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CUET PG Zoology Mock Test - 2 - CUET PG MCQ


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30 Questions MCQ Test - CUET PG Zoology Mock Test - 2

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CUET PG Zoology Mock Test - 2 - Question 1

Following statements are with reference to oogonia. Which of these is incorrect?

Detailed Solution for CUET PG Zoology Mock Test - 2 - Question 1

The correct answer is Number of oogonia grows especially during puberty.

  • Oogonia are the earliest stage of female germ cells in the ovaries that eventually develop into mature oocytes or eggs.
  • They are crucial for female fertility and are formed during fetal development.
  • Oogonia undergo mitotic divisions to increase their number before entering meiosis to become primary oocytes.

Option 1: Common fate of oogonia is death by apoptosis.

  • A significant number of oogonia undergo apoptosis (programmed cell death) during fetal development and after birth. This natural process ensures that only a small number of viable oocytes remain in the ovaries.

Option 2: Oogonia are able to undergo mitotic division.

  • Oogonia undergo several rounds of mitotic division to increase their number before transitioning into primary oocytes.

Option 3: Number of oogonia grows especially during puberty.

  • This statement is incorrect because the maximum number of oogonia is reached during fetal development, not puberty. By the time of birth, the number of oogonia has already started to decline due to apoptosis.
  • After birth, no new oogonia are formed. At puberty, the process that resumes is the maturation of the remaining primary oocytes, not an increase in the number of oogonia.
  • The number of oocytes a female is born with declines steadily throughout her life until menopause.

Option 4: Oogonia are diploid cells.

  • Oogonia are diploid (2n) cells, meaning they have two sets of chromosomes.
CUET PG Zoology Mock Test - 2 - Question 2

According to 2014 IUCN Red List, which of the following vertebrate classes has the largest percentage of threatened species?

Detailed Solution for CUET PG Zoology Mock Test - 2 - Question 2

The correct answer is Amphibians

Concept:

  • The IUCN Red List of Threatened Species, established in 1964, is the world's most comprehensive inventory of the global conservation status of biological species.
  • It uses a set of criteria to evaluate the extinction risk of thousands of species and subspecies.
  • The main goal is to convey the urgency of conservation issues to the public and policy makers, as well as to help the international community reduce species extinction.

Explanation:

  • Amphibians: Amphibians have the largest percentage of threatened species according to the 2014 IUCN Red List. This is due to various factors such as habitat loss, pollution, climate change, and diseases like chytridiomycosis. Amphibians are highly sensitive to environmental changes, making them more vulnerable to threats.
  • Mammals: While many mammal species are also threatened, the percentage is lower compared to amphibians. Major threats to mammals include habitat destruction, hunting, and climate change.
  • Birds: Birds face threats such as habitat loss, climate change, and invasive species. Conservation efforts have helped some species recover, but they still do not have the highest percentage of threatened species among vertebrates.
  • Reptiles: Reptiles are threatened by habitat destruction, climate change, and the pet trade. However, the percentage of threatened reptile species is not as high as that of amphibians.
CUET PG Zoology Mock Test - 2 - Question 3

Which of the options correctly matches the proteins involved in transcription (Column A) with the DNA binding domains they carry (Column B)?

Detailed Solution for CUET PG Zoology Mock Test - 2 - Question 3

The correct answer is Option 4 i.e. A - ii, B - iii, C - iv, D - i
TFIIIA : Zinc finger

  • Zinc fingers are small protein domains that bind to DNA via zinc ions.
  • TFIIIA is a transcription factor that binds to the promoter of the 5S rRNA gene in eukaryotes. It is essential for the initiation of transcription of the 5S rRNA gene.
MyoD : Helix-loop-helix
  • Helix-loop-helix domains are protein domains that consist of two alpha helices connected by a loop.
  • MyoD is a transcription factor that is involved in the regulation of muscle gene expression. It binds to DNA via its helix-loop-helix domain.
Jun : Leucine zipper
  • Leucine zippers are protein domains that consist of a series of leucine residues that form a coiled-coil structure.
  • Jun is a transcription factor that is involved in the regulation of cell growth and differentiation. It binds to DNA via its leucine zipper domain.
Cro: Helix-turn-helix
  • Helix-turn-helix domains are protein domains that consist of two alpha helices connected by a short turn of amino acids.
  • Cro is a transcription factor that is involved in the regulation of the lytic-lysogenic switch in bacteriophage lambda. It binds to DNA via its helix-turn-helix domain.
Conclusion- Therefore, the correct answer is A - ii, B - iii, C - iv, D - i.
CUET PG Zoology Mock Test - 2 - Question 4

Given below are larval stages and phylum to which they belong. Select the INCORRECT combination

Detailed Solution for CUET PG Zoology Mock Test - 2 - Question 4

The correct answer is Tornasia and Axolotl- Phylum Hemichordata.

1. Parenchymula and Amphiblastula - Phylum Porifera

  • Parenchymula and Amphiblastula are larval stages of sponges that belong to the phylum Porifera.
  • Parenchymula is a solid larval form, and Amphiblastula is a hollow larval form characteristic of sponges.
  • Therefore, this combination is correct.

2. Bipinnaria and Auricularia - Phylum Echinodermata

  • Bipinnaria and Auricularia are both larval stages in the phylum Echinodermata.
  • Bipinnaria is a larval stage of starfish (Asteroidea), and Auricularia is a larval stage of sea cucumbers (Holothuroidea).
  • Therefore, this combination is correct.

3. Tornaria and Axolotl - Phylum Hemichordata

  • Tornaria is the larval stage of Hemichordata, particularly of acorn worms (Enteropneusta). However, Axolotl is not a larval form related to Hemichordata.
  • Axolotl is a larval stage of certain species of salamanders, which belong to the class Amphibia in the phylum Chordata.
  • Therefore, this combination is incorrect.

4. Planula and Ephyra - Phylum Cnidaria

  • Planula and Ephyra are both larval stages of Cnidaria.
  • Planula is a free-swimming larva characteristic of many Cnidarians, and Ephyra is a juvenile form of jellyfish (class Scyphozoa) in the same phylum.
  • Therefore, this combination is correct.

Conclusion:
The incorrect combination is Tornaria and Axolotl - Phylum Hemichordata. Axolotl is not a Hemichordate larva but an amphibian larva, making this combination incorrect.

