Nucleic Acids

1. Overview of nucleic acids

Definition

Nucleic acids are large biological molecules made of nucleotide monomers. They store, transmit, and help express genetic information necessary for cellular function, growth, development, and reproduction.

The two major types are:

• DNA (deoxyribonucleic acid): Stores hereditary information.

• RNA (ribonucleic acid): Functions in gene expression, protein synthesis, regulation, and other cellular processes.

Major functions

How are nucleic acids different from other biomolecules?

• Carbohydrates: Primarily provide energy, store energy, and contribute to structural support.

• Lipids: Store energy, form membranes, insulate, and participate in signaling.

• Proteins: Perform most enzymatic, structural, transport, and regulatory functions.

• Nucleic acids: Store, transmit, and help express genetic information.

One important distinction: these functions overlap. For example, proteins help replicate DNA, and DNA contains the instructions for making those proteins.

Where are nucleic acids found?

• In eukaryotic cells, most DNA is located in the nucleus.

• Additional DNA occurs in mitochondria and, in plants, chloroplasts.

• RNA is found in both the nucleus and cytoplasm, including ribosomes.

• In prokaryotes, DNA is located primarily in the nucleoid region, not inside a membrane-bound nucleus.

Correction to your professor's note: DNA does not only become accessible when a cell undergoes mitosis. DNA is continually used during normal cellular activity, including transcription and DNA replication.

2. The basic structure of a nucleotide

A nucleotide consists of three components:

1. Nitrogenous base

Carries the variable genetic information.

2. Pentose sugar

A five-carbon sugar: ribose or deoxyribose.

3. Phosphate group

Connects sugars in the nucleic acid backbone.

Important terminology

• Nitrogenous base: The base alone.

• Nucleoside: Base + sugar.

• Nucleotide: Base + sugar + one or more phosphate groups.

• Nucleic acid: A polymer of nucleotides connected by phosphodiester bonds.

A nucleotide's phosphate group is derived from phosphoric acid. It contains phosphorus bonded to oxygen atoms; the P–O bonds are not all equivalent in bond order or resonance description.

Memory trick:

• Base + sugar = nucleoside.

• Nucleoside + phosphate = nucleotide.

• Many nucleotides linked together = nucleic acid.

3. Nitrogenous bases: purines vs. pyrimidines

Nitrogenous bases are nitrogen-containing ring structures that form part of nucleotides.

They are divided into two groups based on their ring structures.

Purines

A, G

• Adenine

• Guanine

Two fused rings: one six-membered and one five-membered.

Pyrimidines

C, T, U

• Cytosine

• Thymine

• Uracil

One six-membered ring containing two nitrogen atoms.

Which bases occur in DNA and RNA?

Memory trick:

• Purines = A and G — Pure As Gold.

• Pyrimidines = C, T, U.

Important structural differences

• Adenine: Contains an amino group.

• Guanine: Contains an amino group and a carbonyl group.

• Cytosine: Contains an amino group and a carbonyl group.

• Thymine: Contains two carbonyl groups and a methyl group.

• Uracil: Contains two carbonyl groups but lacks thymine's methyl group.

The functional groups influence hydrogen bonding, which helps determine complementary base pairing.

Thymine vs. uracil

Thymine is essentially 5-methyluracil.

In other words, thymine has an additional methyl group (–CH₃) at carbon 5 compared with uracil.

Your professor's comment that uracil is a precursor to thymine is a useful way to think about their structural relationship, but the biochemical pathway requires several reactions. In DNA synthesis, thymine is incorporated as part of the nucleotide thymidine triphosphate's DNA precursor, dTTP.

4. Pentose sugars: ribose vs. deoxyribose

DNA and RNA contain different five-carbon sugars.

Ribose — RNA

• Has an –OH group at C2′.

• Has an –OH group at C3′.

• Makes RNA more susceptible to certain forms of chemical degradation.

Deoxyribose — DNA

• Has H instead of –OH at C2′.

• Still has an –OH group at C3′.

• Contributes to DNA's greater chemical stability.

Why is the difference important?

