Year 12

AQA A level Biology Nucleic Acids and Protein Synthesis Notes

Nucleic Acids & Protein Synthesis

— Biology · AS Level (7402) · Chapter 6 —
★ Exam Tip: A very important topic for both AS and A2. Protein synthesis is a frequently tested topic — ma'am wants it crystal clear. The Meselson–Stahl graph often comes as an MCQ; learn the exact wording of the definitions (gene, codon, mutation types).
Chapter map
  • ① nucleotide → polynucleotide (DNA / RNA structure)
  • ② DNA replication (semi-conservative) · Meselson–Stahl
  • ③ the genetic code (5 features) · ATP structure
  • ④ protein synthesis: transcription → RNA processing → translation
  • ⑤ gene mutations → sickle-cell anaemia

1. Nucleic Acids & the Nucleotide

Nucleic acids come in two types: DNA (deoxyribonucleic acid) and RNA (ribonucleic acid). Both are polymers — more precisely polynucleotides (poly = many; many nucleotides joined together).

Remember: macromolecule = any large molecule. Polymer = large molecule built from a repeat unit (monomer). Like polysaccharide = many saccharides (sugars).

Nucleotide = the monomer / basic unit of nucleic acids. DNA strand = polymer of nucleotides, RNA strand = polymer of nucleotides.

A nucleotide has 3 parts:

  1. Pentose sugar (5-carbon sugar)
  2. Nitrogenous base — adenine, thymine, cytosine, guanine, uracil
  3. Phosphate group (at least one)
Common mistake: it is nitrogenOUS base, not “nitrogen base”. Bases are molecules made of several elements, nitrogen being one of them.

The 2 pentoses (learnt in biological molecules): deoxyribose and ribose. A DNA nucleotide and an RNA nucleotide differ in the sugar and in one base:

OPBASEHOH1′2′3′4′5′deoxyribose
deoxyribonucleotide (DNA) — C2 has H
bases: A, T, G, C
OPBASEOHOH1′2′3′4′5′ribose
ribonucleotide (RNA) — C2 has OH
bases: A, U, G, C
  • Ring = O + 4 carbons (C1′–C4′). There is no room in the ring for the 5th carbon → C5′ sits outside the ring.
  • Base on C1′ · phosphate on C5′ · OH on C3′ (used later to make the bond).
  • de-oxy = “oxygen missing”: at C2′ only, OH becomes just H. The other OH groups keep their oxygen. (Two different pentoses — not the same sugar with an O removed from everywhere.)
  • Only ONE base is attached to each nucleotide — any one of the allowed four.
Nucleotide name Sugar Possible bases
in DNA deoxyribonucleotide deoxyribose A · T · G · C
in RNA ribonucleotide ribose A · U · G · C

Uracil replaces thymine in RNA ⇒ in total 5 bases: A, T, G, C, U.

Two ways to classify the bases

① by which nucleic acid they are in (above). ② by size = number of rings:

PURINESdouble ringed · largerPYRIMIDINESsingle ringed · smallerAdenine (A)Guanine (G)Cytosine (C)Uracil (U)Thymine (T)“Pure As Gold”Pure·A·Gold → Purine·A·G“CUT” = the pyrimidines
size is decided by the number of rings — not by the sugar or phosphate
Tricks: “Pure As Gold” → Purines = Adenine & Guanine. The other three = “CUT” (Cytosine, Uracil, Thymine) = pyrimidines.

Exam style: if a drawn nucleotide shows a double-ringed base on C1′, it is a purine (single ring ⇒ pyrimidine). Purines are larger, pyrimidines smaller.

P= phosphate group= POH⁻OOO –⁻ is a negative CHARGEon oxygen — not a bondbottom O links to C5′
phosphate group (circle P is just shorthand)

2. Joining Nucleotides — the Phosphodiester Bond

One nucleotide = one monomer. To get a dinucleotide and then a polynucleotide, nucleotides join by condensation (a small molecule — water — is released, just like in carbohydrates, proteins, lipids).

The OH on C3′ of one nucleotide + the phosphate (on C5′) of the next nucleotide
⇒ lose H₂O ⇒ phosphodiester bond
OPBASEH OBASEHOH O HO HP⁻OOO H₂OreleasedC3′ –OH of nucleotide ①phosphate on C5′ of ②(loses its OH)nucleotide ①nucleotide ②
CONDENSATION: OH (C3′ of ①) + OH (of phosphate on ②) → H₂O
OPBASEH OBASEHOH OP⁻OOO5′ P free3′ OH freeC – O – P – O – CPHOSPHODIESTER BOND(covalent · strong · solid line)ends = carbons (caps/locks)middle = phosphorusgaps filled by oxygens
dinucleotide: 2 nucleotides joined by ONE phosphodiester bond
Memory trick: C – O – P – O – C. Both ends = carbons (one from each nucleotide, like caps/locks), P in the middle, and the empty spaces are filled by oxygens. Logic, not rote!
  • Phosphodiester bond = covalent = strong (strong bonds = covalent & ionic → drawn as a solid line).
  • Hydrogen bonds = weak → drawn as a dotted line. (Glue analogy: weak bond = temporary glue that peels off; strong bond = permanent adhesive.)
  • Every nucleotide in the chain is held to the next by one phosphodiester bond. More than 2 units ⇒ polynucleotide (e.g. 6 nucleotides = 5 bonds).

A polynucleotide = one DNA (or RNA) strand

sugarAsugarGsugarTsugarCsugarAsugarG P 5′OH3′ 5′3′AGTCAGsugar–phosphate backbone(green = sugars + phosphodiester bonds)read 5′ → 3′ : A G T C A G
one DNA strand: free P at 5′ end, free OH at 3′ end

Why is a free phosphate seen only at ONE end? The phosphate on C5′ of every other nucleotide has lost its OH in condensation — it is busy in a phosphodiester bond. Likewise the 3′-OH of every nucleotide except the last is used up.

