IB Biology A1.2: Nucleic acids worked solutions

Choose A [1]. One sugar, one phosphate and one base make one nucleotide. Guanine can occur in DNA or RNA, so the sugar identity is still needed to distinguish them.

Choose C [1]. Apply C–G and A–T at each position without reversing the top-to-bottom order. Uracil is not a DNA base.

The separated complementary bases no longer hydrogen-bond [1]. Each continuous strand retains its covalent sugar-phosphate backbone, including phosphodiester links [1]. Heating in this model separates strands; it does not cut them into monomers.

Choose B [1]. Ribose retains the carbon-2 hydroxyl group; deoxyribose has hydrogen at that position. Both are five-carbon sugars.

The combination of ribose and uracil identifies RNA [1]. The corresponding DNA sugar is deoxyribose [1]. Strand number supports the usual comparison but should not replace the chemical evidence.

Award one mark each for a phosphate group, ribose, and cytosine correctly labelled and connected [3]. A circle, pentagon and rectangle are suitable symbols. Both phosphate and base join the sugar; a phosphate–base–sugar chain is incorrect. Draw one unit only.

Use ribose in every nucleotide [1] and uracil wherever thymine was shown [1]. Keep covalent sugar-phosphate connections and bases branching from sugars. The chain remains single-stranded.

Thymine is also 17\%, leaving 100-2(17)=66\% for G and C [1]. Equal G and C totals give 66/2=33\% guanine [1]. The whole molecule, not one isolated strand, is the denominator.

A largely conserved codon-to-amino-acid mapping supports inheritance from a shared ancestor [1]. The two organisms can use different base sequences, genes or codon orders while sharing that mapping [1]. The code is nearly universal, with limited exceptions.

The conventional terminal phosphate is associated with sugar carbon 5, so the left end is 5\prime [1]. The free hydroxyl on carbon 3 identifies the right end as 3\prime [1]. These are chemical labels, not fixed left/right page positions.

Correct alternating sugar-phosphate backbones with bases on sugars [1]; A–T, G–C and T–A paired with dashed hydrogen-bond connections [1]; opposing sugar orientations or consistent antiparallel arrows [1]; a bracket enclosing exactly one phosphate, sugar and base [1]. The double helix need not be drawn. Do not attach bases directly to phosphates.

DNA contains deoxyribose whereas RNA contains ribose [1]. DNA uses thymine where RNA uses uracil [1]. Cellular DNA is usually double-stranded whereas RNA is usually single-stranded [1]. State both members of every comparison; one nucleotide itself is not double-stranded.

Each retained A requires T and each G requires C in the replacement DNA, with reciprocal pairings [1]. The retained base order therefore determines the new strand rather than permitting arbitrary additions [1]. A DNA template similarly determines a complementary RNA base order [1], but RNA uses U opposite template A rather than T [1]. This principle explains information transfer; enzyme names are unnecessary.

Two nucleotides per pair give 2(720)=1440 [1]. There are also 264 T, leaving 1440-528=912 G plus C [1]. Hence G is 912/2=456 [1]. The GC fraction is 912/1440\times100=63.333\ldots\%, or 63.3\% [1]. Do not divide the nucleotide count by 720 base pairs.

A covalent backbone helps preserve a long base sequence [1]. Four alternative bases allow different information to be stored at each position [1]. A complementary strand specifies a matching replacement by base pairing [1]. Reversible interstrand hydrogen bonding permits access to templates while leaving each backbone intact [1]. These are separate feature-to-function links, not merely names of DNA components.

Each core contains eight histone proteins [1], so five cores contain 5\times8=40 [1]. Five additional histones give 40+5=45 in total [1]. DNA winding around cores packages a long molecule into a smaller volume and supports further folding [1]. Linker histones are not part of the eight-protein core.

The percentage-point difference is 84-13=71 [1]. The pellet contains sedimented bacteria [1]. The higher fraction of DNA-associated label there supports preferential DNA entry rather than coat-protein entry [1]. Residual external coats, imperfect separation or label specificity can limit this inference [1]. Cell-associated label is evidence of entry only if external contamination has been addressed.

A and T differ by only 0.4 percentage points and their uncertainty intervals overlap [1]. G and C differ by 0.2 points and are also compatible with equal true amounts [1]. The A and T intervals exclude 25\%, as do G and C [1]. The data therefore contradict this equal-composition prediction while remaining consistent with complementary pairing [1]. They do not reveal exact base order or prove the helix.

