Ch 6

When is DNA replicated?

during cell division

how many cells are generated everyday in the human body?

1 billion

mutations/errors in DNA replication can lead to

genetic disorders and diseases.

conservative mode of DNA replication

a method where the original DNA strands remain intact and a completely new double helix is synthesized.

semiconservative mode of DNA replication

a method where each new double helix consists of one original strand and one newly synthesized strand.

dispersive mode of DNA replication

Each strand contains interspersed sections of parental and new DNA

Meselson-Stahl experiment

An experiment conducted by Matthew Meselson and Franklin Stahl in 1958 that demonstrated the semiconservative nature of DNA replication by using isotopes of nitrogen (radiolabeled DNA nitrogenous bases) to label newly synthesized DNA strands and observe the distribution of parental and newly synthesized strands across generations by comparing the mass of DNA strands over 2 generations.

How did they compare the masses of DNA strands?

high-speed centrifugation in graded solution

semiconservative replication over time

The proportion of original strands and newly synthesized strands is maintained, resulting in DNA molecules that contain one old and one new strand for each helix.

each DNA strand can serve as a template for the synthesis of a new _____ strand

complementary

DNA replication

• begins with DNA strands being pulled apart with the aid of initiator proteins

• 
  

initiator proteins

allow the separation of the double-stranded DNA (dsDNA)

replication origins in bacteria vs human genome

bacteria: 1 replication origin

human genome: 10,000 replication origins

are replication origins more likely to be A-T rich or G-C rich?

Replication origins are more likely to be A-T rich due to the presence of two hydrogen bonds between A and T, making them easier to separate compared to G-C pairs which have three hydrogen bonds.

a twisting force that occurs during DNA replication, caused by the unwinding of the double helix. some stress can be relieved by DNA supercoiling

torsional strain

topoisomerase

An enzyme that alleviates torsional strain in DNA during replication by inducing temporary single-strand breaks and allowing the DNA strands to unwind.

How does topoisomerase relieve torsional stress ahead of helicase?

Topoisomerase temporarily breaks one DNA strand. DNA rotates freely around the phosphodiester bond in the opposite, unbroken strand, releasing the twisting stress ahead of helicase. The same topoisomerase then reseals the break.

y-shaped junction where DNA is replicated

replication fork

a group of proteins that are recruited to the site and replicate the DNA after separation of double helix

replication machine

DNA synthesis is ___ to ___

5’ → 3”

nucleotides are added only to the ___ end of a DNA strand

3’

catalyzes new strand synthesis

DNA polymerase

DNA polymerase processivity

clamo proteins anchor the DNA Pol III enzyme to remain clos to the DNA ensure processivity

DNA Pol II can synthesize 1,000 nucleotides/second

processivity

refers to the ability of DNA polymerase to add nucleotides continuously without detaching from the DNA template.

how can the doubling time of E. coli be appx. 15-20 min.? Given that and E. coli genome can be replicated in 50 min..

E. coli starts new rounds of DNA replication before the previous round finishes. This creates multiple replication forks copying DNA simultaneously.

If DNA is synthesized only in the 5’ to 3’ direction, how are both strands replicated at the fork?

Both strands are replicated simultaneously using leading and lagging strands; the leading strand is synthesized continuously, while the lagging strand is synthesized discontinuously in Okazaki fragments.

will synthesize NA primer to begin replication

primase

will extend the okazaki fragment until it runs into the next RNA primer

DNA polymerase

RNA primer will then be degraded by a ____ and a DN polymerase called __________; replaces the RNA with DNA

ribonuclease; repair polymerase (DNA Poly 1)

joins okazaki fragments (5’ Phosphorous to 3’ OH)

DNA ligase

Important replication proteins

initiator proteins, helicase, single-stranded DNA-binding proteins, topoisomerase, primase, DNA Polymerase III, sliding clamp, nuclease, DNA polymerase 1, and DNA ligase.

catalyzed by 3’→5’ exonuclease activity.

polymerase error: error rate is 1 in 10⁷ bp

exonuclease error: error rate is 1 in 10² bp

DNA polymerase proofreading

How does DNA synthesis in the 5’ to 3’ direction facilitate proofreading?

During 5′→3′ synthesis, each incoming nucleotide supplies the energy needed to attach itself to the growing strand.

If proofreading removes an incorrect nucleotide from the 3′ end, it leaves a free 3′-OH. The next correct nucleotide supplies fresh energy, so synthesis can continue.

In hypothetical 3′→5′ synthesis, removing the incorrect nucleotide would also remove the growing strand’s activating triphosphate (energy source), preventing continued synthesis.

Therefore, 5′→3′ synthesis allows mistakes to be removed without losing the ability to keep building DNA.

