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

The correct answer is Homologous recombination and CRISPR-Cas9

Explanation:

• Gene knock-out refers to a genetic technique where a specific gene in an organism is completely inactivated or "knocked out," leading to the loss of its function. This is a powerful tool in molecular biology for studying gene function and plant development.

CRISPR-Cas9:

• The CRISPR-Cas9 system uses a guide RNA (gRNA) to direct the Cas9 nuclease to a specific target sequence in the genome.
• Once the Cas9 protein binds to the target DNA, it introduces double-strand breaks (DSBs) at the specific location.
• In plants, the cell's natural DNA repair mechanisms, such as non-homologous end joining (NHEJ), introduce insertions or deletions (indels) at the DSB site. These mutations often result in a loss of gene function, effectively "knocking out" the target gene.
• CRISPR-Cas9 is highly versatile, efficient, and precise, making it the preferred choice for generating gene knock-outs in plants.
• Homologous recombination: Homologous recombination is a method for precise gene editing, including knock-outs, by replacing a target gene with a desired sequence. 

Other Options (Incorrect):

• Antisense RNA technique: Antisense RNA technology involves introducing a strand of RNA complementary to the target gene's mRNA, which blocks its translation into protein. This technique results in gene silencing but does not lead to a complete knock-out of the gene at the DNA level.
• Activation tagging: Activation tagging is a technique used to increase the expression of a gene rather than knocking it out. This method involves inserting a strong enhancer element near a target gene to boost its transcription. 

The correct answer is 1

Concept:

• Bacteria reproduce by binary fission, where one bacterium divides into two identical daughter cells. Therefore, after one cycle of growth, the number of bacteria doubles.
• Different organisms, or even different types of genes within one organism, can have varying numbers of copies of a specific gene per genome. This is a crucial factor in determining the initial template quantity for PCR.
• PCR is a laboratory technique widely used in molecular biology to make many copies of a specific DNA segment. PCR relies on thermal cycling, consisting of cycles of denaturation, annealing, and extension.
• In an ideal PCR, each DNA template molecule is theoretically doubled in every cycle. Therefore, after 'n' cycles, the number of DNA copies will be 2ⁿ times the initial number of template molecules.

Explanation:

Let N₀ be the initial bacterial count in the sample.

• Prior to DNA extraction, the bacteria were incubated to allow one cycle of growth. This means the initial bacterial population doubled.
  • Number of bacteria after one growth cycle = N₀ x 2 = 2N₀
• Each bacterium contains 3 copies of the target gene. Therefore, the total number of target gene copies extracted and available to serve as the initial template for PCR amplification will be the number of bacteria after growth multiplied by the gene copies per bacterium:
  • Initial PCR template copies = (Number of bacteria after growth) x (Gene copies per bacterium)
  • Initial PCR template copies = 2N₀ x 3 = 6N₀
• PCR Amplification: The PCR was performed for 9 cycles with 100% efficiency. This means the initial template copies were multiplied by 2⁹
  • 2⁹ = 512
• The total number of amplicon copies obtained after 9 cycles is given as 3072. We can set up the equation:
  • Total Amplicon Copies = (Initial PCR template copies) x 2^{Number of PCR cycles}
  • 3072 = 6N₀ x 2⁹
  • 3072 = 6N₀ x 512
  • 3072 = 3072 N₀
  • N₀ = 1

The correct answer is 2

Concept:

• Bacteria, like Escherichia coli, multiply by binary fission. This means that after a certain period (doubling time), a single bacterium divides into two. If the doubling time is 30 minutes, the population of uninfected bacteria will double every 30 minutes.
• Burst Size refers to the number of new phage particles released from a single lysed bacterial cell.
• Multiplicity of Infection (MOI) is the ratio of the number of infectious agents (e.g., phage particles) to the number of target cells (e.g., bacteria) in a given experiment. An MOI of 1 means, on average, one phage particle is added per bacterial cell. Instantaneous absorption means the infection happens immediately upon phage addition.

Explanation:

We need to determine at what time the phage population grows large enough to infect all available E. coli cells. Once all E. coli cells are infected, they will subsequently lyse within the phage life cycle duration.