In X the covalent backbones retain the order of all 12 bases [1]. Breaking hydrogen bonds separates complementary partners without cutting either sequence [1]. Either intact strand constrains its complement and pairing can be restored [1]. In Y, covalent backbone cleavage makes four three-nucleotide fragments per strand [1]. Each fragment retains its internal sequence, but the order of the fragments is no longer physically recorded [1]. Repeated or mutually compatible ends can allow several assemblies, so complete recovery is not guaranteed without reference information [1]. A correct discussion may note that a unique assembly could sometimes be inferred; loss of certainty is the issue.

Pairing in place gives 3\prime–TCAGGTAT–5\prime [1]; reversal gives 5\prime–TATGGACT–3\prime [1]. The original strand has A3, T2, G1 and C2. Adding its partner gives A=T=5 [1] and G=C=3 [1]. Thus GC is 6/16\times100=37.5\% [1]. Complementary equality applies across both strands, so unequal counts on the original strand alone do not contradict pairing [1].

An eight-protein histone core gives 16/8=2 cores [1]. Two separately wrapped DNA segments support that interpretation [1]. Select or colour DNA and protein separately so their association can be followed [1]. Rotate and zoom, or inspect chain identities, to check wrapping around each eight-chain core rather than a misleading overlap [1]. Core-only files omit linker DNA or associated proteins by design [1], so a missing linker histone in the rendered model is not evidence that none occurs in the cell [1]. Artificial colours and a static conformation are additional limitations.

In I, sedimented bacteria contain much more DNA label than protein label [1]. Preferential DNA entry followed by progeny production supports DNA carrying inherited instructions [1]. State a fractionation or labelling limitation, or restrict the conclusion to the tested phage system [1]. In II, a fixed equal four-base repeat would predict 25\% each, inconsistent with 34 and 16 [1]. Purines total A+G=50\% [1]; pyrimidines total T+C=50\%, so the ratio is 1 [1]. I traces which component enters cells; II tests DNA composition, and neither alone determines the detailed sequence or proves every structural feature [1].

Two choices at each of ten positions give 2^{10}=1024 strings [1]. Four choices give 4^{10}=1\,048\,576 [1], a ratio of 1\,048\,576/1024=1024 [1]. Complementarity fixes every partner base, so it is not an additional independent message [1]. The alphabet restriction changes the permitted base sequences [1], whereas the genetic code is the mapping from RNA codons to amino acids or stops, rather than the list of available DNA sequences [1].

R contains ribose and U, identifying RNA rather than DNA [1]. S contains deoxyribose and T, consistent with DNA [1]. Complementary-looking percentages describe composition, not physical arrangement, so they do not prove two strands [1]. A single RNA chain could happen to contain equal A/U and G/C totals [1]. Direct evidence must show whether two separate covalent backbones are associated [1], with complementary bases paired across them rather than merely within a folded single chain [1]. A single chain can fold so that complementary regions hydrogen-bond without creating another covalent backbone. A fully paired RNA duplex is another possible R arrangement; RNA is usually, not invariably, single-stranded.

Interstrand hydrogen bonds can break while covalent backbones remain intact [1], preserving the ordered bases on each original strand [1]. Each original strand is a template [1]; complementary A–T and G–C pairing constrains the new base sequence [1]. Complementarity explains accuracy but cannot certify that no mismatches or damaged bases occurred [1]. A comparison of full daughter sequences with the original, or other appropriate error detection, would be needed to substantiate an error-free claim [1]. Detailed enzyme mechanisms belong to D1.1.

A gene base sequence determines a complementary RNA sequence during transcription [1]. The same codons, read in the same frame with the same assignments, specify the same amino-acid order [1]. This conserved code supports common ancestry [1]. Differences in transcription [1] or translation rates/message persistence [1] can change protein amount despite an identical coding sequence. A base substitution can change a codon and an amino acid, altering folding and function [1], although code degeneracy can make some substitutions silent [1]. Do not assume an identical DNA sequence guarantees identical expression in different cells.

RNA can store information in its nucleotide sequence [1], and complementary interactions can support copying in a suitable system [1]. Catalytic RNA can assist reactions without requiring a pre-existing protein enzyme system [1]. These properties make an early RNA system plausible but do not demonstrate the actual historical sequence of events or solve how the first replicating system formed [1]. RNA viruses are not cellular organisms [1] and their RNA genomes do not contradict cellular organisms using DNA [1]. Catalysis is an emergent property of some folded RNA sequences; it is not a property of every isolated nucleotide.