CUET PG Zoology 2024 Previous Year Question Papers PDF with Solutions Archives - Page 2 of 8 - CSIR NET LIFE SCIENCE COACHING | NTA NET LIFE SCIENCE | CSIR LIFE SCIENCE

An Institute for CSIR NET Life Science- DBT JRF/ M Sc Entrance
letstalkacademy@gmail.com

Q.75 Capacitation is a process in which the – glycoprotein layer, over the acrosome is dissolved. sperm penetrates the zona pellucida. tail of the sperm is activated. tail of the sperm is inactivated.

Capacitation Process in Sperm

Q.74 Select the correct statement about the membrane transport – Channel proteins form open pores, through which suitably sized molecules/ions can cross the membrane. Carrier proteins selectively bind specific molecules, and transport them across the membrane by undergoing a conformation change. Transport of molecules across membranes, through channel proteins and carrier proteins, always involves active transport. Small uncharged molecules can diffuse freely through the phospholipid bilayers. Choose the correct answer from the options given below: (A) and (B) only. (A), (B) and (C) only. (A), (B) and (D) only. (B), (C) and (D) only.

Membrane Transport

Q.73 In a sucrose density gradient, what is the order of organelle sedimentation from lower to higher concentration of sucrose? Golgi, smooth endoplasmic reticulum, rough endoplasmic reticulum. Smooth endoplasmic reticulum, Golgi, rough endoplasmic reticulum. Golgi, rough endoplasmic reticulum, smooth endoplasmic reticulum. Rough endoplasmic reticulum, smooth endoplasmic reticulum, Golgi.

Sucrose Density Gradient Organelle Sedimentation Order

Q.72 A protein coding gene in the mouse was mutated, such that the resultant protein was nonfunctional. The phenotypic effect of the mutation was observed in homozygous mutant mice, in heterozygous mice that inherited the mutation from the mother, and in heterozygous mice, that inherited the mutation from the father. This suggest that - 1. the gene is imprinted. 2. the mutation acts as a dominant negative. 3. the gene possibly has more than two alleles 4. the gene is X-linked.

Dominant Negative Explains Phenotype

Q.71 The square root of the variance is known as - 1. Standard Error (SE) 2. Standard deviation (SD) 3. Standardized coefficients (SC) 4. Protective odds ratio

Square Root of Variance

Q.70 The correlation between the observed and predicted values of the outcome variable is known as - 1. Coefficient of determination (R2) 2. Multiple correlation coefficient 3. Cox regression 4. Multiple linear regression

R² = Correlation(Observed, Predicted)

Q.69 Which of the following molecule is NOT found in the Eosinophilic granule ? 1. Ribonucleases 2. Cytokines 3. Chemokines 4. Histamine

Histamine Absent in Eosinophil Granules

Q.68 Being a medical science student, in your very first anatomical investigation of a female body, arrange the order of appearnce 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: 1. (A), (B), (C), (D). 2. (B), (D), (A), (C). 3. (B), (A), (D), (C). 4. (C), (B), (D), (A).

Immune Organs Anatomical Order

Q.67 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 - 1. mutualism 2. commensalism 3. intraspecific competition. 4. interspecific competition.

Dung Beetle vs Disease Flies

Q.66 Bacteria produce an enzyme methylase, which mainly play a role in - 1. modification of DNA 2. cutting of DNA 3. repair of DNA 4. rearrangement of DNA

Bacterial Methylase Enzyme Role

Q.65 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: 1. (A) and (D) only. 2. (B) and (C) only. 3. (B), (C) and (D) only. 4. (A) and (C) only.

Protostome Cleavage, Flatworms, Ray-Finned Fishes, Lophophore

Q.64 Match List-I with List-II List-I List-II (Taxonomic Unit) (Representative Example) (A) Brachiopods (I) Proboscis worms (B) Ctenophores (II) Lamp shells (C) Nemertea (III) Comb jellies (D) Onychophorans (IV) Velvet worms Choose the correct answer from the options given below: (A) – (I), (B) – (III), (C) – (II), (D) – (IV) (A) – (II), (B) – (III), (C) – (IV), (D) – (I) (A) – (III), (B) – (II), (C) – (IV), (D) – (I) (A) – (II), (B) – (III), (C) – (I), (D) – (IV)

