mince sec 3

DNA/RNA Structure + DNA Replication overall logic

Scientists first had to identify DNA as the hereditary material → then determine DNA’s structure → the structure revealed complementary and antiparallel strands → those structural properties explain how DNA can copy itself → replication must occur before a bacterial cell divides so each daughter receives a genome.

ends the lecture by emphasizing that DNA has to be copied before cell division, and that the next step in genetics is understanding how the information stored in DNA is actually used.

2. How scientists figured out that DNA carries genetic information

evidence of transformation and what it is

Avery, MacLeod, and McCarty: the transforming principle is DNA

Griffith: something can transfer hereditary information

Griffith worked with Streptococcus pneumoniae.

Strain	Appearance	Capsule	Disease
S strain = smooth	Smooth colonies	Capsule present	Virulent
R strain = rough	Rough colonies	No capsule	Nonvirulent

capsule is helps the bacterium evade host defenses, so the encapsulated S strain kills the mice while the nonencapsulated R strain does not.

The important experiment was:

live S → mouse dies
live R → mouse lives
heat-killed S → mouse lives
heat-killed S + live R → mouse dies

The S bacteria were dead, → something from those dead S cells entered the living R bacteria → and caused them to acquire the S phenotype → living bacteria recovered from the dead mouse were now virulent S-type cells → evidence that information was transferred from one population to another→ process is called transformation.

Heat-killed S releases hereditary material → live R takes up that information → R changes into virulent S

Griffith did not know that the transforming substance was DNA.

OpenStax fills in an important step that the lecture moves past quickly.

Avery, MacLeod, and McCarty took material from the virulent S bacteria and selectively destroyed different classes of molecules.

destroy protein → transformation still occurs
destroy RNA → transformation still occurs
destroy DNA with DNase → transformation stops

Therefore:

DNA was required for transformation → DNA was the “transforming principle.”

This is a useful experimental logic pattern:

remove candidate molecule → ask whether biological effect remains

If eliminating the molecule eliminates the effect, that molecule was necessary.

3. Hershey–Chase: what was their experiment

what was the question?

what thing of the virus they awere focused on mainly?

Why use radioactive sulfur and phosphorus?

used T2 bacteriophage, a virus that infects E. coli.

bacteriophage two components for this experiment:

protein coat/capsid + DNA inside

The question was:

Which part enters the bacterium and directs production of new viruses—protein or DNA?

They needed a way to distinguish the two molecular components.

³⁵S labels protein because sulfur is present in sulfur-containing amino acids such as cysteine and methionine, but not in DNA.

³²P labels DNA because DNA has a phosphate-rich backbone, whereas protein generally does not contain phosphorus in its basic structure.

Then:

labeled phage infect bacteria → blender knocks phage coats off bacterial surfaces → centrifuge separates heavy bacterial cells from lighter phage material

After centrifugation:

pellet = bacterial cells

supernatant = detached phage coats/material outside cells

So the prediction is:

If the genetic material enters the bacterium, its radioactivity should be found in the pellet.

The experiment showed:

³²P-DNA → bacterial pellet

³⁵S-protein → supernatant

Therefore:

DNA entered the bacterial cells → DNA directed production of new phages → DNA is hereditary information.

The professor explicitly explains that radioactivity in the bacterial pellet means that labeled material entered the cells. while protein remained outside.

4. From “DNA is genetic material” to “what does DNA look like?”

3 people work, what they did

scientists needed its three-dimensional structure.

Chargaff → base proportions

Franklin/Gosling → X-ray diffraction data revealing helical structure

Watson/Crick → structural model integrating those data

The professor strongly emphasizes Rosalind Franklin’s X-ray crystallography work and how the diffraction pattern revealed DNA’s helical nature.

OpenStax explains that Watson and Crick used Chargaff’s rules plus Franklin/Wilkins X-ray diffraction data to construct the double-helix model.

This matters conceptually because discovering the structure also suggested how DNA could copy itself.

5. Start at the smallest unit: nucleotide

DNA and RNA are nucleic acids.

Like proteins are polymers of amino acids, nucleic acids are polymers of:

Nucleotides

A nucleotide has three components:

phosphate group + 5-carbon sugar + nitrogenous base

For DNA, the sugar is:

Deoxyribose

So DNA nucleotides are properly called:

Deoxyribonucleotides

OpenStax makes an additional terminology distinction:

Nucleoside

sugar + nitrogenous base

Nucleotide

sugar + nitrogenous base + phosphate

That distinction was not emphasized on the slides, but it is proper OpenStax terminology.

6. Nitrogenous bases: purines vs pyrimidines

DNA contains:

A = adenine
G = guanine
C = cytosine
T = thymine

These fall into two structural classes.

Class	Bases	Structure
Purines	A, G	Larger, 2 rings
Pyrimidines	C, T	Smaller, 1 ring

RNA substitutes:

U = uracil for thymine.

So for RNA:

A, G = purines
C, U = pyrimidines

OpenStax explicitly categorizes A/G as purines and C/T as pyrimidines.

7. The DNA backbone:

what is it made of?

bonds differnec between backbone and b/ nucleotides

Nucleotides are not just sitting beside each other. They are covalently connected.

5′–3′ phosphodiester bond

Specifically:

5′ phosphate of one nucleotide ↔ 3′ OH of the next nucleotide

sugar–phosphate–sugar–phosphate–sugar–phosphate

This creates the:

Sugar-phosphate backbone

The backbone is held together by strong covalent bonds.

That is different from the interaction holding the two DNA strands together.

within one DNA strand → covalent phosphodiester bonds

between the two DNA strands → hydrogen bonds between bases

The professor makes this distinction explicitly: backbone = covalent; paired bases = hydrogen bonds.

OpenStax additionally explains that DNA synthesis uses deoxynucleotide triphosphates, dNTPs. When an incoming dNTP is incorporated, terminal phosphates are released as pyrophosphate; hydrolysis of pyrophosphate helps drive polymerization energetically.

incoming dNTP → phosphodiester bond forms → pyrophosphate released → nucleotide becomes part of growing strand

8. Why DNA has a 5′ end and a 3′ end

The numbers refer to carbons in the pentose sugar.

