Cellular Organization MCQ Quiz - Objective Question with Answer for Cellular Organization - Download Free PDF
Last updated on Jul 4, 2025
Latest Cellular Organization MCQ Objective Questions
Cellular Organization Question 1:
Which option(s) correctly match(es) the Antibiotic with their corresponding Target?
Answer (Detailed Solution Below)
Cellular Organization Question 1 Detailed Solution
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.
Cellular Organization Question 2:
Which option(s) correctly match(es) the structures in a bacterial cell (Column I) with their corresponding functions (Column II).
Answer (Detailed Solution Below)
Cellular Organization Question 2 Detailed Solution
The correct answer is P - i; Q - ii; R - iii; S - iv
Explanation:
Bacterial cells have specialized structures that perform distinct functions essential for their survival and interaction with the environment. Each structure in a bacterial cell, such as the cell wall, fimbriae, flagella, and pili, has a specific role that supports various cellular processes like protection, motility, attachment, and genetic exchange.
- Cell wall (P - i): The bacterial cell wall provides structural support and protection to the cell. It prevents the cell from bursting under osmotic pressure, thus offering protection from osmotic stress.
- Fimbriae (Q - ii): Fimbriae are hair-like structures on the bacterial surface that aid in adhesion to surfaces, host tissues, or other cells. They play a crucial role in bacterial colonization and infection.
- Flagella (R - iii): Flagella are whip-like structures that enable bacterial motility by propelling the cell through liquid environments. Motility is essential for bacteria to move toward favorable conditions or away from harmful stimuli.
- Pili (S - iv): Pili are longer appendages than fimbriae and are involved in the transfer of genetic material between bacterial cells during conjugation.
Cellular Organization Question 3:
Which of the following sites is/are the location(s) of ATP generation through oxidative phosphorylation in Escherichia coli?
Answer (Detailed Solution Below)
Cellular Organization Question 3 Detailed Solution
The correct answer is Inner membrane and Mesosome
Explanation:
- In Escherichia coli (a bacterium), which is a prokaryotic organism, there are no membrane-bound organelles like mitochondria.
- In prokaryotes, the processes of cellular respiration, including the electron transport chain and ATP synthesis via oxidative phosphorylation, occur on the cytoplasmic membrane (also known as the inner membrane or plasma membrane)
- Escherichia coli is a Gram-negative bacterium, meaning it has both an inner (cytoplasmic) membrane and an outer membrane. The outer membrane primarily serves as a protective barrier and contains porins, but it does not contain the machinery for oxidative phosphorylation.
Mesosomes are involved in various functions in bacterial cells, including:
- Cell wall formation
- DNA replication
- Cell division
- Cellular respiration/oxidative phosphorylation (acting as an equivalent to mitochondria)
Cellular Organization Question 4:
Which of the following genetic disorders is/are caused by trinucleotide repeat expansions?
Answer (Detailed Solution Below)
Cellular Organization Question 4 Detailed Solution
The correct answer is Option 1 and Option 3
Explanation:
Many trinucleotide repeat expansion diseases in human occur as a result of replication slippage.
- For example, Fragile X-syndrome is a genetic disorder. It occurs due to mutations in the FMR1 gene present on the X-chromosome (Xq27.3).
- The FMR1 gene codes for a protein called FMRP (fragile X-mental retardation protein). This protein plays a role in the development of synapses.
- Nearly all cases of fragile X-syndrome are caused by a mutation in which the expansion of the CGG triplet repeat occurs within the FMR1 gene.
- Normally, this CGG triplet is repeated from 6 to about 50 times in the 5' UTR region of the FMR1 gene.
- In people with fragile X-syndrome, however, the CGG triplet is repeated more than 200 times.
- The abnormally expanded CGG triplet turns off (silences) the FMR1 gene, which prevents the gene from producing FMRP. Males and females with 55 to 200 repeats of the CGG triplet are said to have an FMR1 gene premutation. Most people with a premutation are intellectually normal. In some cases, however, individuals with a premutation have lower than normal amounts of FMRP.