The missing oxygen at C2′ makes DNA less susceptible to backbone cleavage through intramolecular reactions involving the 2′-OH group.

This contributes to DNA's suitability for long-term genetic storage.

RNA's 2′-OH group also affects its structure and flexibility. However, it is not accurate to say that deoxyribose simply has no chirality at C2′: the C2′ carbon is still a stereogenic center in deoxyribose.

Prime notation

The sugar carbons are numbered with primes to distinguish them from the atoms in the nitrogenous base.

• C1′: Attaches to the nitrogenous base.

• C2′: Distinguishes ribose from deoxyribose.

• C3′: Has the –OH used to extend a nucleic acid chain.

• C4′: Part of the sugar ring.

• C5′: Lies outside the ring and is commonly attached to phosphate.

This numbering is essential for understanding nucleotide formation, DNA replication, and RNA transcription.

5. Nucleosides: base + sugar

A nucleoside is a nitrogenous base attached to a pentose sugar, without a phosphate group.

The base connects to C1′ of the sugar through a β-N-glycosidic bond.

The bond uses:

• N9 for purines (adenine and guanine).

• N1 for pyrimidines (cytosine, thymine, and uracil).

This is the standard arrangement in naturally occurring nucleosides.

Nucleoside naming table

Naming rule

• Purines: End in -osine.

• Pyrimidines: End in -idine.

Examples:

• Adenine → adenosine

• Guanine → guanosine

• Cytosine → cytidine

• Uracil → uridine

• Thymine → thymidine

For DNA nucleosides, the prefix deoxy- indicates the deoxyribose sugar, although thymidine conventionally refers to the deoxyribose form.

Correction to your professor's notes

The glycosidic bond in a nucleoside is a C–N bond, not a C–O–C bond. The nitrogenous base is connected to the sugar's C1′ through a nitrogen atom.

Also, a nucleoside is not simply a base "paired with hydrogen." It is a base covalently linked to a sugar.

6. Nucleotides: base + sugar + phosphate

A nucleotide is a phosphorylated nucleoside.

A phosphate group is commonly attached to the sugar's C5′ hydroxyl group, forming a phosphoester bond.

For example:

Adenosine + phosphate → adenosine monophosphate (AMP)

The names indicate how many phosphate groups are present.

For DNA, the corresponding deoxyribonucleotide abbreviations include:

• dAMP

• dGMP

• dCMP

• dTMP

The letter d indicates a deoxyribose-containing nucleotide.

What are the alpha, beta, and gamma phosphates?

In a nucleoside triphosphate, the phosphate groups are named according to their distance from the sugar:

Sugar

α

β

γ

Alpha is closest to the sugar; gamma is farthest away.

• Alpha (α): Closest to the sugar.

• Beta (β): Middle phosphate.

• Gamma (γ): Terminal phosphate.

ATP can be hydrolyzed to ADP and inorganic phosphate, usually by cleavage of the terminal phosphate bond. However, the energy released comes from the overall reaction and the relative stability of its products, not simply from breaking a bond.

What are ATP and GTP used for?

• ATP: Widely used for energy coupling, transport, movement, and biosynthesis.

• GTP: Important in protein synthesis, G-protein signaling, and microtubule dynamics.

ATP and GTP are not just nucleic acid building blocks. They are also essential metabolic molecules.

Correction: nucleotide formation

The addition of a phosphate group to a nucleoside is not the same reaction as joining nucleotides into a nucleic acid chain. These are distinct chemical processes.

7. The three important bonds in nucleic acids

Exam tip: Do not confuse the bond attaching a base to a sugar with the bond connecting one nucleotide to the next.

8. Primary structure of nucleic acids

The primary structure is the linear sequence of nucleotides in a DNA or RNA molecule.

Nucleotides are connected through 3′–5′ phosphodiester bonds.

How does a phosphodiester bond form?

The 3′-OH group of the growing nucleic acid chain reacts with the alpha phosphate of an incoming nucleotide triphosphate. The new nucleotide becomes attached to the chain, and pyrophosphate is released.

