6. Nucleic Acids and Protein Synthesis

A nucleotide is one building block of a nucleic acid. It contains three components: a pentose sugar, a phosphate group and a nitrogenous base. Many nucleotides can join to form a polynucleotide such as DNA or RNA.

• Pentose sugar: DNA nucleotides contain deoxyribose; RNA nucleotides contain ribose. The sugar identity helps distinguish which nucleic-acid context the nucleotide belongs to.
• Phosphate group: The phosphate contributes to the sugar–phosphate backbone when nucleotides join in a nucleic-acid strand. It is one component of a nucleotide, not a whole nucleic acid.
• Nitrogenous base: The base is classified as a purine (adenine or guanine, double-ring) or a pyrimidine (cytosine, thymine or uracil, single-ring). The classification names the base structure; it does not replace the sugar or phosphate components.
• ATP boundary: ATP is a phosphorylated nucleotide: adenine and ribose are attached to three phosphate groups. Its phosphate groups make it suitable for energy transfer in cell processes, but ATP is not a DNA or RNA strand and is not itself a nucleic acid.

The three-part structure supports two different biological roles: repeated sugar–phosphate connections allow nucleotides to build a nucleic-acid chain, while the attached bases provide the base identities used in sequence and pairing rules. ATP uses a nucleotide-like structure with additional phosphate groups for energy transfer, so “nucleotide” is a component-level description rather than a synonym for DNA, RNA or ATP.

A nitrogenous base alone is not a nucleotide, and a nucleotide is not the same as a nucleic-acid polymer. Purine/pyrimidine describes the base; deoxyribose/ribose describes the sugar. ATP is a phosphorylated nucleotide with an energy-transfer role, not a DNA/RNA building strand.

A purine is a nitrogenous base with a fused double-ring structure. In the nucleic acids studied here, the two purines are adenine (A) and guanine (G).

• Where they occur: Adenine and guanine can be the nitrogenous base in nucleotides that contribute to both DNA and RNA. The sugar and phosphate are the other nucleotide components; the base name alone is not a complete nucleotide.
• How the class is used: In a DNA or RNA sequence, A and G identify purine bases. Their complementary partners are pyrimidines: A pairs with T in DNA or U in RNA, while G pairs with C.
• Necessary contrast: Pyrimidines have a single-ring structure. The pyrimidines in this course are cytosine (C), thymine (T) and uracil (U). Purine/pyrimidine is a structural classification, not a synonym for a whole nucleotide or a rule that names the partner without checking the actual base.

The double-ring feature lets a learner classify A and G from the base category, while the surrounding sugar/phosphate context determines whether that base is part of a DNA or RNA nucleotide. Complementary pairing then uses the actual base identity, with one purine pairing with one pyrimidine to maintain the regular width of a nucleic-acid strand pair.

Purine does not mean “the whole nucleotide” and does not mean “a base that can be identified only by its partner”. Learn the two purines directly: adenine and guanine. Do not treat thymine as an RNA purine or uracil as a DNA purine.

A pyrimidine is a nitrogenous base with a single-ring structure. The pyrimidines in this course are cytosine (C), thymine (T) and uracil (U).

• Cytosine: C is used in both DNA and RNA nucleotides.
• Thymine: T is the pyrimidine used in DNA; it pairs with adenine in DNA.
• Uracil: U replaces T in RNA; it can pair with adenine in an RNA sequence.
• Pairing boundary: C pairs with the purine G in both DNA and RNA. A purine–pyrimidine pair maintains the regular spacing across a paired nucleic-acid strand. The ring classification describes the base, not the complete nucleotide.

The DNA/RNA context determines which pyrimidine is present: DNA uses T, whereas RNA uses U, while C is shared. Therefore the same base category can support different nucleic-acid sequences, but a sequence must be read with its molecule type known before assigning the complementary partner.

Uracil is not an additional DNA base in this course, and thymine is not the usual RNA pyrimidine. Pyrimidine means a single-ring nitrogenous base; it does not mean a whole nucleotide or automatically identify the sugar and phosphate attached to it.

