A researcher wants to watch unstained, living fibroblasts crawl across a culture dish and to see their edges and nuclei clearly. The most suitable instrument is

A phase-contrast microscope, because it turns refractive-index differences into brightness differences

A transmission electron microscope, because its resolution is far better than that of any light microscope

A scanning electron microscope, because it shows the three-dimensional surface of each cell

A bright-field microscope, because unstained cells absorb enough visible light to give contrast

A phase-contrast microscope, because it turns refractive-index differences into brightness differences

Which of the following statements about light and fluorescence microscopy are correct?

Immersion oil raises the numerical aperture because it has a higher refractive index than air

In a confocal microscope a pinhole rejects out-of-focus light, so thin optical sections can be recorded

Raising the eyepiece magnification from 10× to 25× on the same objective improves the resolution of the image

Immersion oil raises the numerical aperture because it has a higher refractive index than air

Green fluorescent protein emits light of shorter wavelength than the light used to excite it

In a confocal microscope a pinhole rejects out-of-focus light, so thin optical sections can be recorded

Which cytoskeletal filaments, about 10 nm thick, are built from staggered antiparallel tetramers, have no structural polarity and serve as tracks for no motor protein?

Intermediate filaments, such as keratins and the nuclear lamins

Microtubules, the hollow tracks along which kinesins and dyneins walk

Actin microfilaments of the cell cortex and microvilli

Myosin II thick filaments of skeletal muscle sarcomeres

Dividing animal cells are treated with cytochalasin D just after anaphase begins. The chromosomes still reach the poles and two nuclei re-form, but no cleavage furrow forms and the cells become binucleate. The best conclusion is that

cleavage requires an actin-based contractile ring, while anaphase movement is driven by microtubules

the drug depolymerised the spindle microtubules, so the furrow could not be positioned between the nuclei

separase activity depends on actin filaments, so the sister chromatids were never fully separated

cleavage requires an actin-based contractile ring, while anaphase movement is driven by microtubules

re-forming the nuclear envelopes used up the myosin II that the furrow would have needed

Which of the following statements about microtubules and their motors in a typical interphase animal cell are correct?

Conventional kinesin (kinesin-1) carries vesicles toward the plus ends, which lie near the cell periphery

Cytoplasmic dynein moves cargo toward the minus ends, which are anchored at the centrosome

Conventional kinesin (kinesin-1) carries vesicles toward the plus ends, which lie near the cell periphery

Each centriole is built of nine microtubule doublets surrounding a central pair of single microtubules

Cytoplasmic dynein moves cargo toward the minus ends, which are anchored at the centrosome

Myosin II uses ATP to walk along microtubules and so moves organelles to the cell centre

The N-terminal signal sequence of a secreted digestive enzyme is deleted, and the mutant gene is expressed in pancreatic acinar cells. The mutant protein will most likely

be made on free ribosomes and remain in the cytosol, because SRP never targets it to the ER

enter the ER lumen but stay trapped there, because it can no longer bud into vesicles

be made on free ribosomes and remain in the cytosol, because SRP never targets it to the ER

be secreted normally, because sorting into secretory granules depends on its mature glycans

be delivered to lysosomes, because proteins without a signal follow the default lysosomal route

Fibroblasts from a patient lack the Golgi enzyme that begins adding the mannose-6-phosphate tag to newly made lysosomal hydrolases. Compared with normal fibroblasts in culture, these cells will show

high hydrolase activity in the culture medium and lysosomes swollen with undigested material

hydrolases retained in the ER lumen, because their folding fails without the phosphate tag attached

normal lysosomes, because the hydrolases reach them by default once they pass the Golgi

hydrolases accumulating in the cytosol after they are released from free ribosomes

Which of the following statements about organelles of animal cells are correct?

Mitochondria contain their own circular DNA and ribosomes that translate a few of their proteins

Peroxisomes contain oxidases that produce H₂O₂ and catalase that breaks it down

Mitochondria contain their own circular DNA and ribosomes that translate a few of their proteins

The lysosomal lumen is kept near pH 7.4 by the Na⁺/K⁺-ATPase of the lysosomal membrane

Peroxisomes contain oxidases that produce H₂O₂ and catalase that breaks it down

N-linked glycosylation of secretory proteins begins in the smooth ER, which lacks ribosomes

In skeletal muscle fibres, the Ca²⁺ that triggers contraction is stored mainly in

the sarcoplasmic reticulum, a specialised smooth endoplasmic reticulum

the mitochondrial matrix, from which it leaves through the uniporter

the lysosomes, which release it when their pH rises after stimulation

the Golgi cisternae, which pass it to the sarcolemma in vesicles

the sarcoplasmic reticulum, a specialised smooth endoplasmic reticulum

A cell line expresses a mutant securin that APC/C can no longer ubiquitinate. When these cells enter mitosis and all kinetochores have attached to the spindle, they will most likely

stay at metaphase with sisters joined, because separase remains bound and cohesin is not cleaved

arrest in G1, because Rb stays unphosphorylated and E2F cannot drive the next S phase

separate their sisters normally, because separase is activated directly by kinetochore attachment

stay at metaphase with sisters joined, because separase remains bound and cohesin is not cleaved

re-replicate their DNA at once, because cyclin B is destroyed while the cells are still in mitosis