In biological synthesis, enzymes catalyze this reaction.

The result is a repeating sugar-phosphate backbone with nitrogenous bases extending from the sugars.

Directionality of a nucleic acid strand

A strand has two different ends:

• 5′ end: Usually has a free phosphate group or another group attached at the sugar's 5′ position.

• 3′ end: Usually has a free hydroxyl group at the sugar's 3′ position.

DNA and RNA sequences are conventionally written from 5′ to 3′.

For example:

5′-A-C-G-T-3′

This is a DNA sequence because it contains thymine.

An RNA sequence might be:

5′-A-C-G-U-3′

Why does directionality matter?

DNA polymerases and RNA polymerases extend a nucleic acid chain at its 3′ end, so new nucleotides are added in the 5′→3′ direction.

This is a fundamental rule in DNA replication and transcription.

9. DNA vs. RNA

Important qualification

RNA is often single-stranded, but that does not mean it cannot form base pairs.

RNA can fold back on itself, creating:

• Hairpin loops

• Stems

• Internal loops

• Complex three-dimensional structures

Some viruses also have double-stranded RNA genomes.

10. Secondary structure of DNA: the double helix

DNA commonly exists as a double helix, consisting of two complementary polynucleotide strands.

The classical double-helix model is associated with James Watson and Francis Crick, with crucial experimental evidence provided by Rosalind Franklin and Raymond Gosling, among others.

Rosalind Franklin and Photo 51

Rosalind Franklin was a scientist whose X-ray diffraction work provided important evidence for DNA's helical structure.

Photo 51 is a famous X-ray diffraction image of DNA.

It contributed to the understanding of DNA's helical shape, but the discovery was not the work of one person alone.

Major structural features

1. Two strands: DNA usually contains two complementary polynucleotide chains.

2. Antiparallel arrangement: One strand runs 5′→3′ while the other runs 3′→5′.

3. Sugar-phosphate backbone: Located on the outside of the helix.

4. Nitrogenous bases: Face inward and pair with complementary bases.

5. Hydrogen bonds: Help hold the paired bases together.

6. Base stacking: Additional stabilizing interactions occur between neighboring bases.

Why is DNA's backbone outside?

The sugar-phosphate backbone is hydrophilic and negatively charged, so it interacts favorably with the surrounding aqueous environment.

The bases are relatively less exposed to water and interact through hydrogen bonding and stacking in the interior.

Why does DNA use hydrogen bonds between bases?

Hydrogen bonds provide specificity for complementary base pairing while allowing the two strands to separate when necessary.

However, DNA stability does not depend only on hydrogen bonds. Base stacking and the surrounding ionic environment are also important.

DNA replication requires enzymes and energy, but it is incorrect to say that hydrogen bonds alone make DNA weak or that covalent bonds must be broken continuously during ordinary strand separation.

11. Complementary base pairing

Complementary base pairing follows specific rules.

A — T

2 hydrogen bonds

Adenine pairs with thymine in DNA.

G — C

3 hydrogen bonds

Guanine pairs with cytosine.

In RNA, adenine generally pairs with uracil instead of thymine.

Why do the bases pair specifically?

Their shapes and arrangements of hydrogen-bond donors and acceptors allow the correct partners to interact.

The complementary pairs also have compatible sizes:

• Purine + pyrimidine

• Large double-ring base + smaller single-ring base

This helps maintain a consistent width in the DNA helix.

Important correction to your professor's notes

Hydrogen bonding is not determined simply by whether a molecule is positively or negatively charged.

It depends on the placement of:

• Hydrogen-bond donors

• Hydrogen-bond acceptors

• Functional groups

• The geometry of the interacting bases

Also, uracil can pair with adenine in RNA. Thymine is not required for this pairing.

12. Chargaff's rules

Chargaff's rules describe the base composition of typical double-stranded DNA.

They state:

$\%A=\%T$

$\%G=\%C$

Therefore:

$\%A+\%G=\%T+\%C$

This means the total percentage of purines equals the total percentage of pyrimidines in double-stranded DNA.

