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1
Q

What is the key feature of semi-conservative DNA replication?

A

Each new DNA molecule contains one original strand and one new strand.

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2
Q

During which phase of the cell cycle does DNA replication occur in eukaryotes?

A

It occurs during the S phase.

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3
Q

What factors carry out DNA replication in prokaryotes?

A

Sigma factors.

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4
Q

Are the enzymes involved in DNA replication in eukaryotes similar to those in prokaryotes?

A

Yes, many similar enzymes are required in eukaryotes.

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5
Q

What does semi-conservative DNA replication mean?

A

Each new DNA molecule contains one original (parent) strand and one newly synthesized strand.

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6
Q

How does semi-conservative replication differ from conservative and dispersive replication?

A

Conservative: Keeps the original DNA intact, producing a completely new molecule.
Dispersive: Produces DNA with interspersed original and new segments.
Semi-conservative: One strand of the original DNA is retained in each new molecule.

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7
Q

What key experiment demonstrated the semi-conservative nature of DNA replication?

A

The Meselson-Stahl experiment using isotopes of nitrogen (N-15 and N-14) and density gradient centrifugation.

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8
Q

What is the role of the parent DNA strands during semi-conservative replication?

A

The parent strands act as templates for synthesizing the new complementary strands.

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9
Q

What enzyme is primarily responsible for adding new nucleotides during semi-conservative replication?

A

DNA polymerase.

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10
Q

What is the role of helicase in DNA replication?

A

Helicase unwinds the double helix to separate the two parent strands.

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11
Q

How is the leading strand synthesized during semi-conservative replication?

A

It is synthesized continuously in the 5’ to 3’ direction.

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12
Q

How is the lagging strand synthesized during semi-conservative replication?

A

It is synthesized in short fragments called Okazaki fragments, which are later joined by DNA ligase.

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13
Q

Why is DNA replication considered accurate and reliable?

A

DNA polymerase has proofreading ability, correcting errors during replication.

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14
Q

What stabilizes the single-stranded DNA during replication?

A

Single-stranded binding proteins (SSBs).

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15
Q

What is the role of primase in DNA replication?

A

Primase synthesizes RNA primers, which provide a starting point for DNA polymerase.

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16
Q

Why do eukaryotes use semi-conservative replication?

A

Eukaryotes use semi-conservative replication to maintain a high degree of similarity between generations. With such large genomes, low fidelity could be fatal.

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17
Q

What is semi-conservative replication?

A

Semi-conservative replication is the process where each new DNA molecule consists of one original (parental) strand and one newly synthesized strand.

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18
Q

Why is semi-conservative replication important?

A

It ensures genetic stability by preserving one original DNA strand, minimizing errors during DNA replication, and maintaining fidelity across generations.

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19
Q

How does semi-conservative replication help eukaryotes with large genomes?

A

It maintains high fidelity during DNA replication, preventing errors that could be fatal given the size and complexity of eukaryotic genomes.

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20
Q

What are the key enzymes involved in semi-conservative replication?

A

Key enzymes include DNA helicase (unwinds DNA), DNA polymerase (synthesizes new strand), and DNA ligase (seals fragments).

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21
Q

How was the semi-conservative nature of DNA replication proven?

A

It was proven through the Meselson-Stahl experiment, which used isotopes of nitrogen to demonstrate that new DNA molecules retain one original strand and one new strand.

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22
Q

What could happen if DNA replication had low fidelity in eukaryotes?

A

Low fidelity could lead to mutations and errors in the genome, potentially causing fatal consequences for the organism.

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23
Q

What is genetic integrity, and why is it essential in replication?

A

Genetic integrity ensures the accurate inheritance of nucleic acid sequences, preserving stability and minimizing harmful mutations in both somatic and germ-line cells.

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24
Q

What role do germ-line sequences play in a population?

A

Germ-line sequences introduce genetic variation within a population while mechanisms exist to limit replication errors that could lead to changes in germ-line DNA.

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25
Q

How do errors in replication affect germ-line and somatic cells differently?

A

Errors in germ-line cells can be passed to offspring and affect the entire lineage, while errors in somatic cells only impact the individual.

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26
Q

What mechanisms are in place to maintain genetic integrity during replication?

A

DNA repair systems, proofreading by DNA polymerase, and other cellular mechanisms ensure errors are minimized and genetic integrity is maintained.

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27
Q

How does genetic integrity balance stability and variation?

A

It preserves essential DNA sequences for survival while allowing controlled genetic variation through mutations in germ-line cells, supporting evolution and adaptability.

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28
Q

What happens when there are changes in nucleic acid sequences?

A

Changes in nucleic acid sequences can lead to:
* No changes in the amino acid sequence.
* Changes in amino acid sequences, which may affect protein shape, function, or expression.

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29
Q

What are the possible outcomes of changes in amino acid sequences?

A

Changes in amino acid sequences can:
* Alter protein shape, affecting molecular interactions.
* Change protein function, disrupting metabolic pathways or signaling.
* Modify protein expression levels, leading to overactivity or underactivity.

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30
Q

Why can changes in protein shape be problematic?

A

Altered protein shape affects its ability to interact with other molecules, potentially disrupting biological processes and causing disease.

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31
Q

How do changes in protein function affect the body?

A

Changes in protein function can disrupt critical biological processes such as metabolism, signaling, and immune responses, leading to conditions like cancer or metabolic disorders.

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32
Q

What are the effects of altered protein expression levels?

A

Altered protein expression levels can:
* Cause imbalances in cellular functions.
* Lead to systemic health effects, such as overactive or underactive biological processes.

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33
Q

How can changes in nucleic acid sequences negatively impact the body?

A

Changes in nucleic acid sequences may:
* Alter codons, leading to production of faulty or nonfunctional proteins.
* Introduce premature stop codons, causing truncated proteins.
* Generate proteins with altered functions that disrupt biological pathways.

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34
Q

What are the possible biochemical reasons changes in amino acid sequences can be harmful?

A

Changes in amino acid sequences can:
* Disrupt protein folding, leading to misfolded proteins that cannot function.
* Alter active sites in enzymes, reducing or eliminating catalytic activity.
* Affect binding domains, preventing interactions with other molecules or receptors.

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35
Q

Why can protein misfolding be harmful?

A

Protein misfolding can:
* Lead to aggregation, forming toxic clumps (e.g., amyloids in Alzheimer’s disease).
* Prevent proteins from reaching their functional sites in the cell.
* Trigger cellular stress responses like the unfolded protein response (UPR).

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36
Q

How do changes in protein shape affect biochemical pathways?

A

Altered protein shape can:
* Impair interaction with other proteins or DNA, disrupting complex assembly.
* Prevent recognition by signaling molecules, halting cellular communication.
* Reduce the stability of the protein, increasing degradation rates.

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37
Q

What are the consequences of altered protein function due to amino acid changes?

A

Altered protein function can:
* Cause enzymes to lose specificity, leading to unregulated or inappropriate reactions.
* Create dominant negative effects where the altered protein interferes with normal protein functions.
* Enable proteins to gain toxic functions, damaging cells and tissues.

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38
Q

Why can changes in protein expression be detrimental?

A

Changes in protein expression can:
* Cause an overproduction of proteins, overwhelming cellular systems (e.g., cancer-promoting oncogenes).
* Lead to insufficient production of essential proteins, impairing normal functions (e.g., hormone deficiencies).
* Disrupt cellular balance, affecting energy metabolism and homeostasis.

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39
Q

How do biochemical changes in post-translational modifications affect protein function?

A

Altered post-translational modifications (e.g., phosphorylation, glycosylation) can:
* Impact protein stability and degradation.
* Modify protein localization within the cell.
* Impair or enhance protein interactions, altering cellular signaling networks.

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40
Q

What structures preserve DNA integrity by providing a protective environment for chromosomes?

A

The nuclear envelope.

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41
Q

What mechanism ensures accurate duplication of DNA during cell division?

A

Semi-conservative replication mechanism.

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42
Q

How does chromatin structure contribute to DNA preservation?

A

Chromatin condenses when genes are not active, protecting DNA from damage.