During prophase I of meiosis, crossing over between non-sister chromatids of homologous chromosomes takes place at

Pachytene, when synapsis along the homologues is complete

Zygotene, when homologues first begin to synapse

Pachytene, when synapsis along the homologues is complete

Diplotene, when the homologues pull apart and chiasmata appear

Diakinesis, when the chiasmata slide toward chromosome ends

Which of the following statements comparing mitosis and meiosis in a diploid animal are correct?

Homologous chromosomes pair and separate from each other during meiosis I

Sister chromatids separate at anaphase of mitosis and at anaphase II of meiosis

Homologous chromosomes pair and separate from each other during meiosis I

DNA is replicated a second time between meiosis I and meiosis II

Sister chromatids separate at anaphase of mitosis and at anaphase II of meiosis

Each daughter of a mitotic division receives half the chromosome number of the parent cell

Nuclei are digested briefly with micrococcal nuclease, and the DNA is purified and run on a gel. The DNA appears as a ladder of bands at about 200, 400, 600 bp and so on. This result shows that

DNA is protected in repeating units and cut in exposed linkers between them

histone H1 is required for DNA to wrap around the octamer at all

the enzyme cleaves DNA only at a specific 200 bp-spaced recognition sequence

chromatin is organised as 30 nm fibres, each holding 200 bp of DNA per turn

DNA is protected in repeating units and cut in exposed linkers between them

The giant banded chromosomes of Drosophila larval salivary glands arise by

Repeated rounds of DNA replication without division, with the copies staying aligned

Arrest of oocyte bivalents at diplotene, with their chromatin loops spread out

Fusion of several diploid nuclei from neighbouring cells into a single large syncytium

Repeated non-disjunction in mitosis, which multiplies the number of chromosomes

Cell Biology: Microscopy, Cell Structure, Cytoskeleton, Organelles, Cell Cycle, Cell Division and Chromatin

This chapter answers Section XL4.5 of XL4 Zoology, one of the five optional sections of the Life Sciences (XL) paper: “Basic principles of cellular microscopy, structure of cell, cytoskeletal organization, cellular organelles and their structure and function, cell cycle, cell division, chromosomes and chromatin structure.” It teaches what limits a microscope's resolution and how phase-contrast, fluorescence, confocal and electron microscopes work; the plan of an animal cell and its membrane; the three filament systems and their motors; the organelles and the routes proteins take between them; the cell cycle and its controls; mitosis and meiosis; and how DNA is packed into nucleosomes and chromosomes. The numericals are resolution from wavelength and numerical aperture, DNA content through the cycle and meiosis, phase duration from a mitotic index, and nucleosome and histone counts.

1. Principles of cellular microscopy

A microscope must resolve, not merely magnify. Diffraction spreads each point of the specimen into a small spot, and two points appear separate only if their spots are far enough apart. By the Rayleigh criterion the smallest resolvable distance is d = 0.61λ/NA, where NA = n sin θ is the numerical aperture, n the refractive index of the medium between specimen and objective, and θ the half-angle of the cone of light the objective accepts (some texts use Abbe's d = λ/2NA, so a question must state its formula). Resolution improves with shorter wavelength and larger NA. In air NA stays below 1; immersion oil (n ≈ 1.52) lets it reach about 1.4, so with green light d ≈ 0.61 × 550/1.4 ≈ 240 nm, about 0.2 µm, the practical limit of light microscopy. Magnification beyond about 1,000 × NA is empty magnification: the image grows but shows no new detail.

Microscopes and what each is for
Technique	Principle	Typical use
Bright-field	Stained structures absorb light	Fixed, stained sections
Phase contrast, DIC	Converts phase shifts caused by refractive-index differences into brightness differences	Living, unstained cells
Fluorescence	A fluorophore absorbs a shorter wavelength and emits a longer one; filters pass only the emission	GFP fusions, antibody labelling, DNA stains such as DAPI
Confocal	A scanned laser spot and a pinhole that rejects out-of-focus light give optical sections	Thick specimens, embryos, 3D reconstruction
Transmission EM	Electrons (wavelength in picometres) pass through ultrathin sections stained with heavy metals, in vacuum	Organelle ultrastructure at nanometre resolution; only fixed, dead material
Scanning EM	Electrons scattered from a metal-coated surface are collected as the beam scans	Three-dimensional surface views

⚠️ Exam trap

A smaller d means better resolution. Fluorescence emission is always at a longer wavelength than excitation (GFP is excited by blue light and glows green). Cryo-electron microscopy of vitrified samples and super-resolution light methods that switch fluorophores on and off go beyond these limits.