5′ end = the end with a free (unbonded) phosphate.

3′ end = the end with a free (unbonded) OH.

Rule (convention): always read / write a base sequence from 5′ → 3′. The strand drawn is AGTCAG, not GACTGA.

Backbone: the chain of sugar + phosphate held by phosphodiester bonds = the sugar–phosphate backbone (like our spine). We do not say “deoxyribose-phosphate” because it is the same idea for DNA & RNA.

Nitrogenous bases are NOT part of the backbone and do not join nucleotides together — they stick out and later hydrogen-bond with bases of another strand.

3. Structure of DNA

Classic statement: DNA is a double-helical structure. Break it down: double = made of 2 strands; helical = spiral (braided, like a twisted rope-ladder).

  • ✱ double stranded → 2 sugar–phosphate backbones (each a polynucleotide) — held together by H-bonds between bases
  • ✱ helical
  • ✱ antiparallel

Antiparallel: the two strands are parallel (side by side, not crossing) but run in opposite directions — one is 5′→3′, the other 3′→5′.

Complementary base pairing (like enzyme & substrate being complementary): each base pairs with one particular base only:

Adenine = Thymine  (2 H-bonds)  |  Guanine ≡ Cytosine  (3 H-bonds)
Note: a purine always pairs with a pyrimidine (big + small) → both strands stay the same distance apart.
A T A T G C T A C G G C C G C G T A A T 5′3′3′5′ blue = purines (A, G) · yellow = pyrimidines (C, T)5 purines + 5 pyrimidines in EACH strandpurple dashes = 2 H-bonds (A=T) · red = 3 H-bonds (G≡C)
complementary base pairing & antiparallel strands (teacher's random example)

Getting the other strand: write the complement of each base, then reverse the direction: 5′ AAGTCGCCTA 3′ ⇒ 3′ TTCAGCGGAT 5′ (read 5′→3′ it is TAGGCGACTT).

Chargaff's rule

In one DNA molecule:
% A = % T   and   % G = % C
⇒ % purines = % pyrimidines   (A + G = T + C)

Why? every A in one strand sits opposite a T in the other, every G opposite a C. So if one strand has 6 A, the partner strand has 6 T; 10 G ⇒ 10 C. Add up: the purines of the whole molecule = the pyrimidines.

Check (ma'am's random code above): purines = 3 A + 2 G = 5 in the top strand… and the pyrimidines also 5; same in the bottom strand. ✓ rule holds.

Stability of DNA — 3 reasons

  1. Numerous hydrogen bonds between the complementary base pairs. Each H-bond is weak, but there are so many that together they act like glue holding the 2 strands.
  2. Strong phosphodiester bonds (covalent) within the sugar–phosphate backbone.
  3. Helical structure → provides protection: enzymes cannot easily attack the DNA molecule.
sugar–phosphate backbone= strong phosphodiester bonds base pairs: H-bondsA–T 2 · G–C 3 (many!) helix (spiral) = compact,ends only exposed → protectionThree reasons DNA is stable: ① H-bonds ② phosphodiester bonds ③ helical shape
stability of DNA

Why does the helix protect? Imagine DNA as a plain linear molecule: enzymes could land anywhere and cut it. Twisted into a compact spring-like helix, only the ends are open to enzyme attack → the helix shields it from damage.

✂✂✂LINEAR (if no helix):enzymes can attack at any point ✂✂HELICAL: onlythe two ends areopen to attack
why the helical structure gives protection (teacher's argument)
Not a reason for stability: “ring structure” / “three rings”. Rings (in bases and sugars) only tell the shape/orientation of the molecule. Also never say “3 rings” — purines have 2 rings, pyrimidines 1.

4. RNA and DNA vs RNA

RNA is of 3 types:

mRNAmessenger · linear
tRNAtransfer · clover leaf
rRNAribosomal · linear
  • mRNA messenger RNA · tRNA transfer RNA · rRNA ribosomal RNA (found in ribosomes)
  • Single stranded — ALL RNA is. (tRNA looks double-stranded, but it is one strand folded back on itself.)
  • Pentose = ribose · bases A U G C (U replaces T)
  • Shape: linear (mRNA, rRNA) or clover-leaf (tRNA) · non-helical
  • Shorter than DNA · short-lived · made by transcription
DNA RNA
Sugar deoxyribose ribose
Bases A T G C A U G C
Strands double stranded single stranded
Shape helical non-helical (linear / clover leaf)
Size the larger molecule the smaller molecule
Lifespan long-lived, stable short-lived
Produced by DNA replication (S phase) transcription
Extra: DNA can be circular (bacteria, mitochondria, chloroplasts) or linear (eukaryotic nuclear DNA). Linear DNA is still coiled into a helix.

5. DNA Replication

Replication = making an exact copy of DNA. It happens in the S phase of interphase (before mitosis — this is how the amount of DNA is doubled).

Requirements:

  1. Parental DNA — each strand acts as a template
  2. Enzymes (every reaction in the body is enzyme-catalysed): DNA helicase, DNA polymerase, DNA ligase
  3. Energy in the form of ATP (the energy currency of the cell)
  4. Activated nucleotides

Activated nucleotide — why & what

  • A normal (non-activated) nucleotide has just 1 phosphate — it has no extra energy (like a lazy person); not good enough for the job.
  • Any process needs energy. A nucleotide is activated by adding 2 more phosphate groups (phosphorylated) → 3 phosphates. These phosphates come from ATP (the bonds between phosphates are energy-rich: when a phosphate breaks off, energy is released).
  • Why activated? The new strand is built from DNA monomers = nucleotides; we don't chop the old DNA up — we synthesise from fresh nucleotides.
OBASEHOH1′2′3′4′5′ PPPdeoxyribose 2 ADDED P groups(from ATP)← A / T / G / C
ACTIVATED nucleotide = deoxyribonucleotide with 2 additional P groups (3 P in all)
activated nucleotide = deoxyribonucleotide + 2 additional P groups (also: phosphorylated / energised nucleotide)

Replication is SEMI-CONSERVATIVE

Conserve = preserve unchanged. Semi = half. So semi-conservative = half of the parent molecule is conserved: each new (daughter) DNA = 1 parental strand + 1 newly synthesised strand.