The sugar carbons are numbered:

1′, 2′, 3′, 4′, 5′

The professor spends considerable lecture time on this because replication makes no sense unless she understands 5′ and 3′ orientation. dna replicatio trans

At one end of a DNA strand:

5′ end

There is a free phosphate associated with the 5′ carbon.

At the other end:

3′ end

There is a free 3′ hydroxyl group (-OH).

So a strand has direction:

5′ → 3′

This is not an arbitrary label. It comes directly from the chemical structure of the sugar-phosphate backbone.

9. DNA double helix

what it is

where does phosphate sits

base pairs combo?

DNA is usually composed of two nucleotide strands wrapped around one another.

The sugar-phosphate backbones sit on the outside.

The bases face inward.

The bases form specific complementary pairs:

A ↔ T

G ↔ C

The professor describes complementary as:

not identical, but predictable.

That is a very useful definition.

If one strand is:

5′-A G T C-3′

its complementary DNA sequence is:

3′-T C A G-5′

The lecture explicitly uses AGTC → TCAG to demonstrate complementarity.

10. Hydrogen bonding and why GC is stronger

Base pairs are joined by hydrogen bonds.

A–T = 2 hydrogen bonds

G–C = 3 hydrogen bonds

Therefore:

G–C pairing is stronger than A–T pairing

This becomes important later when considering DNA denaturation.

The professor emphasizes this exact 2-vs-3 distinction

11. Chargaff’s rules

For ordinary double-stranded DNA:

%A = %T

%G = %C

Because every A on one strand must pair with T on the other, and every G must pair with C.

This does not mean all four bases are 25%.

Example:

If:

T = 22%

then:

A = 22%

A + T = 44%.

100 − 44 = 56% remaining for G + C.

G = C, so:

G = 28%, C = 28%.

This is explicitly one of the professor’s critical-thinking questions and she says students should be ready to do it for any single starting base percentage.

• Double-Stranded DNA (Applies Perfectly): In organisms with normal double-stranded DNA (like humans, plants, and most bacteria), the rules are always true. This is because adenine must physically bond with thymine, and guanine must bond with cytosine to form the DNA double helix.

• Single-Stranded DNA (Does Not Apply): Some viruses (like certain bacteriophages or parvoviruses) carry single-stranded DNA (ssDNA). Because there is no complementary matching strand, the amount of A does not have to equal T, and G does not have to equal C.

• RNA (Does Not Apply): RNA is typically single-stranded and uses Uracil (U) instead of Thymine (T). In RNA, Chargaff's 1:1 ratios generally do not hold true.

12. Antiparallel is NOT the same as complementary

These terms answer different questions.

Complementary

Which base belongs opposite another base?

A ↔ T, G ↔ C

Antiparallel

Which direction does each strand run?

If one strand runs:

5′ → 3′

the opposite strand runs:

3′ → 5′

The strands are therefore oriented in opposite directions.

The professor emphasizes that this structural fact becomes the reason the leading and lagging strands behave differently during replication. dna replicatio trans

This is one of the most important connections in the entire chapter:

antiparallel DNA + DNA polymerase can only synthesize 5′→3′ → leading and lagging strands must be synthesized differently

13. DNA major and minor grooves — OpenStax detail

Because the two sugar-phosphate backbones are not spaced perfectly symmetrically around the helix, DNA has:

Major groove: Wider region between backbones.

Minor groove: Narrower region.

Why care?

Proteins can bind DNA through these grooves. Protein binding can influence:

replication, transcription, and DNA structure.

This is a useful structural connection:

DNA sequence → particular chemical pattern exposed in grooves → proteins recognize/bind particular DNA regions → cellular processes can be regulated

This detail comes from OpenStax even though the lecture does not dwell on it.

14. DNA denaturation and reannealing

If heat or certain chemicals disrupt the hydrogen bonds between the strands:

double-stranded DNA → two single strands

This is:

DNA denaturation

Notice what is being broken:

hydrogen bonds between bases

not:

phosphodiester bonds of the backbone

If conditions normalize, complementary strands can find each other and hydrogen-bond again:

Reannealing / renaturation

dsDNA → heat/chemical → ssDNA → cool/remove denaturant → dsDNA again

Why does GC content matter?

GC has 3 H bonds while AT has 2 → high-GC DNA requires more energy to separate → higher-GC DNA is harder to denature.

OpenStax explicitly makes that connection.

15. RNA: similar chemistry, different structure

RNA is also a nucleic acid polymer built from nucleotides.

Each RNA nucleotide contains:

phosphate + ribose + nitrogenous base

The bases are:

A, U, G, C

Major structural differences:

Feature	DNA	RNA
Sugar	Deoxyribose	Ribose
Unique base	Thymine	Uracil
Typical state	Double-stranded	Single-stranded
Stability	More stable	Less stable
Internal base pairing	Between two strands	Often within same molecule

OpenStax points out that the sugar difference contributes to DNA being relatively more stable and therefore well suited for long-term genetic storage, whereas RNA is generally more suited to shorter-term functions.

16. “Single-stranded RNA” does NOT mean RNA cannot base-pair

This was an actual iClicker concept.

RNA is often one strand, but different portions of that same strand can contain complementary sequences.

The strand can fold back on itself:

A ↔ U

G ↔ C

This is:

Intramolecular base pairing / intramolecular hydrogen bonding

The professor uses tRNA as the example. dna replicatio trans

So:

“RNA does not base pair with itself.”

is FALSE.

This folding is not decorative; it determines RNA’s three-dimensional shape and therefore its function.

17. OpenStax: the three major RNAs

The lecture is mainly introducing RNA structure here, but OpenStax expects students to recognize the three major RNA types.

RNA	Structure	Main role
mRNA	Relatively short, unstable, single-stranded	Carries information copied from DNA to the ribosome
rRNA	Larger, stable RNA	Major ribosome component; positions mRNA/tRNA and catalyzes peptide-bond formation
tRNA	Small, ~70–90 nt, folded by extensive intramolecular base pairing	Carries the correct amino acid to the ribosome

OpenStax emphasizes that rRNA is not merely scaffolding: it also has catalytic activity in peptide-bond formation. tRNA has extensive folding because its 3-D shape is essential for matching amino acids to mRNA information.

