Life Sciences MCQ Quiz - Objective Question with Answer for Life Sciences - Download Free PDF

The correct answer is Primitive endoderm (hypoblast)

Explanation:

• During mammalian embryonic development, extraembryonic structures are formed to support the growth and nourishment of the embryo. One of these structures is the yolk sac, which plays a crucial role in early nutrition and hematopoiesis (formation of blood cells).
• The yolk sac is an early extraembryonic structure that originates from the primitive endoderm, also known as the hypoblast.
• It is essential for providing nutrients to the developing embryo before the establishment of placental circulation.
• Primitive endoderm (hypoblast):
  • The yolk sac forms from the primitive endoderm (hypoblast), which is a layer of cells beneath the embryonic epiblast.
  • The hypoblast contributes to the formation of the outer layer of the yolk sac, which is involved in nutrient transfer and initial hematopoiesis.
  • Additionally, the yolk sac serves as a site for the migration of primordial germ cells to the developing gonads.
  • This structure is transient and diminishes in importance as the placenta develops to take over the role of nutrient and gas exchange.

Other Options (Incorrect):

• Syncytiotrophoblast: The syncytiotrophoblast is a layer of the trophoblast that invades the uterine wall to facilitate implantation of the embryo. Its primary function is to form part of the placenta and help in nutrient and gas exchange, but it does not form the yolk sac.
• Amniotic ectoderm: The amniotic ectoderm is derived from the embryonic epiblast and lines the amniotic cavity. Its role is to produce the amniotic fluid that cushions and protects the embryo, but it is not involved in the formation of the yolk sac.
• Embryonic epiblast: The embryonic epiblast is the source of all embryonic germ layers (ectoderm, mesoderm, and endoderm) during gastrulation.

The correct answer is P-3; Q-4; R-2; S-1

Explanation:

• Arabidopsis thaliana is a model organism in plant biology, and its genes play key roles in regulating various biological processes, such as meristem development, floral transition, organ identity, and cell specification.

Explanation:

• (P) CLAVATA3 (3: Meristem size in shoot):
  • CLAVATA3 (CLV3) encodes a signaling peptide that regulates the size of the shoot apical meristem (SAM).
  • It works in conjunction with other genes like CLAVATA1 and WUSCHEL (WUS) to maintain the balance between stem cell proliferation and differentiation in the shoot meristem.
  • Mutations in CLV3 result in an enlarged meristem due to excessive stem cell proliferation.
• (Q) CONSTANS (4: Photoperiodic floral transition):
  • CONSTANS (CO) regulates flowering time in response to photoperiod (day length).
  • It promotes the expression of the FLOWERING LOCUS T (FT) gene, which induces flowering under long-day conditions.
• (R) SCARECROW (2: Cell-type specification in root meristem):
  • SCARECROW (SCR) is involved in cell-type specification and patterning in the root meristem.
  • It is required for the proper development of the endodermis and cortex layers in the root.
  • SCR works with SHORT-ROOT (SHR) to regulate asymmetric cell divisions and maintain the radial organization of the root.
• (S) AGAMOUS (1: Organ identity in flower):
  • AGAMOUS (AG) is a homeotic gene that regulates organ identity in the flower, specifically the development of stamens and carpels (whorls 3 and 4).
  • It belongs to the MADS-box family of transcription factors and is essential for proper floral organ development and termination of floral meristem activity.

The correct answer is Base-excision

Explanation:

• DNA repair systems are critical mechanisms that maintain the integrity of genetic material by correcting damage that occurs due to environmental factors, replication errors, or cellular processes.
• Base-excision repair (BER) is a specialized DNA repair pathway that fixes small, non-helix-distorting base lesions caused by oxidation, alkylation, deamination, or spontaneous hydrolysis.
  • BER is initiated when DNA glycosylases recognize and excise damaged bases, leaving behind an abasic site (also known as an AP site).
  • The AP site is then processed by an enzyme called AP endonuclease, which cuts the DNA backbone at the site.
  • A DNA polymerase fills in the missing nucleotide(s), and DNA ligase seals the nick in the DNA strand, completing the repair.