Invertebrate Phyla Matching

Q.63 Match List-I with List-II List-I List-II Mammalian order Examples (A) Perissodactyla (I) Porcupines (B) Primates (II) Lemurs (C) Lagomorpha (III) Rhinoceroses (D) Rodentia (IV) Hares Choose the correct answer from the options given below: 1. (A) – (II), (B) – (III), (C) – (I), (D) – (IV) 2. (A) – (III), (B) – (IV), (C) – (II), (D) – (I) 3. (A) – (II), (B) – (I), (C) – (IV), (D) – (III) 4. (A) – (III), (B) – (II), (C) – (IV), (D) – (I)

Mammalian Orders Matching

Q.62 The following is the list of steps involved in cloning of a gene. Which option represents the correct sequence of the process ? (A). Transformation with blue and white selection. (B). RFLP analysis. (C). Isolation of plasmid DNA. (D). Ligation of gene, into TA-cloning vector. (E). Amplification of gene by PCR, using Taq DNA polymerase. Choose the correct answer from the options given below: 1. (E), (D), (C), (B), (A). 2. (E), (B), (D), (C), (A). 3. (C), (E), (D), (B), (A). 4. (C), (B), (D), (A), (E).

Gene Cloning Correct Sequence

Q.61 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: 1. (C), (A), (B), (E), (D). 2. (C), (E), (A), (D), (B). 3. (A), (D), (B), (E), (C). 4. (A), (E), (B), (D), (C).

Mastering Generalized Transduction

Q.60 Following statements are with reference to oogonia. Which of these is incorrect? 1. Common fate of oogonia is death by apoptosis. 2. Oogonia are able to undergo mitotic division. 3. Number of oogonia grows especially during puberty. 4. Oogonia are diploid cells.

Oogonia Characteristics

Q.59 During gastrulation in Xenopus frog, the blastocoel 1. becomes the gut 2. is filled with endodermal cells and disappears 3. is filled with mesoderm and disappears 4. is displaced, and its original location becomes an endoderm lined cavity.

Xenopus Gastrulation

Q.58 A student was asked to clone a DNA fragment, flanked by BamHI (GGATCC) sites into a vector, cut with Bglll (AGATCT). The cloning was successful, as these two restriction enzymes are known to create compatible sites. Based on this, what will be the restriction site configurations of these two enzymes (indicates the point of enzymatic cleavage). 1. G/GATCC and AGATC/T 2. G/GATCC and A/GATCT 3. GGATC/C and A/GATCT 4. GGATC/C and AGA/TCT

BamHI and BglII Compatible Restriction Sites

Q.57 Microorganisms that produce antibiotics are unaffected by the same antibiotic. One of the mechanism to explain this is - 1. active pharmacophore of the bactericidal antibiotic, remains bound to antibodies while they are intracellular. 2. antibiotics are always produced in inactive forms, and get activated by the enzymatic action of the target cell. 3. microorganisms export the antibiotic, using efflux pumps, as soon as it exceeds a certain intracellular concentration. 4. antibiotics are made in parts, and are assembled after being transported to the extracellular environment.

Antibiotic-Producing Microbes

Q.56 Which of the following disease in humans, is caused due to expanded microsatellite repeats? 1. Huntington's disease. 2. Cystic fibrosis 3. Tuberculosis 4. Alkaptonuria

Trinucleotide Repeat Expansion Diseases

Q.55 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: 1. (A), (C) and (D) only. 2. (A), (B) and (E) only. 3. (A) and (D) only. 4. (B), (C) and (D) only.

Tiger Reserves India

Q.54 Migratory species pose special preservation challenges, because 1. they are endemic, they are especially susceptible to habitat destruction. 2. their conservation may require international cooperation, when they require habitats in different countries. 3. they are often prone, population number, decline during their long migratory journeys. 4. they reside in biodiversity hot spots, that are most susceptible to habitat degradation.

Why Migratory Species Need International Cooperation

Q.53 The transition zone in biosphere reserve - 1. covers the buffer zone and lies, in between the buffer zone and the core zone. 2. covers the core zone and lies, in between the core zone and the buffer zone. 3. is the inner most zone of the biosphere reserve. 4. is the outer most zone of the biosphere reserve.

Biosphere Reserve Transition Zone

Q.52 Which of the following is a feature of a stable community? 1. Susceptible to invasions by alien species 2. Susceptible to occasional disturbances 3. Resistant to occasional disturbances 4. More variation in the productivity.