Also, RNA can itself be hereditary material in some viruses; DNA is not the universal genome of every virus.

18. Now connect structure to replication

This is where the chapter turns.

Because DNA has:

two complementary strands

each old strand already contains the information needed to construct its partner.

If you separate:

A-T / G-C pairs

then each parental strand can tell the cell exactly which nucleotides belong opposite it.

So:

parental double helix → strands separate → each old strand acts as template → complementary new strand constructed

This is why Watson and Crick’s structural model immediately suggested a copying mechanism.

19. Semiconservative replication

DNA replication is:

Semiconservative

After replication, each daughter DNA molecule contains:

one parental/old strand + one newly synthesized strand

Not:

two old strands

and not:

two new strands

The slide iClicker asks exactly this. DNA_RNA Structure, DNA Replicat…

So:

original DNA = old strand A + old strand B

after replication:

daughter DNA 1 = old A + new complement

daughter DNA 2 = old B + new complement

20. OpenStax: how semiconservative replication was proven

This is an important OpenStax detail absent from the professor’s slides.

Three models had been proposed:

Model	Prediction
Conservative	Original double helix stays completely together; a completely new double helix is made
Semiconservative	Each daughter molecule gets one old + one new strand
Dispersive	Old and new DNA become mixed in pieces throughout each strand

Meselson–Stahl experiment

They grew E. coli in heavy nitrogen:

¹⁵N → DNA becomes “heavy”

Then shifted bacteria to:

¹⁴N → new DNA made with “light” nitrogen

After one generation:

one intermediate-density band

That rules out conservative replication because conservative replication would have produced:

one fully heavy band + one fully light band

After two generations:

one intermediate band + one light band

That rules out dispersive replication.

Therefore:

DNA replication is semiconservative.

OpenStax walks through these density-band predictions explicitly. microbiology_-_WEB

The key logic is not the isotope names themselves; it is:

label old DNA → let cells make new DNA → ask how old and new material are distributed

21. Bacterial DNA replication: where does it begin?

Most bacterial chromosomes are:

Circular

E. coli has one circular chromosome of about 4.6 million base pairs.

Replication begins at a particular DNA sequence:

Origin of replication

In E. coli:

oriC

OpenStax says oriC is about 245 bp and rich in AT sequences. microbiology_-_WEB

Why might an AT-rich area be easier to open?

AT = 2 hydrogen bonds

GC = 3 hydrogen bonds

So AT-rich DNA requires less energy to separate than GC-rich DNA.

That connects DNA structure directly to replication initiation.

22. Bacterial replication

flow

At one bacterial origin: Bidirectional replication

it look like there is one fork, but in bacteria there are actually two forks moving away from the origin.

flow:

single circular chromosome → one origin → replication bubble → dna opens→ two replication forks → forks travel opposite directions around circular chromosome → entire chromosome copied → two daughter chromosomes

E. coli can replicate its ~4.6 Mb chromosome in roughly 42 minutes, with DNA synthesis reaching approximately 1000 nucleotides/sec under the conditions described by OpenStax.

n

23. Opening DNA creates another problem: supercoiling

how to get rid of it?

dna helicase vs dna gyrase

As helicase opens one region, twisting stress builds farther ahead.

Torsional strain / supercoiling

The protein that relieves this in bacteria is:

Topoisomerase II / DNA gyrase

Its job is not simply the same as helicase.

Gyrase : Temporarily cuts/reseals DNA to relieve supercoiling and tension.

Helicase: Separates the two DNA strands by breaking hydrogen bonds between complementary bases.

gyrase relieves torsional stress ahead of fork → helicase breaks H bonds and opens helix

24. what keep the DNA open during replication

is it an enzyme or protein?

Once helicase separates the strands, the complementary bases naturally want to pair back together.

Therefore:

Single-stranded binding proteins (SSBs) bind the exposed parental DNA, so makes it stable, accessible, seperated

Their job:

coat single-stranded DNA → prevent strands from reannealing → keep template accessible

This is a protein rather than an enzyme that synthesizes DNA, but it is part of the replication machinery.

25. Why DNA polymerase cannot simply begin anywhere

The principal bacterial synthesis enzyme is:DNA polymerase III. But DNA pol III cannot start a strand de novo. It needs an existing: Free 3′-OH

Primase

Primase is an RNA polymerase that synthesizes a short:

RNA primer. The primer supplies the free: 3′-OH

OpenStax gives the primer as approximately 5–10 nucleotides.

Then DNA polymerase III can begin adding DNA nucleotides.

flow:

template exposed → primase makes RNA primer → primer provides 3′-OH → DNA pol III attaches to primer → DNA synthesis begins

26. Why DNA polymerase makes DNA only 5′ → 3′

DNA polymerase attaches the incoming nucleotide to the:

3′-OH of the growing DNA strand. Therefore the strand becomes longer at its 3′ end. So every newly synthesized strand grows: 5′ → 3′

DNA polymerase reads its template in the opposite orientation: 3′ → 5′

template read 3′→5′ → daughter synthesized 5′→3′

This directional limitation is what creates the leading/lagging problem.

27. Leading strand: why it can be continuous

Consider a parental strand oriented:3′ → 5′ toward the replication fork

DNA pol III can read that template while synthesizing the new strand: 5′ → 3′ toward the opening fork

So as helicase keeps opening DNA:

fork opens → polymerase moves with it → more DNA continuously added

This is the:

Leading strand

Characteristics:

continuous synthesis

5′→3′ synthesis

toward the opening replication fork

usually one primer for that continuously synthesized stretch

28. Lagging strand: why it MUST be discontinuous

The other parental strand is antiparallel.

So its orientation toward the fork is: 5′ → 3′

DNA polymerase cannot synthesize a daughter strand 3′→5′.

It still has to make DNA:

5′ → 3′

But that forces polymerase to synthesize away from the direction in which the fork is opening.