PARENTAL DNAboth strands old (red) replication daughter DNA 11 parental (red) + 1 NEW (green)daughter DNA 21 parental (red) + 1 NEW (green)semi = half → half of each new molecule is conserved from the parent
semi-conservative replication
  1. Helicase breaks the H-bonds → the 2 strands separate; each is now a template; bases are exposed (unbonded).
  2. Activated nucleotides arrive and pair by complementary base pairing (temporary H-bonds): A–T, G–C.
  3. Phosphodiester bonds form between neighbouring new nucleotides → new sugar–phosphate backbone.
  4. Finally the H-bonds between the template and the new strand hold the new helix together.
Remember: in mitosis “conserved” meant the chromosome number stays the same. Here: half of the DNA is saved.

The mechanism — role of each enzyme

hel pol pol 5′3′3′5′3′5′5′3′3′✱✱LEADING strand: continuous, 5′→3′ towards the fork◀ fork moves this wayDNA helicase:unwinds + unzips (breaks H-bonds)LAGGING strand: discontinuous — short pieces made away from forkOkazaki fragments (green dashes)✱ gaps sealed by DNA LIGASE (joins fragments)DNA polymerase ALWAYS builds the new strand 5′ → 3′ (adds to the 3′ end)black = parental templates · green = new strands · pol = DNA polymerase
replication fork (teacher's drawing, tidied): leading vs lagging strand
Enzyme Role
DNA helicase unwinds & unzips the DNA by breaking H-bonds between the two strands (weak bonds → easy to break); makes 2 template strands
DNA polymerase attaches (at the 3′ end of the template side) and joins complementary activated nucleotides → forms the new strand's sugar–phosphate backbone (phosphodiester bonds). Always 5′ → 3′.
DNA ligase joins the Okazaki fragments on the lagging strand into one continuous strand
Key rule: DNA polymerase can only synthesise a new strand in the 5′ → 3′ direction. It can never make 3′ → 5′.

Leading vs lagging: the two parental strands are antiparallel but polymerase has only one direction.

  • Leading strand = continuous replication: polymerase just follows the helicase as it unzips.
  • Lagging strand = discontinuous replication: here 5′→3′ means working backwards (away from the fork). Polymerase waits until helicase has opened enough, then makes a short fragment, then another … → Okazaki fragments. (It cannot change its method, so it changes its direction.)
  • If both new strands were made the same way they would both be 5′→3′ and the antiparallel principle would break.
  • DNA ligase then joins all the fragments → one solid strand.
  • Memory: “lag” = like a lagging screen / internet — a break in continuity.

Three models of replication

CONSERVATIVE SEMI-CONSERVATIVE ✓ DISPERSIVE one molecule fully old,one fully neweach = 1 old strand+ 1 new strandboth strands = patchesof old + newred = parental · green = newly made. Accepted model = SEMI-conservative.
the three models of DNA replication
  • Conservative: parental DNA stays whole (both old strands together), the other daughter is wholly new.
  • Semi-conservative ✓ (accepted): each daughter = 1 old + 1 new strand.
  • Dispersive: each strand of both daughters is fragmented — some parts old, some new.

Meselson–Stahl experiment — how semi-conservative was proved

  • Two isotopes of nitrogen: N¹⁵ = heavy, N¹⁴ = light (both non-radioactive → harmless). Nitrogen is part of DNA (in the bases!), so growth medium decides which N goes into the DNA.
  • Used E. coli — it divides (one generation) every 20 minutes.
  • Grew E. coli in N¹⁵ medium → all its DNA is N¹⁵ = heavy DNA = generation 0.
  • Moved the bacteria to N¹⁴ medium; sampled every 20 min and checked the DNA.
E. coliN¹⁵ medium (heavy) E. coliN¹⁴ medium (light) E. coliN¹⁴ medium (light) 20 min 20 min GENERATION 0heavy DNA100 % N¹⁵GENERATION 1all HYBRID DNA50 % N¹⁵ : 50 % N¹⁴GENERATION 2hybrid + light DNA25 % N¹⁵ : 75 % N¹⁴ (3:1)blue strand = N¹⁵ (heavy, from original parent) · purple strand = N¹⁴ (light, new)
Meselson–Stahl experiment
  • After 20 min (generation 1): ALL DNA is hybrid — one N¹⁵ strand + one N¹⁴ strand → 50 % N¹⁵ : 50 % N¹⁴.
  • After 40 min (generation 2): hybrid DNA and completely light DNA. Count strands: N¹⁴ : N¹⁵ = 3 : 1 (75 % : 25 %).
  • Why does N¹⁴ rise? The only N in the medium is N¹⁴, so every new strand is N¹⁴; the old N¹⁵ strands are conserved but get diluted.
gen 0 2 strands (2 N¹⁵)100 % N¹⁵ · 0 % N¹⁴gen 1 4 strands (2 N¹⁵)50 % · 50 %gen 2 8 strands (2 N¹⁵)25 % · 75 %gen 3 16 strands (2 N¹⁵)12.5 % · 87.5 %every replication DOUBLES the molecules but the 2 old N¹⁵ strands never disappear —they just get diluted: N¹⁵ % halves each generation (÷2).
strand count per generation (each molecule = pair of bars; blue = N¹⁵, purple = N¹⁴)
Why this proves SEMI-conservative (extra reasoning, not on board): gen 1 = all hybrid ⇒ rules out conservative (that would give separate heavy + light DNA). Gen 2 = hybrid + light (not all intermediate) ⇒ rules out dispersive. Only semi-conservative fits.
100500%time01234N¹⁴ ↑N¹⁵ ↓cross at 50 %= generation 1
N¹⁵ falls, N¹⁴ rises; they meet at 50 % only once (generation 1) — x-axis = generations
gen % N¹⁵ % N¹⁴
0 100 0
1 50 50
2 25 75
3 12.5 87.5
4 6.25 93.75

% N¹⁵ ÷ 2 each generation; % N¹⁴ = 100 − % N¹⁵.