CUET PG Zoology Mock Test - 2 - Question 5

In a sucrose density gradient, what is the order of organelle sedimentation from lower to higher concentration of sucrose?

Detailed Solution for CUET PG Zoology Mock Test - 2 - Question 5

The correct answer is Golgi, smooth endoplasmic reticulum, rough endoplasmic reticulum.
Sucrose density gradient centrifugation is a technique used to separate cellular organelles based on their size and density. In this method, cellular components are placed in a tube with a gradient of sucrose concentrations, and then centrifuged. Organelles will sediment at different rates, allowing them to be separated from each other.

  • When cell homogenates are subjected to centrifugation through a sucrose density gradient, the organelles sediment according to their buoyant densities.
  • The Golgi apparatus, due to its less compact structure and larger vesicles, stays at higher (less dense) regions of the gradient.
  • The smooth endoplasmic reticulum, having fewer ribosomes and a more tubular structure, sediments to the middle.
  • The rough endoplasmic reticulum, with its associated ribosomes, exhibits the highest density among the three organelles given and thus migrates to the regions of the gradient with higher sucrose concentration.

Therefore, the correct order of organelle sedimentation from lower to higher concentration of sucrose is Golgi, smooth endoplasmic reticulum, rough endoplasmic reticulum. The Golgi apparatus has a lower density, followed by the smooth endoplasmic reticulum, and then the rough endoplasmic reticulum which has the highest density due to the presence of ribosomes on its surface.

CUET PG Zoology Mock Test - 2 - Question 6

Which of the following molecule is NOT found in the Eosinophilic granule ?

Detailed Solution for CUET PG Zoology Mock Test - 2 - Question 6

The correct answer is Histamine

  • Eosinophils are a type of white blood cell that play a key role in the immune response. They contain granules filled with various molecules that help them perform their functions.
  • Eosinophilic granules contain a variety of proteins and enzymes, including ribonucleases, cytokines, and chemokines, which contribute to their immune functions.
    • Ribonucleases: These are enzymes found in eosinophilic granules that degrade RNA, playing a role in controlling viral infections and inflammatory responses.
    • Cytokines: These are signaling proteins that are found in eosinophilic granules and are important for cell signaling in the immune system. They help in the communication between cells to trigger the protective defenses of the immune system.
    • Chemokines: These are a subset of cytokines found in eosinophilic granules that induce chemotaxis in nearby responsive cells. They play a crucial role in guiding the movement of immune cells to the site of infection or inflammation.
  • Histamine is not typically found in eosinophilic granules. It is primarily stored in and released by mast cells and basophils, other types of white blood cells involved in the immune response.
CUET PG Zoology Mock Test - 2 - Question 7

Some behaviour patterns appear only after a specific developmental stage or time. This stage/time is called :

Detailed Solution for CUET PG Zoology Mock Test - 2 - Question 7

The correct answer is Maturation
Maturation refers to the process through which behaviors or abilities emerge as a result of the natural aging process, independent of external influences or learning. This concept suggests that certain behaviors or skills will only appear when an individual has reached a certain level of biological or neurological development. Maturation is crucial in understanding developmental milestones in children, such as walking or speaking, which appear as they grow older and their bodies and brains develop.
Incorrect Options:

  • Imprinting:- Imprinting is a form of learning occurring at a particular life stage that is rapid and apparently independent of the consequences of behavior. It's most famously observed in some bird species, where newborn chicks follow the first moving object they see, which is usually their mother.
  • Learning:- Learning involves acquiring new, or modifying existing, knowledge, behaviors, skills, values, or preferences and can happen at any stage in life. It is more dependent on external factors and experiences rather than internal biological growth.
  • Instinct:- Instinct refers to innate, typically fixed patterns of behavior in animals in response to certain stimuli. Instincts are present in an individual from birth, unlike behaviors that appear after a specific developmental stage due to maturation.

In summary, while imprinting, learning, and instinct are all significant in understanding behavior, maturation specifically refers to the developmental stage or time after which certain behavior patterns appear, driven by biological and neurological growth.

CUET PG Zoology Mock Test - 2 - Question 8

Match List-I with List-II

Choose the correct answer from the options given below:

Detailed Solution for CUET PG Zoology Mock Test - 2 - Question 8

The correct answer is (A) - (III), (B) - (IV), (C) - (II), (D) - (I)
Bacterial growth occurs in different phases: Lag phase, Log phase, Stationary phase, and Decline phase. Each phase has distinct characteristics related to cell growth and activity.

  • Lag Phase: During the lag phase, bacteria are adapting to their new environment. They are not immediately dividing as they are synthesizing essential enzymes and molecules, adjusting their metabolic machinery to the conditions of the fresh growth medium. This phase is characterized by metabolic activity without an increase in cell number.
  • Log Phase: The log (logarithmic or exponential) phase is a period of rapid cell division. Binary fission occurs at its maximum rate during this phase, and the population size doubles at a constant rate. This is the phase where bacterial growth is most robust, and it is represented by a straight line when plotted on a logarithmic scale.
  • Stationary Phase: During the stationary phase, the growth rate slows down, and the population reaches a balance where the rate of cell division equals the rate of cell death. This equilibrium is often due to the depletion of nutrients, accumulation of waste products, and limited space. Hence, the overall number of live cells remains constant.
  • Decline Phase: In the decline (or death) phase, the number of dying cells exceeds the number of new cells being produced. As nutrients are exhausted and waste products accumulate to toxic levels, cells cannot sustain their metabolic functions and die at an increasing rate. This phase results in a net decrease in the viable cell population.

CUET PG Zoology Mock Test - 2 - Question 9
Dung beetles live in regions where cattle graze, quickly burying and recycling cattle droppings. Because of the dung beetle's activities, breeding habitats for disease-carrying flies are reduced, and the plants upon which cattle feed are nourished. The relationship between the dung beetle's, and disease-carrying flies is a type of -
Detailed Solution for CUET PG Zoology Mock Test - 2 - Question 9

The correct answer is Interspecific competition

Concept:

  • Interspecific competition refers to the interaction between different species competing for the same resources in an ecosystem. This type of competition can affect the population and distribution of the competing species.
  • In the case of the dung beetle and disease-carrying flies, the dung beetle competes with the flies by quickly burying and recycling cattle droppings, which are essential for the flies' breeding. This reduces the breeding habitats for the flies, hence demonstrating interspecific competition.

