2. Unusual bases pseudouridine and dihydropyridine are most commonly found in the tRNA molecules.

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

Only statement 1 is incorrect,

Five unusual bases, produced by enzymatic modification of the usual bases, may be observed in a tRNA molecule. These are pseudouridine, dihydropyridine, hypoxanthine, thymine and methylguanine. Of these five, unusual bases pseudouridine and dihydropyridine are most commonly found in the tRNA molecules.

CSIR NET Life Sciences Mock Test - 7 - Question 28
Light harvesting complexes are made up of:
Detailed Solution for CSIR NET Life Sciences Mock Test - 7 - Question 28

Light harvesting complexes (LHC) are made up of hundreds of pigment molecules bound to proteins. In LHC, reaction centre is formed by a single chlorophyll a molecule.

CSIR NET Life Sciences Mock Test - 7 - Question 29

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 29

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 30

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 30

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.

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