Graph MCQ: the curves cross at exactly 50 %, only once — at generation 1. N¹⁴ rises, N¹⁵ falls (never quite reaches 0).
Board slip: the notes show “1 : 8” for the later generation — it means 1 in 8 strands = 12.5 % N¹⁵ (as a ratio N¹⁵ : N¹⁴ it is 1 : 7). The middle flask also says “80 %” — it should be 50 %.

6. DNA & the Genetic Code

History: proteins were discovered before nucleic acids. Enzymes (proteins) control every life process, so scientists first assumed proteins = genetic information. Then Watson & Crick discovered DNA, and it became clear that DNA holds the genetic information; proteins carry out metabolism. So: how does DNA decide which protein to make?

Primary structure of a protein = the sequence of amino acids in a polypeptide (e.g. Met – Glu – Ala). Each protein has a unique amino-acid sequence ⇒ unique structure ⇒ unique function.

DNA code = the sequence of bases in DNA, e.g. TAC GAC CGT ACG
① The sequence of DNA bases must be the code for the sequence of amino acids in proteins.
② Each type of protein has a unique amino-acid sequence which determines its structure and hence its function.
③ The sequence of bases in the DNA of a cell = the code for all the proteins of that cell and organism.

(Every body cell has the same DNA as the zygote it came from. Identical twins share the same genotype — but their gene expression/phenotype can still differ.)

What exactly is a gene?

Gene = a section of DNA which codes for a polypeptide.
Better: a particular sequence of DNA nucleotides which codes for a polypeptide.
gene A polypeptide A gene B polypeptide B gene C polypeptide Cone DNA molecule carries many genes — only the section that is needed gets expressedgene = a section of DNA (a particular sequence of DNA nucleotides) that codes for ONE polypeptide
DNA ≠ gene: a gene is just a section of DNA
  • The short form is acceptable for marks, but a gene is NOT the whole DNA — DNA is made of many genes.
  • One gene = one complete polypeptide (insulin gene ⇒ enough nucleotides for all of insulin's amino acids).
  • Gene expression = protein synthesis = when the gene successfully converts into a protein. A gene may sit dormant for years (e.g. a hormone gene switched on only at a certain age) — it was always there, just not expressed.

Note: ma'am quoted “416 amino acids” for an insulin chain but wasn't sure — real human insulin has only 51 (A chain 21 + B chain 30). The idea is what matters.

Understanding the code — 5 features

There are 20 amino acids to be coded for, but DNA has only 4 bases. Which code length works?

bases read at a time possible codes enough for 20 amino acids?
1 (singlet) 4 ✗ only 4 amino acids
2 (duplet) 4² = 4 × 4 = 16 ✗ “cannot code for 20 amino acids”
3 (triplet) 4³ = 64 codons ✓ 64 ≥ 20

(Ma'am's lists: 1 base: A, C, G, T · duplets: AA, AC, AG, AT, CA, CC, CG, CT … → 16 · triplets: AAA, ACC, AGG, ATT, ACA, AGA, ATA … → 64. Letters X, Y, Z… stood for amino acids.)

① Genetic code is a TRIPLET code

One DNA triplet / codon codes for 1 amino acid.
Codon = a sequence of 3 consecutive nucleotides in a DNA or RNA molecule.

The order of triplets decides the order of amino acids in the chain (triplets X, Y, A, Z ⇒ chain X – Y – A – Z). A gene of 45 bases = 15 triplets = 15 codons.

② Genetic code is DEGENERATE (redundant)

64 combinations is more than needed for 20 amino acids ⇒ 44 extra triplets ⇒ each amino acid is coded by more than one triplet.

5554 6671 6781 56714 backup codes →same account opens ✓ 1234✗ not one of the 4 codes → does NOT open(a code for something else / a nonsense code)Genetic code: SEVERAL codons may open the SAME amino acid (degenerate)……but a codon can never code for two different amino acids.
ma'am's passcode analogy for a degenerate code
Examples are made up: the codons for glycine/alanine ma'am used (ACC, GCA, ACG…) were hypothetical. Only AUG = methionine and the stop codons must be remembered.

More than one code for the same amino acid is fine; a codon for a different amino acid, or a nonsense code, would not “open the account”.

③ Genetic code is PUNCTUATED

A sentence with no full stop never ends! The ribosome needs to know where to start and stop: stop codons don't code for any amino acid — they are the “full stop”.

STOP codons (mRNA) mnemonic
UAA “U are away”
UAG “U are gone”
UGA “U go away”
Start codon = AUG (also codes methionine). Ma'am's memory link: AUGur — in a novel she read, augurs were “the start of life”. Make your own link! Recall for 2 days → stuck for life.
Note: mRNA has U, no T (so “TAC” can't be mRNA; “AUG UCU CCG UAC CAC UAA” → the stop is UAA).

④ Genetic code is NON-OVERLAPPING

Adjacent bases / codons do not overlap: each base belongs to only one triplet; each triplet makes its own amino acid without interfering with its neighbour.

NON-overlapping ✓TACGGATGCACC TACGGATGCACCOverlapping ✗TACGGATGCACC each base belongs to ONE triplet only; adjacent codons never share bases
the genetic code is non-overlapping

⑤ Genetic code is UNIVERSAL

All living organisms use similar triplet codons for the same amino acids: AUG codes for methionine in humans, rats, monkeys…. Say “similar”, not “the same” (it is true for most organisms).