Explanation:

  • Mutualism: This is a type of interaction where both species benefit from the relationship. For example, bees and flowers have a mutualistic relationship because bees get nectar from flowers, and flowers get pollinated by bees.
  • Commensalism: This is a type of relationship where one species benefits while the other species is neither helped nor harmed. For example, barnacles attaching themselves to whales.
  • Intraspecific competition: This occurs when members of the same species compete for the same resources. For example, deer competing for food in a forest.
  • Interspecific competition: This correctly describes the relationship between dung beetles and disease-carrying flies. The dung beetles and the flies are different species, and they compete for the same resource - cattle droppings. The flies are negatively impacted because dung beetles bury the droppings, reducing the available breeding habitat for the flies.
CUET PG Zoology Mock Test - 2 - Question 10

The process of synthesis of glucose from a non-carbohydrate source is called:

Detailed Solution for CUET PG Zoology Mock Test - 2 - Question 10
  • Gluconeogenesis is the metabolic pathway through which glucose is synthesized from non-carbohydrate sources.
  • This process is crucial for maintaining blood glucose levels during periods of fasting, intense exercise, or when carbohydrate intake is low.
  • Precursors for gluconeogenesis include lactate, glycerol, and certain amino acids (glucogenic amino acids), which can be converted into intermediates that feed into the gluconeogenesis pathway, ultimately leading to the synthesis of glucose.

Explanation of Other Options:
Glycogenesis:

  • This is the process of synthesizing glycogen from glucose. When glucose levels are high in the bloodstream, such as after a carbohydrate-rich meal, glycogenesis occurs primarily in liver and muscle cells to store excess glucose in the form of glycogen for future use.

Glycogenolysis:

  • This is the breakdown of glycogen into glucose. When the body needs glucose, such as between meals or during physical activity, glycogenolysis occurs in the liver (and to a lesser extent in muscles) to release glucose into the bloodstream.

Lipogenesis:

  • This is the metabolic process of synthesizing fatty acids from acetyl-CoA primarily derived from carbohydrates. Lipogenesis typically occurs in the liver and adipose tissue, and its end products are triglycerides, which are stored in adipose tissue for energy use.
CUET PG Zoology Mock Test - 2 - Question 11
Centrolecithal eggs show
Detailed Solution for CUET PG Zoology Mock Test - 2 - Question 11

Concept:

  • Cleavage is a series of mitotic cell divisions that immediately follows fertilization, which divided a zygote into numerous smaller, nucleated cells.
  • Cleavage sets the groundwork for the next stage of embryo development, i.e., gastrulation.
  • The following are different planes of cleavage:
  1. Meridional plane of cleavage: In this case, the furrow passes through the median axis or centre of the egg and bisects both poles of the egg.
  2. Vertical plane of cleavage: In this case, the furrows pass in any direction except the median axis, from the animal to the opposite pole.
  3. Equatorial plane of cleavage: In this case, the cleavage plane divides the egg only halfway between the animal and vegetal poles. The plane of division is the right angle to the median axis.
  4. Latitudinal plane of cleavage: It is similar to that of the equatorial plane of division and furrows run through the cytoplasm on either side of the equatorial plane.

Important Points

​Option 1: CORRECT

  • In the case of superficial cleavage, nucleus division takes place but cytokinesis does not occur, where the yolk is positioned to the centre of the cell and nuclei migrate to the periphery of the egg.
  • Centrolecithal eggs have yolk at the centre of the egg.
  • Hence, this is the correct answer.

Option 2: INCORRECT

  • In radial cleavage, daughter cells are arranged exactly on top of one another.
  • The cell division is complete, where karyokinesis is followed by cytokinesis.

Option 3: INCORRECT

  • In this case, the plane of the first cleavage furrow determines the plane of bilateral symmetry. The first cleavage bisects the embryo into right and left halves.

Option 4: INCORRECT

  • In discoidal cleavage, the cell division is only restricted to the small disc of cytoplasm at the animal pole, as cleavage furrows cannot penetrate the yolk.

Hence, the correct answer is option 1.

CUET PG Zoology Mock Test - 2 - Question 12

In the following columns, certain terms and their descriptions are given in random order.

Which of the following combination gives correct match for the terms in column I from column II

Detailed Solution for CUET PG Zoology Mock Test - 2 - Question 12

Concept:

  • In animal embryology, the terms Protostome, Deuterostome, Pseudocoely, Schizocoely, and Enterocoely refer to specific developmental processes and characteristics.
  • Protostomes and Deuterostomes represent two major groups of animals based on their embryonic development.
  • Pseudocoely, Schizocoely, and Enterocoely refer to different ways in which the body cavity, or coelom, forms during development.

Explanation:

  • Protostome: In protostomes, the mouth forms from the blastopore during early development. Therefore, the correct match is A - ii (Mouth forming from the blastopore).
  • Deuterostome: In deuterostomes, the mouth forms from a second opening other than the blastopore, and the blastopore becomes the anus. Hence, the correct match is B - iv (Mouth forming from a second opening other than the blastopore).
  • Pseudocoely: Pseudocoely refers to a fluid-filled cavity inside the external body wall that bathes the internal organs but is not entirely lined with mesoderm. Thus, the correct match is C - i (A fluid-filled cavity lying inside the external body wall bathing the internal organs).
  • Schizocoely: Schizocoely is the process of coelom formation by splitting the mesodermal tissue. Therefore, the correct match is D - iii (Coelom formed by splitting the mesodermal tissue).
  • Enterocoely: Enterocoely is the formation of the coelom from pouches pinched off from the digestive tract. Hence, the correct match is E - v (Coelom formed from pouches pinched off from the digestive tract).
CUET PG Zoology Mock Test - 2 - Question 13

Which of the following disease in humans, is caused due to expanded microsatellite repeats?