DNA Priming (leading strand)

• provides starting point DNA polymerase needs to build a new strand since it can’t start from scratch; it needs an existing 3’ OH to add nucleotides

• primase can start from scratch. it joins RNA nucleotides to make a short RNA primer complement to the DNA template

• primer grows 5’→ 3’, and DNA poly adds nucleotides to its 3’ end

DNA Priming (lagging strand)

• primase makes a new RNA primer

• DNA polymerase extends it to form an Okazaki fragment

• the previous RNA primer is removed by nucleases

• a repair DNA polymerase replaces that RNA with DNA

• DNA ligase seals the remaining nick in the backbone

lagging strand needs a new primer for every okazaki fragment

ligation of okazaki fragments

RNA primers removed and replaced with DNA → all nucleotides are present, but the sugar-phosphate backbone isn’t fully connected (nicked)

1. ATP supplies energy to activate 5’ phosphate at the nick

2. ligase forms a phosphodiester bond between 5’ P and adjacent 3’ OH

3. AMP is released, leaving a continuous backbone

draw

end-replication problem

linear chromosomes can get shorter each time DNA is copied

When the last primer at the chromosome’s end is removed, there is no neighboring 3’ OH for DNA polymerase to extend, so that gap can’t be filled. The new strand is shorter at its 5’ end. In the next replication round, that shortened strand serves as a template, causing further loss

solution: telomerase

How does telomerase help elongate chromosome ends?

Telomerase adds repetitive nucleotide sequences to the 3' ends of chromosomes, extending telomeres and preventing loss of genetic information during DNA replication.

when is DNA replicated?

during cell division

permanent changes in DNA sequence; can be deleterious, neutral, or even beneficial (evolution); not always easy to distinguish between deleterious and beneficial; can occur despite proofreading

mutations/ replication errors

an unpaired mismatch can become a permanent ______ after another replication

base-pair substitution

chemical modifications that lead to mutations

deamination: changes C to U; U then pairs with A during replication => substitution after 2 rounds of replication (C-G —deamination changes C into U→ U-G —U pairs with A→ U-A —T is opposite to A→ T-A

depurination: a purine base ( A or G) is lost; the copying machinery skips the missing base, producing deletion

can accumulate mutations over a lifespan that may impact health

somatic cells

excision, replacement, ligation

DNA repair mechanisms

a nuclease will remove the damaged/wrong region

excision

DNA polymerase will synthesize new strand

replacement

the enzyme ligase will connect the newly synthesized nucleotide with the next one (nick repair)

ligation

mutations drive _____ across species

genetic diversity

DNA can help uncover ____ relationships

phylogenetic

_______ mutations that disrupt DNA areplication would be expected to be _____

disease-associated (pathogenic variants); cell lethal

_____ variants likely to be _____ mutations

disease-causing; hypomorphic (partial loss-of-function)

microtia (small ears), distinct facial features, absent or small patellae (kneecaps), short stature, pathogenic variants impair proteins needed to load helicase onto DNA, disrupting replication initiation and normal growth

Meier-Gorlin Syndrome

psychomotor regression (loss of previosuly attained developmental milestones), seizures, liver disease, pathogenic variants in mito polymerase genes, mitochondrial DNA depletion

alpers syndrome

aplastic anemia: underproduction of blood cells, pathogenic variants in telomerase

telomere shortening, cancer, premature aging, alzheimer’s and parkinson’s disease

telomere-related disease

How does nonhomologous end joining (NHEJ) repair a double-strand break?

The broken ends are processed, sometimes removing nucleotides, and DNA ligase joins the ends. No homologous template is required, but repair can cause a deletion.

How does homologous recombination repair a double-strand break?

It aligns the damaged DNA with an undamaged homologous DNA molecule, usually a sister chromatid, and uses it as a template to restore the missing sequence accurately.

What are the steps of homologous recombination shown in the diagram?

1. A nuclease removes DNA from the 5′ ends, exposing 3′ overhangs.

2. A 3′ end invades matching DNA and pairs with its complementary strand.

3. DNA polymerase extends that end using the undamaged DNA as a template.

4. The extended strand returns and pairs with the other broken end.

5. DNA synthesis fills the remaining gaps, and ligase seals the nicks.

When does homologous recombination usually occur, and why?

After DNA replication, when an intact sister chromatid is available as a matching repair template.

What is the main difference between NHEJ and homologous recombination?

NHEJ directly rejoins ends without a homologous template and can lose nucleotides. Homologous recombination copies a matching template and generally restores the sequence accurately.

What could happen if a cell receives foreign DNA with homologous ends?

The matching ends can align with the target DNA, allowing the foreign DNA to serve as a repair template and introduce a specific sequence change through homologous recombination.

How does homology-directed repair (HDR) allow precise genome editing?

Cas9 creates a double-strand break. A supplied DNA repair template contains the desired change between two homology arms, which match DNA on either side of the break. The cell uses this template during HDR to copy the desired change into its chromosome.

How does CRISPR–Cas9 produce different edits through NHEJ versus HDR?

A guide RNA directs Cas9 to a matching DNA sequence, and Cas9 cuts both strands.

• NHEJ: Rejoins the ends without a supplied template and can create insertions or deletions (indels) that disrupt a gene.

• HDR: Uses a supplied matching repair template to introduce a specific sequence change.