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43
Q

What are telomeres, and why are they important for DNA integrity?

A

Telomeres are repeating sequences of DNA at chromosome ends that shorten with age. They are replicated by a mechanism different from the rest of the chromosome to prevent loss of genetic material.

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44
Q

What role do centromeres play in DNA preservation?

A

Centromeres are repetitive sequences of DNA that are highly packaged to provide a robust attachment site to the mitotic spindle during cell division.

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45
Q

Why are telomeres and centromeres critical for chromosomal stability?

A

Telomeres prevent genetic material loss during replication, while centromeres ensure accurate chromosome segregation during cell division.

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46
Q

What happens to telomeres as organisms age?

A

Telomeres shorten with age, potentially impacting the cell’s ability to replicate its DNA accurately.

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47
Q

How does the nuclear envelope help preserve DNA integrity?

A

The nuclear envelope provides a protective environment for chromosomes, shielding DNA from damage caused by external factors such as harmful molecules or physical stress.

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48
Q

How does chromatin condensation preserve DNA?

A

Condensation of chromatin when genes are not active reduces exposure to enzymes and other factors that could damage or alter DNA, ensuring its stability.

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49
Q

How does semi-conservative replication preserve DNA integrity?

A

Semi-conservative replication ensures that each new DNA molecule retains one original strand, providing a template for accurate duplication and reducing errors during replication.

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50
Q

How do telomeres protect chromosome ends during replication?

A

Telomeres are repetitive DNA sequences at the ends of chromosomes that prevent the loss of essential genes during replication by acting as a buffer zone. They also protect DNA ends from being recognized as damage.

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51
Q

How does the unique replication mechanism of telomeres contribute to DNA integrity?

A

Telomeres are replicated by a specialized enzyme, telomerase, which helps maintain their length and prevents chromosome degradation over successive cell divisions.

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52
Q

What role do centromeres play in preserving DNA during cell division?

A

Centromeres serve as the attachment site for the mitotic spindle, ensuring accurate segregation of chromosomes during cell division and preventing loss or duplication of genetic material.

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53
Q

How do telomeres and centromeres together ensure chromosomal stability?

A

Telomeres prevent degradation at chromosome ends, while centromeres ensure proper alignment and separation during cell division, safeguarding the entire chromosome’s structure and function.

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54
Q

Why is chromatin structure critical for DNA preservation?

A

The tightly packed structure of chromatin protects DNA from enzymatic damage and mechanical stress, while also regulating access for replication and repair.

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55
Q

What four histones make up the core histone octamer in a nucleosome?

A

Two molecules each of H2A, H2B, H3, and H4 form the octamer core.

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56
Q

How many times does DNA wrap around the histone core, and how many nucleotides are involved in this wrap?

A

146 nucleotides of DNA wrap 1.65 times around the histone core to form the nucleosome.

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57
Q

Why are histones considered highly conserved across species?

A

Histones are considered highly conserved across species due to their essential role in DNA packaging and regulation of gene expression.

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58
Q

Why are histones considered highly conserved across species?

A

Histones have very similar amino acid sequences in different organisms, reflecting their fundamental role in DNA packaging and gene regulation.

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59
Q

What types of bonds help stabilize the interaction between DNA and the histone octamer?

A

Hydrogen bonds form between the DNA and the histone octamer, helping to maintain the structure of the nucleosome.

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60
Q

How are nucleosomes spaced along DNA, and what is the function of the DNA between nucleosomes?

A

Nucleosomes are separated from each other by up to 80 nucleotides of linker DNA.

The linker DNA connects one nucleosome to the next and provides sites where other proteins can interact or additional regulatory factors can bind.

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61
Q

What role does histone H1 play in nucleosome structure?

A

Histone H1 acts as a clamp, binding to the nucleosome where the DNA leaves and enters, thereby helping stabilize the higher-order structure of chromatin.

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62
Q

What is DNA packaging, and why is it important?

A

DNA packaging is the process of coiling and folding DNA into more compact forms. It is crucial for fitting large amounts of DNA into the limited space of the nucleus and for regulating access to DNA during processes such as transcription and replication.

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63
Q

What happens to the space DNA occupies as packaging becomes more complex?

A

The ‘room’ taken up by DNA gets smaller. However, as DNA becomes more tightly packed, it becomes harder for proteins to access the DNA for transcription and replication.

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64
Q

What does the ‘beads-on-a-string’ form of chromatin refer to?

A

This form represents DNA wrapped around histone protein cores (nucleosomes) connected by short segments of linker DNA, resembling ‘beads’ on a ‘string.’

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65
Q

What is a chromatin fiber of packed nucleosomes?

A

When nucleosomes coil and fold more tightly, they form a thicker, more compact fiber (often referred to as the 30 nm fiber). This additional packaging level further organizes the DNA, reducing its exposed length.

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66
Q

What happens when the chromatin fiber is folded into loops?

A

The 30 nm fiber forms looped domains attached to a scaffold, leading to an even higher level of compaction. These loops help regulate which sections of the DNA are more or less accessible to cellular machinery.

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67
Q

How does the fully condensed, mitotic chromosome relate to the earlier stages of packaging?

A

In preparation for cell division, the looped chromatin undergoes further coiling and compaction to form the fully condensed mitotic chromosomes visible under a microscope. This highest level of compaction ensures equal segregation of genetic material during mitosis.

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68
Q

Why is it harder for proteins to access DNA in highly compacted forms?

A

More compact structures shield DNA from proteins, reducing the likelihood that enzymes or transcription factors can bind. As a result, genes in tightly packed regions are often less actively transcribed or replicated.

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69
Q

How do cells overcome the challenge of tightly packaged DNA when they need to read or replicate it?

A

Cells use chromatin remodeling complexes and histone modifications (e.g., acetylation, methylation) to loosen specific DNA regions. These changes temporarily unwind or relax the chromatin structure to allow proteins to access genes for transcription or replication.

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70
Q

What is the function of DNA helicase in DNA replication?

A

DNA helicase unwinds the double helix by breaking hydrogen bonds between the complementary strands, creating the replication fork.

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71
Q

What role does DNA polymerase play during DNA replication?

A

DNA polymerase synthesizes new DNA strands by adding nucleotides complementary to the template strand. It also proofreads to correct errors in base-pairing.

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72
Q

Why is DNA primase necessary for replication?

A

DNA primase synthesizes short RNA primers that provide a 3ʹ hydroxyl group for DNA polymerase to start adding DNA nucleotides.

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73
Q

What does DNA ligase do in the replication process?

A

DNA ligase seals the nicks (breaks) in the sugar-phosphate backbone of the lagging strand, joining Okazaki fragments into a continuous strand.

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74
Q

How does topoisomerase aid in DNA replication?

A

Topoisomerase relieves the supercoiling and torsional strain that develop ahead of the replication fork by cutting and rejoining the DNA strands.

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75
Q

Why must DNA replication occur before cell division?

A

To ensure each new daughter cell inherits a complete, accurate copy of the genome. Without prior replication, cells would not have the correct genetic material to pass on.

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76
Q

How does the original DNA strand function during replication?

A

The original strand serves as a template to produce a complementary copy of the genomic DNA.

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77
Q

What is the first major step in DNA replication?

A

The double helix is unwound and separated into two strands, forming a replication fork. This step is essential for exposing the bases on each strand, allowing them to serve as templates.

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78
Q

What role does DNA primase play in the replication process?

A

DNA primase synthesizes short RNA primers that provide a starting point for DNA polymerase. On the leading strand, a single primer is typically needed to initiate replication.

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79
Q

How do free nucleotide bases participate in forming the new DNA strand?

A

Free nucleotide bases pair with the exposed template bases according to complementary base-pairing rules (A with T, G with C). DNA polymerase then joins these nucleotides together, synthesizing the new strand.

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80
Q

What happens on the lagging strand during replication?

A

On the lagging strand, DNA primase creates primers every 100–200 bp. DNA polymerase extends these primers, forming Okazaki fragments. DNA ligase then joins the fragments to create a continuous DNA strand.