Detailed Solution for CUET PG Zoology Mock Test - 2 - Question 13

The correct answer is Huntington's disease

  • Huntington's disease (HD) is a progressive neurodegenerative genetic disorder caused by an abnormal expansion of CAG trinucleotide repeats in the HTT gene (huntingtin gene) located on chromosome 4.
  • It is an autosomal dominant condition, meaning that a single copy of the altered gene in each cell is sufficient to cause the disorder.
  • The disease typically affects muscle coordination and leads to cognitive decline and psychiatric problems.
  • Symptoms usually appear between the ages of 30 and 50 and progressively worsen over time.

Normal Function of HTT Gene:

  • The HTT gene encodes for the huntingtin protein, which is involved in various cellular processes, including signaling, transport, and protecting against cell death.
  • In a healthy individual, the HTT gene contains a region where the DNA sequence "CAG" is repeated multiple times. This CAG trinucleotide repeat codes for the amino acid glutamine.

CAG Repeats in HTT Gene:

  • The number of CAG repeats in the HTT gene is variable among individuals.
  • Normal Range: Typically, a healthy person will have 10 to 35 CAG repeats in the HTT gene. This range is considered normal and does not cause Huntington's disease.
  • Intermediate Range: Individuals with 36 to 39 CAG repeats may or may not develop the disease. This intermediate range often leads to a later onset and milder symptoms if the disease does develop.
  • Pathogenic Range: Individuals with 40 or more CAG repeats will develop Huntington's disease. The greater the number of repeats beyond 39, the earlier the onset and often the more severe the disease symptoms.

Mechanism of Disease:

  • In individuals with pathogenic CAG repeat expansions, the HTT gene produces an abnormally long huntingtin protein with an excessively long polyglutamine tract (a series of glutamine amino acids). This mutated protein tends to misfold and aggregate.
  • The misfolded huntingtin protein forms toxic aggregates that accumulate in neurons, particularly in the brain's striatum and cortex regions, which are crucial for movement control and cognitive functions. These aggregates disrupt normal cellular functions, leading to neuronal dysfunction and cell death.

Other Options:

  • Cystic fibrosis: This is a genetic disorder that affects the lungs and digestive system. It is caused by mutations in the CFTR gene, leading to the production of thick and sticky mucus that can clog the airways and lead to respiratory and digestive problems.
  • Tuberculosis: This is an infectious disease caused by the bacterium Mycobacterium tuberculosis. It primarily affects the lungs but can also affect other parts of the body. It is not a genetic disorder and is transmitted through the air from person to person.
  • Alkaptonuria: This is a rare genetic disorder caused by the deficiency of the enzyme homogentisate 1,2-dioxygenase (HGD). It leads to the accumulation of homogentisic acid in the body, which can cause darkening of the urine and damage to cartilage and heart valves. It is not related to expanded microsatellite repeats.
CUET PG Zoology Mock Test - 2 - Question 14

The following is the list of steps, involved in the transfer of bacterial DNA by generalized transduction. Which option represents the correct sequence of steps?
(A) Packaging of host genomic DNA fragment, into P1 phage head.
(B) Injection of DNA of transducing particle, into bacterial host.
(C) Injection of P1 phage DNA, into bacterial host.
(D) Release of transducing particle, into bacterial host.
(E) Fragmentation of host genomic DNA, by phage-encoded endonuclease.
Choose the correct answer from the options given below:

Detailed Solution for CUET PG Zoology Mock Test - 2 - Question 14

The correct answer is (C), (E), (A), (D), (B).

Concept:

  • Generalized transduction is a process by which bacterial DNA is transferred from one bacterium to another by a virus (bacteriophage).
  • This process involves several steps where the bacteriophage mistakenly packages bacterial DNA and transfers it to a new host bacterium.

Explanation:

  • (C) Injection of P1 phage DNA into bacterial host: The process begins with the injection of the P1 phage DNA into the bacterial host, initiating the infection.
  • (E) Fragmentation of host genomic DNA by phage-encoded endonuclease: After the phage DNA enters the bacterial cell, it hijacks the bacterial machinery to replicate its own DNA and produce phage proteins. Phage-encoded endonucleases (enzymes that cut DNA) degrade the bacterial DNA into smaller fragments. This fragmentation ensures that the cell's resources are channeled towards producing new phage particles rather than bacterial proteins.
  • (A) Packaging of host genomic DNA fragment into P1 phage head: During the assembly of new phage particles, the viral machinery occasionally makes a mistake. Instead of packaging phage DNA into the phage heads, fragments of the degraded bacterial DNA are mistakenly packaged. These phage particles containing bacterial DNA are called transducing particles
  • (D) Release of transducing particle into bacterial host: Once the phage assembly is complete, the bacterial cell undergoes lysis (ruptures), releasing new phage particles, including both normal phages with viral DNA and transducing particles with bacterial DNA, into the surrounding environment
  • (B) Injection of DNA of transducing particle into bacterial host: The transducing particles can go on to infect new bacterial cells. When a transducing particle infects a new bacterium, it injects the packaged bacterial DNA into the new host cell. Since this DNA is from the previously infected bacterium, it can recombine with the new host bacterium's genome through homologous recombination, integrating into its genome and potentially altering its genetic makeup.
CUET PG Zoology Mock Test - 2 - Question 15
Individuals occupying a particular habitat and adapted to it phenotypically but not genotypically are known as
Detailed Solution for CUET PG Zoology Mock Test - 2 - Question 15

The correct answer is Ecophenes

Concept:

  • Individuals occupying a particular habitat and adapted to it phenotypically but not genotypically are known as ecophenes.
  • Phenotypic adaptation refers to changes in an organism's behavior, morphology, or physiology in response to their environment, which do not involve changes in the genetic code.
  • Such adaptations are generally reversible and can occur within a single generation.