2. Structure of the animal cell and its membrane

Prokaryotic cells have no nucleus or membrane-bound organelles, a nucleoid usually holding one circular chromosome, 70S ribosomes and, in most, a peptidoglycan wall. Eukaryotic cells, typically 10 to 100 µm across, have a nucleus, an endomembrane system, mitochondria, 80S cytosolic ribosomes and a cytoskeleton. Animal cells lack a wall and plastids, carry centrosomes with centrioles, and are held in tissues by an extracellular matrix (collagen, proteoglycans, fibronectin) and by junctions: tight junctions seal epithelia; adherens junctions and desmosomes link neighbours through cadherins, to actin and to intermediate filaments respectively; hemidesmosomes anchor cells to the basal lamina through integrins; and gap junctions of connexins pass ions and small molecules between cells.

The plasma membrane follows the fluid mosaic model of Singer and Nicolson (1972): a phospholipid bilayer about 5 nm thick in which proteins float. Integral proteins span the bilayer; peripheral proteins sit on its surfaces. Lipids diffuse quickly within a leaflet but flip between leaflets rarely unless a flippase acts, so the leaflets differ: phosphatidylserine faces the cytosol, and glycolipids and glycoproteins face outward as the glycocalyx. Unsaturated fatty acids raise fluidity and cholesterol buffers it. When Frye and Edidin fused mouse and human cells, the two sets of surface proteins intermixed within about 40 minutes at 37 °C, showing that membrane proteins move laterally.

3. Cytoskeletal organisation

The three filament systems
Feature	Microfilaments	Microtubules	Intermediate filaments
Subunit	G-actin, binds ATP	α/β-tubulin dimer, binds GTP	Keratins, vimentin, neurofilament proteins, nuclear lamins
Diameter	About 7 nm	About 25 nm; hollow, 13 protofilaments	About 10 nm
Polarity	Plus (barbed) and minus (pointed) ends	Plus end grows; minus end anchored at the centrosome	None (antiparallel tetramers)
Motors	Myosins	Kinesins (mostly to the plus end), dyneins (to the minus end)	None
Main roles	Cortex, microvilli, crawling, contractile ring, muscle contraction	Spindle, vesicle transport, cilia and flagella	Tensile strength, nuclear lamina
Drugs	Cytochalasin caps plus ends; latrunculin sequesters G-actin; phalloidin stabilises	Colchicine and nocodazole block assembly; taxol stabilises	None in routine use

Actin and tubulin polymers are dynamic because nucleotide hydrolysis follows assembly. Microtubules show dynamic instability: a growing plus end carries a cap of GTP-tubulin, and if hydrolysis catches up with the end the GDP lattice peels apart and the microtubule shrinks fast (catastrophe) until rescued. Actin filaments treadmill, adding ATP-actin at the plus end while ADP-actin leaves the minus end. In animal cells microtubules grow from the centrosome: two centrioles, each of nine triplet microtubules, in pericentriolar material whose γ-tubulin ring complexes nucleate minus ends. Motile cilia and flagella have a 9 + 2 axoneme, nine outer doublets around a central pair; dynein arms slide neighbouring doublets and cross-links turn the sliding into bending. Their basal bodies have the nine-triplet centriole structure.

🎯 Immotile cilia

In primary ciliary dyskinesia the dynein arms are defective: cilia and sperm tails cannot beat, so patients suffer chronic airway infections and male infertility, and about half show reversed organ asymmetry, because cilia in the early embryo set up left–right asymmetry.

4. Organelles and protein sorting

The nucleus is bounded by a double envelope continuous with the ER and pierced by nuclear pore complexes; small molecules diffuse through, but large proteins need a nuclear localisation signal and importins. The nucleolus assembles ribosomal subunits around the rRNA genes. Proteins reach their compartments by signals. A hydrophobic N-terminal signal sequence is bound by the signal recognition particle (SRP), which docks the ribosome on the rough ER so that the protein enters the lumen as it is made; there it folds, forms disulfides and receives N-linked oligosaccharides, while residents are retrieved by a C-terminal KDEL signal. Vesicles carry cargo to the Golgi (cis to trans), which trims and adds sugars and sorts proteins at the trans-Golgi network to the plasma membrane, secretory granules or lysosomes. Lysosomal hydrolases are tagged with mannose-6-phosphate; without the enzyme that adds it (I-cell disease) they are secreted instead. Proteins with no signal stay in the cytosol.