Huntington's Disease is caused by expansion of CAG (cytosine-adenine-guanine) repeats in the HTT gene on chromosome 4.
- Normal Range: ≤ 35 repeats
- Disease Range: ≥ 36 repeats
The expanded polyglutamine (polyQ) tract in the huntingtin protein causes neurodegeneration, especially in the basal ganglia.
β-thalassemia is caused by point mutations or deletions in the HBB gene (β-globin gene) on chromosome 11. These mutations reduce or abolish β-globin chain production.
Cystic fibrosis is caused by mutations in the CFTR gene, most commonly the ΔF508 mutation (a 3-bp deletion causing loss of phenylalanine at position 508). Although it's a 3-base deletion, it is not a repeat expansion. It affects chloride channels, leading to thick mucus in lungs, pancreas, etc
Cellular Organization Question 5:
Phospholipid vesicles prepared in 50 𝑚𝑀 KCl were diluted in water. Based on this information, statements P and Q are made.
P: The diluted vesicles will develop a membrane potential.
Q: There is a K+ concentration difference across the vesicular membrane.
Which one of the following options is correct?
Answer (Detailed Solution Below)
Cellular Organization Question 5 Detailed Solution
The correct answer is P is false but Q is true.
Concept:
- Phospholipid vesicles are bilayer structures that can encapsulate an aqueous solution. When these vesicles are prepared in a specific ionic solution (e.g., 50 mM KCl), they trap that solution inside their lumen.
- When such vesicles are diluted in water, the external environment changes, while the internal environment remains as it was during preparation. This sets up a concentration gradient for ions (e.g., K+ and Cl-) across the vesicle membrane.
- Membrane potential refers to the electrical potential difference across a membrane, typically created due to selective ion permeability and ion gradients.
Explanation:
- Statement P: "The diluted vesicles will develop membrane potential." This is false.
- Although a concentration gradient for K+ exists across the vesicular membrane after dilution, phospholipid bilayers are not inherently permeable to ions like K+ or Cl- without the presence of ion channels or transporters.
- In the absence of selective ion permeability, no net charge separation occurs across the membrane, and thus no membrane potential develops.
- Statement Q: "There is a K+ concentration difference across the vesicular membrane." This is true.
- When vesicles prepared in 50 mM KCl are diluted in water, the internal K+ concentration remains at 50 mM, but the external K+ concentration decreases significantly due to dilution.
- This creates a K+ concentration gradient across the vesicular membrane, even if no membrane potential develops.
Top Cellular Organization MCQ Objective Questions
Which one of the following combinations represents the major protein or protein complex involved in chromatin condensation in yeast and human, respectively?
Answer (Detailed Solution Below)
Cellular Organization Question 6 Detailed Solution
Download Solution PDFConcept:
- 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:
- Heterochromatin - transcriptionally inactive part due to dense chromatin condensation.
- Euchromatin - transcriptionally active part due to comparatively loose chromatin condensation or presence of expanded DNA regions for transcription.
Heterochromatin |
Euchromatin |
Found only in eukaryotes |
Found in both prokaryotes and eukaryotes |
Stains dark with DNA staining dye |
Stains light with DNA staining dye |
Replication is slow due to dense DNA packaging |
Replication is faster due to loose DNA packaging |
Constitutes 97 to 98% of the genome |
Constitutes only 2-3% of the genome |
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.
Progression across G1/S boundary followed by entry into S‐phase is promoted by the activation of which one of the following protein complexes?
Answer (Detailed Solution Below)
Cellular Organization Question 7 Detailed Solution
Download Solution PDFThe 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.
In eukaryotic cells, covalently attached lipids help to anchor some water soluble proteins to the plasma membrane. One group of cytosolic proteins are anchored to the cytosolic face of membrane by a fatty acyl group (e.g. myristate or palmitate). These groups are generally covalently attached to which amino acids present at the N‐terminus of the polypeptide chain?
Answer (Detailed Solution Below)
Cellular Organization Question 8 Detailed Solution
Download Solution PDF
Concept:
- Transmembrane proteins are characterized by having transmembrane-spanning segments.