Explanation:

  • Ecophenes: These are individuals of the same species that show phenotypic variations due to environmental influences. These variations do not involve any genetic changes, meaning the genetic makeup of ecophenes remains unchanged.
  • Ecotypes: These are populations of a species that have genetically adapted to specific environmental conditions. Unlike ecophenes, the adaptations in ecotypes involve genetic changes, making them distinct in terms of genotype.
  • Ecospecies: This term refers to a group of populations that can interbreed and are adapted to a particular set of environmental conditions. Ecospecies include genetic adaptations that help them thrive in their specific habitats.
  • Coenospecies: This term is used to describe a group of related species that can interbreed to some extent but are typically reproductively isolated. They share common characteristics and may have a common ancestry.
CUET PG Zoology Mock Test - 2 - Question 16

Being a medical science student, in your very first anatomical investigation of a female body, arrange the order of appearance of following organs of immunological importance, in anterior to posterior direction in the body.
(A) Thymus
(B) Adenoids
(C) Peyer's patches
(D) Spleen
Choose the correct answer from the options given below:

Detailed Solution for CUET PG Zoology Mock Test - 2 - Question 16

The correct answer is (B), (A), (D), (C)

  • Adenoids (B): The adenoids, also known as pharyngeal tonsils, are located in the roof of the nasopharynx, which is in the upper part of the throat behind the nose.
  • Thymus (A): The thymus is located in the anterior part of the upper chest, just behind the sternum (breastbone) and in front of the heart.
  • Spleen (D): The spleen is located in the left upper quadrant of the abdomen, behind the stomach and under the diaphragm.
  • Peyer’s patches (C): Peyer's patches are found in the ileum of the small intestine, more posteriorly and inferiorly in the abdominal cavity.

Thus, the order of appearance of these organs from anterior to posterior direction is Adenoids (B) → Thymus (A) → Spleen (D) → Peyer's patches (C)

CUET PG Zoology Mock Test - 2 - Question 17

The increasing order of pKa of the following amino acids in aqueous solution is:
Gly, Asp, Lys, Arg

Detailed Solution for CUET PG Zoology Mock Test - 2 - Question 17

Concept:
Order of acidic strength:
 
Aspartic Acid is a non-essential amino acid in humans, Aspartic Acid has an overall negative charge and plays an important role in the synthesis of other amino acids and in the citric acid and urea cycles. Asparagine, arginine, lysine, methionine, isoleucine, and some nucleotides are synthesized from aspartic acid.
Glycine is an amino acid, a building block for protein. It is not considered an "essential amino acid" because the body can make it from other chemicals. A typical diet contains about 2 grams of glycine daily. The primary sources are protein-rich foods including meat, fish, dairy, and legumes.
Amino acid arginine changes into nitric oxide (NO).
Nitric oxide is a powerful neurotransmitter that helps blood vessels relax and also improves circulation. Some evidence shows that arginine may help improve blood flow in the arteries of the heart.
Amino acids are organic compound containing amine (-NH2) and carboxyl (-COOH) functional groups, along with a side chain (R group) specific to each amino acid. The key elements of an amino acid are carbon (C), hydrogen (H), oxygen (O), and the nitrogen (N) although other elements are found in the side chains of certain amino acids.
The amino acids are classified according to the core structural functional groups location as alpha (α), beta (β), gamma (ϒ) or delta (δ).
The increasing order of pKa of the following amino acids in the aqueous solution is, Asp < Gly < Lys < Arg

Nature of ‘R’ group will determine the basicity (hence, pKa) of an amino acid.

CUET PG Zoology Mock Test - 2 - Question 18

Which one of the following combinations represents the major protein or protein complex involved in chromatin condensation in yeast and human, respectively?

Detailed Solution for CUET PG Zoology Mock Test - 2 - Question 18

Concept:

  • Chromatin condensation is a process by which chromatin gets densely packaged and reduced in volume for the broader purpose of gene regulation.
  • Subsets of chromatins are:
    1. Heterochromatin - transcriptionally inactive part due to dense chromatin condensation.
    2. Euchromatin - transcriptionally active part due to comparatively loose chromatin condensation or presence of expanded DNA regions for transcription.

Explanation:
HP1 -

  • HP1 is a family of non-histone chromosomal proteins found in mammals.
  • HP1 has three paralogs: HP1alpha, HP1 beta and HP1 gamma.
  • HP1 belongs to the heterochromatin protein 1 family, which binds to methylated histone H3 at the lysine 9 position and represses DNA transcription of the region.

SIR Complex-

  • SIR (silent information regulator) proteins are nuclear proteins found in budding yeast (Saccharomyces cerevisiae).
  • These proteins form specialized chromatin structures that resemble heterochromatin of higher eukaryotes.
  • SIR-3 is known to be the primary structural component of SIR proteins of heterochromatin condensation.
  • SIR 2-4 complex helps in the recruitment of other SIR proteins.

Su(var) -

  • The role of Su(var) heterochromatin protein is seen in Drosophila only.
  • It controls position effect variegation in Drosophila by methylation at H3-K9 position.

Hence, the correct option is option 2.

CUET PG Zoology Mock Test - 2 - Question 19

Select the correct statement about the membrane transport -
(A) Channel proteins form open pores, through which suitably sized molecules/ions can cross the membrane.
(B) Carrier proteins selectively bind specific molecules, and transport them across the membrane by undergoing a conformation change.
(C) Transport of molecules across membranes, through channel proteins and carrier proteins, always involves active transport.
(D) Small uncharged molecules can diffuse freely, through the phospholipid bilayers.
Choose the correct answer from the options given below:

Detailed Solution for CUET PG Zoology Mock Test - 2 - Question 19

The correct answer is (A), (B) and (D) only

  • Membrane transport refers to the movement of substances across the cell membrane, either into or out of the cell.
  • There are different types of membrane transport mechanisms, including passive transport (such as diffusion and facilitated diffusion) and active transport.

Proteins play a crucial role in facilitating the transport of molecules that cannot freely diffuse through the lipid bilayer.

  • (A). Channel proteins form open pores, through which suitably sized molecules/ions can cross the membrane: This statement is correct. Channel proteins provide a passageway for specific ions or molecules to pass through the membrane. They can open or close in response to certain stimuli, allowing selective permeability.
  • (B). Carrier proteins selectively bind specific molecules, and transport them across the membrane by undergoing a conformation change: This statement is also correct. Carrier proteins bind to specific molecules on one side of the membrane, undergo a conformational change, and release the molecule on the other side. This can happen through facilitated diffusion or active transport.
  • (D). Small uncharged molecules can diffuse freely, through the phospholipid bilayers: This statement is correct. Small, nonpolar, and uncharged molecules, such as oxygen and carbon dioxide, can diffuse directly through the phospholipid bilayer without the need for transport proteins.
  • (C). Transport of molecules across membranes, through channel proteins and carrier proteins, always involves active transport: This statement is incorrect. Transport through channel proteins and carrier proteins can involve both passive (facilitated diffusion) and active transport. Facilitated diffusion does not require energy, while active transport does.
CUET PG Zoology Mock Test - 2 - Question 20

Which of the following statements are INCORRECT?
(A) Many animals with protostome development, undergo spiral cleavage.
(B) Flat worms exhibit, radial symmetry
(C) Ray-finned fishes are also called Sarcopterygians.
(D) Lophophore is a crown of ciliated tentacles, that function in feeding
Choose the correct answer from the options given below:

Detailed Solution for CUET PG Zoology Mock Test - 2 - Question 20

The correct answer is (B) and (C) only.