Other organelles at a glance
Organelle	Structure	Main functions
Smooth ER	Tubules without ribosomes	Lipid and steroid synthesis, detoxification by cytochrome P450, Ca²⁺ store (sarcoplasmic reticulum of muscle)
Lysosome	Single membrane; lumen near pH 5, kept acidic by a V-type H⁺-ATPase	Acid hydrolases digest endocytosed and autophagic material
Peroxisome	Single membrane; imports proteins with a C-terminal SKL signal	Oxidases make H₂O₂ and catalase destroys it; β-oxidation of very-long-chain fatty acids
Mitochondrion	Double membrane, inner one folded into cristae; matrix holds circular DNA and bacteria-like ribosomes	Krebs cycle, oxidative phosphorylation, fatty-acid oxidation, apoptosis via cytochrome c release; maternally inherited

ℹ️ Note

Mitochondria descend from an endosymbiotic α-proteobacterium: they divide by fission, keep their own genome and ribosomes, and their inner membrane resembles a bacterial membrane. Most of their proteins are now encoded in the nucleus and imported.

5. The cell cycle and its control

The cycle is interphase (G1, S, G2) followed by M phase (mitosis and cytokinesis). In a cultured human cell dividing every 24 h, M lasts about an hour; G1 is the most variable phase, and cells that stop dividing leave G1 for G0, as neurons do. DNA content is counted in C-values, 1C being the DNA of one haploid set. A diploid G1 cell has 2C; DNA doubles in S, so G2 and mitotic cells have 4C until anaphase, and each daughter returns to 2C. The chromosome number does not change in S: a replicated chromosome is still one chromosome, with two sister chromatids held by cohesin. Because cells in an asynchronous population are spread around the cycle, the fraction in a phase estimates that phase's share of the cycle: a mitotic index of 5% in a 20 h cycle means mitosis lasts about 1 h.

Progression is driven by cyclin-dependent kinases (CDKs), active only with a cyclin whose level rises and falls. Cyclin D–CDK4/6 phosphorylates Rb, releasing E2F to transcribe S-phase genes; passing this restriction point commits the cell. Cyclin E–CDK2 triggers S, cyclin A–CDK2 runs it, and cyclin B–CDK1 (maturation-promoting factor, MPF) drives entry into mitosis. Checkpoints guard each transition: at G1/S, DNA damage stabilises p53, which induces the CDK inhibitor p21 and arrests the cell or triggers apoptosis; at G2/M, damaged or unreplicated DNA keeps CDK1 off; and the spindle assembly checkpoint holds metaphase until every kinetochore is attached. Then the APC/C ubiquitinates securin and cyclin B: freed separase cleaves cohesin, sisters part at anaphase, and falling CDK1 activity allows exit from mitosis. Cell-fusion experiments showed the logic: an S-phase cell drives a G1 nucleus into DNA synthesis, but a G2 nucleus does not re-replicate.

6. Mitosis and meiosis

In mitosis, prophase condenses the chromatin and the duplicated centrosomes move apart to build the spindle; in prometaphase the nuclear envelope breaks down (lamins are phosphorylated) and microtubules capture the kinetochores at the centromeres; at metaphase the chromosomes align at the equator with sister kinetochores attached to opposite poles; in anaphase A the chromatids move poleward as kinetochore microtubules shorten, and in anaphase B the poles separate; in telophase the envelopes re-form and the chromosomes decondense. Animal cytokinesis pinches the cell in two with a contractile ring of actin and myosin II placed under the old metaphase plate.

Meiosis makes four haploid cells from one round of replication and two divisions. Meiosis I is reductional: homologues pair, recombine and separate. Its long prophase I runs through leptotene (chromosomes condense), zygotene (homologues synapse through the synaptonemal complex), pachytene (synapsis complete; crossing over), diplotene (homologues pull apart except at chiasmata; mammalian oocytes arrest here, often for years) and diakinesis. At metaphase I bivalents align with both sister kinetochores of each homologue facing the same pole; at anaphase I homologues part while sisters stay joined, because cohesin at the centromeres is protected. Meiosis II is equational, like mitosis. Crossing over and the independent orientation of bivalents create variety; non-disjunction yields aneuploid gametes.

DNA content through meiosis: if a G1 diploid cell holds 6 pg, the column reads 6, 12, 6 and 3 pg
Cell in spermatogenesis	Chromosomes	Chromatids per chromosome	DNA
Spermatogonium in G1	2n	1	2C
Primary spermatocyte (after S)	2n	2	4C
Secondary spermatocyte	n	2	2C
Spermatid	n	1	1C

7. Chromosomes and chromatin structure

Eukaryotic DNA is packed with histones into chromatin. The nucleosome core particle is 147 bp of DNA wrapped in about 1.65 left-handed turns around an octamer of two each of H2A, H2B, H3 and H4, small basic proteins rich in lysine and arginine. Linker DNA joins the cores and histone H1 binds where DNA enters and leaves; the repeat length (core plus linker) varies between cell types, commonly near 200 bp. Light digestion with micrococcal nuclease cuts the linkers and gives a ladder of fragments in multiples of the repeat, while exhaustive digestion leaves the 147 bp core. Nucleosomes form the 10 nm “beads on a string” fibre, which folds further into loops anchored by cohesin and CTCF and into chromosome territories in the interphase nucleus; a metaphase chromosome is roughly ten-thousand-fold shorter than its DNA. Counting example: a 3 Mb domain with a 200 bp repeat carries 3 × 10⁶/200 = 15,000 nucleosomes, and so 30,000 molecules of each core histone.