Stable Community Features

Q.51 Which organism's species have greater diversity in the Western Ghats than the Eastern Ghats? 1. Mammals 2. Amphibians 3. Reptiles 4. Pisces

Western Ghats vs Eastern Ghats

Q.50 Which of the following statements made with regards to plasmid incompatibility are true ? (A). Plasmids having the same origin of replication, belong to the same incompatibility group. (B). Plasmids having different origins of replication, belong to the same incompatibility group. (C). Plasmids of the same incompatibility group are compatible. (D). Plasmids of the same incompatibility group are incompatible. Choose the correct answer from the options given below: 1. (A) and (C) only. 2. (A) and (D) only. 3. (B) and (C) only. 4. (B) and (D) only.

Plasmid Incompatibility

Q.49 Which of the following statements about cortisol are correct? (A). It decreases the circulating lymphocytes. (B). It increases the circulating eosinophils. (C). It decreases the production of prostaglandins (D). It inhibits the production of fibroblasts (E). It alters immune system responses and suppresses the digestive and the reproductive systems. Choose the correct answer from the options given below: 1. (A), (B) and (D) only. 2. (B), (C) and (D) only. 3. (A), (C), (D) and (E) only. 4. (A), (B), (C) and (E) only.

Cortisol Statements Correct?

Q.48 Estrous and menstrual cycle are two types of cycles, that occur in female mammals. Both cycles begin after sexual maturity in females. Reproductive hormones induce these cycles in females. These cycles happen in order to prepare female mammals, for holding a child during pregnancy. The primary difference between estrous and menstrual cycles is - 1. The endometrium shed by the uterus, during the estrous cycle, is reabsorbed, but the shed endometrium is excreted from the body, during the menstrual cycles. 2. Behavioural changes, during estrous cycles are much less pronounced effects, on estrous cycle, than they do on menstrual cycles. 3. Season and climate have less pronounced effects on estrous cycle, than they do on menstrual cycles. 4. Most estrous cycles are of much longer duration, compared to menstrual cycles.

Estrous vs Menstrual Cycle Primary Difference

Q.47 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: 1. (E), (D), (C), (B), (A). 2. (B), (A), (C), (E), (D). 3. (E), (B), (C), (D), (A). 4. (B), (C), (D), (E), (A).

Blood Clot Formation Steps

Q.46 Match List-I with List-II List-I (Structure) List-II (Function) (A) Ascending loop of Henle (I) Reabsorption of glucose (B) Descending loop of Henle (II) Sodium is actively thrown out (C) Proximal tubule (III) Diffusion of water (D) Collecting duct (IV) Passive entry of sodium Choose the correct answer from the options given below: (A) – (II), (B) – (IV), (C) – (I), (D) – (III) (A) – (II), (B) – (IV), (C) – (III), (D) – (I) (A) – (II), (B) – (III), (C) – (I), (D) – (IV) (A) – (IV), (B) – (II), (C) – (III), (D) – (I)

Loop of Henle Matching

Q.45 Match List-I with List-II List-I (Technique) List-II (Connecting / Purpose) (A) DNA footprinting (I) Protein–protein interaction (B) Yeast two hybrid system (II) VNTR (C) DNA fingerprinting (III) DNA binding protein (D) SAGE (IV) Transcriptome analysis Choose the correct answer from the options given below: (A) – (I), (B) – (II), (C) – (IV), (D) – (III) (A) – (III), (B) – (I), (C) – (II), (D) – (IV) (A) – (III), (B) – (IV), (C) – (I), (D) – (II) (A) – (IV), (B) – (II), (C) – (I), (D) – (III)

DNA Footprinting & SAGE Matching

Q.44 Match List-I with List-II List-I (Phase) List-II (Characteristics of the phase) (A) Lag Phase (I) Dying cells are higher than the dividing cells. (B) Log Phase (II) Number of new cells equal to dying cells. (C) Stationary Phase (III) Adaptation to the nutrient environment. (D) Decline Phase (IV) Binary fission is at maximum rate. Choose the correct answer from the options given below: (A) – (I), (B) – (III), (C) – (IV), (D) – (II) (A) – (IV), (B) – (I), (C) – (III), (D) – (II) (A) – (II), (B) – (III), (C) – (I), (D) – (IV) (A) – (III), (B) – (IV), (C) – (II), (D) – (I)

Bacterial Growth Curve Phases Matching

Q.43 Which of the following statements is true concerning infrared spectroscopy ? 1. Functional groups can be identified by looking in the fingerprint region of the spectrum. 2. It shines infrared light on a compound and records the position, where the light is blocked by the compound. This results in the spectrum peaks. 3. It is useful in determining the size and shape of a compound's carbon skeleton. 4. When the infrared light frequency matches the frequency of bond vibration in a molecule, a peak is recorded on the spectrum.