✓ Genetic code in one line each
  • Triplet — 3 bases = 1 codon = 1 amino acid (4³ = 64)
  • Degenerate — 64 − 20 = 44 extra → >1 codon per amino acid
  • Punctuated — start AUG; stop UAA, UAG, UGA
  • Non-overlapping — adjacent codons don't share bases
  • Universal — (similar) in all organisms

7. Structure of ATP

ATP = adenosine triphosphate (A from adenosine, T = “tri”, P = phosphate).

OadenineOHOH1′2′3′4′5′ PPPribose(pentose) 2 extra P groups(energy-rich bonds)← base = adenine
ATP = adenosine triphosphate = ribose + adenine + 3 phosphate groups
  • Adenosine = ribose (pentose) + adenine (base). Triphosphate = 3 phosphate groups (mono = 1, di = 2, tri = 3).
  • So ATP = a modified nucleotide = activated nucleotide = phosphorylated nucleotide (phosphorylation = adding phosphate groups): a ribonucleotide (A) with 2 extra P groups.
  • AMP (adenosine mono-phosphate, 1 P) = just a simple nucleotide; ADP = 2 P; ATP = 3 P.
  • The bonds between the phosphate groups are energy-rich (energy is stored there; released when a phosphate breaks off).
Careful: a random nucleotide (e.g. deoxyribose + thymine) with extra phosphates is an activated nucleotide but NOT ATP. ATP = ribose + adenine + 3 P.

8. Protein Synthesis — Overview

Protein synthesis is an active process = energy-consuming → needs ATP (ATP is the cell's energy currency — like only Pakistani rupees work in Pakistan, only ATP energy works in the cell).

Two stages (each followed by a modification):

  • (i) TRANSCRIPTION — in the nucleus: synthesis of mRNA using a gene; then post-transcriptional modification (RNA processing).
  • (ii) TRANSLATION — at the ribosomes: synthesis of a polypeptide using mRNA; then post-translational modification (= protein folding).
“DNA makes RNA (mRNA) and RNA makes protein”
NUCLEUS DNA (gene)template strand primary transcript(immature mRNA) MATURE mRNAcap · tail · no introns transcriptionRNA processing exits vianuclear poreCYTOPLASM RIBOSOME(free or on RER) polypeptide(1° structure) PROTEINfolded 2° 3° 4° translationfolding in RER(post-translational modification)
flow of information: nucleus → cytoplasm

Why do we need mRNA?

  • All genetic information is in the DNA, which cannot leave the nucleus; but proteins cannot be made inside the nucleus (too little space) → the information must be carried out to the cytoplasm.
  • mRNA = the translator / messenger between the language of DNA and the language of protein (like an interpreter between an Urdu speaker and an English speaker).
  • Only the gene needed at that moment is unzipped and expressed. If the whole DNA were transcribed, hundreds of unneeded polypeptides would be made together → protein levels rise → harmful, the cell could burst.

9. Transcription

Transcription = copying the genetic information in a molecule of DNA into a complementary strand of mRNA; a single strand of the DNA is used as the template.
  • Site: nucleus · result: synthesis of mRNA using a gene.
  • Why only one DNA strand? RNA is single stranded, so one template is enough.
  • Template = a pre-existing design (like a Canva template): the strand that gets copied = template strand = transcribed strand.

Requirements:

  1. Activated ribonucleotides — ribo, not deoxyribo (deoxyribonucleotides would build DNA, we want RNA!); activated by ATP phosphates.
  2. Template strand of the gene (of the DNA) — only the gene needed, not the whole DNA.
  3. RNA polymerase — joins the monomers (poly-mer-ase!) = the ribonucleotides complementary to the template.
  4. Energy in the form of ATP

Gene structure & the 3 stages

promoter exon intron exon intron exon terminatorRNA pol promoter = start:initiation (RNA pol binds)terminator = end:termination ELONGATION — phosphodiester bonds join ribonucleotidesexons = coding (carry the message)introns = non-coding (no message) — still transcribed!
a gene (eukaryote): promoter · exons / introns · terminator — e.g. 45 bases = 15 triplets
  • Promoter = start point of a gene; the site to which RNA polymerase binds for transcription.
  • Terminator (transcription terminator sequence) = marks the end of transcription.
  • Exons = coding part (carry a message). Introns = non-coding (no code, no message — like the appendix: present, but no role).
① Initiation
RNA polymerase binds to the promoter.
③ Termination
Transcription ends when RNA polymerase reaches the terminator.
② Elongation
  • H-bonds between the DNA strands break ⇒ unzipping — only the relevant gene unzips.
  • One strand = the template strand.
  • RNA polymerase joins activated ribonucleotides (complementary to the template) via phosphodiester bonds, in the 5′ → 3′ direction (all polymerases!).
non-transcribed(coding) strandmRNA copytemplate =transcribed strandGTTAAACTTGGGGUUAAACUUGGGCAATTTGAACCC 5′3′5′3′3′5′ RNA polymerase builds 5′→3′purple dashes = TEMPORARY H-bonds (DNA template ↔ mRNA); break once mRNA is completemRNA = same sequence as the NON-transcribed strand (U instead of T); complementary to the template
transcription: one DNA strand is the template
  • Before the phosphodiester backbone is made, the mRNA is held to the template by H-bonds. These are temporary: when mRNA is complete they break — mRNA must leave the nucleus and attach to a ribosome.
  • The product is the primary transcript (immature mRNA): a complementary copy of the whole gene including intron sequences — not yet able to bind a ribosome.
ATGGCGGGACTCTGCTAG TACCGCCCTGAGACGATC5′…3′3′…5′template strand = “TAC strand” (read it: TAC → AUG) transcriptionAUGGCGGGACUCUGCUAG 1 2 3 4 5 6STARTSTOPMetAlaGlyLeuCys—4 triplets between start & stop → 5 amino acids (Met-Ala-Gly-Leu-Cys);the stop codon codes for no amino acid
from gene to codons to amino acids

Which strand was transcribed? The mRNA is complementary to the template: AUG… ⇒ template TAC… (the “TAC strand”). The ATG-strand is the non-transcribed one. AUG = start, UAG = one of the 3 stop codons.