- They contain a stretch of 21 to 26 hydrophobic amino acid residues coiled into an alpha-helix that is believed to facilitate the spanning of a lipid bilayer.
- In a few membrane proteins, transmembrane portions comprise beta-barrel made up of antiparallel beta strands.
Explanation:
- Membrane proteins are covalently bound to lipids molecules and are called lipid-linked or lipid-anchored proteins.
- They form covalent attachments with three classes of lipids- compounds formed from isoprene units such as farnesyl and geranylgeranyl residues, fatty acids such as myristic acid and palmitic acid, and glycosylated phospholipid.
- Proteins that are covalently attached with isoprenoid compounds such as farnesyl (15-carbon compound) and geranylgeranyl (20-carbon compound) are termed prenylated proteins. In these proteins, isoprenoid compounds are covalently linked to a cysteine residue at C-terminal via thioether linkage.
- Proteins covalently attached with fatty acids such as palmitic acid and myristic acid are termed fatty acylated proteins. Myristic acid is a 14-carbon molecule that is attached to a protein through an amide linkage to the alpha-amino group of an N-terminus Glycine residue (myristoylation).
- Palmitic acid is attached to cysteine residue close to N or C-terminus via amide linkage (palmitoylation).
- A glycophosphatidylinositol molecule (GPI) attaches at the C terminal amino acid via an amide linkage.
Lipid anchor |
Protein |
Attachment site |
Subcellular location |
Lipids built from isoprene units |
Prenylated protein |
Cys residue at C-terminal |
Intracellular |
Myristic acid |
Fatty acylated protein |
Gly residue at N-terminus |
Intracellular |
Palmitic acid |
Fatty acylated protein |
Cys residue near N or C terminus |
Intracellular |
GPI |
GPI-linked protein |
Various residues at the C terminus |
Cell surface |
So, the correct answer is option 1.
Porins, which are normally present on the outer mitochondrial membrane, reach their destination by
Answer (Detailed Solution Below)
Cellular Organization Question 9 Detailed Solution
Download Solution PDFThe correct answer is "synthesis in the cytosol, import by TOM complex and insertion from the inter-mitochondrial membrane space".
Explanation-
Porins are present in the outer mitochondrial membrane, and their synthesis typically occurs in the cytosol. The TOM complex, located in the outer mitochondrial membrane, facilitates the import of precursor proteins into the mitochondria. After entering the inter-membrane space of mitochondria, the precursor proteins are translocated across the outer mitochondrial membrane.
Synthesis in the Cytosol: Like most other proteins in the cell, mitochondrial porins, also known as Voltage Dependent Anion Channels (VDACs), are synthesized in the cytosol from mRNA translated by free ribosomes—not by the endoplasmic reticulum (ER) or by mitochondrial ribosomes.
Import by TOM Complex: The Translocase of the Outer Membrane (TOM) complex facilitates the transport of these porins from the cytosol across the outer mitochondrial membrane. The TOM complex forms a general entry gate for almost all mitochondrial precursor proteins that are synthesized in the cytosol.
Insertion in the Membrane & Folding: Once in the intermembrane space (the space between the inner and outer mitochondrial membranes), the proteins have to be inserted into the outer mitochondrial membrane. This task is accomplished by the SAM (sorting and assembly machinery) complex. The SAM complex inserts the precursor proteins into the outer membrane and assists in their folding and assembly to form functional porin channels.
The TIM (translocase of the inner mitochondrial membrane) complex is not involved in this process. This complex targets proteins to the inner mitochondrial membrane, the intermembrane space, or the matrix of mitochondria but does not affect proteins destined for the outer mitochondrial membrane like the porins.
The movement of proteins bigger than 50 kDa across the nuclear envelope requires:
Answer (Detailed Solution Below)
Cellular Organization Question 10 Detailed Solution
Download Solution PDFConcept:
- Transporters are membrane proteins or carrier proteins that span the membrane and assist in the movement of ions, molecules, small peptides, and certain macromolecules.
- Transport across the membrane can occur via simple diffusion, facilitated diffusion, osmosis, or active transport.