So:

fork opens a little → primase places primer → DNA pol III makes short 5′→3′ piece away from fork → fork opens farther → new primer placed closer to fork → another short piece made

Those short pieces are:

Okazaki fragments

This is the:

Lagging strand

The professor repeatedly returns to this because it is one of the concepts students find hardest. dna replicatio trans

So the causal chain she needs to be able to explain is:

DNA is antiparallel → DNA polymerase only works 5′→3′ → one new strand can follow the fork continuously → the opposite strand cannot → it is synthesized in separate Okazaki fragments

Do not memorize “lagging = fragments” without that reasoning.

29. Each Okazaki fragment needs a primer

Every new fragment has to begin somewhere.

DNA pol III cannot create its own starting 3′-OH.

Therefore:

each new Okazaki fragment → new RNA primer made by primase → DNA pol III extends it

This is why the lagging strand needs many primers, whereas the leading strand needs far fewer.

30. What happens to the RNA primers afterward?

The finished chromosome should be DNA, not DNA with RNA pieces left throughout it.

In bacteria:

DNA polymerase I

removes the RNA primers using exonuclease activity and replaces the removed RNA with DNA.

This is distinct from:

DNA polymerase III

The main enzyme performing most new DNA synthesis.

So:

DNA pol III = major builder

DNA pol I = primer removal/replacement + repair roles

OpenStax makes this enzyme distinction explicit.

31. DNA ligase: “making ends meet”

Even after DNA pol I replaces the RNA primers with DNA, the DNA pieces are not yet covalently continuous.

There are remaining:

Nicks

between adjacent DNA fragments.

DNA ligase

seals those nicks by making the missing phosphodiester bond between the fragments.

So on the lagging strand:

primase → RNA primer → DNA pol III → Okazaki fragment → DNA pol I removes primer/replaces it with DNA → ligase seals final nick

The professor’s cartoon says ligase is “making ends meet,” which is actually a useful memory device. The lecture states that ligase connects the Okazaki fragments after synthesis.

32. Sliding clamp — OpenStax machinery detail

Sliding clamp

This ring-shaped protein holds DNA polymerase III onto DNA while polymerase adds nucleotides.

Its purpose is to improve processivity—polymerase can stay associated with the template instead of repeatedly falling off.

You do not need to confuse it with an enzyme performing chemistry.

sliding clamp = holds polymerase in place

33. The replication machinery table she should actually know

Factor	Precise function
DNA gyrase / topoisomerase II	Relieves supercoiling/torsional strain ahead of fork
Helicase	Breaks H bonds between bases; separates strands
SSB proteins	Stabilize separated parental strands; prevent reannealing
Primase	Synthesizes RNA primers
RNA primer	Provides free 3′-OH
DNA polymerase III	Main bacterial DNA synthesis enzyme; adds DNA 5′→3′
Sliding clamp	Holds DNA pol III on template
DNA polymerase I	Removes RNA primers and replaces them with DNA
DNA ligase	Seals remaining phosphodiester-bond nicks
Topoisomerase IV	Separates interlocked circular daughter chromosomes after replication

34. mega flow Full bacterial replication mechanism in one continuous flow

circular bacterial chromosome → oriC recognized → DNA gyrase relieves supercoiling → helicase breaks H bonds and opens DNA → two replication forks form → SSB proteins stabilize exposed strands → primase makes RNA primers → DNA pol III extends from each primer 5′→3′ → leading strand made continuously toward fork while lagging strand made discontinuously as Okazaki fragments → DNA pol I removes RNA primers and fills gaps with DNA → DNA ligase seals nicks → forks eventually complete replication → topoisomerase IV separates interlocked daughter chromosomes → cell can divide

Termination of bacterial replication — OpenStax detail

When replication of the circular chromosome finishes, the two new circular DNA molecules can be:

Concatenated

Interlocked like links in a chain.

Topoisomerase IV

temporarily breaks DNA, allows the circles to separate, and reseals them.

OpenStax also notes that bacterial DNA gyrase/topoisomerase IV differ enough from eukaryotic enzymes that they can be targeted by quinolone antibiotics. microbiology_-_WEB

That gives a nice microbiology connection:

bacterial DNA replication machinery differs from host machinery → bacterial-specific enzymes can become antimicrobial targets

36. Bacteria vs eukaryotes

Bacteria	Eukaryotes
Usually one circular chromosome	Multiple linear chromosomes
Usually one origin per chromosome	Many origins per chromosome
Two forks from that origin	Many replication bubbles/forks
~1000 nt/s cited for E. coli	~100 nt/s cited by OpenStax
No telomere end problem on circular chromosome	Linear chromosomes have chromosome-end problem
DNA pol I removes bacterial primers	Eukaryotes use other machinery such as RNase H for primer removal

The professor specifically explains why eukaryotes need multiple origins: there is too much DNA to efficiently copy from only one starting point.

OpenStax adds that linear eukaryotic chromosome ends are:

Telomeres

repetitive noncoding DNA that protects coding regions.

Telomerase

extends telomeric DNA in certain cells.

This is lower priority for this lecture because the professor says she is not focusing on eukaryotic replication, but it is part of the OpenStax section

37. One OpenStax topic the lecture does not emphasize: rolling-circle replication

OpenStax §11.2 also discusses rolling-circle replication, used by some plasmids and viruses.

Basic idea:

one strand of circular DNA is nicked → DNA polymerase extends from the exposed 3′-OH around the circle → old strand is displaced → displaced strand can then be copied to produce another double-stranded circular molecule

I would put this below the professor’s leading/lagging/bacterial chromosome material in priority, because it is not emphasized in the lecture deck, but it belongs to the assigned OpenStax replication section.

mega connection

DNA’s bases contain information → complementary pairing means each strand contains enough information to reconstruct the other → antiparallel orientation means the two strands run opposite directions → polymerase chemistry permits synthesis only 5′→3′ → one strand can therefore be copied continuously while the other must be copied in Okazaki fragments → bacterial enzymes coordinate opening, copying, primer replacement, joining, and chromosome separation → two equivalent DNA molecules are produced before binary fission

1. Label sugar, phosphate, bases and complete pairing

On a nucleotide diagram:

pentagon = deoxyribose sugar

phosphate group = phosphate

A/T/G/C structure = nitrogenous base

The repeating outer chain:

sugar → phosphate → sugar → phosphate = sugar-phosphate backbone

Bases point inward.