DNA is a double helix made from two polynucleotide strands. Each strand has a deoxyribose–phosphate backbone on the outside, while the nitrogenous bases face inward and pair by complementarity.

• Backbone and direction: Covalent phosphodiester bonds link deoxyribose sugars and phosphate groups along each strand. The two strands run antiparallel: one is 5′→3′ while the other is 3′→5′.
• A–T comparison: Adenine (A), a purine, pairs with thymine (T), a pyrimidine, using two hydrogen bonds.
• C–G comparison: Cytosine (C), a pyrimidine, pairs with guanine (G), a purine, using three hydrogen bonds.
• Shared consequence: Each pair combines one purine with one pyrimidine, helping maintain a regular distance between the backbones. Hydrogen bonds hold the strands together but can be separated during replication; phosphodiester bonds keep each individual backbone intact.

The outward-facing sugar–phosphate backbones provide continuous covalent support, while inward-facing complementary bases hold the two strands together through hydrogen bonds. Antiparallel direction and specific pairing mean that the sequence on one strand determines the complementary sequence on the other, giving DNA both stability and a usable template for copying.

Hydrogen bonds join complementary bases across the two strands; phosphodiester bonds join adjacent nucleotides within one backbone. The strands are antiparallel, not parallel. The A–T and C–G comparison is a bond-and-pairing aid, not a complete replication mechanism; replication details belong to card 4574. Staff-only visual brief retained in the Topic Blueprint: show two antiparallel backbones, inward bases, bond types and 2-vs-3 hydrogen-bond contrast; no image is generated here.

Semi-conservative DNA replication produces two DNA molecules from one original molecule. Each product contains one original template strand and one newly synthesised complementary strand.

1. Separate the templates: Hydrogen bonds between complementary bases break, so the double helix opens and the two original antiparallel strands are exposed. Each original strand remains intact and acts as a template.
2. Match free nucleotides: Free DNA nucleotides align with exposed template bases by complementary pairing: A with T and C with G. This selects the sequence of each new strand.
3. Extend new strands: DNA polymerase joins adjacent nucleotides by forming phosphodiester bonds. It can extend a new strand only in the 5′→3′ direction, so the two templates are copied differently.
4. Account for leading and lagging synthesis: The leading strand is made continuously in the direction of the replication fork. The lagging strand is made as short Okazaki fragments because polymerase still works only 5′→3′; DNA ligase joins the fragments into a continuous strand.
5. Check the outcome: The original molecule has become two DNA molecules, each with one old strand and one new complementary strand: each product = 1 original strand + 1 new strand.

Complementary base pairing copies the information, while the fixed 5′→3′ direction of DNA polymerase explains why one new strand is continuous and the other is assembled in fragments. Ligase completes the lagging-strand backbone; the semi-conservative result follows because each original strand is retained as a template in one product.

Semi-conservative means that each complete DNA product keeps one whole original strand; it does not mean random pieces of old DNA are mixed into both strands. Hydrogen bonds open between the templates, whereas phosphodiester bonds form the new backbones. Staff-only visual brief retained in the Topic Blueprint: show one replication fork, antiparallel templates, continuous leading synthesis, Okazaki fragments and ligase joining; no image is generated here.

RNA is a nucleic acid made from nucleotides and is typically a single polynucleotide strand. Messenger RNA (mRNA) is an RNA transcript copy of a gene that carries information from DNA to a ribosome.

• Sugar: DNA nucleotides contain deoxyribose; RNA nucleotides contain ribose.
• Bases: DNA uses adenine, thymine, cytosine and guanine (A, T, C, G). RNA uses adenine, uracil, cytosine and guanine (A, U, C, G); U replaces T in RNA.
• Strands: DNA is typically a double-stranded antiparallel helix. RNA molecules are typically single-stranded, so an RNA molecule is not automatically a second DNA-like helix.
• Stability and role: DNA is relatively stable and suited to retaining genetic information. mRNA is a more temporary, mobile transcript that can be read by a ribosome; this lets information be used without moving the DNA molecule.
• Shared structure: Both are polynucleotides with sugar–phosphate backbones joined by phosphodiester bonds and nitrogenous bases projecting from the backbone.