- Two distinct translocation complexes that mediate translocation are situated in the outer and inner mitochondrial membrane.
Important Points
Sec 61 -
- Nearly every newly synthesized polypeptide translocation to the endoplasmic reticulum occurs via a translocon protein.
- This protein is present in the ER membrane of all nucleated cells.
- Translocon contains sec 61 channel protein along with other protein complexes.
- Sec 61 transports proteins to the endoplasmic reticulum in eukaryotes and out of the cell in prokaryotes.
TOM -
- TOM complex (translocase of outer membrane) consists of receptor proteins (Tom20, Tom22, and Tom70), channel-forming proteins (Tom40), and three small Tom proteins (Tom5, Tom6, and Tom7).
- TOM 20 is a mitochondrial import receptor.
- It is the translocase in the outer mitochondrial membrane.
Importin -
- Importin is a type of karyopherin (protein transporter for transporting molecules between cytoplasm and nucleus).
- It is found in eukaryotic cells. Importin beta specifically transports proteins inside the nucleus.
- Importin beta must associate with the nuclear pore complexes to deliver cargo protein into the nucleus.
- This is accomplished by binding with the nuclear pore complex.
- It transports proteins bigger than 50 kDa across the nuclear membrane.
Tim 44 -
- Tim 44 (translocase inner membrane 44) is located in the mitochondrial matrix and is also peripherally attached to the inner membrane.
So, the correct answer is option 3.
So, the correct answer is option 3.
Which type of haploid unicellular eukaryote cells has a diameter of ∼10 μm, and about half of their volume is occupied by cup-shaped chloroplasts?
Answer (Detailed Solution Below)
Cellular Organization Question 11 Detailed Solution
Download Solution PDFThe correct answer is Chlamydomonas.Key Points
- Chlamydomonas is a type of haploid unicellular eukaryote cell that has a diameter of approximately 10 μm.
- Half of its volume is occupied by cup-shaped chloroplasts, which are responsible for photosynthesis.
- Chlamydomonas is found in freshwater environments and is capable of both sexual and asexual reproduction.
- It is commonly used as a model organism in genetic and biochemical research.
Additional Information
- Hydrodictyon is a genus of green algae that forms a net-like structure.
- It is commonly known as the water net.
- Ulva is a genus of green algae that is commonly known as sea lettuce.
- It is often found in marine environments.
- Oedogonium is a genus of filamentous green algae.
- It is commonly found in freshwater environments.
Which one of the following activities is associated with Mitochondria-associated ER membranes (MAM)?
Answer (Detailed Solution Below)
Cellular Organization Question 12 Detailed Solution
Download Solution PDFThe correct answer is Phospholipid metabolism
Concept:
Mitochondria-associated ER membranes (MAM) are specialized regions where the endoplasmic reticulum (ER) is closely associated with mitochondria. This association facilitates various important cellular functions, particularly related to lipid metabolism, calcium signaling, and communication between the ER and mitochondria.
- Phospholipid metabolism: MAM plays a critical role in the synthesis and metabolism of phospholipids. The close proximity of the ER and mitochondria at MAM allows for the transfer of lipids and coordination of lipid synthesis, which is essential for maintaining membrane integrity and function.
Explanation:
-
1) Protein glycosylation: This process primarily occurs in the ER, where proteins are modified (glycosylated) as they are synthesized. While MAM may have some role in protein processing, it is not specifically known for glycosylation.
-
2) ATP synthesis: This is a primary function of mitochondria themselves, specifically in the inner mitochondrial membrane, where ATP synthase operates to produce ATP during oxidative phosphorylation.
-
4) Iron-sulfur cluster assembly: This process primarily occurs in mitochondria and involves specific mitochondrial proteins. While there may be some interactions related to iron-sulfur clusters at the MAM, it is not the main associated activity.
Therefore, phospholipid metabolism is the key activity associated with Mitochondria-associated ER membranes (MAM).
Hyperlinks can be
(A). Text
(B). Drawing objects
(C). A group of cells
(D). Pictures
Choose the correct answer from the options given below:
Answer (Detailed Solution Below)
Cellular Organization Question 13 Detailed Solution
Download Solution PDFThe correct answer is (A), (B), (C), and (D)
Key Points
- Hyperlinks are versatile and can be applied to various elements within a document.