Example 1: If adenine is 15%

If a double-stranded DNA sample contains 15% adenine:

• A = 15%

• T = 15%

• Remaining percentage = 70%

• G = 35%

• C = 35%

Example 2: If cytosine is 20%

• C = 20%

• G = 20%

• Remaining percentage = 60%

• A = 30%

• T = 30%

Important correction

Chargaff's rules do not mean that any deviation automatically proves a mutation. The rules describe the expected composition of ordinary double-stranded DNA. Single-stranded nucleic acids do not necessarily follow these equality rules.

Why do species have different base percentages?

Different species can have different overall proportions of A–T and G–C base pairs.

The percentages of A and T match each other, as do G and C, but the overall GC content can vary among species.

13. Constructing a complementary DNA strand

Given the original strand:

5′-A-G-T-C-C-A-A-T-C-3′

The complementary strand is:

3′-T-C-A-G-G-T-T-A-G-5′

Notice that the two strands are antiparallel.

How to solve these questions

Use the base-pairing rules:

• A ↔ T

• G ↔ C

Then make sure the two strands run in opposite directions.

If your instructor requests the complementary strand written in the conventional 5′→3′ direction, reverse the complementary sequence:

5′-G-A-T-T-G-G-A-C-T-3′

Exam tip: Complement the bases first, then check strand orientation. Many students get the base pairs right but reverse the sequence incorrectly.

14. DNA replication

DNA replication is the process of copying DNA before cell division.

It allows each daughter cell to receive genetic information.

Semiconservative replication

DNA replication is called semiconservative because each daughter DNA molecule contains:

• One original parental strand

• One newly synthesized strand

The two resulting DNA molecules have the same sequence information as the original DNA, assuming no replication errors or mutations.

Main stages

1. The double helix is opened.

2. Each parental strand serves as a template.

3. Complementary nucleotides are incorporated.

4. The newly synthesized strands are extended.

5. The new DNA molecules are formed and proofread.

Major enzymes and proteins

These components work together at a region called the replication fork.

Replication fork

A replication fork is a Y-shaped region where parental DNA is unwound and new DNA strands are synthesized.

Replication generally proceeds bidirectionally from an origin in many cellular DNA molecules.

15. Leading and lagging strands

DNA polymerase can extend a strand only in the 5′→3′ direction.

Because the two parental DNA strands are antiparallel, replication occurs differently on the two templates.

Leading strand

Synthesized continuously in the 5′→3′ direction toward the advancing replication fork.

Lagging strand

Synthesized discontinuously in short pieces called Okazaki fragments. Each fragment is still synthesized 5′→3′, but away from the advancing fork.

Okazaki fragments

These are short DNA segments synthesized on the lagging strand.

After the RNA primers are removed and replaced with DNA, DNA ligase seals the remaining nicks in the sugar-phosphate backbone.

Important correction

The lagging strand is not synthesized 3′→5′. Each new DNA fragment is synthesized 5′→3′. The overall direction of fragment production relative to the fork is what differs.

16. Energy in DNA replication

Incoming DNA building blocks are called deoxyribonucleoside triphosphates (dNTPs):

• dATP

• dGTP

• dCTP

• dTTP

Each incoming nucleotide carries three phosphate groups.

During DNA synthesis, DNA polymerase forms a phosphodiester bond between the growing strand and the incoming nucleotide. Pyrophosphate is released, and its subsequent hydrolysis helps drive the overall reaction forward.

This means the incoming nucleotide supplies the chemical potential needed for polymerization.

Important distinction

DNA replication uses energy from nucleotide triphosphates. It does not require the cell to repeatedly break the covalent backbone of the existing DNA strand just to separate its two strands. The strands are separated mainly by disrupting noncovalent interactions, including base pairing and stacking.

17. Central dogma of molecular biology

The central dogma describes the main flow of genetic information:

DNA

Genetic information

Transcription

RNA

Transcribed information

Translation

Protein

Functional product

DNA replication

DNA can also be copied to DNA.

Three essential processes

Replication: DNA → DNA

Copies genetic information.