Infrared Spectroscopy

Q.42 Cytoskeleton of eukaryotic cells consists of (A). Peptidoglycans (B). Microtubules (C). Chitin (D). Microfilaments (E). Cellulose Choose the correct answer from the options given below: 1.(B) and (D) only 2. (A), (B) and (C) only 3. (A), (B) and (D) only. 4 (B),(D) and (E) only

Cytoskeleton of Eukaryotic Cells

Q.41 Which two embryological structures together form the bilaminar disc? Cytotrophoblast and Hypoblast Epiblast and Hypoblast Cytotrophoblast and Epiblast Syncytiotrophoblast and Embryoblast

Which Two Embryological Structures Form the Bilaminar Disc?

Q.34 Match List-I with List-II List-I (Structure) List-II (Details relating to structure) (A) Diaphysis (I) The regions between the diaphysis and epiphysis (B) Epiphysis (II) A thin membrane that lines the internal bone surface (C) Metaphysis (III) The proximal and distal end of the bone (D) Endosteum (IV) The long cylindrical main portion of the bone Choose the correct answer from the options given below: (A) – (I), (B) – (II), (C) – (IV), (D) – (III) (A) – (I), (B) – (III), (C) – (II), (D) – (IV) (A) – (IV), (B) – (III), (C) – (I), (D) – (II) (A) – (IV), (B) – (I), (C) – (III), (D) – (II)

Diaphysis Epiphysis Metaphysis Endosteum Matching

Q.40 Which of the following statements are true? (A) Differential gene expression can occur only at the level of transcription (B) RNA polymerase binds to the promoter region to initiate transcription (C) Enhancers regulate transcription (D) Transcription factors can recognize specific sequences of DNA in heterochromatin region Choose the correct answer from the options given below: (A), (B) and (D) only. (B) and (C) only. (B), (C) and (D) only. (A) and (C) only.

Gene Expression and Transcription

Q.39Arrange the following evolutionary events in the history of horse evolution chronologically – (A) Miohippus (B) Orohippus (C) Hyracotherium (Eohippus) (D) Merychippus Choose the correct answer from the options given below: (A), (B), (C), (D) (C), (B), (D), (A) (B), (A), (D), (C) (C), (B), (A), (D)

Horse Evolution Chronological Order

Q.38 Arrange the following animals on the basis of their phyla, according to increasing level of complexity: Scolopendra Nereis Dentalium Euplectella Physalia Choose the correct answer from the options given below: (D), (E), (B), (C), (A) (D), (E), (C), (B), (A) (E), (D), (A), (B), (C) (E), (D), (B), (C), (A)

Arrange Animals by Phyla Increasing Complexity

Q.37 Arrange the following in the correct sequence of DNA packaging – (A) DNA scaffold (B) Nucleosome formation (C) Chromatosome formation (D) Histone dimerization Choose the correct answer from the options given below: (D), (A), (C), (B) (D), (C), (B), (A) (B), (A), (D), (C) (B), (C), (D), (A)

DNA Packaging Sequence

Q.36 Arrange the events in correct order of sequence: Cleavage Cortical rotation Mid-blastula transition Gastrulation Choose the correct sequence answer from the options given below: (A), (B), (C), (D) (C), (B), (A), (D) (B), (A), (C), (D) (B), (C), (A), (D)

Cleavage Mid-Blastula Transition Gastrulation Sequence

Q.35 Match List-I with List-II List-I (Species) List-II (Identifying Characters of species) (A) Allopatric species (I) Species inhabiting the same geographical area (B) Sympatric species (II) Species inhabiting different geographical areas (C) Sibling species (III) Species occupying separate areas that share a common boundary (D) Parapatric species (IV) Species which are morphologically identical but reproductively isolated Choose the correct answer from the options given below: (A) – (II), (B) – (I), (C) – (IV), (D) – (III) (A) – (I), (B) – (II), (C) – (III), (D) – (IV) (A) – (I), (B) – (III), (C) – (IV), (D) – (II) (A) – (II), (B) – (IV), (C) – (I), (D) – (III)