10. Post-transcriptional Modification (RNA processing)

Site: nucleus. The primary transcript is modified so that it can survive in the cytoplasm and attach to a ribosome after leaving the nucleus. Also called RNA processing. 3 steps:

(1) Splicing

  • Removal of introns (non-coding) and joining of exons (coding).
  • Enzyme: spliceosome — a “genetic scissor”.
  • Non-coding ≠ no bases. Introns are copied into the primary transcript; their base sequence just doesn't code for any amino acid.
Spelling: it is SPLICING, not “slicing” — students lose marks for this!

(2) 5′ capping   (3) Polyadenylation at 3′

  • 5′ capping = addition of a modified guanine at the 5′ end (like putting a cap on a toothpaste tube — protects the contents).
  • Polyadenylation at 3′ = poly (many) + adenylation (adenine nucleotides added): about 150–200 adenine nucleotides at the 3′ end (the poly-A tail).
  • Cap + tail protect both ends from intracellular enzymes that would attack the mRNA.
exon 1 intron exon 2 intron exon 3① primary transcript (immature) — has introns5′3′✂✂SPLICING by spliceosome (genetic scissors): introns cut out, exons joined exon 1 exon 2 exon 3② introns removed exon 1 exon 2 exon 3③ + 5′ cap + 3′ poly-A tail = MATURE mRNA G A A A A … (150–200 A)5′ cap = modified G
RNA processing in the nucleus (post-transcriptional modification)
✓ Significance of RNA processing
  • ① Removal of introns
  • ② Protection of mRNA from degradation (cap + poly-A tail)
  • ③ Enables mRNA to attach to ribosomes

After splicing + cap + tail: mature mRNA (the earlier primary transcript = immature / unprocessed).

11. tRNA and Ribosomes

tRNA — the amino-acid carrier

  • Single stranded · bases A U G C (common to all RNA).
  • Clover-leaf shaped — looks double-stranded but is ONE strand folded (5′ end and 3′ end are both on top).
  • Intramolecular complementary base pairing with H-bonds between the bases — within the same molecule (A–U, G–C), which holds the fold.
  • Anticodon: each tRNA has a specific anticodon, and is specific for carrying one particular amino acid. The anticodon determines which amino acid attaches to the tRNA = its “badge / ID card”. One tRNA carries one amino acid (alanine-tRNA → alanine, glycine-tRNA → glycine …).
  • The amino acid attaches at the 3′ end (opposite the anticodon). The 3′ end always ends with CCA (read 5′→3′; drawn top-down as A-C-C).
5′3′ amino acid ends in …CCAD loopT loopanticodonloop U A C3′5′anticodon AUG 5′3′codon AUG (mRNA)anticodon is alwaysOPPOSITE the 3′ endfolded by intramolecularH-bonds (A–U, G–C)
tRNA — ONE folded single strand (clover leaf)
Codon = triplet on DNA or mRNA.  Anticodon = the complementary triplet on tRNA.
e.g. codon UCA ⇒ anticodon AGU  ·  codon AUG ⇒ anticodon UAC (Met-tRNA).

Board slip: the notes write the anticodon of UCA as “A G A” — the complement is A G U.

Drawing tRNA quickly: draw a cross; the long line is the 3′ end and the anticodon sits opposite to it.

3′ endtRNAanticodon+ amino acid enzymereversible amino acidCHARGED tRNA= aminoacyl-tRNA complex
tRNA picks up its specific amino acid at the 3′ end

tRNA + its amino acid = charged tRNA = aminoacyl-tRNA complex. The attachment is reversible (an enzyme is involved — details not needed): what attaches can detach.

Ribosome

  • Two sub-units: a large and a small sub-unit, made of protein + rRNA (ribosomal RNA).
  • The large sub-unit contains the enzyme peptidyl transferase (makes peptide bonds).
  • The two sub-units assemble (ribosome assembly — possibly helped by Mg²⁺ ions) when needed and disassemble afterwards.
LARGE sub-unitpeptidyl transferaseSMALLmRNAenzyme in the largesub-unit = PEPTIDYLTRANSFERASEmRNA binds thesmall sub-unitribosome = protein + rRNA; only 2 codons are exposed inside it at a time
eukaryotic ribosome

12. Translation

Translation = synthesis of a polypeptide using mRNA. Energy-consuming.
Site: ribosomes — free in the cytoplasm or on the RER.

Requirements: ① ATP ② ribosomes (the site) ③ mRNA (the molecule being translated; it left the nucleus through a nuclear pore) ④ amino acids ⑤ tRNA (carrying specific amino acids).

Divided (like transcription) into initiation → elongation → termination. The mRNA of the example has start AUG, 4 triplets in between, and a stop codon (UAG): AUG GCG GGA CUC UGC UAG.