Euchromatin is less condensed, gene-rich and largely active; heterochromatin stays condensed in interphase and is mostly silent. Constitutive heterochromatin, such as satellite repeats around centromeres, is condensed in every cell; facultative heterochromatin is silenced only in some cells, like the inactive X of female mammals, seen as the Barr body. Each chromosome has a centromere, marked by the histone H3 variant CENP-A, on which the kinetochore assembles, and two telomeres, tandem TTAGGG repeats in vertebrates, capped by proteins so that the ends are not treated as breaks. By centromere position chromosomes are metacentric, submetacentric, acrocentric or telocentric.

Two giant chromosomes are classic zoological material. Polytene chromosomes of Drosophila larval salivary glands arise by repeated replication without division (endoreplication): about a thousand aligned chromatids, with homologues synapsed, give a band pattern used to map genes and rearrangements, and puffs (Balbiani rings in Chironomus) mark active transcription, appearing in sequence after the moulting hormone ecdysone. Lampbrush chromosomes of amphibian and bird oocytes are diplotene bivalents whose loops are extended and heavily transcribed as the oocyte stocks up RNA.

Key takeaways

Resolution d = 0.61λ/NA with NA = n sin θ: oil immersion and short wavelengths give about 0.2 µm for light microscopes, and magnification beyond about 1,000 × NA is empty. Phase contrast shows living unstained cells, fluorescence emits at a longer wavelength than excitation, confocal pinholes give optical sections, and EM resolves nanometres in fixed material.
The fluid mosaic membrane allows fast lateral diffusion but rare flip-flop, so its leaflets are asymmetric. Animal cells are joined by tight, adherens, desmosome and gap junctions and anchored to the matrix through integrins.
Actin (7 nm, ATP, myosins), microtubules (25 nm, GTP, 13 protofilaments, kinesin to the plus end and dynein to the minus end, dynamic instability) and intermediate filaments (10 nm, no polarity, no motors). Cytochalasin and phalloidin act on actin; colchicine and taxol on microtubules.
A signal sequence and SRP send proteins into the rough ER, then through the Golgi; mannose-6-phosphate routes hydrolases to lysosomes (pH about 5). Peroxisomes make and destroy H₂O₂; mitochondria keep their own DNA and ribosomes and are maternally inherited.
G1 is 2C, S doubles DNA, G2 and M are 4C. Cyclin–CDKs drive the cycle; Rb–E2F sets the restriction point, p53–p21 arrests damaged cells, and APC/C destroys securin so separase can cleave cohesin. Meiosis I separates homologues (crossing over at pachytene, chiasmata at diplotene); meiosis II separates sisters.
A nucleosome is 147 bp around two each of H2A, H2B, H3 and H4, with H1 on the linker and a repeat near 200 bp. Heterochromatin is constitutive or facultative; centromeres carry CENP-A and vertebrate telomeres TTAGGG repeats; polytene and lampbrush chromosomes show active genes directly.