The sugar and base differences distinguish the two nucleic acids, while strand arrangement supports their typical roles: DNA provides a stable information store, whereas a single-stranded mRNA copy can carry a selected sequence to a ribosome. “Typically” matters—RNA structure is not being defined as an absolute rule that every RNA molecule must be single-stranded or equally short-lived.

mRNA is not DNA with thymine: it contains ribose and uracil. A DNA coding strand and its mRNA transcript can have a related sequence, but the RNA uses U where DNA uses T. This card establishes RNA structure and mRNA’s transcript role; detailed transcription/translation steps belong to the next protein-synthesis topic.

A gene is a defined sequence of nucleotides in a DNA molecule that contains the information for producing a specific polypeptide. In the wider idea of gene expression, a DNA sequence is used to make an RNA message and, for a protein-coding gene, that message specifies an amino-acid sequence.

• Information source: The ordered DNA bases in the gene carry the sequence information; the gene is a section of a DNA molecule, not the whole chromosome.
• Expression chain: DNA base sequence → transcribed RNA information → codons read in order → amino-acid sequence → polypeptide.
• Product boundary: The polypeptide sequence is the immediate product specified by this syllabus objective. Its folding and interactions can give the final protein its structure and function.
• Expression boundary: Having a gene in the DNA does not mean it is being used at the same level in every cell or situation; gene expression refers to when and how the information is transcribed and used.

Because the DNA base order is copied into an RNA message and decoded into amino-acid order, changing the gene sequence can change the instructions available for the polypeptide. The gene supplies information; it is not itself the RNA message, amino-acid chain or finished protein.

A gene is not the entire chromosome, all of the DNA in a cell, or a protein molecule. It is a nucleotide sequence within DNA that can be expressed to specify a polypeptide; the later transcription and translation steps explain how that information reaches the product.

A codon is a triplet of bases on mRNA read during translation. Each codon specifies one amino acid or acts as a start or stop signal; it does not directly name a complete protein.

• Triplet and continuity: The mRNA sequence is read three bases at a time from a defined start point. The codons are read consecutively and do not overlap, so the reading frame determines which bases belong to each codon.
• Mapping: A codon maps to an amino acid, and the ordered amino acids form the polypeptide. A start codon establishes where the coding sequence is read; a stop codon signals termination and does not add an amino acid.
• Degeneracy: More than one codon can specify the same amino acid. This is degeneracy, not ambiguity: a given codon has one assigned meaning in the code.
• Near-universality: Almost all organisms use the same codon assignments, which is why the code is described as universal at A-Level. No special exception is needed to apply the core mapping here.

The triplet rule provides enough combinations to assign amino acids, while the fixed reading frame prevents the message from being regrouped at every step. Translation therefore follows: mRNA codons → amino-acid sequence → polypeptide, with start and stop signals defining the usable coding run.

A codon is read on mRNA, not as an untranslated DNA triplet, and it specifies an amino acid or signal rather than an entire protein. A stop codon ends translation but is not incorporated as an amino acid. Do not shift the reading frame or treat overlapping groups as the standard code.

Protein synthesis converts genetic information into an amino-acid sequence through two linked stages: transcription makes an mRNA copy from a DNA template, and translation reads the mRNA to assemble a polypeptide.

1. Transcription — DNA template to mRNA: The relevant DNA region is used as a template. RNA polymerase joins complementary RNA nucleotides to make a single-stranded mRNA molecule containing codons. The DNA sequence remains the information source; mRNA is the transportable message.
2. Message to ribosome: The mRNA leaves the nucleus and attaches to a ribosome. The ribosome reads the mRNA codons in order from the defined start point.
3. tRNA matching: Each tRNA carries a specific amino acid and has an anticodon. Its anticodon pairs with the complementary mRNA codon, bringing the corresponding amino acid to the ribosome.
4. Translation — chain assembly: The ribosome positions successive tRNAs so peptide bonds form between adjacent amino acids. The chain lengthens in the order specified by the mRNA codons.
5. Completion: Translation stops when a stop codon is reached. The completed amino-acid chain is released and can fold into the particular polypeptide structure specified by its sequence.