① INITIATION AUGGCGGGACUCUGCUAG MetmRNA attaches to the SMALL sub-unit;tRNA carrying methionine (anticodon UAC)pairs with the START codon AUG② ELONGATION AUGGCGGGACUCUGCUAG Ala Gly Met peptide bondcodon sequence → amino-acid sequencepeptidyl transferase makes peptide bondsbetween neighbouring amino acids;tRNA leaves, ribosome moves along③ TERMINATION CUCUGCUAG UAG = stop Met–Ala–Gly–Leu–Cys stop codon (UAA / UAG / UGA)reached — no tRNA for it;ribosome disassembles (large + smallsub-units separate) & the polypeptideis released into the cytoplasm
translation in three steps (mRNA read 5′→3′, ribosome moves along it)
① Initiation
  • mRNA attaches to the small sub-unit of the ribosome.
  • The start codon AUG codes for methionine, so the first tRNA (carrying Met, anticodon UAC) pairs up with it — temporary H-bonds; no peptide bond yet.
  • The first amino acid of every chain is Met. (Not every finished polypeptide contains Met — it can be removed later.)
② Elongation
  • The sequence of codons on mRNA is translated into the sequence of amino acids of the polypeptide chain.
  • 2 codons / 2 tRNAs are exposed in the ribosome at one time. The enzyme peptidyl transferase (large sub-unit) forms peptide bonds between the neighbouring amino acids (covalent bonds of the primary structure).
  • Once the peptide bond is made, the tRNA detaches (it was only a carrier) — codon–anticodon H-bonds are weak; the ribosome moves on to the next codon.
③ Termination
  • Occurs when the ribosome reaches a stop codon (UAA, UAG or UGA — any one).
  • The ribosome disassembles (large + small sub-unit separate) and releases the polypeptide chain into the cytoplasm.
5′3′ ribosomes move along the mRNA →short chainLONGEST chainPOLYSOME (polyribosome) = many ribosomes on ONE mRNA → many copies of the polypeptide at once(fast & efficient — cell saves time)
polysome
Polysome: one mRNA with many ribosomes attached = polysome / polyribosome → many polypeptide chains made from a single mRNA at the same time.

Post-translational modification = protein folding

Takes place in the rough ER: the polypeptide enters the RER sacs. Primary structure (from translation) → secondary (α-helices, β-pleated sheets) → tertiary (R-group interactions: H-bonds, hydrophobic interactions, ionic bonds, disulfide links) → quaternary (when > 1 polypeptide chain interact).

Transcription Translation
Where nucleus ribosomes (free / RER)
Makes mRNA (using a gene) polypeptide (using mRNA)
Enzyme RNA polymerase peptidyl transferase
Bond made phosphodiester peptide
Start / end promoter / terminator AUG / stop codon
Then RNA processing protein folding (RER)

13. Gene Mutation

Mutation = a change in the genetic information (DNA = our hereditary information bank). Two kinds:

Gene mutation
random
change in the base sequence in part of the DNA molecule ⇒ alteration in the sequence of nucleotides which may alter the sequence of amino acids in a polypeptide chain
Chromosome mutation
change in the structure and/or number of chromosomes (e.g. a lost arm; Down syndrome = 3 copies of chromosome 21 instead of 2 — not studied in detail)
Key word “MAY”: not every mutation changes the amino-acid sequence — if the same amino acid is still coded, the protein is unchanged.

3 types of gene mutation: base substitution · base addition (insertion) · base deletion.

Base substitution — a base is replaced by a different base

original MET GLU SER VAL SILENT MET GLU SER VAL no changedespite mutationMISSENSE MET PRO SER VAL same length,1 amino acid wrongNONSENSE MET STOPchain ends abruptly(truncated protein)
effects of a base substitution on the polypeptide
Type Definition Why (analogy)
Silent no overall change in the polypeptide chain despite gene mutation / base substitution degenerate code: new codon codes the same amino acid
Missense a mutation that changes an amino acid in a polypeptide chain miscommunication; complete chain, same length, but wrong amino acid
Nonsense a mutation that changes a codon to a stop codon, so protein synthesis (the polypeptide chain) ends abruptly makes no sense; chain incomplete
Chinese whisper: original “Are you going to China?” → last person says “He is going to China.” = missense (complete sentence, different meaning). If it arrives as “Are you going to…” = nonsense (abrupt stop, incomplete).
DNA tripletsTACMetGGTProAGTSerCAAValACT(stop)after mutation (A → T in 3rd base of 4th triplet):TACMetGGTProAGTSerCATValSAME ✓ACT(stop)CAA and CAT both code valine (degenerate code) → no overall change = SILENT
silent mutation (template-strand triplets shown)
original mRNAAUGUACUCACCGACAUAGMetTyrSerProThrSTOPC → G in the 3rd codon (UCA → UGA):AUGUACUGACCGACAUAGMetTyrSTOP!(never read)chain stops after 2 amino acids → incomplete, usually useless protein = NONSENSE
nonsense mutation (illustration with made-up sequence)

Base addition (insertion) & base deletion

Addition / removal of one or more nucleotides. This may lead to a FRAMESHIFT mutation. (Think photo frame: add or remove something and the frame shifts.)

Frameshift mutation = a gene mutation caused by insertion or deletion of one or more nucleotides, resulting in incorrect reading of the sequence of triplets in the genetic code.
originalAUGMetGCAAlaUUCPheGAAGluinsert GAUGMetGGCGlyAUUIleCGAArgAdelete CAUGMetGAUAspUCGSerAAinsert 3 (GGG)AUGMetGGGGlyGCAAlaUUCPheGAAGlu↑ ONE extra base: every triplet after it is re-read differently↑ ONE missing base: same — whole downstream sequence changes↑ delete/insert a MULTIPLE OF 3 → reading frame kept (only an amino acid gained/lost)frame = the way the bases are grouped in 3s. Shift the frame → wrong amino acids from that point on.
frameshift: insertion / deletion (made-up sequence; real codon table)

Board example (codons): UAC UGG CUA GUA CAU = Tyr–Trp–Leu–Val–His. Insert one C → UAC UCG GCU AGU ACA U = Tyr–Ser–Ala–Ser–Thr: one C, and everything after it is read wrongly.

Board slip: its first line is printed “TAT TGG…” — it should be TAC TGG… for the inserted C to give the second line.

Not every insertion/deletion shifts the frame: adding or removing a multiple of 3 nucleotides (a whole triplet) keeps the frame — the code is a triplet code. Leftover 1 or 2 bases = the problem.