  • (A) Many animals with protostome development undergo spiral cleavage: This statement is correct. Protostomes typically exhibit spiral and determinate cleavage during embryonic development.
    • Spiral cleavage is a type of embryonic development where cells divide at oblique angles, creating a helical arrangement.
    • This mode of development is characteristic of many protostomes, such as annelids, molluscs, and some other invertebrates.
    • Protostomes are a major group of animals defined by their embryonic development; the mouth forms from the initial opening (blastopore) in the embryo.
  • (B) Flatworms exhibit radial symmetry: This statement is incorrect. Flatworms (Platyhelminthes) exhibit bilateral symmetry, their body can be divided into mirror-image halves along one plane.
    • Symmetry is a fundamental characteristic used to classify animals.
    • Radial symmetry means the body plan is arranged around a central axis, and it can be divided into similar halves by multiple planes through the central axis (e.g., cnidarians like jellyfish).
    • Bilateral symmetry, on the other hand, means the body has a distinct left and right side and can only be divided into mirrored halves along one plane (e.g., flatworms, humans).
  • (C) Ray-finned fishes are also called Sarcopterygians: This statement is incorrect. Ray-finned fishes are called Actinopterygians. Sarcopterygians are lobe-finned fishes, which include coelacanths and lungfish.
    • Fish are classified based on the structure of their fins.
    • Ray-finned fishes (Actinopterygii) have fins supported by long, flexible rays, whereas lobe-finned fishes (Sarcopterygii) have fleshy, lobed fins with a central bony axis.
    • Sarcopterygians include the ancestors of terrestrial vertebrates (tetrapods). This distinction is significant in vertebrate evolutionary histor
  • (D) Lophophore is a crown of ciliated tentacles that function in feeding. This statement is correct. Lophophores are found in lophophorate animals like brachiopods and bryozoans, aiding in feeding by filtering food particles from the water.
    • The lophophore is a unique feeding structure found in certain groups of invertebrates, such as bryozoans, brachiopods, and phoronids.
    • It consists of a crown of ciliated tentacles that effectively capture and move food particles toward the mouth. ​
CUET PG Zoology Mock Test - 2 - Question 21

Sequentially arrange the following steps involved in blood clot formation -
(A) Conversion of fibrinogen to fibrin
(B) Conversion of prothrombin to thrombin
(C) Adhesion and aggregation of platelets on damaged vessels
(D) Prothrombinase formed by extrinsic or intrinsic pathway
(E) Reduction of blood loss by initiation of a vascular system
Choose the correct answer from the options given below:

Detailed Solution for CUET PG Zoology Mock Test - 2 - Question 21

The correct answer is (E), (D), (C), (B), (A)
Blood clot formation, also known as coagulation, is a complex process that prevents excessive bleeding when a blood vessel is injured. It involves multiple steps that are initiated when a blood vessel is damaged, leading to the formation of a stable blood clot that seals the injury.

  • Step (E) - Reduction of blood loss by initiation of a vascular system: When a blood vessel is injured, the first response is vasoconstriction, which reduces blood flow to the area and minimizes blood loss.
  • Step (D) - Prothrombinase formed by extrinsic or intrinsic pathway: The damaged tissue exposes collagen and tissue factors that activate the extrinsic and intrinsic pathways, leading to the formation of prothrombinase.
  • Step (C) - Adhesion and aggregation of platelets on damaged vessels: Platelets adhere to the exposed collagen and aggregate to form a temporary platelet plug at the site of injury.
  • Step (B) - Conversion of prothrombin to thrombin: Prothrombinase converts prothrombin (a plasma protein) into thrombin, an enzyme that plays a critical role in the next step of coagulation.
  • Step (A) - Conversion of fibrinogen to fibrin: Thrombin converts fibrinogen (a soluble plasma protein) into fibrin strands, which weave through the platelet plug to form a stable and insoluble clot.
CUET PG Zoology Mock Test - 2 - Question 22

Match List-I with List-II

Choose the correct answer from the options given below:

Detailed Solution for CUET PG Zoology Mock Test - 2 - Question 22

The correct answer is (A) - (III), (B) - (I), (C) - (II), (D) - (IV)

  • DNA footprinting (A) - DNA binding protein (III): DNA footprinting is a technique used to identify the specific DNA sequences that are bound by proteins. This helps in understanding the DNA-protein interactions and identifying regulatory regions in the DNA.
  • Yeast two-hybrid system (B) - Protein-protein interaction (I): The yeast two-hybrid system is a molecular biology technique used to study protein-protein interactions. It involves the use of yeast cells to test the interaction between two proteins, helping in the identification of protein functions and interaction networks.
  • DNA fingerprinting (C) - VNTR (II): DNA fingerprinting is a method used to identify individuals based on their unique genetic makeup. This technique often uses Variable Number Tandem Repeats (VNTRs), which are highly polymorphic regions in the DNA, to differentiate between individuals.
  • SAGE (D) - Transcriptome analysis (IV): Serial Analysis of Gene Expression (SAGE) is a technique used to analyze the expression of genes in a genome-wide manner. It helps in studying the transcriptome, which is the complete set of RNA transcripts produced by the genome under specific conditions.
CUET PG Zoology Mock Test - 2 - Question 23

Which of the following protected areas are declared Tiger Reserves in India?
(A) Bandipur
(B) Bhitarkanika
(C) Manas
(D) Sunderbans
(E) Kaziranga
Choose the correct answer from the options given below:

Detailed Solution for CUET PG Zoology Mock Test - 2 - Question 23

The correct answer is (A), (C) and (D) only
Tiger Reserves are protected areas that are specifically designated for the conservation of tigers in India. These reserves are part of Project Tiger, which was launched in 1973 to save the declining population of tigers in the country.