Pair only:

A ↔ T

G ↔ C

2. Label 5′ and 3′ ends

Determine one strand first.

If:

top strand = 5′ → 3′

then the complementary strand must be:

3′ → 5′

because the strands are antiparallel.

Remember the chemistry:

5′ end = free phosphate side

3′ end = free 3′-OH side

3. Draw RNA with the same sequence as the DNA strand

The wording says same sequence, not complementary RNA.

Example:

DNA:

5′-A T G C C T-3′

RNA with the same sequence:

5′-A U G C C U-3′

Replace:

T → U

and draw RNA as a single strand

4. Three structural differences between DNA and RNA

Do not give functional differences because she specifically says structure.

DNA has deoxyribose; RNA has ribose

DNA uses thymine; RNA uses uracil

DNA is typically double-stranded; RNA is typically single-stranded

5. Four different enzyme jobs during DNA replication

A strong answer would give the job, not merely the enzyme name.

separate DNA strands by breaking H bonds → helicase

relieve supercoiling → DNA gyrase/topoisomerase

make RNA primers → primase

synthesize new DNA → DNA polymerase III

Other acceptable jobs:

remove RNA primers/replace with DNA → DNA pol I

seal fragments → ligase

6. Why do Okazaki fragments form?

DNA strands are antiparallel, but DNA polymerase can synthesize DNA only 5′→3′ by adding nucleotides to a free 3′-OH. On the leading-strand template, polymerase can synthesize continuously toward the opening replication fork. On the opposite template, polymerase would have to synthesize 3′→5′ to follow the fork, which it cannot do. Instead, primase repeatedly places new primers and DNA polymerase synthesizes short 5′→3′ pieces away from the fork. These are Okazaki fragments, which are later joined by ligase.

That is much better than:

7. DNA has 22% thymine

T = 22% → A = 22%

Remaining:

100 − 44 = 56%

Split equally:

G = 28%

C = 28%

Final:

A 22%, T 22%, G 28%, C 28%

8. Why is high-GC DNA harder to denature?

G–C pairs have 3 hydrogen bonds; A–T pairs have 2.

higher GC content → more hydrogen bonding → more energy required to separate strands → harder/higher temperature needed to denature

9. P³²/S³⁵ virus question

The reasoning comes directly from Hershey-Chase.

Why P³²?

Nucleic acids contain phosphate.

So if viral nucleic acid is labeled with P³²:

expose cell types to labeled virus → allow infection → remove extracellular virus → detect which cells retain intracellular P³²

Cells containing the radioactive nucleic-acid signal are the cells in which viral material entered.

For SARS-CoV-2 specifically, its genome is RNA, but RNA also contains a phosphate backbone, so the underlying P³² labeling logic still concerns nucleic acid.

Why not S³⁵ for the same purpose?

S³⁵ labels:

Protein

not nucleic acid.

Therefore S³⁵ would track viral protein rather than specifically tracking the viral genome.

The professor is explicitly asking students to apply the Hershey-Chase distinction rather than merely recall the experiment.

main things

The professor’s highest-emphasis connections are:

DNA hereditary evidence → Hershey-Chase labeling logic

nucleotide structure → sugar/phosphate/base

phosphodiester backbone vs hydrogen-bonded bases

complementary vs antiparallel

5′/3′ chemistry

A-T 2 H bonds / G-C 3 H bonds

RNA can intramolecularly base-pair

semiconservative replication

origin → two forks → bidirectional bacterial replication

gyrase vs helicase vs SSB

primase/primer → free 3′-OH

DNA pol III = 5′→3′ synthesis

antiparallel + 5′→3′ restriction → leading vs lagging

Okazaki fragments → DNA pol I primer replacement → ligase joining

replication finishes before cell division

That is the level of continuity I think the earlier notes were missing.

The big picture: gene expression

DNA by itself is the cell’s stored information/blueprint. A cell actually needs proteins because proteins can become cellular structures and enzymes that carry out cellular work. Therefore, the information stored in a gene has to be expressed.

DNA gene → transcription by RNA polymerase → mRNA copy of gene → translation by ribosome → amino-acid sequence/polypeptide → protein folds/functions → phenotype/cellular activity

This overall information flow is the central dogma:

DNA → RNA → Protein

The chromosome contains the cell’s genetic information; specific DNA segments are genes. The collection of genetic information is the genotype, while the traits/functions actually produced from expressed genes contribute to the phenotype. The professor uses examples such as genes for superoxide dismutase, catalase, flagella, chemotaxis, adhesion, and bacterial toxins to show that differences in DNA can lead to different cellular capabilities.

2. Transcription: DNA → RNA

Term	Where it is / comes from	Exact job	What happens next
Gene	Region of DNA	Contains information for an RNA/protein product	RNA polymerase transcribes it
Promoter	DNA sequence before/upstream of gene	Binding/start-recognition region for RNA polymerase	RNA polymerase binds → transcription begins
RNA polymerase	Enzyme in cell	Opens a small DNA region and builds RNA 5′→3′ complementary to template	Moves along gene until termination signal
Template strand = antisense strand	One DNA strand of that gene	Actual strand RNA polymerase reads 3′→5′	Determines complementary RNA sequence
Coding strand = nontemplate/sense strand	Other DNA strand	Same sequence as RNA except DNA T ↔ RNA U	Useful for predicting mRNA sequence
Transcription bubble	Small locally opened region of DNA	Allows RNA polymerase access to template	Moves with polymerase; DNA rewinds behind it
RNA transcript / mRNA	Made from template DNA	Carries gene information to ribosome	Used during translation
Terminator	DNA sequence	Signals RNA polymerase to stop transcription	Polymerase + RNA release; DNA rewinds

The slides define the promoter as where RNA polymerase binds and the terminator as where RNA synthesis stops. .

Template vs coding strand

Suppose:

Coding DNA: 5′-ATG GAA TTC-3′
Template DNA: 3′-TAC CTT AAG-5′
mRNA: 5′-AUG GAA UUC-3′

Notice:

mRNA is complementary to the template strand

but

mRNA matches the coding strand except U replaces T

The professor explicitly stresses this relationship.