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81
Q

Why is replication said to be ‘based upon the formation of base-pairs’?

A

DNA replication relies on complementary base pairing (A-T, G-C). Each strand of the original double helix acts as a template for generating its complementary strand.

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82
Q

What is the first step of DNA replication?

A

The double helix must be unwound and separated into two strands, generating a replication fork where the bases become exposed.

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83
Q

What is the second step of DNA replication?

A

DNA primase synthesizes short RNA primers, which provide a starting platform (3′ hydroxyl group) for DNA polymerase to begin DNA synthesis.

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84
Q

What is the third step of DNA replication?

A

DNA polymerase recognizes the exposed bases on the template strands and adds complementary nucleotides, building the new DNA strand in the 5′ to 3′ direction.

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85
Q

How does replication differ on the lagging strand (fourth step)?

A

On the lagging strand, DNA primase lays down multiple RNA primers every 100–200 nucleotides. DNA polymerase extends these segments (Okazaki fragments). DNA ligase then joins the fragments, forming one continuous strand.

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86
Q

What must happen before DNA replication can begin?

A

The double helix must be separated into two template strands. This is done by DNA helicase, which unwinds the helix.

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87
Q

How does DNA helicase separate the two strands of DNA?

A

DNA helicase binds to one strand of DNA and spins around it. Using energy from ATP hydrolysis, it breaks the hydrogen bonds between complementary bases.

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88
Q

What is the energy source for DNA helicase activity, and why is it needed?

A

ATP hydrolysis provides the energy. This energy is needed for DNA helicase to physically unwind the double helix and move along the DNA strand.

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89
Q

Why is the action of DNA helicase essential for replication?

A

By unwinding and separating the strands, it exposes the bases on each template strand. This exposure is required for the next steps in replication, where DNA polymerase can begin synthesizing new strands.

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90
Q

What happens to the single-stranded DNA immediately after the double helix is unwound?

A

Single-stranded binding (SSB) proteins bind to the separated DNA strands. This stabilizes the single strands and prevents the formation of secondary structures (e.g., hairpins).

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91
Q

Why are single-stranded binding proteins especially important for the lagging strand?

A

The lagging strand is synthesized in short fragments, leading to more frequent exposure of single-stranded DNA. SSB proteins maintain the strands in an extended form so replication enzymes can access them efficiently.

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92
Q

How do single-stranded binding proteins help replication proceed smoothly?

A

By preventing single-stranded DNA from re-annealing or forming loops, SSB proteins ensure that DNA polymerase and other replication factors can quickly bind and replicate the exposed strands.

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93
Q

What is ‘cooperative binding’ in the context of single-stranded binding (SSB) proteins?

A

Cooperative binding means that the binding of one SSB protein to single-stranded DNA increases the likelihood that additional SSB proteins will bind nearby. This cooperative effect helps stabilize the DNA in an extended conformation, preventing secondary structures and ensuring efficient replication.

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94
Q

What is the replication fork, and how is it formed?

A

The replication fork is a Y-shaped region created when the DNA double helix is unwound by helicase. This unwinding exposes the individual template strands, allowing replication enzymes to access and copy the DNA.

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95
Q

Why does DNA replication need an RNA primer?

A

DNA polymerases can only add nucleotides to an existing 3′ hydroxyl (3′-OH) group. RNA primers, made by DNA primase, provide the initial 3′-OH so that DNA polymerase can start synthesizing the new DNA strand.

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96
Q

How long are the RNA primers synthesized by DNA primase, and what do they do?

A

The RNA primers are typically about 10 nucleotides in length. They base-pair with the DNA template at the replication fork, creating a short stretch of double-stranded nucleic acid for DNA polymerase to bind and initiate DNA synthesis.

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97
Q

What is the role of DNA primase in the initiation stage of replication?

A

DNA primase recognizes the exposed bases at the replication fork. It then synthesizes the short RNA primers needed for DNA polymerase to begin elongation of the new DNA strand.

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98
Q

Why is this primer-initiated process referred to as ‘initiation’?

A

It’s called ‘initiation’ because laying down the RNA primer is the critical first step in starting the synthesis of the new DNA strand, enabling other replication proteins to join and continue the process.

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99
Q

What is the overall role of DNA polymerase in DNA replication?

A

DNA polymerase adds deoxyribonucleotides to the growing DNA strand, using the existing template strand and an RNA primer. It forms phosphodiester bonds between the 3′ hydroxyl (3′-OH) of the primer and the 5′ phosphate (5′-P) of the incoming nucleotide.

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100
Q

In which direction does the newly synthesized DNA strand grow?

A

The new DNA strand always elongates in the 5′ to 3′ direction. DNA polymerase attaches each new nucleotide to the 3′-OH end of the growing chain.

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101
Q

How does the incoming nucleotide supply energy for bond formation?

A

The incoming nucleotide is a deoxyribonucleoside triphosphate (dNTP) (e.g., dATP, dTTP, dCTP, dGTP). When a nucleotide is added, pyrophosphate (PPi) is released. The hydrolysis of pyrophosphate provides energy that drives the formation of the phosphodiester bond.

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102
Q

What ensures the correct nucleotide is added during DNA synthesis?

A

Complementary base pairing (A–T, G–C) ensures the correct match. DNA polymerase also has a ‘checking’ mechanism (often proofreading) that helps correct mispaired bases.

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103
Q

What is meant by ‘phosphodiester bonds’ in DNA synthesis?

A

A phosphodiester bond is the linkage between the 3′ hydroxyl group of the existing strand and the 5′ phosphate of the incoming nucleotide. It creates the sugar-phosphate backbone of DNA.

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104
Q

What is the significance of the polymerase ‘hand’ shape shown in the diagram?

A

DNA polymerase is often depicted as a right-handed shape. The ‘palm’ region is where the catalytic site resides, binding DNA and checking base pairing. The ‘fingers’ help position the incoming dNTP, and the ‘thumb’ helps secure the DNA in place.

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105
Q

How does DNA polymerase proceed along the template once a nucleotide is incorporated?

A

After each correct nucleotide is added and the bond is formed, DNA polymerase translocates forward. It repositions so the next 3′-OH is aligned in the active site, ready to bind the next incoming nucleotide.

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106
Q

What happens to pyrophosphate after it is released?

A

Pyrophosphate is typically hydrolyzed into two phosphate ions (Pi). This hydrolysis makes the DNA synthesis reaction effectively irreversible under cellular conditions.

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107
Q

Why does the lagging strand require multiple primers during replication?

A
  • DNA polymerase synthesizes DNA only in the 5′ to 3′ direction.
  • On the lagging strand, the replication fork opens in the opposite direction, so DNA primase must repeatedly lay down new RNA primers to initiate each Okazaki fragment.
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108
Q

What are Okazaki fragments, and how are they formed?

A
  • Okazaki fragments are short segments of newly synthesized DNA on the lagging strand.
  • DNA polymerase uses each new RNA primer as a starting point, extending the DNA until it reaches the previous primer.
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109
Q

Once an Okazaki fragment is completed, what happens to the RNA primer?

A
  • The RNA primer is removed (often by an exonuclease activity or specialized enzyme) and replaced with DNA by DNA polymerase.
  • This replacement ensures the final strand is entirely made of DNA.
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110
Q

How do the individual Okazaki fragments become one continuous strand?

A
  • DNA ligase forms a phosphodiester bond between the adjacent 3′ hydroxyl (from the newly synthesized DNA) and the 5′ phosphate (remaining from the replaced primer region).
  • This “glues” all the fragments together into a continuous strand.
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111
Q

Why is the lagging strand referred to as ‘lagging’?

A
  • Because it’s synthesized in a discontinuous manner, in short segments (Okazaki fragments), it lags behind the continuous synthesis on the leading strand.
  • Nonetheless, both strands complete replication around the same time.
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112
Q

What is the significance of ‘3′ to 5′ direction’ limitation for DNA polymerase?

A
  • DNA polymerase cannot synthesize DNA in the 3′ to 5′ direction.
  • Therefore, the lagging strand must adopt a discontinuous approach, repeatedly using primers to allow 5′ to 3′ extension in small sections.
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113
Q

How does the process on the lagging strand ensure fidelity (accuracy) despite multiple steps?