Practice questions (18)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

An objective lens accepts a cone of light of half-angle 60° and is used with an immersion medium of refractive index 1.50. Using d = 0.61λ/NA with NA = n sin θ, the smallest resolvable distance for light of wavelength 500 nm is ____ nm (to the nearest whole number).
Numerical answer — type the value.
Show answer
Answer: 235
NA = 1.50 × sin 60° = 1.50 × 0.866 = 1.299. d = 0.61 × 500/1.299 = 305/1.299 = 234.8 nm, which rounds to 235 nm. Leaving out the refractive index (NA = 0.866, d = 352 nm) or using Abbe's λ/2NA (192 nm) are the usual slips; the stem fixes the Rayleigh form.
A researcher wants to watch unstained, living fibroblasts crawl across a culture dish and to see their edges and nuclei clearly. The most suitable instrument is
1. A transmission electron microscope, because its resolution is far better than that of any light microscope
2. A scanning electron microscope, because it shows the three-dimensional surface of each cell
3. A bright-field microscope, because unstained cells absorb enough visible light to give contrast
4. A phase-contrast microscope, because it turns refractive-index differences into brightness differences
Show answer
Answer: D — A phase-contrast microscope, because it turns refractive-index differences into brightness differences
Living, unstained cells are nearly transparent: they shift the phase of light rather than absorb it, and phase-contrast optics convert those phase shifts into visible contrast. Both electron microscopes need fixed, dehydrated specimens in vacuum, so the cells could not crawl. Bright-field is the tempting choice, but unstained cells absorb very little visible light and give almost no contrast.
Which of the following statements about light and fluorescence microscopy are correct?
1. Raising the eyepiece magnification from 10× to 25× on the same objective improves the resolution of the image
2. Immersion oil raises the numerical aperture because it has a higher refractive index than air
3. Green fluorescent protein emits light of shorter wavelength than the light used to excite it
4. In a confocal microscope a pinhole rejects out-of-focus light, so thin optical sections can be recorded
Show answer
Answer: B — Immersion oil raises the numerical aperture because it has a higher refractive index than air; D — In a confocal microscope a pinhole rejects out-of-focus light, so thin optical sections can be recorded
NA = n sin θ, so oil (n ≈ 1.52) raises NA and with it resolution, and the confocal pinhole blocks light from planes above and below focus, giving optical sections. Eyepiece magnification only enlarges the image the objective has already resolved, so it adds no detail (empty magnification). A fluorophore loses some energy before emitting, so emission is at a longer wavelength than excitation: GFP absorbs blue and emits green.
Which cytoskeletal filaments, about 10 nm thick, are built from staggered antiparallel tetramers, have no structural polarity and serve as tracks for no motor protein?
1. Intermediate filaments, such as keratins and the nuclear lamins
2. Microtubules, the hollow tracks along which kinesins and dyneins walk
3. Actin microfilaments of the cell cortex and microvilli
4. Myosin II thick filaments of skeletal muscle sarcomeres
Show answer
Answer: A — Intermediate filaments, such as keratins and the nuclear lamins
Intermediate filaments assemble from coiled-coil dimers that pair antiparallel into tetramers, so the filament has no polarity and no motor can move directionally along it; their job is tensile strength, and lamins form the nuclear lamina. Microtubules (25 nm) and actin filaments (7 nm) are polar and carry motors. Myosin thick filaments are motor assemblies, not part of the three cytoskeletal filament systems, and are bipolar rather than built of nonpolar tetramers.
Dividing animal cells are treated with cytochalasin D just after anaphase begins. The chromosomes still reach the poles and two nuclei re-form, but no cleavage furrow forms and the cells become binucleate. The best conclusion is that
1. the drug depolymerised the spindle microtubules, so the furrow could not be positioned between the nuclei
2. separase activity depends on actin filaments, so the sister chromatids were never fully separated
3. cleavage requires an actin-based contractile ring, while anaphase movement is driven by microtubules
4. re-forming the nuclear envelopes used up the myosin II that the furrow would have needed
Show answer
Answer: C — cleavage requires an actin-based contractile ring, while anaphase movement is driven by microtubules
Cytochalasin D caps actin filament plus ends and blocks polymerisation. Chromosome segregation, which depends on spindle microtubules, went ahead normally, but the actin–myosin II contractile ring could not assemble, so cytokinesis failed. The drug does not act on microtubules, and normal segregation into two nuclei rules out the claims that the spindle was destroyed or that sisters failed to separate; nuclear envelope assembly does not use myosin II.
Which of the following statements about microtubules and their motors in a typical interphase animal cell are correct?
1. Conventional kinesin (kinesin-1) carries vesicles toward the plus ends, which lie near the cell periphery
2. Each centriole is built of nine microtubule doublets surrounding a central pair of single microtubules
3. Cytoplasmic dynein moves cargo toward the minus ends, which are anchored at the centrosome
4. Myosin II uses ATP to walk along microtubules and so moves organelles to the cell centre
Show answer
Answer: A — Conventional kinesin (kinesin-1) carries vesicles toward the plus ends, which lie near the cell periphery; C — Cytoplasmic dynein moves cargo toward the minus ends, which are anchored at the centrosome