Transcription: DNA → RNA

Produces RNA from a DNA template.

Translation: mRNA → protein

Uses the mRNA sequence to assemble amino acids into a polypeptide.

Important qualification

The diagram is a simplified representation of the usual flow of genetic information. Some viruses, for example, use RNA-dependent RNA replication or reverse transcription. Proteins are not normally translated directly back into DNA.

Your professor's comment that proteins can produce DNA appears to confuse the role of enzymes with the flow of information. Proteins such as DNA polymerases help synthesize DNA, but they do not reverse-translate a protein's amino acid sequence into its original DNA sequence.

18. The three major types of RNA

RNA is involved in several processes, especially protein synthesis.

A. Messenger RNA (mRNA)

mRNA carries the information copied from a protein-coding gene.

Its nucleotide sequence is read in groups of three called codons.

Each codon specifies an amino acid or a translation stop signal.

B. Transfer RNA (tRNA)

tRNA serves as an adaptor between the nucleotide language of mRNA and the amino acid language of proteins.

Each tRNA has:

• An amino acid attachment site at its 3′ end

• An anticodon that recognizes a complementary mRNA codon

• A folded structure maintained by internal base pairing

C. Ribosomal RNA (rRNA)

rRNA combines with proteins to form ribosomes.

It has both structural and catalytic roles. The ribosome's rRNA participates directly in forming peptide bonds during protein synthesis.

Relative abundance

Your notes list approximate proportions of 80% rRNA, 15% tRNA, and 5% mRNA. These are useful as rough classroom estimates, but the exact proportions vary by cell type, physiological state, and measurement method.

For many cells, rRNA is the most abundant class, tRNA is next, and mRNA makes up a smaller fraction.

19. Structure of tRNA

Transfer RNA has a characteristic cloverleaf secondary structure in two-dimensional diagrams.

Its three-dimensional shape is more compact and is often described as an L-shaped structure.

The main regions are:

• Acceptor stem: Where the amino acid attaches.

• Anticodon loop: Contains the anticodon that recognizes an mRNA codon.

• D loop: Helps with tRNA recognition and folding.

• TΨC loop: Helps interactions with the ribosome and other translation components.

• Variable loop: Varies in size among different tRNAs.

Amino acid attachment

The amino acid attaches to the terminal adenosine at the tRNA's 3′ end, which has the conserved sequence CCA.

The attachment is covalent and is catalyzed by an enzyme called an aminoacyl-tRNA synthetase.

Codon vs. anticodon

• Codon: A three-nucleotide sequence on mRNA.

• Anticodon: A complementary three-nucleotide sequence on tRNA.

For example, if an mRNA codon is 5′-AUG-3′, the complementary tRNA anticodon can be written 3′-UAC-5′.

This matching helps the ribosome incorporate the correct amino acid.

20. Transcription: DNA to RNA

Transcription is the synthesis of RNA using DNA as a template.

The enzyme responsible is RNA polymerase.

General stages

1. Initiation: RNA polymerase and associated factors recognize a promoter.

2. Elongation: RNA polymerase reads the DNA template and synthesizes RNA.

3. Termination: Transcription ends through mechanisms specific to the organism and gene.

Direction of transcription

RNA polymerase reads the DNA template strand in the:

3′→5′ direction

The RNA transcript is synthesized in the:

5′→3′ direction

Transcription base-pairing rules

RNA contains uracil (U) instead of thymine (T).

Example

DNA template:

3′-C-T-A-A-G-G-5′

RNA transcript:

5′-G-A-U-U-C-C-3′

Template strand vs. coding strand

This distinction is important for exams.

• Template strand: The DNA strand read by RNA polymerase.

• Coding strand: The opposite DNA strand, whose sequence matches the RNA transcript except that DNA uses T where RNA uses U.

For the example above:

• Template: 3′-CTAAGG-5′

• Coding strand: 5′-GAT TCC-3′

• RNA: 5′-GAUUCC-3′

The coding strand and RNA have the same sequence orientation, with T replaced by U.

