IB Biology A2.2: Cell structure worked solutions

Choose B [1]. Ribosomes translate mRNA in both groups. A nucleus is absent from bacteria; a wall supports rather than assembles proteins; Golgi is absent from bacteria.

DNA lies in an unenclosed bacterial nucleoid but within a nuclear envelope in the animal cell [1]. Bacterial cytoplasmic ribosomes are 70S, compared with animal cytoplasmic 80S [1]. Both sides of each comparison are needed.

Choose C [1]. A chitin-containing wall is diagnostic here; chloroplast absence alone would not separate an animal cell from a fungus.

Identify Golgi apparatus [1]. State modification or sorting of proteins/lipids for onward packaging [1]. A stack alone is insufficient to call it rough ER when attached ribosomes are absent.

Choose D [1]. Carbon dioxide is a product of metabolism. Its removal is excretion; egestion concerns undigested food.

A skeletal muscle fibre or aseptate fungal hypha supplies a multinucleate example [1]. A mature mammalian erythrocyte or mature phloem sieve tube element supplies an enucleate example [1]. Maturity and mammalian identity must be retained where relevant.

Convert 32\,\mathrm{mm} to 32\,000\,\mu\mathrm m [1]. Divide image length by actual length: 32\,000/8.0=4000, reported as 4000\times without a length unit [1].

Use the length ratio 21/12 [1]. Multiply by the bar value: (21/12)(4.0)=7.0\,\mu\mathrm m [1]. The image units cancel; the bar provides the physical unit.

The digital operation increases displayed magnification [1]. No new spatial information was recorded, so close points that were not originally separated remain unresolved [1].

Predict that the organism consists of one or more cells [1]. Examine it microscopically for membrane-enclosed cellular units [1]. Do not predict a nucleus or a fixed number of cells for every living organism.

Use a thin layer laid flat in water [1]. Lower a coverslip at an angle to displace air [1]. A suitable stain increases contrast, although it may prevent live observation [1]. Start at low power because the wider field makes the cells easier to locate and centre [1]. Use fine focus at high power to avoid a large lens movement into the slide [1].

c_X=90/60=1.5\,\mu\mathrm m per division [1]. Cell length =28(1.5)=42\,\mu\mathrm m [1]. At Y, c_Y=45/60=0.75\,\mu\mathrm m per division [1], so the unchanged cell spans 42/0.75=56 divisions [1]. The objective changes calibration, not actual size.

Draw two envelope outlines with a gap through both at a pore [1]. Preserve a large closed profile elsewhere and internal chromatin threads [1]. Place three accurate label lines on envelope, pore and chromatin [1]. Annotate the envelope as separating nuclear and cytoplasmic contents, with exchange through pores [1]. Annotate chromatin as DNA carrying genetic instructions, or the pore as a route for regulated exchange [1]. Names alone do not earn the annotation marks.

Freeze fracture exposes internal membrane faces [1]. Immunofluorescence locates the antigen [1]. Cryo-EM preserves rapidly frozen particles without requiring crystals [1]. An antibody supplies target recognition and is detected with a fluorescent label [1]; fluorescence alone does not establish antigen specificity.

R is plant because the cellulose wall and plastids support that identity [1]. S is fungal because its wall contains chitin [1]. Non-photosynthetic plant plastids explain why chloroplast absence cannot prove animal identity [1]. A mature sieve tube element loses its nucleus during differentiation; this does not make it a bacterium [1].

Different genes are expressed in the two cells [1]. Expression produces different proteins, here transport versus contractile proteins [1]. Those products enable different specialised activities without requiring loss of unused genes or a sequence change [1].

Circular organellar DNA resembles bacterial chromosomes [1]. Bacterial-type ribosomes resemble bacterial protein-synthesis machinery [1]. Replication from existing chloroplasts is consistent with retained cellular ancestry [1]. Modern chloroplasts still depend on host resources and nuclear gene products, so replication is not proof of independent life [1].

Drawing length =24(1500)=36\,000\,\mu\mathrm m [1] =36\,\mathrm{mm} [1]. An 8.0\,\mu\mathrm m bar requires 8.0/0.25=32 pixels [1]. Keep the bar attached and resize it by the same factor as the photograph [1]. Its label remains 8.0\,\mu\mathrm m.

The no-primary control tests background from the detection system and cells without the target-recognising reagent [1]. The target-negative control tests signal when the antigen is absent [1]. Difference =120-24=96 units [1]. The stronger complete signal in positive cells is consistent with target-associated labelling [1]. No spread or uncertainty is given, so statistical reliability cannot be judged [1]. Fluorescence intensity does not itself establish finer resolving power, molecular binding in every bright pixel, or a live-cell result [1].

Same tested sequence plus different products is consistent with altered gene expression rather than required gene deletion [1]. A can attach neighbours [1]; T can enable material exchange [1], supporting cooperation and possible division of labour [1]. A group of attached cells alone does not demonstrate distinct specialised cell types [1]. Remove or block S, A or T separately in matched viable cultures and measure adhesion/exchange [1]. Retain appropriate untreated/vehicle controls and replication to distinguish each causal role from correlated expression or damage [1].

Stacked thylakoids within a double envelope support chloroplast identity [1]. Actual length =(360/120)(2.5)=7.5\,\mu\mathrm m [2: ratio and answer with unit]. The folded-inner-membrane profile supports mitochondrion identity [1]. A drawing must preserve two envelope membranes [1] and the stacks versus folds that distinguish the profiles [1]. One missing profile in a thin section cannot establish absence from the entire cell.

Nuclei and mitochondria support eukaryotic identity [1]. Fungi and non-photosynthetic plant cells also lack chloroplasts, so the observation does not establish animal identity [1]. Bacterial genetic affinities are consistent with mitochondrial descent from bacteria [1]; organellar circular DNA or bacterial-type ribosomes supply a further independent line of evidence [1]. Differentiated cells can divide tasks and cooperate [1], allowing a larger body while cells remain small enough for exchange [1].

A suitable light microscope can observe the living filling/emptying cycle [1]. Expelling excess water contributes to homeostasis [1]. Water enters by osmosis down a water-potential gradient through the selectively permeable membrane [1]. The supplied inhibition should reduce effective water expulsion [1]. Water may accumulate and increase cell volume or cause rupture because a wall is absent [1]. Ion transport and selective exchange belong to the membrane, not a hypothetical wall [1].

Different genes are transcribed, giving different mRNA abundances [1]. Ribosomes translate the mRNAs into the corresponding proteins [1]. Different products support export and contraction as specialised activities [1]. The broadly shared genome can remain unchanged [1]. Alter a relevant regulatory signal in matched cultures and measure the mRNA followed by protein [1]; include a control culture and repeated independent samples to test the proposed expression link [1].

Cellulose walls and plastids support plant identity in both tissues [1]. Chloroplasts account for light-dependent carbon fixation in L [1]. Mitochondria support aerobic respiration in both, so absence of chloroplasts does not remove that pathway in R [1]. Circular organellar DNA and bacterial-type ribosomes are two linked observations supporting bacterial ancestry [2]. Retained organelles depend on host/nuclear products and cannot be inferred to be independently living bacteria [1].