The information is converted rather than moved unchanged: DNA base sequence → mRNA codons → tRNA anticodon matching → amino-acid order → polypeptide. Transcription separates the protected DNA information source from the message, while translation converts the message into a peptide-bonded chain at the ribosome.

Transcription produces RNA, not a polypeptide; translation reads mRNA, not DNA directly. tRNA brings amino acids and uses anticodons, whereas mRNA carries codons. Staff-only visual brief retained in the Topic Blueprint: show DNA → mRNA → ribosome, tRNA delivery and chain growth; no image is generated or bound here.

During transcription, RNA is made from only one DNA strand. The template (transcribed) strand is read to build a complementary mRNA sequence; the non-template (coding) strand is not transcribed and has the same base sequence as the mRNA when T in DNA is replaced by U in RNA.

1. Identify the template: The gene region unwinds and the hydrogen bonds between the DNA strands break. The transcribed/template strand is the one whose exposed bases are used to build the RNA.
2. Build the RNA: RNA polymerase joins complementary RNA nucleotides to the template. The template is read 3′→5′ while the mRNA is built 5′→3′.
3. Use the direction cue: Template DNA 3′–TAC–5′ → mRNA 5′–AUG–3′.
4. Check the coding strand: The non-template/non-transcribed strand is written 5′→3′ and matches the mRNA sequence except that DNA has T where mRNA has U: Coding DNA 5′–ATG–3′ ↔ mRNA 5′–AUG–3′.
5. Place the product: The completed mRNA can leave the nucleus through a nuclear pore and carry the transcript to a ribosome; this is the end of the transcription-focused sequence, before translation is treated in the neighbouring process card.

Only the template strand is complementary to the new RNA, so it determines the mRNA sequence. The coding strand is a useful check because its 5′→3′ sequence matches the mRNA apart from T/U; labelling strand identity and direction prevents complementing the wrong strand twice.

Both DNA strands are present, but only one is transcribed for a given gene. Do not call the coding/non-template strand the template, and do not use T in the mRNA. This card resolves strand identity and direction; the complete DNA→RNA→polypeptide process belongs to card 4578, while intron removal and exon joining belong to card 4580. Staff-only transcription visual brief retained in the Topic Blueprint; no image is generated or bound here.

In a eukaryotic cell, the first RNA made from a gene is a primary transcript. It contains both the gene’s coding exons and non-coding introns, so it must be processed before it becomes mature mRNA ready to leave the nucleus.

1. Transcribe the whole gene region: Transcription produces a primary RNA transcript containing the exon and intron sequences that were present in the transcribed gene region.
2. Identify the sections: Exons are the coding sequences retained for the message; introns are non-coding sequences that are not to be translated into the polypeptide.
3. Remove introns: The intron sections are cut out of the primary transcript. This processing changes the RNA molecule, not the DNA gene itself.
4. Join exons: The remaining exon sections are joined to form one continuous mature mRNA molecule. This joining step is splicing.
5. Export the message: The mature mRNA leaves the nucleus through a nuclear pore and becomes available for the later translation stage. Translation is the next use of the processed message, not part of this card’s processing sequence.

Processing converts a mixed primary transcript into a continuous message: primary transcript (exons + introns) → introns removed → exons joined → mature mRNA → export from the nucleus. Without intron removal and exon joining, the RNA would not present the intended continuous coding sequence for later use.

Introns are removed from the RNA transcript, not deleted from the DNA template, and mature mRNA is not a protein. The primary transcript and mature mRNA are different RNA forms; the detailed codon/anticodon and peptide-bond process belongs to card 4578, not this post-transcriptional processing card.

A gene mutation is a change in the DNA base or base-pair sequence of a gene. It changes the stored sequence information; its biological consequence must be traced rather than assumed from the word “mutation”.