In interphase animal cells microtubule minus ends sit at the centrosome and plus ends point outward, so kinesin-1 moves cargo outward and cytoplasmic dynein moves it inward. A centriole has nine triplets and no central pair; the 9 + 2 pattern of doublets is the motile axoneme. Myosins are actin-based motors and do not walk on microtubules.
The N-terminal signal sequence of a secreted digestive enzyme is deleted, and the mutant gene is expressed in pancreatic acinar cells. The mutant protein will most likely
1. enter the ER lumen but stay trapped there, because it can no longer bud into vesicles
2. be made on free ribosomes and remain in the cytosol, because SRP never targets it to the ER
3. be secreted normally, because sorting into secretory granules depends on its mature glycans
4. be delivered to lysosomes, because proteins without a signal follow the default lysosomal route
Show answer
Answer: B — be made on free ribosomes and remain in the cytosol, because SRP never targets it to the ER
Entry into the secretory pathway starts when SRP binds the emerging signal sequence and docks the ribosome on the rough ER. Without the signal the protein is completed on a free ribosome and stays in the cytosol, so it never reaches the ER, the Golgi or the granules, and it acquires no N-linked glycans. Lysosomal delivery needs ER entry plus the mannose-6-phosphate tag; the cytosol, not the lysosome, is the default for signal-less proteins.
Fibroblasts from a patient lack the Golgi enzyme that begins adding the mannose-6-phosphate tag to newly made lysosomal hydrolases. Compared with normal fibroblasts in culture, these cells will show
1. high hydrolase activity in the culture medium and lysosomes swollen with undigested material
2. hydrolases retained in the ER lumen, because their folding fails without the phosphate tag attached
3. normal lysosomes, because the hydrolases reach them by default once they pass the Golgi
4. hydrolases accumulating in the cytosol after they are released from free ribosomes
Show answer
Answer: A — high hydrolase activity in the culture medium and lysosomes swollen with undigested material
Mannose-6-phosphate receptors in the trans-Golgi network capture tagged hydrolases and divert them to lysosomes. Untagged enzymes follow the default route to the cell surface and are secreted, so the medium (and, in patients, the blood) is rich in lysosomal enzymes while lysosomes, lacking them, fill with undigested material: I-cell disease. The enzymes fold and leave the ER normally, they are made on ER-bound ribosomes so they never sit in the cytosol, and the default route out of the Golgi is secretion, not the lysosome.
Which of the following statements about organelles of animal cells are correct?
1. Mitochondria contain their own circular DNA and ribosomes that translate a few of their proteins
2. The lysosomal lumen is kept near pH 7.4 by the Na⁺/K⁺-ATPase of the lysosomal membrane
3. Peroxisomes contain oxidases that produce H₂O₂ and catalase that breaks it down
4. N-linked glycosylation of secretory proteins begins in the smooth ER, which lacks ribosomes
Show answer
Answer: A — Mitochondria contain their own circular DNA and ribosomes that translate a few of their proteins; C — Peroxisomes contain oxidases that produce H₂O₂ and catalase that breaks it down
Mitochondria keep a small circular genome and their own ribosomes, which make a handful of respiratory-chain subunits, and peroxisomes pair H₂O₂-producing oxidases with catalase. The lysosomal lumen is acidic, near pH 5, kept so by a V-type H⁺-ATPase, which suits its acid hydrolases. N-linked glycosylation starts co-translationally in the rough ER lumen, where the oligosaccharide is transferred from dolichol to asparagine.
In skeletal muscle fibres, the Ca²⁺ that triggers contraction is stored mainly in
1. the mitochondrial matrix, from which it leaves through the uniporter
2. the lysosomes, which release it when their pH rises after stimulation
3. the Golgi cisternae, which pass it to the sarcolemma in vesicles
4. the sarcoplasmic reticulum, a specialised smooth endoplasmic reticulum
Show answer
Answer: D — the sarcoplasmic reticulum, a specialised smooth endoplasmic reticulum
The sarcoplasmic reticulum is the smooth ER of muscle. It pumps Ca²⁺ in with a Ca²⁺-ATPase and releases it through ryanodine receptor channels when an action potential travels down the T-tubules; Ca²⁺ then binds troponin C. Mitochondria take up Ca²⁺ but are not the trigger store, and lysosomes and the Golgi play no part in excitation–contraction coupling.
In a mammal, a diploid somatic cell in G1 contains 8 pg of nuclear DNA. Most mammalian oocytes are ovulated arrested at metaphase II. The nuclear DNA content of such an ovulated oocyte (excluding any polar body) is ____ pg.
Numerical answer — type the value.
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Answer: 8
G1 diploid = 2C = 8 pg, so 1C = 4 pg. The primary oocyte replicates to 4C (16 pg). Meiosis I separates homologues, so the secondary oocyte has n chromosomes, each still with two chromatids: 2C = 8 pg, and it stays at 2C while arrested at metaphase II. Answering 4 pg treats it as already haploid in DNA (1C), which is true only after meiosis II; 16 pg forgets that meiosis I has been completed.
In an asynchronously dividing population of cells with a cycle time of 24 h, 50 out of 1,200 cells scored are in mitosis. Assuming the fraction of cells in a phase equals the fraction of the cycle spent in it, the duration of mitosis is ____ minutes.
Numerical answer — type the value.
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Answer: 60
Mitotic index = 50/1,200 = 0.0417. Duration of M = 0.0417 × 24 h = 1.0 h = 60 min. Expressing the answer in hours (1) instead of minutes, or dividing 1,200 by 50 and multiplying by 24, are the slips to avoid.