Other Options:

• Direct repair:
  • Direct repair is a mechanism that directly reverses specific types of DNA damage without removing the damaged nucleotide or disrupting the DNA backbone.
  • For example, photolyase enzymes repair UV-induced thymine dimers through photoreactivation, and O₆-methylguanine-DNA methyltransferase (MGMT) repairs alkylated guanine bases by transferring the alkyl group to itself.
• Mismatch repair (MMR):
  • MMR corrects replication errors such as mispaired bases or small insertion/deletion loops that escape proofreading by DNA polymerase.
  • It involves proteins like MutS and MutL that recognize mismatches, recruit endonucleases to excise the erroneous DNA segment, and allow DNA polymerase to resynthesize the correct sequence.
  • MMR does not involve DNA glycosylases, as it addresses replication errors rather than damaged bases.
• Nucleotide-excision repair (NER):
  • NER repairs bulky, helix-distorting lesions such as thymine dimers and large chemical adducts caused by UV radiation or chemical exposure.
  • The pathway involves the excision of a short single-stranded DNA segment containing the damage, followed by resynthesis using the complementary strand as a template.
  • NER does not require DNA glycosylases, as it removes lesions through a different mechanism involving excision and resynthesis.

The correct answer is Predatory spiders.

Key Points

• Predatory spiders have a high assimilation efficiency (over 90%) due to their ability to effectively convert prey biomass into their own biomass.
• Spiders are efficient predators, consuming prey with minimal energy wastage, which allows them to achieve high energy transfer rates.
• They utilize external digestion by secreting digestive enzymes, which enhances nutrient assimilation.
• Predatory spiders are important in regulating prey populations and maintaining ecological balance.

Additional Information

• Secondary carnivores: These are organisms that feed on primary carnivores. While they are efficient predators, their assimilation efficiency is generally lower than predatory spiders as they consume organisms with more complex energy pathways.
• Herbivores: Herbivores feed on plant material, which is difficult to digest due to the presence of cellulose and lignin. Their assimilation efficiency is typically lower (around 20–50%) compared to carnivores.
• Social insects: Social insects like ants and bees have specialized roles within colonies. Their efficiency depends on their dietary habits, but they generally do not achieve assimilation efficiencies as high as predatory spiders.

The correct answer is The presence of a network of water-filled tubes for movement.

Explanation:

• The absence of sexual reproduction: This is not a distinguishing feature because both Echinoderms and Cnidarians are capable of sexual reproduction. While there are variations in their reproductive strategies, both groups utilize sexual reproduction as a means to propagate their species.
• The presence of radial symmetry: This feature is shared by both Echinoderms and Cnidarians. Cnidarians (such as jellyfish and sea anemones) exhibit radial symmetry, typically in the form of a basic radiate symmetry. Echinoderms (such as starfish and sea urchins) also exhibit a type of radial symmetry, usually pentaradial symmetry in adulthood.
• Having a central cavity involved in digestive processes: Both Echinoderms and Cnidarians have a cavity that plays a crucial role in their digestive system. In Cnidarians, the gastrovascular cavity (sometimes simply called the coelenteron) is a central cavity in which digestion takes place, and it also helps distribute nutrients throughout the organism's body. It has a single opening that functions as both mouth and anus. Echinoderms, while having a more complex digestive system including separate mouth and anus in many forms, also possess internal cavities that serve various functions including digestion. Their digestive system includes a stomach and intestine, and they have internal water vascular systems that support their physiological processes, which includes nutrient distribution.
• The presence of a network of water-filled tubes for movement: This feature is exclusive to Echinoderms. Echinoderms possess a unique water vascular system, which is a network of water-filled canals that aid in locomotion, feeding, and gas exchange. The system includes structures such as tube feet (or podia), ampullae, and the madreporite. This system allows for complex movements and is a defining characteristic of the phylum Echinodermata. Cnidarians do not have such a system; they primarily rely on simple muscle contractions and passive drifting for movement.

The correct answer is Option 4 i.e.Nus A is necessary for intrinsic transcription termination.