Promoters and the TATA box

A promoter is a DNA region where transcription machinery assembles.

Some eukaryotic promoters contain a TATA box, often represented by a sequence resembling TATAAA. However, not every gene has a TATA box, and bacterial promoters use different recognition sequences.

21. Eukaryotic mRNA processing

In eukaryotic cells, the initial RNA transcript of a protein-coding gene is generally called pre-mRNA.

It must be processed before it becomes mature mRNA.

1. 5′ capping

A modified guanine nucleotide is added to the 5′ end.

Functions include:

• Protecting the RNA from degradation

• Helping with nuclear export

• Supporting translation initiation

2. RNA splicing

Noncoding introns are removed, and exons are joined together.

• Exons: Sequences retained in the mature RNA.

• Introns: Sequences removed during splicing.

Important nuance: Exons are not always entirely protein-coding; some exonic sequences form untranslated regions.

3. Polyadenylation

A poly-A tail is added to the 3′ end of most eukaryotic mRNAs.

This helps with:

• RNA stability

• Nuclear export

• Translation efficiency

4. Nuclear export

After processing, mature mRNA can pass through nuclear pores into the cytoplasm, where ribosomes translate its coding sequence.

Memory trick: Capping → Splicing → Poly-A tail → Export.

22. Translation: RNA to protein

Translation occurs on ribosomes and converts the nucleotide sequence of mRNA into an amino acid sequence.

Basic steps

1. Initiation: The ribosome assembles on the mRNA and identifies the start site.

2. Elongation: tRNAs deliver amino acids according to mRNA codons, and peptide bonds join the amino acids.

3. Termination: A stop codon signals the release of the completed polypeptide.

Start and stop codons

The common start codon in mRNA is:

AUG — methionine

The three standard stop codons are:

• UAA

• UAG

• UGA

Stop codons are recognized by release factors, not by ordinary tRNAs carrying amino acids.

Why is a codon three bases long?

There are four standard RNA bases. If each codon contains three bases, there are:

\[ 4^3=64 \]

possible codons.

These encode 20 standard amino acids plus stop signals.

Several codons can specify the same amino acid. This property is called the degeneracy of the genetic code.

Synonymous mutations

A synonymous mutation changes a codon without changing the encoded amino acid.

However, not every mutation is synonymous. Other mutations can change an amino acid, introduce a premature stop codon, or affect RNA processing and gene expression.

23. Gene regulation and the lac operon

The lac operon is a classic example of gene regulation in bacteria, especially Escherichia coli.

It controls the expression of genes involved in lactose utilization.

Main components

What do the structural genes do?

lacZ — β-galactosidase

Breaks lactose into glucose and galactose. It can also help produce allolactose, the natural inducer of the lac operon.

lacY — lactose permease

Helps lactose enter the cell.

lacA — thiogalactoside transacetylase

Encodes an enzyme involved in acetylating certain galactosides. It is not the primary enzyme for breaking down lactose.

24. Lac operon: when lactose is absent

When lactose is absent:

1. The lacI gene produces the lac repressor.

2. The active repressor binds to the operator.

3. Repressor binding interferes with transcription.

4. Expression of the structural genes remains very low.

The cell avoids wasting resources on enzymes that are not currently needed.

Important detail

The lacI regulatory gene has its own regulatory region. Its transcription does not depend on the lac operon being turned on.

25. Lac operon: when lactose is present

When lactose enters the cell, some of it is converted to allolactose, which acts as the natural inducer.

The sequence is:

1. Allolactose binds the lac repressor.

2. The repressor changes shape.

3. Its affinity for the operator decreases.

4. The operator becomes available.

5. RNA polymerase can transcribe the structural genes.

The resulting mRNA is translated into the proteins needed for lactose utilization.

A crucial correction: lactose alone does not determine the highest expression

The lac operon is also regulated by glucose availability.

When glucose is low, intracellular cAMP generally rises. The cAMP–CAP complex can help recruit RNA polymerase and increase transcription.

Thus, the strongest expression generally occurs when:

• Lactose is available.