A
  • DNA polymerase has proofreading capability, removing incorrectly paired bases.
  • Enzymes that remove RNA primers and DNA ligase also check and correct any structural or bonding errors during fragment joining.
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114
Q

Why are single-stranded binding (SSB) proteins necessary during DNA replication?

A
  • SSB proteins prevent secondary structures (such as hairpin loops) from forming in single-stranded DNA.
  • By keeping the strands in an extended conformation, they ensure replication enzymes have unobstructed access.
115
Q

How do SSB proteins help maintain an ‘open’ DNA structure?

A
  • SSB proteins bind cooperatively to single-stranded DNA, covering exposed regions so they can’t fold back on themselves.
  • This keeps the DNA straightened and prevents base-pairing within the same strand.
116
Q

What role do RNA primers play in the formation of secondary structures, and how do SSB proteins mitigate this?

A
  • RNA primers might inadvertently enhance base-pairing between complementary regions on the same strand.
  • SSB proteins override this tendency by continuously binding along the ssDNA, blocking such intramolecular pairing.
117
Q

Why is preventing secondary structures critical for replication fidelity?

A
  • Secondary structures can stall or impede the progress of DNA polymerase and other replication enzymes.
  • By avoiding these structures, SSB proteins help maintain a smooth, efficient replication process, reducing errors.
118
Q

What does the term ‘cooperative binding’ mean in the context of SSB proteins?

A
  • It means once an SSB protein binds to DNA, it increases the likelihood that additional SSB proteins will bind nearby.
  • This cooperative effect ensures rapid coverage of ssDNA regions and reinforces a stable, open conformation.
119
Q

What does it mean that replication forks are ‘bidirectional’?

A
  • From each origin of replication, two replication forks form and move in opposite directions.
  • This allows DNA to be copied more quickly than if replication proceeded in only one direction.
120
Q

Why is each replication fork described as having an ‘asymmetric’ structure?

A
  • The leading strand is synthesized continuously in the 5′ to 3′ direction.
  • The lagging strand is synthesized discontinuously as Okazaki fragments, because DNA polymerase can only add nucleotides in the 5′ to 3′ direction.
121
Q

What is the leading strand, and how is it synthesized?

A
  • The leading strand is the newly made DNA strand that can be elongated in the same direction as the replication fork opens.
  • DNA polymerase synthesizes it continuously, adding nucleotides as the fork progresses.
122
Q

What is the lagging strand, and why does it form Okazaki fragments?

A
  • The lagging strand is the DNA strand oriented in the opposite direction of the fork’s opening.
  • Because DNA polymerase must work 5′ to 3′, it synthesizes short segments (Okazaki fragments) that are later joined by DNA ligase.
123
Q

Which accessory proteins help maintain the structure of replication forks?

A
  • Key accessory proteins include the sliding clamp, which secures DNA polymerase to the DNA, and topoisomerase, which relieves supercoiling and torsional stress ahead of the replication fork.
124
Q

How does the sliding clamp enhance DNA replication efficiency?

A
  • The sliding clamp encircles the DNA and holds DNA polymerase in place, preventing it from dissociating.
  • This increases processivity, allowing polymerase to synthesize longer stretches of DNA without stopping.
125
Q

Why is topoisomerase important at replication forks?

A
  • As the replication fork advances, twisting tension (supercoiling) accumulates ahead of the fork.
  • Topoisomerase temporarily cuts one or both DNA strands, relieves this tension, then re-ligates the DNA to allow replication to proceed smoothly.
126
Q

What is the sliding clamp, and why is it needed in DNA replication?

A
  • The sliding clamp is a ring-shaped protein complex that encircles the DNA.
  • It keeps DNA polymerase firmly attached to the DNA template, enabling the polymerase to synthesize long stretches of DNA without falling off (enhancing processivity).
127
Q

What problem arises if DNA polymerase is not attached by the sliding clamp?

A
  • DNA polymerase can only stay bound to DNA for 50–200 nucleotides before dissociating.
  • The sliding clamp prevents this premature dissociation, ensuring continuous and efficient DNA synthesis.
128
Q

How does the sliding clamp get onto the DNA?

A
  • A clamp loader complex (which binds and hydrolyzes ATP) opens the sliding clamp ring.
  • The loader then threads the clamp around the DNA strand, and upon ATP hydrolysis, locks the clamp shut and releases it onto DNA.
129
Q

What is the function of the clamp loader, and how does it operate?

A
  • The clamp loader binds to both parts of the sliding ring.
  • It controls the opening/closing of the ring using energy from ATP hydrolysis.
  • After loading the clamp onto DNA, it detaches, leaving the clamp in place.
130
Q

How does the sliding clamp interact with DNA polymerase?

A
  • One segment of the ring binds to the back of DNA polymerase.
  • This physical attachment prevents the polymerase from sliding off the DNA and thus maintains contact with the template strand.
131
Q

Why is the clamp complex described as having three parts?

A
  • The ring itself is formed by two subunits that hinge open and close.
  • The clamp loader is the third component, which attaches to these subunits to load and unload the clamp around DNA.
132
Q

What is the initial action of the clamp loader in the sliding clamp mechanism?

A
  • The clamp loader binds ATP, which causes it to open the sliding clamp ring.
  • This allows the DNA strand to be threaded through the opening of the clamp.
133
Q

How does the clamp loader position the sliding clamp onto DNA?

A
  • With the ring open, the clamp loader moves the clamp onto the single-stranded DNA region (adjacent to where DNA synthesis will occur).
  • The clamp loader precisely aligns the clamp so it encircles the DNA in the correct orientation.
134
Q

What triggers the clamp loader to release the clamp onto the DNA?

A
  • ATP hydrolysis triggers the clamp loader to close the clamp ring around the DNA and release the clamp from the loader.
  • This ensures the clamp is now securely latched around the DNA.
135
Q

Once the sliding clamp is on the DNA, how does DNA polymerase attach?

A
  • DNA polymerase binds to the back face of the sliding clamp.
  • This attachment locks the polymerase in close proximity to the DNA, allowing highly processive DNA synthesis.
136
Q

What happens to the clamp loader after it has placed the clamp?

A
  • After releasing the clamp, the clamp loader is free to pick up another clamp (if necessary).
  • It can repeat the cycle whenever a new primer or Okazaki fragment requires the sliding clamp.
137
Q

What happens when ATP binds to the clamp loader in the first step?

A
  • ATP binding causes the clamp loader to open the sliding clamp.
  • The open ring of the clamp can then be threaded around the DNA strand.
138
Q

How does the clamp loader move the opened sliding clamp onto the DNA?

A
  • With the clamp ring held open, the clamp loader positions it so that the DNA helix passes through the opening.
  • This step aligns the clamp in the correct orientation for replication.
139
Q

What triggers the clamp loader to release the clamp?

A
  • ATP hydrolysis (splitting ATP into ADP and Pi) triggers the clamp loader to close the sliding clamp around the DNA.
  • The clamp loader then releases from the clamp, leaving it locked in place.
140
Q

After the clamp is locked onto DNA, how does DNA polymerase interact with the sliding clamp?

A
  • DNA polymerase binds to the outer surface of the clamp.
  • This ensures the polymerase remains firmly attached to the DNA, improving its processivity.
141
Q

What is meant by the ‘recycling’ of the released clamp loader?

A
  • Once the clamp loader has closed the clamp and detached, it can bind another clamp.
  • The clamp loader then repeats the cycle whenever a new clamp is needed on the lagging strand or for other replication tasks.
142
Q

Why does the unwinding of DNA during replication cause torsional stress?

A
  • As the replication fork moves and the two strands are pulled apart, the tightly coiled DNA ahead of the fork cannot rotate freely.
  • This leads to a buildup of torsional (twisting) strain, which can kink or supercoil the DNA.
143
Q

What role does topoisomerase play in preventing replication ‘catastrophe’?