Allopatric Sympatric Sibling Parapatric Species Matching

Q.33 Identify the correct statements related to sex-linked inheritance in humans: Males are considered to be hemizygous for genes present on the X-chromosome. Majority of sex-linked traits are carried on the X-chromosome. Genes that are not sex-linked are present in somatic cells and absent in germ cells. 1:1 ratio of males and females in humans is the result of meiosis in males. The single X-chromosome present in the father is inherited by the son. Choose the correct answer from the options given below: (A), (B) and (E) only (B) and (D) only (A) and (B) only (B), (C) and (E) only

Sex-Linked Inheritance in Humans

Q.32 Arrange the following in order of their action – (A) G1 cyclin CDK complex (B) APC–Cdh1 (C) S phase cyclin CDK complex (D) APC–Cdc20 Choose the correct sequence answer from the options given below: (B), (C), (A), (D) (A), (C), (D), (B) (B), (A), (D), (C) (A), (D), (B), (C)

Cell Cycle Regulators Order

Q.31 N-linked glycosylation in proteins of eukaryotes is initiated in the: Lumen of the endoplasmic reticulum Lumen of the Golgi apparatus Inside the nucleus Within endosomes

N-Linked Glycosylation Initiation in Eukaryotes

$Q.30 Rifampicin, an inhibitor of RNA polymerase, is added to a bacterial culture. The culture is immediately divided into three fractions, to which radioactive thymine or uracil or methionine are added, respectively. Five minutes later the cells are lysed and the radioactive label in all macromolecules is monitored. What will you observe? Thymine and methionine incorporated will be greater than uracil. The labels will be equally incorporated. Uracil incorporated is similar to methionine but less than thymine. Uracil incorporated will be greater than methionine but less than thymine.$

Rifampicin RNA Polymerase Inhibition Experiment

Q.29 The Asian mongoose and the American skunk evolved independently of each other, but show similar structures and features (for example, the ability to spray their attacker with musk). The similarities between the two organisms are most likely the result of: Genetic drift Divergent evolution Allopatric speciation Convergent evolution

Asian Mongoose American Skunk Similarities

Q.28 Which of the following is NOT TRUE with respect to control of respiration? Peripheral chemoreceptors respond to changes in O2 and CO2 concentration in the blood. Respiration responds to smaller changes, in the blood concentration of O2 than CO2. Chemoreceptors in the CNS respond to changes in CO2 concentration in the blood. Central chemoreceptors are sensitive to changes in pH of the cerebrospinal fluid

Control of Respiration

Q.27 Place these steps in temporal order, first to last: Exocytosis of synaptic vesicles. Binding of neurotransmitter molecules to its receptors. Opening of voltage-gated Ca2+ channels. Opening of Na+ channels and influx of sodium ions. Generation of post synaptic potential. Choose the correct answer from the options given below: (C), (A), (B), (D), (E) (B), (D), (C), (E), (A) (B), (A), (D), (E), (C) (C), (B), (D), (A), (E)

Synaptic Transmission Steps Temporal Order

Q.26 Match List-I with List-II List-I List-II Hormone Function (A) Leptin (I) Conversion of glycogen to glucose-1 phosphate (B) Pancreatic polypeptide (II) Suppresses appetite (C) Parathyroid hormone (III) Inhibits somatostatin secretion (D) Epinephrine in skeletal muscle (IV) Promotes formation of calcitriol Choose the correct answer from the options given below: (A) – (IV), (B) – (I), (C) – (II), (D) – (III) (A) – (I), (B) – (III), (C) – (II), (D) – (IV) (A) – (I), (B) – (II), (C) – (IV), (D) – (III) (A) – (II), (B) – (III), (C) – (IV), (D) – (I)

Leptin Parathyroid Hormone Epinephrine Matching

Q.25 Which one of the following are correct characteristic features set for bony fishes? Heterocercal caudal fin; 5–7 pairs of gill slits; exoskeleton is made up of cycloid scales. Homocercal caudal fin; 4 pairs of gill slits; ventrally placed mouth. Homocercal caudal fin; 5–7 pairs of gill slits; presence of air bladder. Homocercal caudal fin; 4 pairs of gill slits; presence of operculum.