  • Bandipur: Located in Karnataka, Bandipur National Park is one of the earliest Tiger Reserves in India and is part of the Nilgiri Biosphere Reserve.
  • Manas: Situated in Assam, Manas National Park is a UNESCO World Heritage Site and a designated Tiger Reserve, known for its rich biodiversity.
  • Sunderbans: The Sunderbans in West Bengal is the largest estuarine mangrove forest in the world and is also a Tiger Reserve, known for its unique ecosystem and tiger population.

Other Options:

  • Bhitarkanika: Located in Odisha, Bhitarkanika National Park is known for its mangrove forests and saltwater crocodiles.
  • Kaziranga: Also in Assam, Kaziranga National Park is famous for its one-horned rhinoceros population.
CUET PG Zoology Mock Test - 2 - Question 24

Microorganisms that produce antibiotics are unaffected by the same antibiotic.
One of the mechanism to explain this is -

Detailed Solution for CUET PG Zoology Mock Test - 2 - Question 24

The correct answer is microorganisms export the antibiotic, using efflux pumps, as soon as it exceeds a certain intracellular concentration.
Microorganisms that produce antibiotics have developed mechanisms to avoid self-destruction by their own antibiotics. One of the key mechanisms is the use of efflux pumps, which actively transport the antibiotic out of the cell, preventing it from reaching toxic concentrations within the microorganism. Efflux pumps are proteins located in the cell membrane that act as transporters, expelling unwanted toxic substances, including antibiotics.

Option 1: Active pharmacophore of the bactericidal antibiotic remains bound to antibodies while they are intracellular.

  • This option is incorrect because antibiotics do not bind to antibodies inside the producing microorganisms. Antibodies are part of the immune system of higher organisms, not bacteria or fungi that produce antibiotics.

Option 2: Antibiotics are always produced in inactive forms, and get activated by the enzymatic action of the target cell.

  • This option is incorrect because while some antibiotics may require activation, this is not a general mechanism for all antibiotics. Many antibiotics are produced in their active forms.

Option 3: Microorganisms export the antibiotic, using efflux pumps, as soon as it exceeds a certain intracellular concentration.

  • This option is correct. Efflux pumps help the microorganism to expel the antibiotic, thereby preventing it from accumulating to toxic levels inside the cell. This mechanism is well-documented and effective in protecting the antibiotic-producing microorganism.

Option 4: Antibiotics are made in parts, and are assembled after being transported to the extracellular environment.

  • This option is incorrect because antibiotics are synthesized as complete molecules within the cell and are not assembled extracellularly. The biosynthetic pathways for antibiotics take place within the producing microorganism.
CUET PG Zoology Mock Test - 2 - Question 25

Which two embryological structures together form the bilaminar disc?

Detailed Solution for CUET PG Zoology Mock Test - 2 - Question 25

The correct answer is Epiblast and Hypoblast

  • The bilaminar disc is an essential structure during early embryonic development. It forms during the second week of development and consists of two distinct layers: the epiblast and the hypoblast.
  • The epiblast and hypoblast are derived from the inner cell mass of the blastocyst and play crucial roles in the establishment of the body plan and the formation of extraembryonic structures.
  • The epiblast is the upper layer, and the hypoblast is the lower layer of the bilaminar disc.
  • The epiblast gives rise to all three germ layers (ectoderm, mesoderm, and endoderm) during gastrulation, while the hypoblast contributes to the formation of the yolk sac.

Fig:Bilaminar embryo disc
Other Options:

  • Cytotrophoblast: The cytotrophoblast is part of the trophoblast, which contributes to the formation of the placenta, not the bilaminar disc.
  • Syncytiotrophoblast and Embryoblast: The syncytiotrophoblast is another part of the trophoblast, which also contributes to placenta formation. The embryoblast gives rise to both the epiblast and hypoblast but isn't itself a part of the bilaminar disc. Hence, this combination is incorrect.
CUET PG Zoology Mock Test - 2 - Question 26

Match the RNAs in Group I with their corresponding function in Group II

Detailed Solution for CUET PG Zoology Mock Test - 2 - Question 26

The correct answer is Option 4

  • snRNA (small nuclear RNA): Involved in RNA splicing, which is the process of removing introns from pre-mRNA to produce the mature messenger RNA that will be translated into protein.
  • piRNA (PIWI-interacting RNA): Plays a crucial role in germ line cells, primarily through the silencing of transposable elements to protect the integrity of the genetic material.
  • snoRNA (small nucleolar RNA): Involved in the chemical modification and processing of rRNA (ribosomal RNA), which is essential for the proper assembly and function of ribosomes.
  • siRNA (small interfering RNA): Functions in the RNA interference (RNAi) pathway, where it binds to complementary mRNA molecules leading to their degradation and thus regulating gene expression and also protecting the genome from viruses and other genomic parasites.

Therefore, the correct match is P-2, Q-4, R-1, S-3

CUET PG Zoology Mock Test - 2 - Question 27

Which one of the following statements is TRUE for positive-frequency dependent selection?

Detailed Solution for CUET PG Zoology Mock Test - 2 - Question 27

The correct answer is Option 2 i.e. Fitness of a genotype increases as it becomes more common.
In evolutionary biology, "fitness" refers to an organism's ability to survive and reproduce in its environment. The concept of frequency-dependent selection involves situations where the fitness of a phenotype (observable traits) or genotype (genetic makeup) is directly related to its frequency within the population.
Positive-Frequency Dependent Selection
Positive-frequency dependent selection occurs when the fitness of a phenotype or genotype increases as it becomes more prevalent in the population. This means that common traits become even more advantageous as they spread, leading to an acceleration in their uptake within the population until they potentially become fixed (found in all individuals).
Mechanisms and Examples
One classic example of positive-frequency dependent selection is seen in warning coloration among certain species, a phenomenon known as "Müllerian mimicry." Predators learn to avoid prey with specific warning colors after bad experiences. As a result, the more common a warning color becomes in a population (shared by multiple species), the more likely predators are to recognize and avoid it, increasing the survival rate of individuals displaying that coloration.
Negative-Frequency Dependent Selection
Negative-frequency dependent selection occurs when the fitness of a phenotype or genotype increases as it becomes rarer in the population. This mechanism helps maintain genetic diversity within the population, as it prevents any one allele or trait from becoming overly dominant. An example of negative-frequency dependent selection is seen in host-parasite interactions, where rare genotypes in a host population may be less susceptible to common parasite strains, giving these rare genotypes a reproductive advantage.
Conclusion:
Therefore, the correct answer is Option 2

CUET PG Zoology Mock Test - 2 - Question 28

Which of the following is NOT a premating mechanism of isolation?