A
  • Topoisomerase relieves the torsional stress by creating temporary nicks (cuts) in one or both strands of the DNA helix.
  • After unwinding or relaxing the supercoils, it re-seals the breaks, allowing replication to continue smoothly.
144
Q

What would happen if torsional stress is not alleviated?

A
  • Excessive stress can lead to DNA breaks, stalling of the replication fork, or entanglement of the DNA strands.
  • Without topoisomerase, replication could be severely hindered or the DNA could become damaged.
145
Q

How does topoisomerase differ from helicase?

A
  • Helicase unwinds the DNA helix by breaking hydrogen bonds between bases, separating the two strands.
  • Topoisomerase manages the torsional strain caused by this unwinding.
  • Both are essential, but they address different replication challenges.
146
Q

Why can topoisomerase be considered a ‘relaxation enzyme’?

A
  • By cutting, twisting, and re-ligating DNA strands, topoisomerase effectively ‘relaxes’ the supercoiled regions.
  • This prevents knots and tangles, allowing the replication machinery to work uninterrupted.
147
Q

Why does DNA unwinding during replication generate torsional stress?

A
  • As the replication fork progresses, the DNA helix ahead of the fork becomes tightly twisted.
  • If it cannot freely rotate, torsional stress (supercoiling) builds up, potentially halting replication.
148
Q

How does topoisomerase alleviate torsional stress?

A
  • Topoisomerase cuts one or both DNA strands, relieves the supercoil by allowing rotation, and then reseals the break.
  • This prevents the DNA from becoming overly wound or tangled.
149
Q

What is the difference between Type I and Type II topoisomerases?

A
  • Type I Topoisomerases: Create a single-strand nick, allowing the helix to unwind (rotate) around that nick.
  • Type II Topoisomerases: Create a double-strand break, allowing one helix to pass through another; they’re essential when two double helices cross.
150
Q

How do Type I topoisomerases work at the molecular level?

A
  • They form a covalent bond with one strand of DNA, break its phosphodiester bond, and allow the strand to rotate around the uncut strand.
  • After unwinding, they reseal the nick, restoring DNA integrity.
151
Q

How do Type II topoisomerases manage more complex DNA crossings?

A
  • They bind to and cut both strands of one DNA helix, forming covalent bonds with them.
  • Another DNA helix can then pass through the break, resolving knots or tangles.
  • The enzyme finally re-ligates the cut strands.
152
Q

Why is topoisomerase often called the ‘catastrophe preventer’ during replication?

A
  • Without topoisomerase, excessive supercoiling would cause replication forks to stall, break, or collapse.
  • By relieving these strains, topoisomerase ensures smooth replication progression.
153
Q

How does topoisomerase restore DNA integrity after relieving torsional stress?

A
  • After creating a controlled “nick” (or break) in one or both DNA strands and allowing rotation, topoisomerase then re-ligates (reseals) the broken phosphodiester bond(s).
  • This “ligation” step ensures the DNA double helix is returned to a continuous, intact form once the strain is relieved.
  • topoisomerase has its own ligation activity
154
Q

Where does DNA replication start?

A

It starts at specific DNA sequences called origins of replication (ori).

155
Q

Where are these origins of replication located in eukaryotes?

A

Eukaryotic chromosomes have multiple origins of replication dispersed either side of the centromere. These sites are recognized by the origin recognition complex (ORC), which recruits other factors to initiate replication.

156
Q

Why do eukaryotic chromosomes contain multiple origins of replication?

A

Eukaryotic chromosomes are large, so having multiple origins allows simultaneous replication at multiple sites. This significantly reduces the total time needed to replicate the entire genome.

157
Q

How many replication origins can be found in a eukaryotic genome, and do all cell types use the same origins?

A

There can be thousands of replication origins per chromosome. Different cell types can activate different sets of origins, allowing for specialized or regulated replication programs.

158
Q

What advantage do ‘many potential origins’ provide to the cell?

A

They permit complex regulation of replication timing, ensuring certain regions replicate first or later. They also act as a back-up system in case some origins fail or are blocked.

159
Q

How do replication forks move once they are initiated at these origins?

A

Replication forks typically form in pairs at each origin. Each fork moves in opposite directions along the chromosome, thereby replicating the DNA bidirectionally.

160
Q

What is the relationship between replication and the cell cycle in eukaryotes?

A

DNA replication occurs during the S phase (synthesis phase) of the cell cycle. This timing is tightly synchronized with other cell cycle events, ensuring chromosomes are fully replicated before cell division.

161
Q

In the diagram, what do the colored bands on the chromosomes represent?

A

The blue regions at the ends are telomeres. The red region in the center is the centromere. The yellow marks indicate replication origins where the replication forks initiate.

162
Q

Why do the replicated chromosomes appear ‘double’ after S phase but before mitosis?

A

Each chromosome has been copied into two sister chromatids, held together at the centromere. They remain attached until they are separated during mitosis, ensuring each daughter cell inherits one chromatid (complete copy).

163
Q

Why do eukaryotic chromosomes have multiple replication origins?

A

Eukaryotic chromosomes are large, so having multiple origins lets replication start at many points simultaneously, dramatically speeding up the overall replication process.

164
Q

How many replication origins can a eukaryotic chromosome have, and do all cells use the same ones?

A

There can be thousands of replication origins. Different cell types or conditions can activate different subsets of these origins, allowing for complex regulation and ensuring flexibility in the replication program.

165
Q

What advantages do multiple ‘potential origins’ offer to eukaryotic cells?

A

They permit complex regulation of replication timing (certain regions can be replicated earlier or later). They serve as a backup system if some origins fail to fire or become blocked, ensuring complete genome duplication.

166
Q

How do replication forks move once initiated at each origin?

A

Two forks form at each origin (they move in opposite directions), creating bidirectional replication. The forks typically move at the same speed, working in pairs to replicate DNA efficiently.

167
Q

Why is replication said to be synchronized with the cell cycle?

A

DNA replication occurs during the S phase (synthesis phase) of the eukaryotic cell cycle. This ensures the genome is fully replicated before the cell proceeds into mitosis (cell division).

168
Q

What does the diagram show regarding interphase and mitosis?

A

Interphase: Replication origins (yellow) fire, creating replication forks. The chromosome eventually becomes fully duplicated, forming sister chromatids. Mitosis: The sister chromatids are separated and equally distributed to daughter cells.

169
Q

What are the key labeled regions on the chromosome in the diagram?

A

Telomeres (blue): Protective ends of chromosomes. Replication origins (yellow): Sites where replication begins. Centromere (red): The region where sister chromatids connect and where the mitotic spindle attaches during cell division.

170
Q

How do these multiple replication origins affect the overall replication process?

A

Multiple origins enable multiple replication forks to work simultaneously. This reduces total replication time and ensures complete replication within the timeframe of S phase before cell division.

171
Q

What does the slide illustrate about eukaryotic DNA replication origins?

A

Eukaryotic chromosomes have multiple replication origins per chromosome, each giving rise to two replication forks that progress bidirectionally. This setup allows large eukaryotic chromosomes to be fully replicated in a reasonable amount of time.

172
Q

How do prokaryotes differ from eukaryotes in terms of replication origins?

A

Prokaryotes (such as bacteria) typically have one replication origin on their circular chromosome. Eukaryotes have linear chromosomes and many replication origins distributed along each chromosome.

173
Q

Why are multiple replication origins necessary in eukaryotes?

A

Eukaryotic genomes are much larger and more complex, so multiple origins shorten the total replication time. Each origin creates two forks moving in opposite directions, allowing simultaneous replication along the length of the chromosome.

174
Q

What is the significance of having only one origin in prokaryotes?

A

Prokaryotic chromosomes are typically smaller and circular, making a single origin sufficient. The replication forks move around the circle and meet on the opposite side, completing replication efficiently for smaller genomes.

175
Q

How does the cell cycle coordination differ between prokaryotes and eukaryotes?

A

Eukaryotes replicate DNA only during the S phase of the cell cycle and have checkpoints to ensure accurate replication before mitosis. Prokaryotes do not have a defined S phase; replication can be continuous or overlapping with cell division, depending on growth conditions.