Characteristic Features of Bony Fishes

Q.24 Select the vegetation type having highest productivity per unit weight (g leaf biomass-1 year-1) from the following sites: Desert Deciduous forest Coniferous forest Grassland

Vegetation Type with Highest Productivity per Unit Weight

Q.23 Select the plant or ecosystem known to have highest energy conversion efficiency: Tropical forest plantation Sugar cane Cornfield Open sea

Highest Energy Conversion Efficiency

Q.22 Papain digestion of IgG antibody molecule results into – 2 Fab fragments and 1 Fc fragment 2 Fc fragments and 1 Fab fragment 4 Fab fragments and 2 Fc fragments 1 Fab and 1 Fc fragment

Papain Digestion of IgG Antibody

Q.21 In an experiment, two fly populations are separately maintained for many generations. Population A contains closely related individuals, whereas Population B contains a set of unrelated individuals. Over many generations, it was observed that the fitness of individuals in Population A is lower than the fitness of those in Population B, because: Close relatives will not mate with one another. Close relatives compete for common resources more than unrelated individuals. The offspring of close relatives harbour more deleterious recessive mutations. Unrelated individuals are likely to contain new high-fitness genes.

Inbreeding Depression in Fly Populations

Q.20 In the process of exploring eukaryotic 80S ribosome, the following components may be observed: 28S rRNA, 5.8S rRNA 5S rRNA and 34 proteins 23S rRNA 5S rRNA and 49 proteins Choose the correct answer from the options given below: (A) and (D) only (A) and (B) only (A), (C) and (D) only (A), (B) and (C) only

Eukaryotic 80S Ribosome Components

Q.19 Match List-I with List-II List-I List-II (Hormone) (Site of synthesis) (A) Progesterone (I) Adrenal cortex (B) Testosterone (II) Corpus luteum (C) Estrone (III) Testis (D) Aldosterone (IV) Ovary Choose the correct answer from the options given below: (A) – (I), (B) – (III), (C) – (II), (D) – (IV) (A) – (II), (B) – (III), (C) – (I), (D) – (IV) (A) – (II), (B) – (III), (C) – (IV), (D) – (I) (A) – (I), (B) – (III), (C) – (IV), (D) – (II)

Match List-I with List-II

Q.18 Which of the following statements related to transcription in bacteria/eukaryotes are correct? During concurrent promoter sequence recognition and melting, melting commences with base flipping, where two bases are flipped out into pockets of the primary sigma factor. Binding of α-amanitin to RNA polymerase II permits entry of nucleotides into RNA polymerase II active site and synthesis of RNA, but prevents translocation. RNA polymerase I cannot bind to upstream promoters. FACT is associated with RNA polymerase during transcriptional elongation and helps displace one histone octamer during transcription. Choose the correct answer from the options given below: (A), (B) and (D) only (A), (B) and (C) only (B), (C) and (D) only (B) and (D) only

Bacterial Eukaryotic Transcription

Q.17 In the process of protein sequencing, trypsin and chymotrypsin are used for generating fragments. Select tryptic peptide from the list – Try Phe Trp Lys

Trypsin Cleavage Sites in Protein Sequencing

Q.16 After electrophoresis, the gel containing the separated proteins is immersed in an acid alcoholic solution of the dye. This: Denatures the proteins Fixes them in the gel Ensures they do not wash out Allows dye to bind them Choose the correct answer from the options given below: (A), (C) and (D) only (A), (B) and (C) only (A), (B), (C) and (D) (B), (C) and (D) only

Protein Electrophoresis Gel Staining

Q.15 Following are few statements with reference to Biostatistics. Choose the statement/s which are incorrect. (A) Type II error : α - error (B) Type I error : β - error (C) Correct null hypothesis is rejected in β - error (D) Wrong null hypothesis is accepted in α - error Choose the correct option from the options given below: (A) and (B) only. (A), (B) and (C) only. (A), (B), (C) and (D). (B), (C) and (D) only.

Biostatistics Type I Error and Type II Error

Q.14 Calculate the harmonic mean of the numbers 4, 5, and 10 collected in one experiment. 6.83 5.45 7 6

Harmonic Mean of 4, 5, and 10

Q.13 Match List-I with List-II List-I List-II (Statistical Terminology) (Definition) (A) Median (I) The reciprocal of the average of reciprocals of the values of items of a series. (B) Mean (II) The value of the middle item of series when it is arranged in ascending or descending order of magnitude. (C) Mode (III) The value obtained by dividing the total of the values of various given items in a series by the total number of items. (D) Harmonic mean (IV) The most commonly or frequently occurring value in a series. Choose the correct answer from the options given below: (A) – (I), (B) – (II), (C) – (III), (D) – (IV) (A) – (II), (B) – (III), (C) – (IV), (D) – (I) (A) – (I), (B) – (II), (C) – (IV), (D) – (III) (A) – (III), (B) – (IV), (C) – (I), (D) – (II)