So if you are given the template DNA, complement it:

DNA A → RNA U
DNA T → RNA A
DNA C → RNA G
DNA G → RNA C

If you are given the coding DNA, just copy its sequence into RNA and replace:

T → U

Also remember direction:

RNA polymerase reads template DNA 3′→5′ → synthesizes RNA 5′→3′

Transcription mechanism — one continuous process

Promoter on DNA → bacterial RNA polymerase recognizes/binds promoter → DNA opens locally to form a transcription bubble → one DNA strand becomes the template → RNA polymerase reads template 3′→5′ while adding complementary ribonucleotides to the RNA 5′→3′ → phosphodiester bonds join RNA nucleotides → DNA immediately rewinds behind RNA polymerase → RNA chain elongates until polymerase encounters a termination sequence → RNA polymerase stalls/releases → completed RNA transcript separates → DNA fully rewinds → mRNA becomes available for translation

The professor organizes transcription as initiation → elongation → termination. transcription and translation t…

OpenStax adds an important detail: unlike DNA polymerase, RNA polymerase does not need a primer/free preexisting 3′-OH to begin a new RNA molecule. In bacteria, RNA polymerase includes a core enzyme plus a sigma (σ) factor; σ helps the polymerase recognize specific promoters, then dissociates as elongation proceeds. microbiology_-_WEB

The three transcription stages

Stage	What specifically happens
Initiation	RNA polymerase recognizes promoter, binds, locally opens DNA, identifies template, begins first RNA nucleotides
Elongation	Polymerase moves along template 3′→5′, RNA grows 5′→3′; DNA opens ahead and rewinds behind
Termination	Terminator sequence causes polymerase to stop/release → RNA transcript released → DNA closes

A useful detail from OpenStax is that bacterial promoter recognition involves σ factor. OpenStax discusses conserved promoter regions upstream of the +1 transcription start site, including -10/-35 regions. That is useful for understanding how RNA polymerase knows where a gene starts, although the professor is not emphasizing detailed promoter sequences in this lecture.

transcription vs tarnslation signals

Signal	Process	Molecule it is on	Meaning
Promoter	Transcription	DNA	RNA polymerase binds/start region
Terminator	Transcription	DNA	RNA polymerase stops
Start codon AUG	Translation	mRNA	Ribosome starts protein synthesis; codes Met/fMet
Stop/nonsense codon UAA, UAG, UGA	Translation	mRNA	Translation stops; no amino acid encoded

RNA polymerase binds PROMOTER, not start codon.

Ribosome stops at STOP CODON, not terminator.

That distinction is tested directly in the lecture iClickers.

The RNA molecules needed for translation

RNA	Where it comes from / structure	Information or cargo	Exact job → what happens next
mRNA = messenger RNA	Transcribed from a DNA gene by RNA polymerase	Contains codons	Goes to ribosome → codons determine amino-acid order
tRNA = transfer RNA	RNA that folds by intramolecular base pairing; has anticodon on one end and amino-acid attachment site at 3′ end	Carries one specific amino acid	Anticodon pairs with mRNA codon → delivers correct amino acid to ribosome
rRNA = ribosomal RNA	Made by transcription; combines with proteins to form ribosome	Does not carry protein code	Structural + catalytic part of ribosome; helps align mRNA/tRNAs and catalyzes peptide bonds

The professor specifically says rRNA contributes both structure and enzymatic function, while tRNA brings amino acids to the ribosome. transcription and translation t…

OpenStax makes the tRNA structure more precise: the amino acid attaches to the CCA amino-acid-binding end at the tRNA’s 3′ end, while the anticodon lies at the opposite functional region and base-pairs with the mRNA codon.

Genetic code: how RNA letters specify amino acids

Proteins use 20 amino acids, but RNA has only four bases:

A, U, G, C

One base would provide:

4¹ = 4 possibilities → not enough

Two-base combinations:

4² = 16 → still not enough

Three-base combinations:

4³ = 64 → enough

Therefore the genetic code is read in:

Codons = groups of 3 mRNA nucleotides

The slides explicitly derive the 64 possible codons this way. Transcription and Translation

Codon vs anticodon

Codon = three bases on mRNA

Anticodon = complementary three bases on tRNA

Example:

mRNA codon: 5′-AUG-3′

pairs with:

tRNA anticodon: 3′-UAC-5′

That tRNA carries the amino acid specified by AUG.

The professor’s examples connect:

AUG ↔ UAC tRNA → methionine

UUC ↔ AAG tRNA → phenylalanine

AAA ↔ UUU tRNA → lysine transcription and translation t…

Start and stop codons

AUG = start codon → methionine; in bacteria the initiator amino acid is N-formylmethionine (fMet)

UAA, UAG, UGA = stop/nonsense codons → no amino acid

Her lecture explicitly expects students to recognize AUG and all three stop codons. transcription and translation t…

Redundancy / degeneracy of the genetic code

There are 64 codons but only 20 amino acids, so multiple codons can specify the same amino acid.

Example from lecture:

UUU and UUC → phenylalanine

This is called redundancy or, in OpenStax terminology, degeneracy of the genetic code.

This explains the iClicker:

CAU → CAC

Both encode histidine → no amino-acid change.

So a nucleotide change does not automatically mean the protein changes.

The professor says she will provide a codon chart if needed; she does not expect memorization of every codon, but she does expect students to know how to read the chart. transcription and translation t…

Important: codon charts are read using the mRNA codon, not the DNA template and not the tRNA anticodon.

Before a tRNA can work, it has to be “charged”

A tRNA cannot simply grab any amino acid.

A specific enzyme:

Aminoacyl-tRNA synthetase

recognizes a tRNA and attaches its proper cognate amino acid.