• Initial State (Time = 0 minutes):
  • Initial E. coli cells (N₀) = 2 x 10⁷
  • Initial T7 phage (PFU) = 5000
  • Phage absorption is instantaneous at MOI 1. So, 5000 E. coli cells are infected immediately.
  • Number of E. coli cells remaining uninfected = 2 x 10⁷ - 5000. This is approximately 2 x 10⁷ (since 5000 is negligible compared to 2 x 10⁷).
  • The 5000 infected cells will lyse after 20 minutes (phage life cycle).
• First Phage Burst (Time = 20 minutes):
  • The 5000 E. coli cells infected at t=0 lyse.
  • Number of new phages released = (Number of cells lysed) x (Burst size) = 5000 x 200 = 1,000,000 = 1 x 10⁶ phages.
  • During these 20 minutes, the uninfected E. coli cells have been growing. The growth factor for 20 minutes (given a 30-minute doubling time) is 2^(20/30) = 2^{2/3}  =1.587.
  • Approximate number of uninfected E. coli cells available for infection at t=20 min = 2 x 10⁷ x 2^{2/3} = 3.174 x 10⁷ cells. (The 5000 initial infected cells are no longer part of the viable population.)
  • At t=20 min, the 1 x 10⁶ newly released phages infect 1 x 10⁶ additional E. coli cells (since 1 x 10⁶ < 3.174 x 10⁷). These newly infected cells will lyse at t = 20 + 20 = 40 minutes.
• Second Phage Burst (Time = 40 minutes):
  • The 1 x 10⁶ E. coli cells infected at t=20 min lyse.
  • Number of new phages released = 1 x 10⁶ x 200 = 200 x 10⁶ = 2 x 10⁸ phages.
  • During the period from t=20 min to t=40 min (another 20 minutes), the *remaining uninfected* E. coli cells continued to grow. The approximate uninfected population at t=20 was 3.174 x 10⁷ - 1 x 10⁶ = 3.074 x 10⁷ cells.
  • Number of uninfected E. coli cells available for infection at t=40 min = (3.074 x 10⁷) x 2^{2/3} = 3.074 x 10⁷ x 1.587 = 4.877 x 10⁷ cells.
  • At t=40 min, the 2 x 10⁸ newly released phages are available for infection.
  • Since 2 x 10⁸ (available phages) is greater than 4.877 x 10⁷ (available E. coli cells), all the remaining uninfected E. coli cells can be infected by these phages.
  • These newly infected cells at t=40 min will lyse after 20 minutes, i.e., by t = 40 + 20 = 60 minutes.
• Total Time to Complete Lysis:
  • The entire E. coli culture will be completely lysed by 60 minutes, as all cells that were not lysed earlier would have been infected at the 40-minute mark and would lyse by the 60-minute mark.
• Convert Time to Bacterial Division Cycles:
  • E. coli doubling time = 30 minutes.
  • Number of full cycles of bacterial division = Total lysis time / Doubling time
  • Number of full cycles = 60 minutes / 30 minutes/cycle = 2 cycles.

The correct answer is P - iv; Q - i; R - ii; S - iii and P - iv; Q - i; R - ii; T - iii

Explanation:

• P. Penicillin: Penicillin is a beta-lactam antibiotic. It inhibits cell wall synthesis in bacteria by interfering with the transpeptidation step (cross-linking of peptidoglycan chains) catalyzed by penicillin-binding proteins (PBPs), which are enzymes involved in bacterial cell wall formation.
• Q. Kanamycin: Kanamycin belongs to the aminoglycoside class of antibiotics. Aminoglycosides bind to the 30S ribosomal subunit of bacteria, interfering with protein synthesis (translation) by causing misreading of mRNA and premature termination of translation.
• R. Rifampicin: Rifampicin is an antibiotic used to treat various bacterial infections, including tuberculosis. Its primary target is bacterial RNA polymerase, which it binds to and inhibits, thereby blocking transcription (RNA synthesis).
• S. Nalidixic acid: Nalidixic acid is an older quinolone antibiotic. Its primary targets are bacterial DNA gyrase (topoisomerase II) and topoisomerase IV. These enzymes are crucial for DNA replication, repair, and transcription by managing DNA supercoiling. Inhibition leads to DNA damage and bacterial cell death.
• T. Ciprofloxacin: Ciprofloxacin is a newer, broader-spectrum fluoroquinolone antibiotic. Like nalidixic acid, its primary targets are bacterial DNA gyrase (topoisomerase II) and topoisomerase IV. It also inhibits DNA replication and repair.

The correct answer is Option 1 and Option 2

Explanation:

• The enzyme aldolase (specifically Fructose-1,6-bisphosphate aldolase, but also other aldolases) catalyzes aldol condensation or retro-aldol cleavage reactions.
• Its primary role in glycolysis is the reversible cleavage of fructose 1,6-bisphosphate.
• However, aldolases are also involved in other metabolic pathways, such as the Calvin cycle and the pentose phosphate pathway, where they catalyze similar reactions involving different sugar phosphates.

Dihydroxyacetone phosphate + Glyceraldehyde-3-phosphate → Fructose 1,6-biphosphate:

• This is the synthesis of fructose 1,6-bisphosphate from two triose phosphates. This is the reverse reaction of a key step in glycolysis and is directly catalyzed by aldolase.
• (3-carbon DHAP + 3-carbon G3P → 6-carbon FBP)

Dihydroxyacetone phosphate + Erythrose-4-phosphate → Sedoheptulose-1,7-biphosphate:

• This reaction occurs in the Calvin cycle (in plants) and also in the non-oxidative phase of the pentose phosphate pathway.
• An aldolase catalyzes the condensation of the 3-carbon dihydroxyacetone phosphate with the 4-carbon erythrose-4-phosphate to form the 7-carbon sedoheptulose-1,7-bisphosphate. This reaction is indeed catalyzed by aldolase.
• (3-carbon DHAP + 4-carbon E4P → 7-carbon SBP)

(GAP = glyceraldehyde 3-phosphate, PRPP = 5-O-phosphono- ribose 1-diphosphate, R5P = ribose 5-phosphate, Ru5P = ribulose 5-phosphate, RuBP = ribulose 1,5-bisphosphate, S7P = sedoheptulose 7-phosphate, SBP = sedoheptulose 1,7-bisphosphate, SBPase = sedoheptulose 1,7-bisphosphatase, and Xu5P = xylulose 5-phosphate)

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.