Statistical Terminology Match

Q.12 Identify the rule that suggests that the endothermic animals from cold climates tend to have shorter extremities (ears and limbs) compared with animals from warmer climate. Bergmann's rule Allen's rule Bottleneck effect rule Survival of fittest

Endothermic Animals Cold Climates Shorter Extremities

Q.11 Match List-I with List-II List-I List-II (Pollutant) (Characteristics) (A) Sulphur dioxide (I) Colourless gas with an offensive ‘rotten egg’ odor, slightly water soluble (B) Nitrogen dioxide (II) Colourless, heavy, water soluble gas, with a pungent odor (C) Ozone (III) Reddish-brown gas, slightly water soluble (D) Hydrogen sulphide (IV) Pale-blue gas, water soluble, unstable, sweetish odor Choose the correct answer from the options given below: (A) – (I), (B) – (II), (C) – (III), (D) – (IV) (A) – (I), (B) – (III), (C) – (II), (D) – (IV) (A) – (II), (B) – (III), (C) – (IV), (D) – (I) (A) – (III), (B) – (IV), (C) – (I), (D) – (II)

Pollutant Matching List-I List-II

Q.10 Which of the following statements are correct regarding Immunoglobulin A (IgA)? IgA is the main class of antibody in secretions such as tears and saliva. IgA is also found in secretions of lungs and intestine. IgA is the first line of defense against infection at these sites. IgA is the main immunoglobulin in the bloodstream late in the primary immune response. Choose the correct answer from the options given below: (A), (B) and (D) only (A), (B) and (C) only (A), (B), (C) and (D) (B), (C) and (D) only

Immunoglobulin A IgA Functions

Q.9 Match List-I with List-II List-I List-II (Vitamin precursor) (Deficiency disease) (A) Vitamin B1 (I) Pernicious anemia (B) Vitamin B2 (II) Beriberi (C) Vitamin B6 (III) Growth retardation (D) Vitamin B12 (IV) Dermatitis Choose the correct answer from the options given below: (A) – (I), (B) – (II), (C) – (III), (D) – (IV) (A) – (II), (B) – (III), (C) – (IV), (D) – (I) (A) – (I), (B) – (II), (C) – (IV), (D) – (III) (A) – (III), (B) – (IV), (C) – (I), (D) – (II)

Vitamin Deficiency Matching

Q.8 Match List-I with List-II List-I List-II (Coenzyme) (Precursor) (A) Coenzyme A (I) Vitamin B1 (B) FAD, FMN (II) Pantothenic Acid (C) NAD+, NADP+ (III) Vitamin B2 (D) Thiamine pyrophosphate (IV) Niacin Choose the correct answer from the options given below: (A) – (I), (B) – (II), (C) – (III), (D) – (IV) (A) – (I), (B) – (III), (C) – (II), (D) – (IV) (A) – (II), (B) – (III), (C) – (IV), (D) – (I) (A) – (III), (B) – (IV), (C) – (I), (D) – (II)

Coenzyme Precursor Matching

Q.7 Match List-I with List-II List-I List-II Enzyme Type of reaction catalyzed (A) Alcohol Dehydrogenase (I) Bond formation coupled to ATP hydrolysis (B) Trypsin (II) Transfer of electrons (C) Hexokinase (III) Hydrolysis reactions (D) Pyruvate carboxylase (IV) Transfer of functional groups Choose the correct answer from the options given below: (A) – (I), (B) – (II), (C) – (III), (D) – (IV) (A) – (I), (B) – (III), (C) – (III), (D) – (IV) (A) – (II), (B) – (III), (C) – (IV), (D) – (I) (A) – (II), (B) – (III), (C) – (I), (D) – (IV)

Enzyme Matching List-I List-II

Q.6 Read the following structure which is a combination of: Clinical Syndrome : Karyotype : Chromosome Formula : Estimated Frequency at Birth (A) Down Syndrome : 47, +13 : 2n+1 : 1/20,000 (B) Patau Syndrome : 47, +21 : 2n+1 : 1/2,000 (C) Edward Syndrome : 47, +18 : 2n+1 : 1/8000 (D) Klinefelter Syndrome : 47, XXY : 2n+1 : 1/500 male births Choose the correct answer for the above statements from the options given below: (A), (B) and (D) only. (C) and (D) only. (A), (B), (C) and (D). (B), (C) and (D) only.