A cell line expresses a mutant securin that APC/C can no longer ubiquitinate. When these cells enter mitosis and all kinetochores have attached to the spindle, they will most likely
1. arrest in G1, because Rb stays unphosphorylated and E2F cannot drive the next S phase
2. separate their sisters normally, because separase is activated directly by kinetochore attachment
3. stay at metaphase with sisters joined, because separase remains bound and cohesin is not cleaved
4. re-replicate their DNA at once, because cyclin B is destroyed while the cells are still in mitosis
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Answer: C — stay at metaphase with sisters joined, because separase remains bound and cohesin is not cleaved
Securin is an inhibitory partner of separase. At the metaphase–anaphase transition APC/C–Cdc20 ubiquitinates securin for proteasomal destruction, which frees separase to cleave cohesin. A non-degradable securin keeps separase inactive, so sisters stay joined and the cells sit at metaphase even though the checkpoint is satisfied. Kinetochore attachment only silences the checkpoint; it does not activate separase directly. Rb–E2F acts in G1, and no re-replication can occur without passing through G1 licensing.
During prophase I of meiosis, crossing over between non-sister chromatids of homologous chromosomes takes place at
1. Zygotene, when homologues first begin to synapse
2. Pachytene, when synapsis along the homologues is complete
3. Diplotene, when the homologues pull apart and chiasmata appear
4. Diakinesis, when the chiasmata slide toward chromosome ends
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Answer: B — Pachytene, when synapsis along the homologues is complete
Crossing over is completed at pachytene, when the synaptonemal complex holds the homologues in full register. Diplotene is the tempting answer because chiasmata, the visible results of earlier crossovers, first appear then as the synaptonemal complex dissolves. Zygotene is when synapsis begins, and diakinesis is the final stage of further condensation before metaphase I.
Which of the following statements comparing mitosis and meiosis in a diploid animal are correct?
1. Homologous chromosomes pair and separate from each other during meiosis I
2. DNA is replicated a second time between meiosis I and meiosis II
3. Sister chromatids separate at anaphase of mitosis and at anaphase II of meiosis
4. Each daughter of a mitotic division receives half the chromosome number of the parent cell
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Answer: A — Homologous chromosomes pair and separate from each other during meiosis I; C — Sister chromatids separate at anaphase of mitosis and at anaphase II of meiosis
Meiosis I is reductional because homologues synapse and then segregate, and sister chromatids separate only when cohesin is cleaved along their whole length, at mitotic anaphase and at anaphase II. There is no S phase between the two meiotic divisions, which is exactly why the second division halves the DNA again. Mitosis keeps the chromosome number constant: each daughter is diploid like the parent.
A 1.0 Mb stretch of chromatin is fully packed into regularly spaced nucleosomes, each with a 147 bp core and an average linker of 53 bp. The number of histone H3 molecules in the nucleosome cores of this stretch is ____.
Numerical answer — type the value.
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Answer: 10000
Repeat length = 147 + 53 = 200 bp. Number of nucleosomes = 1,000,000/200 = 5,000. Each histone octamer contains two H3, so H3 = 2 × 5,000 = 10,000. Dividing by 147 instead of the full repeat, or forgetting that the octamer has two copies of H3, are the usual slips.
Nuclei are digested briefly with micrococcal nuclease, and the DNA is purified and run on a gel. The DNA appears as a ladder of bands at about 200, 400, 600 bp and so on. This result shows that
1. histone H1 is required for DNA to wrap around the octamer at all
2. the enzyme cleaves DNA only at a specific 200 bp-spaced recognition sequence
3. chromatin is organised as 30 nm fibres, each holding 200 bp of DNA per turn
4. DNA is protected in repeating units and cut in exposed linkers between them
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Answer: D — DNA is protected in repeating units and cut in exposed linkers between them
Micrococcal nuclease cuts accessible DNA. Partial digestion cuts some linkers and not others, releasing mono-, di- and trinucleosomes whose DNA lengths are multiples of the repeat (about 200 bp), which revealed the nucleosome as a repeating unit; longer digestion trims down to the 147 bp core. The enzyme has no sequence-specific recognition site, the ladder says nothing about higher-order 30 nm folding, and core particles wrap DNA even without H1.
The giant banded chromosomes of Drosophila larval salivary glands arise by
1. Repeated rounds of DNA replication without division, with the copies staying aligned
2. Arrest of oocyte bivalents at diplotene, with their chromatin loops spread out
3. Fusion of several diploid nuclei from neighbouring cells into a single large syncytium
4. Repeated non-disjunction in mitosis, which multiplies the number of chromosomes
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Answer: A — Repeated rounds of DNA replication without division, with the copies staying aligned
Polytene chromosomes form by endoreplication: the DNA is copied about ten times without separation, so about a thousand chromatids stay side by side (with homologues synapsed) and their condensed regions line up as bands. Arrest of oocyte bivalents at diplotene describes lampbrush chromosomes, the tempting confusion. Nuclear fusion and non-disjunction would change chromosome number, not create single giant aligned chromosomes.