Concept:

• Bacterial transcription termination serves two important purposes:
  • regulation of gene expression
  • recycling of RNA polymerase (RNAP)
• Bacteria have 2 major modes of termination of bacterial RNA polymerase (RNAP):
  • Intrinsic (Rho-independent)
  • Rho-dependent

Intrinsic Termination - 

• Intrinsic termination occurs by the specific sequences present in the mRNA sequence itself. 
• These RNA sequences form a stable secondary hairpin loop-type structure signaling for termination. 
• The base-paired region called the stable 'stem' consists of 8-9 'G' and 'C' rich sequences.
• The stem is followed by 6-8 ‘U’ rich sequences.
• Intrinsic transcription terminators consist of an RNA hairpin followed by Uridine-rich nucleotide sequences.
• Intrinsic termination needs two major interactions: 1) nucleic acid elements with 2) RNAP.
• Additional interacting factors like Nus A, could enhance the efficiency of termination but not necessary for intrinsic termination.

Rho-dependent Termination - 

• Rho-dependent termination on the other hand requires Rho protein which is an ATP-dependent RNA hexamer translocase (or helicase). 
• Rho protein binds with ribosome-free mRNA and 'C' rich sites on the mRNA (Rut site).

Explanation:

Option 1: Some terminator sequences require Rho protein for termination

• Since Rho protein is required for termination this option is correct. 

Option 2: Inverted repeat and ‘T’ rich non‐ template strand define intrinsic terminators.

• The image given below represents a pre-requisite template for the intrinsic terminator.
• We can find a T-rich sequence on the non-template DNA strand.
• The inverted repeat sequence is also present and helps in the formation of the hairpin loop (as shown in the image).
• Hence, the statement is correct.

Option 3: Rho-dependent terminators may possess inverted repeat elements.

• In some cases, Rho-dependent terminators could possess inverted repeat elements, but Rho proteins do not rely on these inverted repeat elements for their action.
• Hence the statement is correct.

Option 4: Nus A is necessary for intrinsic transcription termination.

• NusA is not a necessary element for intrinsic transcription termination.
• It might enhance transcription termination in some cases but only as an accessory element.
• Hence, this option is incorrect.

Additional Information

Other mode of termination -

• It is reported in bacteria and is Mfd dependent.
• Mfd-dependent termination occurs with the help of Mfd protein which is a type of DNA translocase and requires ATP for its action just like Rho.

Hence, the correct answer is option 4.

The correct answer is Option 3 i.e.Cdc45

Concept:

• DNA replication in eukaryotes could be divided into three major parts:
  • Initiation
  • Elongation
  • Termination.
• DNA replication initiation could be divided into:
  •  pre-replicative complex
  •  initiation complex.
• Pre-replicative complex majorly consists of
  • ORC (origin recognition complex)+ Cdc6 + Cdt + MCM complex (mini-chromosome maintenance complex)
• Initiation complex consists of 
  • Cdc45 + MCM 10 + GINS + DDK and CDK kinase + Dpb11, Sld3, Sld2 protein complex.

Explanation:

• All the proteins given in the options belong to eukaryotic cells and so we must consider only eukaryotic DNA replication here.

Option 1: ORC - INCORRECT

• DNA replication is initiated from the origin of replication, having specific sequences to initiate replication.
• The ORC is a hexameric DNA binding complex that binds with the origin of replication followed by the recruitment of Cdc6 protein followed by Cdt1.
• ORC dephosphorylates and becomes inactivated before the elongation process.
• Hence, this option is incorrect. 

Option 2: Geminin - INCORRECT

• It binds to cdt1 to prevent the re-initiation of DNA replication and hence it works as a regulator/inhibitor rather than an initiator of DNA replication.
• It is an inhibitor of Cdt1.

Option 3: Cdc45 - CORRECT

• Cdc refers to the cell division control proteins that are involved in various steps of DNA replication process.
• Cdc45 remains with MCM complex and GINS to work as a helicase.
• Thus, it helps in the initiation of DNA replication as well as advancement of the replication fork.

​Option 4: Cdc6 - INCORRECT

• It helps in the assembling of the pre-replicative complexes and interacts with the ORC.
• Cdc6 degrades before initiation of the replication fork.
• The concentration of both cdc6 and cdt1 declines before DNA elongation starts.

​Hence, the correct option is option 3.

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.

The correct answer is  Pseudopodia.

Key Points

• The change in shape in amoeba which facilitates movement is due to Pseudopodia.
• The pseudopodia extends from the two sides of the food molecule and surrounds it and finally engulfs the food.
• Pseudopodia is used in movement and as a tool to capture prey or obtain required nutrition.