• Glucose is low.

This allows bacteria to preferentially use glucose when it is readily available.

Quick comparison

These are simplified descriptions of the regulatory states.

26. Nucleic acid metabolism and hydrolysis

Your professor mentioned breaking the phosphodiester backbone, which is an important biochemical concept.

Hydrolysis

Hydrolysis uses water to break a chemical bond.

In nucleic acids, enzymes called nucleases catalyze the cleavage of phosphodiester bonds.

Two major types are:

• Endonucleases: Cut bonds within a nucleic acid chain.

• Exonucleases: Remove nucleotides from the ends of a chain.

Why is this important?

Nucleic acid cleavage is involved in:

• DNA repair

• DNA replication processing

• RNA degradation

• Recycling nucleotides

• Laboratory techniques used to analyze and manipulate DNA

Restriction endonucleases, for example, recognize specific DNA sequences and cut DNA at or near those sequences.

However, DNA editing is not simply a matter of cutting any phosphodiester bond. The cut must be made at the intended site, and the cell or laboratory system must repair or modify the DNA appropriately.

27. DNA mutations and genetic variation

A mutation is a change in the nucleotide sequence of genetic material.

Mutations can occur because of DNA replication errors, DNA damage, or other biological processes.

Major types

Why can mutations affect proteins?

Protein-coding sequences are read in groups of three nucleotides.

An insertion or deletion can alter how those groups are read, potentially changing many amino acids downstream.

For example:

AUG-AAA-GGC-UGA

If a nucleotide is inserted into the coding region, the grouping of subsequent bases may change.

However, not every mutation is harmful. Some have little or no effect, some are harmful, and some can be beneficial in a particular environment.

Important correction

A break in the DNA backbone does not automatically mean complementary base pairing disappears or that replication becomes impossible. Cells have DNA repair mechanisms that can restore many forms of DNA damage. Unrepaired damage can nevertheless interfere with replication and contribute to mutations or cell death.

28. Common misconceptions from your professor's notes

Here are the statements I would be especially careful about when studying.

29. High-yield exam review

These are the facts I would prioritize for memorization.

Revision checklist

0 of 21

Nucleotide structure

Nucleoside = base + sugar

Nucleotide = base + sugar + phosphate

Purines = adenine and guanine

Pyrimidines = cytosine, thymine, and uracil

DNA has deoxyribose; RNA has ribose

DNA structure

A–T forms 2 hydrogen bonds

G–C forms 3 hydrogen bonds

DNA strands are antiparallel

Nucleic acid sequences are written 5′→3′

DNA backbone contains 3′–5′ phosphodiester bonds

Replication and gene expression

DNA replication is semiconservative

DNA polymerase synthesizes DNA 5′→3′

Lagging strand contains Okazaki fragments

Transcription makes RNA from DNA

Translation uses mRNA to synthesize proteins

RNA and regulation

mRNA carries coding information

tRNA carries amino acids and has an anticodon

rRNA forms the catalytic and structural core of ribosomes

Start codon: AUG

Stop codons: UAA, UAG, UGA

Allolactose induces the lac operon

Reset checklist

30. Final memory map

NUCLEIC ACIDS

Polymers of nucleotides

DNA

• Deoxyribose

• A, T, G, C

• Genetic storage

• Replication

RNA

• Ribose

• A, U, G, C

• Gene expression

• Protein synthesis

Nucleotide → Nucleic acid

Base + Sugar = Nucleoside

Nucleoside + Phosphate = Nucleotide

Nucleotides linked by phosphodiester bonds = Nucleic acid

Central dogma

DNA

Transcription

RNA

Translation

Protein

My biggest recommendation is to learn the structures and the processes together. If you understand how the sugar's 3′-OH, the 5′ phosphate, and complementary base pairing work, DNA replication and transcription become much easier to understand instead of feeling like separate lists to memorize.

And remember this especially for practical identification: a base alone is not a nucleoside, a nucleoside is not a nucleotide, and a nucleotide is not yet a nucleic acid polymer. Those distinctions are easy exam points to secure.