Detailed Solution for CUET PG Zoology Mock Test - 2 - Question 28

The correct answer is Zygotic mortality

  • Zygotic mortality refers to the failure of the zygote to develop properly, leading to its death. This occurs after mating has taken place, making it a postmating mechanism of reproductive isolation rather than a premating one. It prevents species that have already mated from producing viable offspring, thus maintaining species boundaries.

Overview of Incorrect Options:

  • Temporal Isolation: This occurs when two species mate or flower at different times of the day, season, or even years. Since they do not overlap in their periods of sexual activity or reproduction, they cannot mate with each other. This is a premating mechanism because it prevents mating from occurring in the first place.
  • Behavioural Isolation: This involves differences in mating rituals or behaviors that prevent members of different species from recognizing each other as suitable mates. For example, different bird species may have unique songs or dances that are attractive only to their own species. This is also a premating mechanism because it stops different species from attempting to mate.
  • Mechanical Isolation: This occurs when differences in the physical structures (morphology) of the reproductive organs of two species prevent them from successfully mating. For instance, if the genitalia of two species do not fit together properly, mating cannot occur. This is a premating mechanism as it physically prevents successful copulation or pollination.
CUET PG Zoology Mock Test - 2 - Question 29

Given below are a few statements regarding the rate of glycolysis, gluconeogenesis and glycogen metabolism.
A. Increased blood glucose would decrease gluconeogenesis and increase glycogen synthesis.
B. Increased levels of fructose-1, 6-bisphosphate inhibits glycolysis.
C. Increased blood glucagon inhibits glycogen synthesis and stimulates glycogen break down.
D. Increase in AMP levels inhibits glycolysis and stimulates gluconeogenesis.
Which one of the following options represents INCORRECT statements?

Detailed Solution for CUET PG Zoology Mock Test - 2 - Question 29

The correct answer is B and D
Explanation:
A. Increased blood glucose would decrease gluconeogenesis and increase glycogen synthesis: This statement is correct. When blood glucose levels are high, the body tends to store glucose as glycogen and reduce the production of glucose through gluconeogenesis. Insulin is released, which promotes glycogen synthesis (glycogenesis) and decreases gluconeogenesis.
B. Increased levels of fructose-1, 6-bisphosphate inhibit glycolysis: This statement is incorrect. Fructose-1,6-bisphosphate is an intermediate in glycolysis and generally acts to stimulate the process rather than inhibit it. In glycolysis, fructose-1,6-bisphosphate is formed from fructose-6-phosphate and later split into two three-carbon sugars that continue through the glycolytic pathway.
C. Increased blood glucagon inhibits glycogen synthesis and stimulates glycogen breakdown: This statement is correct. Glucagon is a hormone that is released when blood glucose levels are low. It signals the liver to break down glycogen into glucose (glycogenolysis) and to decrease glycogen synthesis. This increases blood glucose levels.
D. Increase in AMP levels inhibits glycolysis and stimulates gluconeogenesis: This statement is incorrect. AMP (adenosine monophosphate) is an indicator of low energy status in the cell. High levels of AMP activate glycolysis to produce ATP and inhibit gluconeogenesis, which consumes ATP.
Table: Regulators of Glycolysis

Conclusion: Therefore, the statements that are incorrect are B and D.

CUET PG Zoology Mock Test - 2 - Question 30

Consider the following statements regarding the Cell organelles:

  1. Microtubules help in the spindle and astral ray formation during cell division and form the cytoskeleton of cilia and flagella.
  2. Golgi apparatus helps in the recycling of broken plasma membrane during endocytosis.
  3. Rough Endoplasmic Reticulum is involved in fat and steroidal hormone synthesis.

Which of the statements given above is/are correct?

Detailed Solution for CUET PG Zoology Mock Test - 2 - Question 30

The correct answer is 1 and 2 only.
Key Points

  • Cytoskeleton
    • Microtubules, Microfilaments, and intermediate filaments are proteinaceous structures to support the extensive system of membranes that collectively form the cytoskeleton of a cell.
    • The microtubules occur in cilia, flagella, centrioles and basal bodies, mitotic apparatus, etc.
    • Microtubules help in the spindle and astral ray formation during cell division and form the cytoskeleton of cilia and flagella. Hence statement 1 is correct.
    • The microfilaments are mainly composed of the globular protein actin but have filamentous protein myosin also.
    • Microfilaments are involved in cytoplasmic streaming and amoeboid movements.
  • Golgi Complex
    • Golgi complex is in the form of a parallel arranged interconnecting system of a cluster of smooth membranes having a central stack of flattened sacs or cisternae.
    • It has two distinct faces i.e, forming faces (cis region) and a maturing face (trans region).
    • Glycosidation of lipids and glycosylation of proteins to form glycolipids and glycoproteins.
    • Golgi apparatus helps in the recycling of broken plasma membrane during endocytosis. Hence statement 2 is correct.
  • Endoplasmic Reticulum
    • They divide the intercellular space into two components i.e luminal (inside ER) and extraluminal (cytoplasm). These can extend from nuclear envelope to plasma membrane.
    • The endoplasmic reticulum is composed of the following three kinds of structures: cisternae which occur in those cells which have synthetic roles, tubules that are involved in lipid and sterol synthesis and vesicles that are membrane-bound vacuolar structures.
    • The Endoplasmic Reticulum of two types: Smooth Endoplasmic Reticulum consists mainly of tubules and vesicles and the Rough Endoplasmic Reticulum consists of cisternae, it has ribosomes attached to its cytoplasmic surface.
    • The Rough Endoplasmic Reticulum provides a surface for protein synthesis and Smooth Endoplasmic Reticulum is involved in fat and steroidal hormone synthesis. Hence statement 3 is incorrect.
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