Structure of Amoeba:

Additional Information 

The correct answer is Option 4 i.e.Brugiya malayi

Concept:

• Filariasis is a parasitic disease caused by the spread of roundworms belonging to the Filarioidea type.
• These parasites are spread via mosquitoes or other blood-feeding insects.
• This disease is found in subtropical regions (hot, humid, and damp regions) such as South Asia, South Africa, the South Pacific, and parts of South America.
• Humans are their definitive hosts.

• Depending on the major affected areas of the human body, this disease is categorized into:
  • Lymphatic filariasis
  • Subcutaneous filariasis
  • Serous cavity filariasis

Explanation:

Option 1: Listeria monocytogenes

• It is a pathogenic bacteria that causes listeriosis.
• It is usually transmitted by contaminated food.
• It causes serious infection and severely affects pregnant women and older people.
• Hence, this option is incorrect.

Option 2: Cryptococcus neoformans

• This is an encapsulated yeast-like fungus that causes cryptococcal meningitis.
• It is life-threatening only in immunocompromised patients like AIDS patients.
• Hence, this option is incorrect.

Option 3: Francisella tularensis

• It is a gram-negative bacteria that causes tularemia.
• It is a zoonotic pathogen that causes febrile conditions in affected person.
• In this disease, the affected person suffers from respiratory troubles like cough and breathing problems.
• Hence, this option is incorrect.

Option 4: Brugiya malayi

• It is one of the filarial nematodes that causes lymphatic filariasis in humans.
• The Mansonia and Aedes mosquitoes are the known vectors for this nematode species.
• They are exclusively found in south-east Asia.
• Hence, this option is correct.

Hence, the correct answer is Brugiya malayi.

The correct answer is Option 4 i.e.Opines are a source of nitrogen for Agrobacterium cells.

Concept:

• Agrobacterium tumefaciens and A. rhizogenes are two species of Agrobacterium that are gram-negative soil bacteria.
• Ti plasmid is a plasmid that is present in the A. tumefaciens.
• Agrobacterium are also called natural genetic engineers because of their ability to naturally transformed plants.
• A. tumefaciens causes crown gall disease in some plants.

Ti plasmid - 

• It is a large-sized tumour-inducing plasmid. 
• A. tumifaciens infect wounded or damaged plant tissue and cause the formation of Crown gall disease. 
• Following are three important regions present in Ti-plasmid.

1. T-DNA region - 
   • This region has genes for the synthesis of auxin, cytokinins, and opine. It is flanked by the left and right borders of the T-DNA region.
   • It contains a set of 24-kb sequences that is flanked on either side of the T-DNA region. 
   • Right border is more critical in transfer of Ti plasmid in plant.  
2. Virulence region - 
   • The genes that help in the transfer of the T-DNA in the plant is located outside T-DNA. 
   • At least, 9 different vir genes are identify in the plant. 
3. Opine catabolism region - 
   • This region contains genes that are involved in uptake and metabolism of opine.

Important Points

Option 1: INCORRECT

• The virulence region of the Ti plasmid contains 9 vir genes.
• Out of 9 vir genes, vir A and vir G are the only two vir genes that are constitutively expressed. 

Option 2: INCORRECT

• The integration of T-DNA into the plant genome is based on the specific DNA sequence that is present at the right border of T-DNA.
• If any gene or DNA sequence is present in this T-DNA region then it is also transferred and gets integrated into the nuclear genome of the plant. 
• The integration of the T-DNA region in the nuclear genome of a plant occurs at an approximate random location through non-homologous recombination. 

Option 3: INCORRECT

• Host plants encode proteins that play any role in the Agrobacterium-mediated transfer of T-DNA into plant cells and integration of T-DNA region in the plant genome.

Option 4: CORRECT

• Ti plasmid contains the different types of opine genes i.e., nopaline, octopine, and atropine. 
• These opines are condensation products of either amino acids or keto acids or amino acids and sugar. 
• These opines are used as a source of nitrogen and carbon for Agrobacterium.

Hence, the correct answer is Option 4.

The correct answer is Option 4 i.e. Asialoglycoprotein

Concept:

Cytoplasmic domains of any receptor bind with different proteins and signals cell for specific functions accordingly.

