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

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 2: Higher affinity for regulatory region of f than that of c

Concept:

• Gene expression in different cell types is regulated by morphogen concentration gradients. Morphogens are signaling molecules that determine cell fate based on their local concentrations.
• The transcription factor activated by the morphogen binds to the regulatory regions of specific genes to activate or repress their expression.
• The affinity of the transcription factor for the regulatory regions of various genes determines which genes are expressed in response to varying morphogen concentrations.
  • High affinity: The transcription factor can bind and activate gene expression even at low concentrations.
  • Low affinity: The transcription factor requires a high concentration to bind effectively and activate gene expression.

Explanation:

• The transcription factor activated by the morphogen has higher affinity for the regulatory region of gene f than that of gene c. This means gene f is expressed at lower morphogen concentrations compared to gene c. The spatial pattern of gene expression is determined by this differential affinity, ensuring that specific genes are activated in different cell types based on morphogen concentration.
• The higher affinity for f's regulatory region implies that f is expressed earlier (or in cells exposed to lower morphogen concentrations) than c, which requires higher morphogen levels for activation.

Other Options:

• Higher affinity for regulatory region of a than that of d. This is incorrect because 'a' requires higher morphogen (and thus more activated transcription factor) to be expressed than 'd', the transcription factor must have a lower affinity for 'a's regulatory region compared to 'd's.
• Same affinity for regulatory regions of a and b. This is incorrect as morphogen signaling typically induces differential gene expression, meaning the transcription factor does not have identical affinity for two distinct genes.
• Lower affinity for regulatory region of m than that of c. This is incorrect because Gene 'm' is expressed in Cell Types 1, 2, and 3 (all morphogen levels). Gene 'c' is expressed only in Cell Type 1 (highest morphogen). Since 'm' is expressed at lower morphogen levels, the transcription factor has a higher affinity for 'm's regulatory region. Since 'c' requires the highest morphogen level, the transcription factor has a lower affinity for 'c's regulatory region.

The correct answer is It is unaffected by the evenness of species abundances.

Concept:

• The Shannon-Wiener diversity index is a widely used measure for quantifying species diversity in an ecosystem. It accounts for both species richness (the number of species present) and species evenness (how evenly the individuals are distributed among the species).

•  The formula for the Shannon index (H) is often expressed as:

  \(H' = -\sum_{i=1}^{S} p_i \ln(p_i) \)

• In the Shannon index, p is the proportion (n/N) of individuals of one particular species found (n) divided by the total number of individuals found (N), ln is the natural log, Σ is the sum of the calculations, and s is the number of species. 
• The higher the Shannon index, the greater the diversity in a community.
• This index gives higher values when there is greater species richness and evenness, indicating a more diverse ecosystem.

Explanation of the Correct Answer:

• It is unaffected by the evenness of species abundances is incorrect because the Shannon-Wiener diversity index incorporates species evenness into its calculations. Evenness refers to how similar the abundance of different species is within the ecosystem. 
• If abundances are very uneven (e.g., one species is super abundant and others are rare), the p_i(lnp_i)​ term for the dominant species will heavily influence the sum, leading to a lower diversity value.
• The index is sensitive to both richness and evenness. Ecosystems where species are more evenly distributed yield higher index values compared to those where one or a few species dominate.

Brief Overview of Other Options:

• Option 1: "It increases as species richness increases." This is correct because the Shannon-Wiener index rises when more species are present. Greater species richness contributes positively to the index value.
• Option 2: "It is maximized when all species have equal abundances." This is correct because the index reaches its maximum value when all species in the ecosystem have equal proportions (maximum evenness). This state represents the highest level of diversity.
• Option 4: "A low index value indicates dominance of one or a few species." This is correct because ecosystems with low diversity (where one or a few species dominate) yield low Shannon-Wiener index values. Such dominance reduces both richness and evenness.

The correct answer is A, B, and D

Concept:

• A promoter is a DNA sequence that facilitates the initiation of transcription by recruiting transcription machinery, such as RNA polymerase II and transcription factors.
• The TATA box is a highly conserved sequence found in many eukaryotic promoters. It is recognized and bound by the TATA-binding protein (TBP), which is a critical component of the transcription factor complex.
• If the TATA box is mutated and cannot bind TBP, the transcription process may still occur but with reduced efficiency, as other promoter elements and transcription factors may partially compensate for the loss of TBP binding.

Explanation:

• Option A: An mRNA will be generated with an alternate reading frame.
  This is incorrect. A mutation in the promoter or TATA box affects the transcription initiation process, not the reading frame of the mRNA. The reading frame is determined by the coding sequence of the gene and the proper start codon during translation, which remains unaffected by the promoter mutation.
• Option B: The mRNA will be transcribed by RNA polymerase I instead of RNA polymerase II.
  This is incorrect. RNA polymerase I is responsible for transcribing ribosomal RNA (rRNA) genes, while RNA polymerase II transcribes protein-coding genes. A mutation in the TATA box does not alter the specificity of RNA polymerase recruitment. The gene will still be transcribed by RNA polymerase II but with reduced efficiency.
• Option C: Transcription may occur with a reduced efficiency.
  This is correct. The TATA box plays a key role in recruiting TBP and the transcription machinery. If the TATA box is mutated, the binding of TBP is impaired, leading to decreased transcription efficiency. However, transcription may still occur through the activity of other promoter elements or compensatory mechanisms.
• Option D: Transcription may occur but will always result in the formation of a non-functional mRNA.
  This is incorrect. A mutation in the TATA box impacts transcription initiation but does not inherently affect the functionality of the mRNA. The mRNA produced may still be functional if it is correctly transcribed from the gene's coding sequence.

The correct answer is A and B only

Explanation:

• Post-translational modifications (PTMs) are chemical changes to proteins that occur after translation. These modifications play a crucial role in regulating protein function, localization, stability, and interactions.
• Phosphorylation (A):
  • Phosphorylation is the addition of a phosphate group to an amino acid residue, typically catalyzed by kinases.
  • The most commonly phosphorylated amino acids are serine, threonine, and tyrosine. However, histidine can also be phosphorylated, though it is less common and often overlooked in studies.
• Ubiquitination (B):
  • Ubiquitination involves the attachment of ubiquitin, a small protein, to lysine residues on target proteins. This modification regulates protein degradation, signaling, and trafficking.
  • Although lysine is the predominant target for ubiquitination, N-terminal methionine can also be modified under specific circumstances.
• O-linked glycosylation (C):
  • O-linked glycosylation is the attachment of sugar molecules to the hydroxyl group of serine or threonine residues.
  • The table incorrectly lists "Asparagine" as the target residue for O-linked glycosylation. Asparagine is involved in N-linked glycosylation, not O-linked glycosylation.
  • Thus, the match for "O-linked glycosylation" is incorrect.
• Hydroxylation (D):
  • Hydroxylation is the addition of a hydroxyl group (-OH) to amino acid residues, typically catalyzed by hydroxylases. It is commonly seen in proline and lysine residues, particularly in collagen proteins.
  • The table incorrectly lists "Cysteine" as a residue subject to hydroxylation. Cysteine is not typically hydroxylated..

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.