This process is:

tRNA charging

The relationship is:

uncharged tRNA + correct amino acid + ATP → aminoacyl-tRNA synthetase uses energy to attach amino acid → charged aminoacyl-tRNA → travels to ribosome → delivers amino acid during translation → becomes uncharged → leaves ribosome → gets charged again

The professor emphasizes that charging costs energy and describes ATP being used when the amino acid is loaded onto its tRNA. transcription and translation t…

OpenStax adds the mechanistic terminology: the amino acid is activated during charging, then transferred to the appropriate tRNA; there is at least one aminoacyl-tRNA synthetase type corresponding to each amino acid class. microbiology_-_WEB

So the tRNA is basically an adapter:

anticodon side identifies the mRNA instruction ↔ amino-acid side carries the physical amino acid that instruction represents

That is why tRNA is what actually connects the “language” of nucleotides to the “language” of amino acids.

The bacterial ribosome and its three sites

The professor focuses heavily on what happens at A, P, and E.

OpenStax adds that a bacterial ribosome is:

30S small subunit + 50S large subunit → 70S ribosome

The numbers are Svedberg sedimentation units, so they are not mathematically additive in the ordinary sense.

The small subunit binds/positions mRNA; the large subunit contains the catalytic machinery and accommodates tRNAs. Ribosomes contain both protein and rRNA, and rRNA performs an important catalytic role. microbiology_-_WEB

A–P–E sites

Site	Full name	What is there	What happens next
A site	Aminoacyl / acceptor site	Incoming charged tRNA carrying next amino acid	Codon–anticodon match confirmed → peptide bond forms
P site	Peptidyl site	tRNA holding the growing polypeptide	Chain is transferred toward amino acid in A site
E site	Exit / eject site	Uncharged tRNA after losing amino acid	tRNA leaves ribosome → can be recharged

Normal movement:

A → P → E → out

That is exactly the answer to the professor’s iClicker about tRNA movement. Transcription and Translation

One very important exception

The first initiator tRNA does NOT enter through A.

During initiation:

initiator fMet-tRNA goes directly into the P site

This leaves the A site empty and ready for the second aminoacyl-tRNA. OpenStax emphasizes this exception.

Translation: mRNA → polypeptide

Just like transcription, translation is divided into:

initiation → elongation → termination

But these are completely different events from transcription initiation/elongation/termination.

Component	Where it comes from / where found	Exact job	Where it goes/what happens next
mRNA	Produced by transcription	Provides codon sequence	Ribosome reads it 5′→3′
Small ribosomal subunit	rRNA + protein	Binds/positions mRNA	Helps establish start site
Large ribosomal subunit	rRNA + protein	Contains A/P/E sites; peptide-bond catalytic activity	Joins initiation complex
Initiator tRNA-fMet	tRNA charged with formylmethionine in bacteria	Pairs with AUG at start	Sits directly in P site
Charged aminoacyl-tRNA	tRNA charged by aminoacyl-tRNA synthetase	Delivers next amino acid	Enters A → P → E
Peptidyl transferase	Catalytic rRNA in large subunit	Catalyzes peptide bond	Polypeptide lengthens
Release factor	Protein	Recognizes stop codon in A site because no stop-codon tRNA exists	Releases polypeptide + dismantles translation complex

Translation initiation — full connected mechanism

mRNA becomes available → small 30S ribosomal subunit binds mRNA near the start region → bacterial Shine-Dalgarno ribosome-binding sequence helps correctly position the ribosome → AUG start codon is placed in the P-site position → initiator tRNA carrying N-formylmethionine (fMet) base-pairs with AUG → initiation factors and GTP help assemble the complex → large 50S subunit joins → complete 70S initiation complex forms with fMet-tRNA already in P site and A site empty → elongation can begin

OpenStax identifies the bacterial Shine-Dalgarno sequence as the ribosome-binding site upstream of AUG and describes the special fMet initiator tRNA. microbiology_-_WEB

For the professor’s level, the indispensable pieces are:

ribosomal subunits + mRNA + AUG + initiator fMet-tRNA → initiation complex → first tRNA starts in P site

Her transcript explicitly emphasizes that the first methionine-containing tRNA starts in the P site.

Translation elongation — follow one complete cycle

This is the section where people usually memorize A/P/E without understanding what is physically moving.

Assume fMet-tRNA is already sitting in the P site.

next mRNA codon exposed in A site → matching charged tRNA enters A site by complementary codon–anticodon pairing → its amino acid is positioned beside growing chain held in P site → peptidyl transferase forms a peptide bond → growing polypeptide becomes attached to the tRNA in A site → ribosome translocates one codon along mRNA → former A-site tRNA carrying the chain moves into P site → old uncharged P-site tRNA moves into E site → uncharged tRNA exits → next charged tRNA enters A site → cycle repeats

The professor walks through this with methionine → leucine → glycine and shows the tRNAs shifting through the sites. transcription and translation t… transcription and translation t…

OpenStax adds two useful mechanisms:

peptidyl transferase is actually catalytic rRNA—a ribozyme—in the 50S subunit, and GTP supplies energy for several elongation events, including tRNA delivery/translocation. microbiology_-_WEB

So remember:

ATP is especially important for charging tRNAs

while

GTP is used during translation machinery movements/initiation/elongation

11. Translation termination

Eventually the ribosome reaches:

UAA, UAG, or UGA

There is no tRNA with an amino acid corresponding to a stop codon.

Therefore:

stop codon enters A site → no aminoacyl-tRNA can bind → release factor recognizes stop codon → finished polypeptide is released from P-site tRNA → ribosomal subunits separate → mRNA and tRNAs are released/recycled

The professor describes the stop/nonsense codon as the signal that the complex should come apart and notes that the released tRNA can be recharged and reused. transcription and translation t…

OpenStax explicitly identifies release factors, which are important terminology missing from the simpler slide diagram.

What happens to the protein afterward?

The product directly leaving the ribosome is a:

Polypeptide

A polypeptide may not yet be a fully functional protein.

The lecture notes that the protein can fold and may receive help from chaperonins/chaperone proteins. transcription and translation t…

OpenStax expands this into post-translational processing, which can include protein folding, assembly with other subunits, cleavage of certain sequences, or chemical modifications. microbiology_-_WEB

So:

translation → linear amino-acid chain/primary structure → folding ± modification → functional protein

13. Why multiple ribosomes can use one mRNA

One mRNA does not necessarily make only one protein.