A cell has two types of receptors:

1. Cytoplasmic receptors -
   • These are found in the cytoplasm of a cell and respond to hydrophobic ligands.
   • They are also known as internal or intracellular receptors.
2. Transmembrane receptors -
   • These are membrane-anchored receptors that are also known as integral membrane proteins.
   • Transmembrane receptors bind to extracellular signals and transmit them into the intracellular environment.

Tyrosine kinase receptors (RTK) - 

• They belong to the high-affinity cell surface receptor category and aid in the binding of many growth factors, cytokines and hormones.
• RTKs possess intrinsic cytoplasmic enzymatic activity which catalyzes the transfer of phosphate from ATP to a tyrosine residue in protein substrates.
• Epidermal growth factor, platelet-derived growth factor, and insulin function as a protein that binds to tyrosine kinase receptors.

Explanation:

Option 1: Epidermal growth factor (EGF)

• EGF receptor (EGFR) is a transmembrane protein that binds to EGF.
• EGFR contains a cytoplasmic tyrosine kinase active site.
• It is expressed in the human body at multiple locations like gums, placenta, vulva, superficial temporal artery, human penis, urethra, mouth cavity, etc.
• Hence, this option is incorrect.

Option 2: Platelet-derived growth factor (PDGF)

• The receptors of PDGF belongs to family of cell surface tyrosine kinase receptors.
• These function for cell proliferation, cellular growth and differentiation.
• Hence, this option is incorrect.

Option 3: Insulin

• The insulin receptors are heterotetrameric transmembrane glycoproteins.
• It contains 2 α-subunits and 2 β-subunits.
• They have one transmembrane domain and one tyrosine-kinase cytoplasmic domain.
• Hence, this option is incorrect.

Option 4: Asialoglycoprotein

• Asialogycoprotein and glycoprotein binds to asialoglycoprotein receptors (ASGPR).
• ASGPR are transmembrane receptors which are specifically present on hepatocytes (liver cells) and thus also called as hepatic lectin.
• The human ASGPR has 4 functional domains:
  • Cytoplasmic domain
  • Transmembrane domain
  • Stalk
  • Carbohydrate recognition domain (CRD)
• The cytoplasmic or cytosolic domain does not act as a tyrosine kinase here.
• Hence, this option is correct. 

Thus, the correct answer is Asialoglycoprotein.

The correct answer is Protoderm.

Key Points

• Protoderm is the outermost primary meristem in plants.
• In roots, it differentiates to form the epidermis.
• Produces epidermal cells, including root hairs.
• Root hairs play a crucial role in water and nutrient absorption.
• Acts as a protective barrier.
• Facilitates root-soil interactions.

Additional Information

• Axillary buds, located at leaf-stem junctions, can grow into branches or flowers, influenced by hormonal signals and environmental factors. 
• A leaf primordium is the initial embryonic tissue from which a leaf develops, found at the shoot apex or growing tip.
• Vascular tissue in plants consists of xylem and phloem.
• Xylem transports water and minerals from roots, providing structural support with tracheids, vessels, and fibers. 
• Phloem transports sugars and nutrients throughout the plant.   

The correct answer is Cdk2/Cyclin E

Concept:

• Cell cycle is a highly regulated and ordered series of events. The engines that derive the progression from one step of the cell cycle to the next are cyclin-CDK complexes.
• These complexes are composed of two subunits- cyclin and cyclin-dependent protein kinase. Cyclin is a regulatory protein whereas CDK is a catalytic protein and acts as serine/threonine protein kinase.
• Cyclins are so named as they undergo a cycle of synthesis and degradation in each cycle.   
• Humans contain four cyclins- G1 cyclins, G1/S cyclins, S cyclins, and M cyclins.

Explanation:

• Cyclin-CDK complexes trigger the transition from G1 to the S phase and from G2 to the M phase by phosphorylating a distinct set of substrates.
• According to the classical model of cell cycle control, D cyclins and CDK4/CDK6 regulate events in the early G1 phase. Cyclin E-CDK2 regulates the completion of the S-phase.
• The transition from G2 to M is driven by sequential activity of cyclin A-CDK1 and cyclin B-CDK1 complexes.

So, the correct answer is Option 2.