176
Q

What are other key differences between eukaryotic and prokaryotic replication?

A

Chromosome Structure: Eukaryotes have linear chromosomes with telomeres; prokaryotes have circular chromosomes without telomeres. Speed & Complexity: Eukaryotic replication forks are often slower and involve more complex protein machinery (e.g., multiple polymerases, histone handling). Prokaryotes use a simpler set of proteins, but replication rates can be faster due to fewer regulatory hurdles.

177
Q

Why is replication considered ‘synchronized’ with the cell cycle in eukaryotes?

A

In eukaryotes, replication occurs during S phase and must finish before the cell enters mitosis. This ensures that each daughter cell inherits a complete set of chromosomes once cell division occurs.

178
Q

In the diagram, what are the labels (telomere, replication origin, centromere) highlighting about eukaryotic chromosomes?

A

Telomeres: Protective ends of linear chromosomes. Replication origins: Multiple sites where replication begins. Centromere: The region for sister chromatid attachment and spindle binding during mitosis.

179
Q

What is the primary role of DNA polymerase α (Pol α) in eukaryotic DNA replication?

A

DNA polymerase α forms a multi-subunit complex where one subunit has primase activity. It synthesizes a short RNA–DNA hybrid primer to initiate new DNA strands on both the leading and lagging strands. After laying down this primer, Pol α typically hands off synthesis to the main replicative polymerases (e.g., Pol δ).

180
Q

How does DNA polymerase δ (Pol δ) function during replication, especially on the lagging strand?

A

Pol δ is the major lagging-strand polymerase in eukaryotes. It extends from the RNA–DNA primer laid down by Pol α, synthesizing Okazaki fragments (up to ~200 bp in length). Pol δ associates with the sliding clamp (PCNA), improving its processivity so it can replicate long stretches of DNA without dissociating.

181
Q

What is DNA polymerase η (Pol η), and why is it important for genomic stability?

A

Pol η (eta) is a translesion synthesis (TLS) polymerase that can bypass certain DNA lesions, especially UV-induced thymine dimers, without stalling the replication fork. By performing relatively accurate bypass of these lesions, Pol η prevents mutations and replication collapse. A defect in Pol η leads to xeroderma pigmentosum variant (XP-V), highlighting its role in protecting against UV-induced DNA damage.

182
Q

What is 3′→5′ exonucleolytic proofreading in DNA replication, and why is it important?

A

3′→5′ exonucleolytic proofreading is the mechanism by which many DNA polymerases can reverse direction and remove a wrongly inserted nucleotide. This proofreading corrects base-pairing errors immediately as they occur, reducing mutation rates and ensuring high-fidelity DNA replication.

183
Q

Which structures in eukaryotic chromosomes prevent the loss of genome integrity?

A

Centromeres: Ensure proper segregation of chromosomes during cell division. Telomeres: Protect the ends of linear chromosomes from degradation and fusion.

184
Q

Which enzymes are crucial in preventing genome instability, and what do they do?

A

DNA Polymerase (with proofreading): Detects and corrects base-pairing errors during replication. Topoisomerase: Relieves supercoiling and torsional stress as the DNA helix unwinds.

185
Q

What is one key mechanism that helps maintain genomic integrity?

A

DNA Repair Pathways (e.g., mismatch repair, excision repair): Constantly scan for and fix DNA damage or replication errors, preventing mutations and preserving the genome.

186
Q

Why is the centromere region more about chromatin structure rather than a single DNA sequence?

A

In higher eukaryotes, there is no strict consensus sequence for the centromere. Instead, epigenetic factors (e.g., specialized histones like CENP-A) and repetitive DNA (e.g., α-satellite repeats) define the centromere’s identity and function.

187
Q

How large can centromere regions be in complex eukaryotes, and what are they typically composed of?

A

Centromeric regions can span several million base pairs in length. They often contain repetitive DNA called α-satellite DNA, which assembles into higher-order repeat structures.

188
Q

What is α-satellite DNA, and why is it important for centromere function?

A

α-satellite DNA consists of repeating units (171 bp monomers) arranged in large tandem arrays. This repetitive sequence forms a foundation for centromeric chromatin and helps recruit kinetochore proteins essential for chromosome segregation.

189
Q

How do centromeres ensure proper chromosome segregation during cell division?

A

Centromeres are the binding site for kinetochore complexes, which attach to microtubules. This ‘strong attachment’ enables sister chromatids to be pulled apart accurately during mitosis and meiosis.

190
Q

Why is the centromeric region often considered ‘heterochromatin’?

A

The DNA in centromeres is tightly packed with distinctive histone modifications and repetitive sequences. This heterochromatic state helps stabilize centromere function and maintains the specialized kinetochore structure.

191
Q

What happens when replication forks converge and finish replicating most of the chromosome?

A

As forks ‘run into one another,’ the helicases become ubiquitinated, leading to their rapid dissociation. DNA ligase and DNA polymerase then fill in any remaining gaps to complete replication.

192
Q

Why must replication stop at the end of a chromosome, and what is this region called?

A

Linear chromosomes terminate in specialized repetitive regions called telomeres. The normal replication machinery cannot fully replicate the very ends, so telomeres protect chromosome ends and prevent the cell from mistaking them for DNA damage.

193
Q

What is the general makeup of telomeres in humans, and how do they help the cell?

A

Telomeres consist of many tandem repeats of a short sequence (in humans, it’s typically TTAGGG repeated). These repeats form a protective ‘cap’ that stops the cell from recognizing the chromosome end as a DNA break, thus preventing unnecessary repair responses.

194
Q

How do telomeres form a T-loop, and why is it significant?

A

The 3′ overhang folds back on itself to form a T-loop, which helps protect the chromosome end from degradation and prevents the activation of DNA damage responses.

195
Q

What is the general makeup of telomeres in humans, and how do they help the cell?

A
  • Telomeres consist of many tandem repeats of a short sequence (in humans, it’s typically TTAGGG repeated).
  • These repeats form a protective “cap” that stops the cell from recognizing the chromosome end as a DNA break, thus preventing unnecessary repair responses.
196
Q

How do telomeres form a T-loop, and why is it significant?

A
  • The 3′ overhang folds back and invades the double-stranded telomeric DNA, creating a T-loop.
  • This loop masks the chromosome end, further safeguarding it from being labeled damaged and helping maintain chromosome stability.
197
Q

Which enzyme recognizes telomere sequences and helps maintain telomere length?

A
  • Specific DNA-binding proteins recognize telomere repeats and recruit the enzyme telomerase.
  • Telomerase, a specialized reverse transcriptase, extends the 3′ end of telomeres, compensating for the incomplete replication and preventing gradual telomere shortening.
198
Q

What happens when two replication forks meet (“run into one another”) near the end of replication?

A
  • The helicases on those forks are tagged with ubiquitin, which causes them to dissociate rapidly.
  • After helicase dissociation, DNA ligase and DNA polymerase fill any remaining gaps in the newly synthesized DNA.
199
Q

Once forks converge and helicases dissociate, how are the final stretches of DNA replication completed?

A
  • DNA polymerase extends the remaining short segments, and DNA ligase seals the nicks in the sugar-phosphate backbone.
  • This process ensures the chromosome is completely replicated, except for the extreme ends (the telomeres).
200
Q

Why must replication machinery also stop at the chromosome ends, and what is this region called?

A
  • Linear chromosomes have ends known as telomeres.
  • The standard replication enzymes cannot fully copy these very ends, so the cell relies on special mechanisms (like telomerase) to maintain them.
201
Q

What are telomeres typically composed of in humans, and why is this repetitive structure important?

A
  • Human telomeres consist of many tandem repeats of a short sequence (often cited as GGGTT(A) or TTAGGG).
  • These repeats prevent the cell from mistaking the natural chromosome end for DNA damage, thus avoiding erroneous repair events.
202
Q

How does the cell prevent the very end of the chromosome from being recognized as broken DNA?