• Substitution: one base pair is replaced by another. This may leave the encoded amino acid unchanged or may alter a codon, depending on the position and the new base.
• Insertion: one or more base pairs are added to the sequence. The addition can change how downstream bases are grouped into codons, especially when the number added is not a multiple of three.
• Deletion: one or more base pairs are removed. Like an insertion, a non-triplet change can shift the downstream reading frame; a triplet-sized change can have a different reach.
• Consequence boundary: DNA change → possible mRNA/codon change → possible amino-acid or polypeptide change. The result depends on where the change occurs, how many bases are affected, and whether the relevant sequence is read or expressed.

A mutation matters through the expression chain, not by definition: a DNA sequence change may be silent, may alter one codon, or may regroup many downstream codons if the reading frame changes. A change outside the relevant coding information can have a different outcome, so mutation type alone is not enough to predict a polypeptide or phenotype.

Substitution, insertion and deletion describe different edits to DNA; they do not automatically describe the final protein effect. Not every mutation is harmful or changes an amino acid, and not every insertion/deletion causes a frameshift. Do not name a disease or phenotype without evidence for the specific sequence and expression context.

Substitution, insertion and deletion are three ways a gene’s DNA base sequence can change. Their different effects on the triplet grouping explain why some changes are local while others affect many downstream codons and the polypeptide produced.

• Substitution — swap one base: One DNA base is replaced by another. It changes the triplet at that position but is not a frameshift, so it does not regroup all downstream triplets. Because the genetic code is degenerate, the amino-acid sequence may stay the same or may change at that triplet.
• Insertion — add base(s): One or more bases are added to the DNA sequence. The added base changes the local triplet grouping and can shift the reading frame, changing downstream triplets and potentially many amino acids.
• Deletion — remove base(s): One or more bases are removed. Like an insertion, a non-triplet change can shift the reading frame and alter downstream amino-acid instructions.
• Comparison boundary: Substitution is local with respect to the reading frame; insertion/deletion can have a downstream frameshift effect. The actual polypeptide consequence still depends on the sequence position and how the changed codons are read.

The causal chain is DNA edit → altered triplet grouping or codon → possible amino-acid sequence change → possible polypeptide shape/function change. Insertions and deletions can propagate the change through later triplets, whereas a substitution does not automatically do so; the genetic code’s degeneracy means even a changed base need not change the polypeptide.

An insertion or deletion is not automatically the same as a substitution: check whether the reading frame is shifted. A mutation type predicts a mechanism of sequence change, not a guaranteed disease or protein outcome. This card compares the edits; the broader definition of mutation is card 4581 and the context-dependent polypeptide effect belongs to card 4583.

The effect of a gene mutation on a polypeptide must be traced through gene expression. A DNA change may alter the mRNA message, the amino-acid sequence or the amount of product, but it may also have little or no effect.

1. DNA information: Start with the changed base sequence and ask whether it lies in the part of the gene that is read and expressed.
2. RNA message: Transcription copies the relevant information into RNA; processing can produce the mature mRNA message used for expression.
3. Codon reading: A ribosome reads the mRNA codons while tRNA anticodons bring amino acids. A changed codon or a shifted reading frame can change the amino-acid instructions, but the degenerate code means some base changes do not change the amino acid.
4. Polypeptide outcome: The resulting chain may be unchanged, may contain a limited amino-acid change, or may be substantially altered/shortened when many downstream codons are affected.
5. Function boundary: A changed amino-acid sequence can alter folding, shape or function, but a changed DNA base does not guarantee a changed polypeptide or phenotype. The final claim depends on the mutation’s position, reading context and the protein’s role.

The complete reasoning path is DNA change → RNA message/processing → codon and tRNA matching at the ribosome → amino-acid sequence → polypeptide shape and function. This chain explains both possibilities: a mutation can be buffered by the code or location, or it can propagate through the reading frame and produce a markedly different polypeptide.

Do not jump directly from “mutation” to “disease” or “non-functional protein”. Explain each link that is supported: DNA sequence, mRNA/codon, amino-acid chain, then shape/function. Cards 4581 and 4582 define and compare mutation edits; this card is the high-level completion and consequence check, not a repeat of their classifications.