Clinical Syndromes Karyotype and Chromosome Formula

Q.5 People with hemophilia are unable to produce a factor needed for blood clotting; the cuts, bruises, and wounds of hemophiliacs continue to bleed and, if not stopped by transfusion with clotting factor, can cause death. Select the correct statement about hemophilia: (A) The principal type of hemophilia in humans is due to a recessive X-linked mutation. (B) These males have inherited the mutation from their heterozygous mothers. (C) Affected males never transmit the mutant allele to their sons. (D) Nearly all the individuals who have hemophilia are male. Choose the correct answer from the options given below: (A), (B) and (D) only. (A), (B) and (C) only. (A), (B), (C) and (D). (A), (B), (C) and (D)

Hemophilia X-Linked Inheritance

Q.4 Match List-I with List-II List-I (Organism) List-II (Number of Chromosome) (A) Fruit fly (Drosophila melanogaster) (I) Diploid number 40 (B) Nematode (Caenorhabditis elegans) (II) Haploid number 06 (C) Human (Homo sapiens) (III) Haploid number 23 (D) Mouse (Mus musculus) (IV) Diploid number 08 Choose the correct answer from the options given below: 1. (A) – (I), (B) – (II), (C) – (III), (D) – (IV) 2. (A) – (II), (B) – (IV), (C) – (III), (D) – (I) 3. (A) – (IV), (B) – (II), (C) – (III), (D) – (I) 4. (A) – (I), (B) – (IV), (C) – (III), (D) – (II)

Fruit Fly Chromosome Number

Q.3 Gluconeogenesis synthesizes glucose from non-carbohydrate precursors. (A) The main site of gluconeogenesis is the liver. (B) Far lesser extent of gluconeogenesis occurs in kidneys. (C) Very little gluconeogenesis occurs in brain or muscle. (D) The second main site of gluconeogenesis is the heart. Choose the correct answer from the options given below: (A), (B) and (D) only. (A), (B) and (C) only. (A), (B), (C) and (D). (B), (C) and (D) only.

Gluconeogenesis Main Site

Q.2 Match List-I with List-II List-I List-II (Amino Acid) (pK value) (A) α-Carboxyl groups (I) 1.8 – 2.9 (B) α-Amino groups (II) 8.8 – 10.8 (C) Acidic Amino Acid – Asp (III) 12.5 (D) Basic Amino Acid – Arg (IV) 3.9 Choose the correct answer from the options given below: (A) – (I), (B) – (II), (C) – (III), (D) – (IV) (A) – (I), (B) – (II), (C) – (IV), (D) – (III) (A) – (I), (B) – (III), (C) – (IV), (D) – (II) (A) – (III), (B) – (IV), (C) – (I), (D) – (II)

Amino Acid pKa Values Matching

Q.1 Sickle cell anemia is a disease characterised by the patient's erythrocyte, having a characteristic sickle or crescent shape. Select the correct statement hut the disease - l. Deletion of glutamic acid from the 6th position of ß-chain. 2 Substitution of hydrophobic residue by a polar one. 3. Substitlon of polar residue by a hydrophobic one 4 Substitution of valine residue by a glutamic acid one.

Sickle Cell Anemia Mutation

Latest Courses

IIT JAM / CUET- PG

Are you aspiring to crack the IIT JAM Biotechnology exam with flying colors? Choosing the right coaching institute is crucial for your success. Let's Talk Academy (LTA) has established itself as the best coaching institute for IIT JAM Biotechnology, helping students achieve their dreams through expert guidance, structured study material, and personalized mentorship.

CSIR UGC NET Life Science Course

Let's Talk Academy is the number one CSIR NET Life Science coaching in Jaipur for Life Science. At Let's Talk Academy, you can be sure to get the right coaching strictly in adherence to the CSIR NET Life Science course. CSIR NET Life Science is an exam that requires in-depth knowledge, consistent practice, and the right guidance to crack. If you are looking for a place where you can get the best CSIR NET Life Science coaching, you are at the right place.

ICMR JRF Life Science

If you are serious about cracking the ICMR Life Science JRF exam, Let’s Talk Academy is your ultimate destination. Join now and take the first step toward a successful research career!