A
  • The telomeric repeats form a protective “cap” (sometimes looping back on itself as a T-loop) so that the chromosome end is shielded and not flagged as a DNA break.
203
Q

Which proteins recognize telomeric DNA, and what do they do next?

A
  • Sequence-specific DNA-binding proteins bind the repeated telomeric sequences.
  • They recruit telomerase, the enzyme that can add more telomeric repeats to the 3′ overhang, thereby compensating for the incomplete replication of chromosome ends.
204
Q

Why is telomerase crucial for chromosome-end maintenance?

A
  • Normal DNA polymerases cannot fully replicate the 5′ ends of linear DNA (the “end-replication problem”).
  • Telomerase extends the 3′ overhang by adding repeated sequences, preventing progressive shortening of chromosomes over multiple cell divisions.
205
Q

Why can’t the final RNA primer on the lagging strand be replaced by DNA polymerase at the very end of a chromosome?

A

Because there is no upstream primer to provide the necessary 3′-OH group for DNA polymerase.
* Standard DNA replication machinery therefore cannot fill in that last stretch of DNA on the lagging strand, which would otherwise result in chromosome shortening each cell cycle.

206
Q

How does telomerase solve this end-replication problem?

A

Telomerase binds to the 3′ overhang at the chromosome end.
* It carries its own RNA template, which it uses for RNA-templated DNA synthesis, extending the 3′ end in a 5′→3′ direction and providing additional sequence to protect the end of the chromosome.

207
Q

What is the essential enzymatic activity of telomerase, and how does it function?

A

Telomerase has a reverse transcriptase domain (often called TERT) that synthesizes DNA from its internal RNA.
* By repeatedly adding short telomeric repeats (e.g., “TTGGGG” or “TTAGGG” in humans) to the 3′ end, telomerase extends the chromosome end without needing a conventional RNA primer.

208
Q

Once telomerase extends the 3′ end, how is the complementary strand (on the lagging side) completed?

A

The newly elongated 3′ end provides a template that can then be filled in by standard DNA polymerases (using a new RNA primer made by primase, if necessary).
* DNA ligase seals any remaining nicks, ensuring both strands of the telomere are replicated.

209
Q

Why is telomerase activity crucial for cells with linear chromosomes?

A

Each round of replication would otherwise leave the chromosome ends shorter, eventually threatening the integrity of coding regions.
* Telomerase maintains telomere length, delaying or preventing the senescence that occurs when telomeres become too short.

210
Q

What does the diagram illustrate about the sequential steps of telomerase action?

A
  1. Telomerase binds to the 3′ overhang of the telomere.
  2. It extends the 3′ end using its RNA template (reverse transcription).
  3. DNA polymerase later fills in the complementary (lagging) strand.
  4. This process is repeated as needed, preserving telomere length.
211
Q

How does the cell recognize the newly formed telomeric repeats and prevent unwanted DNA repair at the chromosome ends?

A

Telomere-binding proteins (and associated complexes) specifically bind the repeated telomeric sequences.
* These proteins form a protective cap (sometimes a T-loop structure), preventing the cell’s DNA damage response from mistakenly treating chromosome ends as double-strand breaks.

212
Q

Why must DNA polymerase have proofreading ability when copying DNA?

A

Proofreading ensures a high-fidelity copy of the original DNA.
* It detects and corrects base-pairing errors immediately as they occur, thereby dramatically reducing the mutation rate.

213
Q

What are the two discrete sites in DNA polymerase, and what are their functions?

A

P site (Polymerizing site): Adds nucleotides to the growing DNA strand based on the template sequence.
* E site (Editing site): Detects and removes incorrectly paired nucleotides via exonuclease activity.

214
Q

How does the DNA polymerase editing (E) site remove an incorrectly paired nucleotide?

A

The E site’s exonuclease activity breaks the phosphodiester bond of the mismatched nucleotide.
* After removal, the polymerase “slides” the DNA backbone back to the P site, where the correct nucleotide can be inserted.

215
Q

What does the figure show about the polymerizing (P) and editing (E) sites?

A

The P site is where the incoming dNTP is added to the 3′ end of the newly synthesized strand.
* The E site is spatially separate but connected; when an error is detected, the DNA shifts to the E site for excision of the mismatch before returning to the P site to continue synthesis.

216
Q

Why is it beneficial for DNA polymerase to have both polymerizing and editing sites in the same enzyme?

A

This integrated proofreading allows immediate error correction, enhancing the overall accuracy of replication.
* It saves time and resources by preventing errors from going undetected and potentially requiring more extensive repair later.

217
Q

What happens if proofreading during DNA replication fails to remove a mismatched base?

A

An incorrectly paired base (mismatch) remains in the DNA sequence.
* This mismatch is subsequently detected by the cell’s repair mechanisms, which then attempt to correct it.

218
Q

What determines which repair pathway is used to fix a DNA error?

A

The type of error (e.g., mismatched base, UV-induced dimer, deaminated base) influences which specific pathway is activated.
* Cells have multiple, specialized DNA repair mechanisms.

219
Q

What are some common DNA repair mechanisms, and what errors do they address?

A
  1. Nucleotide Excision Repair (NER): Removes bulky lesions like pyrimidine dimers caused by UV light.
  2. Base Excision Repair (BER): Corrects single-base errors (e.g., deamination of cytosine to uracil).
  3. Mismatch Repair (MMR): Fixes errors that escape replication proofreading (e.g., replication errors leading to mismatched bases).
220
Q

Why is DNA repair crucial for maintaining genome integrity?

A

Unrepaired mismatches or lesions can lead to mutations.
* Over time, these mutations can accumulate and potentially cause cell dysfunction, disease, or cancer.

221
Q

How does mismatch repair (MMR) specifically address replication errors?

A

MMR enzymes scan newly synthesized DNA for mispaired bases.
* Once a mismatch is found, the incorrect base is excised and replaced with the correct nucleotide, using the parental strand as the template.

222
Q

What is a mismatch error, and how does it occur?

A

A mismatch error happens when an incorrect nucleotide is incorporated opposite the template strand during replication (e.g., a G mistakenly paired with T).
* If not corrected by proofreading or Mismatch Repair (MMR), it becomes a permanent base substitution in the DNA.

223
Q

What are pyrimidine dimers, and what typically causes them?

A

Pyrimidine dimers form when UV light induces a covalent bond between adjacent pyrimidine bases (often two thymines, called a “thymine dimer”).
* This bulky lesion distorts the DNA helix and can block replication if not removed by Nucleotide Excision Repair (NER).

224
Q

What is deamination, and which repair pathway usually fixes it?

A

Deamination is a chemical reaction where an amine group is removed from a base (e.g., cytosine can become uracil).
* Base Excision Repair (BER) typically corrects these single-base alterations by removing the damaged base and restoring the correct one.

225
Q

Why are these errors (mismatches, dimers, deaminations) critical to fix?

A

Each type of error, if left uncorrected, can lead to variants—changes in the DNA sequence that may alter gene function.
* Accumulated variants contribute to genetic diseases, cancer, or loss of cellular viability.

226
Q

Which repair mechanisms address these specific errors?

A

Mismatch Repair (MMR): Fixes replication errors like mismatched pairs.
* Nucleotide Excision Repair (NER): Removes bulky lesions such as pyrimidine dimers.
* Base Excision Repair (BER): Corrects single-base modifications (e.g., deaminated bases, oxidized or alkylated bases).

227
Q

What is depurination and how does it affect DNA?

A

Depurination is the loss of a purine base (adenine or guanine) from the DNA backbone.
* This creates an “abasic” site (missing a base), which can lead to a base deletion in the daughter strand after replication.

228
Q

Why can depurination lead to a loss of a nucleotide pair in the replicated DNA?

A

When the template strand is missing a base (due to depurination), DNA polymerase may skip it during replication.
* This results in one less base pair in the newly synthesized strand—a deletion variant.

229
Q

What types of effects can a deletion caused by depurination have on a protein sequence?

A

A frameshift variant could occur if the deleted base is within a coding region.
* This frameshift might lead to:
* A premature stop codon (truncating the protein).
* A drastic change in the amino acid sequence from that point onward.
* Possibly altered splicing signals if the deletion affects regulatory regions.