- These elements include text, drawing objects, groups of cells, and pictures.
- Using hyperlinks enhances the interactivity and navigation within and outside the document.
- Hyperlinks can lead to external websites, email addresses, or other sections within the same document, making it easier to access related information.
Additional Information
- Apply Hyperlinks: To insert a hyperlink, select the text or object, right-click, and choose 'Hyperlink' from the context menu. This allows linking to various destinations.
- Edit Hyperlinks: Existing hyperlinks can be modified by right-clicking the hyperlink and selecting 'Edit Hyperlink'.
- Remove Hyperlinks: Hyperlinks can be removed by right-clicking the hyperlink and selecting 'Remove Hyperlink'.
Some features mentioned below are important for the segregation of homologous chromosomes in meiosis I.
A. Synaptonemal complex formation between homologous chromosomes.
B. Degradation of cohesins at the chromosome arms.
C. Retention of cohesins at the centromeres.
D. Bi-orientation of kinetochores of sister chromatids.
Which one of the following options has all correct features?
Answer (Detailed Solution Below)
Cellular Organization Question 14 Detailed Solution
Download Solution PDFThe correct answer is A, B, C only
Explanation:
A. Synaptonemal complex formation between homologous chromosomes.
- Correct. The synaptonemal complex is a protein structure that forms between homologous chromosomes during prophase I of meiosis. It facilitates the pairing (synapsis) of homologs, allowing for crossover events, which are critical for genetic recombination and proper segregation.
B. Degradation of cohesins at the chromosome arms.
- Correct. During meiosis I, cohesins, which hold sister chromatids together, are degraded at the arms of the chromosomes. This degradation allows homologous chromosomes to separate while sister chromatids remain connected at the centromeres, ensuring proper segregation.
C. Retention of cohesins at the centromeres.
- Correct. While cohesins are degraded along the arms of the chromosomes, they are retained at the centromeres. This retention is crucial because it keeps the sister chromatids together until anaphase II, ensuring that they are properly segregated in the second meiotic division.
D. Bi-orientation of kinetochores of sister chromatids.
- Incorrect. Bi-orientation refers to the alignment of kinetochores of sister chromatids in opposite directions (one facing each pole).
- In meiosis I, the kinetochores of sister chromatids do not bi-orient as they do in mitosis or meiosis II.
- Instead, the kinetochores of homologous chromosomes are oriented toward opposite poles, ensuring the separation of the homologs.
Thus, the correct option that includes all correct features is A, B, C only.
Cell lysate in 1% TX100 was purified over an affinity column to isolate a complex with certain enzymatic activity. The purified enzyme complex was separated on a 10-50% continuous sucrose gradient. Shown below are the UV spectra using an absorbance filter at 280 nm or 260 nm.
Which one of the following combinations of molecules would generate the spectra shown above?
Answer (Detailed Solution Below)
Cellular Organization Question 15 Detailed Solution
Download Solution PDFThe correct answer is Protein-RNA
Concept:
UV Absorbance Characteristics:
- Proteins primarily absorb at 280 nm due to the presence of aromatic amino acids (such as tryptophan, tyrosine, and phenylalanine). The strong absorbance at 280 nm indicates the presence of proteins.
- Nucleic acids (RNA/DNA) absorb strongly at 260 nm due to the presence of nucleotide bases. If the spectrum shows significant absorbance at 260 nm, it suggests the presence of nucleic acids (RNA or DNA).
Explanation:
- The two spectra shown likely correspond to absorbance at 260 nm and 280 nm (one for each filter).
- If one of the spectra shows a peak at 260 nm, it strongly suggests the presence of RNA (or DNA), as nucleic acids absorb more at this wavelength.
- If another peak appears at 280 nm, it indicates the presence of protein.
- The combination of peaks at 260 nm and 280 nm suggests a complex of protein and RNA. This could be a riboprotein complex or a protein that interacts with RNA, leading to the combined absorbance pattern.