14. Sickle-Cell Anaemia

  • An inherited (genetic) disorder caused by base substitution — a classic missense mutation.
  • Homozygous recessive: the child inherits the recessive allele from both parents.
  • Affects the RBCs — which are mostly haemoglobin (Hb). Hb = 4 polypeptide chains: 2 α-globin + 2 β-globin. The disorder is due to the β-globin chain (made by one gene).
Genotype Allele Result
HbA HbA HbA = normal / dominant normal
HbA HbS HbS = abnormal / recessive carrier (sickle-cell trait) — no disease
HbS HbS both abnormal sickle-cell DISEASE

What goes wrong?

β-globin gene, 6th triplet: CTC → CAC (T replaced by A) = BASE SUBSTITUTION → MISSENSE mutation 6th amino acid: glutamic acid (hydrophilic, soluble)→ VALINE (hydrophobic, insoluble) β-globin chain of Hb is LESS SOLUBLE→ forms precipitates (lumps) in the RBC lumps polymerise → long fibres pull on the RBC wall→ SICKLE shape ↓ O₂-carrying capacity · fewer RBCs · vessels blocked
sickle-cell: from one base to a sick cell
  • Normally the 6th triplet of the β-globin gene is CTC → glutamic acid (hydrophilic = soluble). After substitution: CAC → valine (hydrophobic = insoluble). T is replaced by A.
  • The β-globin chain, so haemoglobin, becomes less soluble → precipitates (small lumps — like lumps in a flour/besan mix) inside the RBC. Lumps polymerise into long fibres which pull the RBC wall inward → sickle (crescent) shape, smaller capacity.
  • RBCs have no nucleus/mitochondria so as to carry the max. of Hb and O₂ — so any loss of capacity matters a lot.
NORMAL RBCbiconcave disc, lots of Hbcarries O₂ well SICKLE-SHAPED RBClong Hb S fibres pulled it out of shape↓ O₂ capacity · blocks vessels
normal vs sickled red blood cell

Symptoms & treatment

reduced O₂-carrying capacity of the blood ⇒

  1. Fatigue — less O₂ ⇒ less aerobic respiration ⇒ less ATP ⇒ tires quickly
  2. Swelling / pain in hands & feet
  3. Reduced number of RBCs (anaemia — RBCs are smaller, wear out faster)
  4. Frequent infections
  5. Stroke / heart attack — rigid sickle cells block blood vessels (obstruction → blood cannot reach tissues)
Treatment: bone marrow transplant — the only definitive cure. Stem cells in the bone marrow divide and differentiate into all blood cells; the patient's stem cells make faulty RBCs; a healthy donor's stem cells make normal ones.
Symptoms only are controlled by: antibiotics, vaccinations (frequent infections), analgesics / painkillers (swelling & pain).

15. Full Chapter Cheat-Sheet

✓ Nucleic acids
  • Nucleotide = pentose + nitrogenous base (on C1′) + phosphate (on C5′). Deoxyribose (C2′ = H) in DNA; ribose (C2′ = OH) in RNA.
  • Bases: DNA A T G C · RNA A U G C. Purines (A, G) double ring “Pure As Gold”; pyrimidines (C, U, T) single ring “CUT”.
  • Condensation (C3′–OH + phosphate) → phosphodiester bond C–O–P–O–C (covalent) → sugar–phosphate backbone. 5′ end = free P; 3′ end = free OH; read 5′→3′.
  • DNA: double-stranded, helical, antiparallel; A=T (2 H), G≡C (3 H); Chargaff: %A=%T, %G=%C, purines = pyrimidines.
  • Stability: many H-bonds · strong phosphodiester bonds · helix (protection from enzymes).
  • RNA: single stranded, ribose, AUGC, shorter, short-lived, non-helical; mRNA / tRNA / rRNA; made by transcription (DNA by replication).
✓ Replication
  • S phase; needs parental DNA, helicase / polymerase / ligase, ATP, activated nucleotides (deoxyribonucleotide + 2 P).
  • Semi-conservative: each daughter = 1 old + 1 new strand. Polymerase builds 5′→3′ only → leading (continuous) / lagging (Okazaki fragments, joined by ligase).
  • Meselson–Stahl: N¹⁵ → N¹⁴, E. coli, 20 min; gen 1 all hybrid (50:50), gen 2 hybrid + light (N¹⁴:N¹⁵ = 3:1); % N¹⁵ ÷ 2 each generation; lines cross at 50 % at generation 1.
✓ Code & protein synthesis
  • Gene = particular sequence of DNA nucleotides coding for a polypeptide; gene expression = protein synthesis.
  • Code: triplet (64), degenerate (44 extra), punctuated (start AUG; stop UAA UAG UGA), non-overlapping, universal.
  • ATP = ribose + adenine + 3 P = modified / activated / phosphorylated nucleotide.
  • Transcription (nucleus): template strand + activated ribonucleotides + RNA polymerase + ATP; initiation (polymerase binds promoter) → elongation → termination (terminator); product = primary transcript.
  • RNA processing: splicing (introns out; spliceosome) · 5′ cap (modified G) · 3′ poly-A tail (150–200 A) → mature mRNA (protection + ribosome attachment).
  • tRNA: clover leaf, intramolecular H-bonds, anticodon, aa at 3′ end (CCA). Ribosome: large + small sub-unit (protein + rRNA); peptidyl transferase in large.
  • Translation (ribosomes): initiation (mRNA to small sub-unit; Met-tRNA pairs with AUG) → elongation (peptide bonds) → termination (stop codon; ribosome disassembles, chain released). Polysome = many ribosomes on 1 mRNA. Then folding in RER.
✓ Mutation & sickle cell
  • Gene mutation (random): substitution (silent / missense / nonsense), addition & deletion → may cause frameshift.
  • Sickle cell: homozygous recessive HbSHbS; β-globin 6th triplet CTC→CAC; glutamic acid (hydrophilic) → valine (hydrophobic) → precipitates → fibres → sickle RBC → ↓ O₂ capacity; treatment = bone marrow transplant.