Once one ribosome moves far enough down the mRNA, another ribosome can bind behind it.

Therefore:

one mRNA + many ribosomes = polyribosome / polysome

Each ribosome makes its own copy of the same polypeptide.

So:

one gene → multiple mRNAs can be transcribed → each mRNA can be translated by multiple ribosomes → many copies of protein rapidly

OpenStax explicitly calls an mRNA containing multiple translating ribosomes a polyribosome/polysome and notes how this permits rapid protein production.

14. The major bacterial advantage: transcription and translation can be coupled

Prokaryotes

No nucleus separates DNA from ribosomes.

Therefore both processes happen in the cytoplasm.

That means a ribosome can begin translating an mRNA before RNA polymerase has even finished transcribing that mRNA.

DNA being transcribed → 5′ end of mRNA emerges → ribosome binds emerging mRNA → translation begins while RNA polymerase continues transcription

The professor emphasizes that multiple RNA polymerases can simultaneously produce RNA from a gene while ribosomes are already translating the emerging transcripts. transcription and translation t…

OpenStax calls this coupled/concurrent transcription and translation and connects it to rapid bacterial responses. microbiology_-_WEB

Eukaryotes

These processes are spatially separated:

DNA in nucleus → transcription → pre-mRNA processing → mature mRNA exits nuclear pore → translation in cytoplasm

The slides specifically contrast prokaryotic simultaneous transcription/translation with eukaryotic mRNA needing to leave the nucleus. Transcription and Translation

For this course, the professor is not emphasizing detailed eukaryotic processing, and she explicitly says not to worry about details such as caps/poly-A tails from the eukaryotic animation. transcription and translation t…

OpenStax does explain the full eukaryotic sequence for context:

pre-mRNA → 5′ cap + poly-A tail + intron removal/splicing → mature mRNA → export to cytoplasm → translation. microbiology_-_WEB

For her exam, I would prioritize the bacterial pathway unless the professor directly asks for the comparison.

15. Transcription vs translation — same stage names, completely different meanings

	TRANSCRIPTION	TRANSLATION
Goal	DNA → RNA	mRNA → polypeptide
Main machine	RNA polymerase	Ribosome + tRNAs
Starting signal	Promoter	AUG start codon
Template	Template DNA strand	mRNA
Building blocks	Ribonucleotides A/U/G/C	Amino acids
Product	RNA transcript	Polypeptide
Initiation	RNA polymerase binds promoter	Ribosome assembles at AUG with initiator fMet-tRNA
Elongation	RNA polymerase adds RNA nucleotides 5′→3′	Repeated A→P→E tRNA cycle; peptide bonds form
Termination signal	DNA terminator	UAA/UAG/UGA stop codon
Termination result	RNA transcript released	Polypeptide released + ribosome dissociates

16. One worked sequence connecting EVERYTHING

Start with DNA:

Coding DNA: 5′-ATG TTT GGC TAA-3′
Template DNA: 3′-TAC AAA CCG ATT-5′

Transcription

RNA polymerase uses the template:

3′-TAC AAA CCG ATT-5′ → 5′-AUG UUU GGC UAA-3′

Notice mRNA matches the coding strand except:

T → U

Translation

Break mRNA into codons:

AUG | UUU | GGC | UAA

Codon chart:

AUG = Met/fMet
UUU = Phe
GGC = Gly
UAA = Stop

Therefore:

DNA → mRNA → fMet–Phe–Gly → STOP

And at the tRNA level:

AUG codon ↔ UAC anticodon carrying fMet → UUU codon ↔ AAA anticodon carrying Phe → GGC codon ↔ CCG anticodon carrying Gly → UAA has no tRNA → release factor terminates

That one example connects coding strand, template strand, transcription, codons, anticodons, tRNA, amino acids, start and stop.

17. MEGA FLOWCHART

Bacterial chromosome contains genes → promoter marks where a gene can begin transcription → σ-containing RNA polymerase recognizes/binds promoter → DNA opens locally into transcription bubble → RNA polymerase uses one DNA strand as template and reads it 3′→5′ → complementary RNA is synthesized 5′→3′ using A/U/G/C → mRNA sequence therefore matches coding DNA except U replaces T → DNA rewinds behind polymerase → polymerase reaches terminator → RNA transcript is released → mRNA codons are read in groups of 3 → AUG establishes start and bacterial initiator tRNA carrying fMet pairs with AUG in ribosomal P site → meanwhile aminoacyl-tRNA synthetases use ATP to attach the correct amino acids to their tRNAs → each charged tRNA uses its anticodon to recognize a complementary mRNA codon → after initiation, new charged tRNAs enter ribosomal A site → rRNA/peptidyl transferase catalyzes peptide bond → growing chain transfers to A-site tRNA → ribosome translocates one codon using energy → tRNA carrying growing chain shifts A→P while old uncharged tRNA shifts P→E→exits → emptied tRNA returns to aminoacyl-tRNA synthetase to be recharged → A→P→E cycle repeats while ribosome reads mRNA 5′→3′ and polypeptide grows → UAA/UAG/UGA reaches A site → no corresponding tRNA exists → release factor binds → completed polypeptide released → ribosomal subunits separate/recycle → polypeptide folds ± chaperones/modifications into functional protein → in bacteria multiple ribosomes can translate one mRNA and translation can begin before transcription finishes because both occur in cytoplasm → one gene can therefore rapidly produce many copies of a protein

Professor’s iClickers / things she clearly expects

Question	Answer + reason
RNA polymerase begins transcription by binding what?	Promoter
Protein synthesis stops at what?	Nonsense/stop codon, not terminator
Order a normal elongating tRNA moves through ribosome?	A → P → E
Exception to A→P→E?	Initiator fMet-tRNA begins directly in P
What does ACA code for?	Threonine, using provided mRNA codon table
CAU → CAC gives amino-acid change?	No; both encode histidine because genetic code is redundant
Start codon?	AUG
Stop codons?	UAA, UAG, UGA
Codon is located where?	mRNA
Anticodon is located where?	tRNA
Which strand matches mRNA?	Coding/nontemplate strand, except T→U
Which strand is actually read?	Template/antisense DNA strand