230
Q

Is it possible that a depurination event has no significant effect on the protein?

A

Yes, in rare cases:
1. If the deletion occurs in a non-coding region with no regulatory effect.
2. If it’s in a region where a frameshift or changed codon doesn’t affect overall protein function.
* However, most often it’s detrimental if it happens in a critical coding or regulatory region.

231
Q

Why might depurination-induced base loss be “unrepairable” in some instances?

A

If the DNA repair machinery does not recognize or fix the abasic site before replication, the polymerase skips that position.
* Once replication occurs, the missing base is “locked in” as a permanent deletion.

232
Q

Which DNA repair mechanism typically handles abasic (AP) sites if detected in time?

A

Base Excision Repair (BER) can recognize an AP site.
* An AP endonuclease can cut the backbone, and DNA polymerase + ligase fill and seal.
* But if replication happens first, the deletion is propagated.

233
Q

What happens if the DNA backbone is broken in only one strand?

A

A single-strand break (SSB) occurs, which is usually repaired by dedicated single-strand break repair mechanisms.

234
Q

What if both DNA strands are broken?

A

A double-strand break (DSB) forms. Cells must use either non-homologous end joining (NHEJ) or homologous recombination (HR) to fix it.

235
Q

What is non-homologous end joining (NHEJ)?

A

An error-prone DSB repair method where the broken ends of DNA are directly ligated without using a template.

236
Q

Why is NHEJ considered more error-prone?

A

Processing of the DNA ends can remove nucleotides, leading to small deletions.

Often used outside S-phase when a sister chromatid is not available as a template.

237
Q

How does homologous recombination (HR) differ from NHEJ?

A

HR uses a homologous DNA sequence (often the sister chromatid) as a template, making it more accurate than NHEJ.

238
Q

Why is HR considered the less error-prone pathway?

A

By using a matching chromosome or sister chromatid, it can resynthesize the correct sequence without losing nucleotides.

239
Q

Why is double-strand break repair critical for cells?

A

DSBs are highly dangerous lesions. If not repaired, or repaired incorrectly, they can cause genome instability, chromosome rearrangements, or cell death.

240
Q

When do cells typically use each repair pathway?

A

NHEJ: Common in all cell-cycle phases but predominantly outside of S-phase (no sister chromatid).

HR: Generally occurs in S and G2 phases when a homologous template (sister chromatid) is available.

241
Q

What repair mechanism is initiated if only one strand of the DNA helix is broken?

A

Single Strand Break Repair is initiated for single-strand breaks (details not covered on this slide).

242
Q

What happens when both strands of the DNA helix are broken?

A

Double Strand Break Repair is initiated, requiring either Non-Homologous End Joining (NHEJ) or Homologous Recombination (HR).

243
Q

What is Non-Homologous End Joining (NHEJ) and when does it typically occur?

A

NHEJ directly rejoins the broken ends and is more error-prone.

It commonly occurs outside of S-phase, where no sister chromatid is available as a template.

244
Q

What is Homologous Recombination (HR) and what does it use as a template?

A

HR is a more accurate repair pathway that uses a second, homologous chromosome (or sister chromatid) as a template.

This approach avoids losing nucleotides at the break site.

245
Q

When does crossing over occur during meiosis?

A

Crossing over occurs during Prophase I of meiosis.

246
Q

What do double-stranded breaks allow during meiosis?

A

Double-stranded breaks allow the translocation of DNA from one homologous chromosome to the other, generating chiasma.

247
Q

What is the result of the formation of chiasma in gametes?

A

When gametes are formed, chiasma contain genetic sequences from both parents, contributing to genetic diversity.

248
Q

Does crossing over always occur at the end of a gene?

A

No, crossing over between chromosomes does not always occur at the end of a gene. However, it is more likely to occur at the end of a gene.

249
Q

What is the role of chiasma in genetic diversity?

A

Chiasma contribute to genetic diversity by exchanging DNA between homologous chromosomes, which combines genetic material from both parents.

250
Q

What happens during metaphase I of meiosis?

A

During metaphase I, homologous chromosome pairs align at the cell’s equator.

251
Q

What occurs during anaphase I of meiosis?

A

During anaphase I, homologous chromosomes are pulled to opposite poles of the cell.

252
Q

What happens during metaphase II of meiosis?

A

During metaphase II, the chromosomes align at the equator of the two haploid cells.

253
Q

What happens during anaphase II of meiosis?

A

During anaphase II, the sister chromatids are pulled apart and move toward opposite poles of the cell.

254
Q

What is the final outcome of meiosis?

A

The final outcome of meiosis is the formation of four haploid daughter nuclei, each with half the number of chromosomes.

255
Q

What is independent assortment in genetics?

A

Independent assortment is the principle that traits located on different chromosomes do not influence each other’s inheritance when passed to offspring.

256
Q

How does independent assortment affect genetic traits?

A

Independent assortment means that the inheritance of one trait does not influence the inheritance of another trait, as long as the traits are located on different chromosomes.

257
Q

Who is credited with discovering the principle of independent assortment?

A

The principle of independent assortment was discovered by Gregor Mendel.

258
Q

What is the significance of the formation of gametes in the context of independent assortment?

A

During gamete formation, the alleles for different traits (such as B and E) are inherited independently of each other, creating genetic diversity.

259
Q

What are linked genes?

A

Linked genes are genes located physically close to one another on the same chromosome, and they are part of a linkage group.

260
Q

What is a linkage group?

A

A linkage group is a set of genes that are inherited together because they are located close to each other on the same chromosome.

261
Q

How does the distance between genes affect their separation during crossing over?

A

The closer the genes are to one another, the lower the chance they will be separated during crossing over.

262
Q

What happens when genes are part of a linkage group?

A

When genes are part of a linkage group, they are inherited together, and Mendel’s principles of independent assortment do not apply to them.

263
Q

Can crossing over separate linked genes?

A

Yes, crossing over can allow linked genes to be separated from one another and inherited in a different pattern.

264
Q

What happens during chromosome duplication and meiosis for linked genes?

A

During chromosome duplication and meiosis, chromosomes undergo crossing over, and linked genes may be separated and undergo gene conversion.

265
Q

What does the diagram illustrate about chromosome duplication?

A

The diagram shows chromosome duplication, where homologous chromosomes duplicate before undergoing meiosis, leading to the formation of sister chromatids.

266
Q

What is the site of crossover in the diagram?

A

The site of crossover in the diagram refers to the location on homologous chromosomes where genetic material is exchanged between chromatids, leading to genetic recombination.

267
Q

What happens at the site of gene conversion in the diagram?

A

The site of gene conversion in the diagram indicates where one chromatid may change its genetic sequence, a process that alters the gene’s expression due to the crossover event.

268
Q

How are linked genes separated during meiosis, according to the diagram?

A

The diagram shows that during meiosis, crossing over can separate linked genes from one another. The crossover event occurs between chromatids of homologous chromosomes, leading to the inheritance of genes in different combinations.

269
Q

What happens after meiosis, as shown in the diagram?

A

After meiosis, the result is the production of haploid cells with chromosomes that have undergone crossing over and gene conversion, which contributes to genetic diversity.

270
Q

How does crossing over affect the genetic material in gametes?

A

Crossing over exchanges genetic material between homologous chromosomes, resulting in gametes that carry a mix of genetic information from both parents.

271
Q

What is the effect of crossing over events on genes located on the same chromosome?

A

Crossing over events (recombination) can separate genes that are located on the same chromosome, creating new combinations of alleles.

272
Q

What are recombination events in genetics?

A

Recombination events refer to the process of crossing over, where homologous chromosomes exchange genetic material, resulting in new combinations of alleles.

273
Q

What is the meaning of ‘P generation’ in the context of genetic inheritance?

A

The ‘P generation’ refers to the parental generation, which is the first set of parents in a genetic cross.

274
Q

What is the meaning of ‘F1 generation’ in the context of genetic inheritance?

A

The ‘F1 generation’ refers to the first progeny (offspring) resulting from the cross between the P generation.

275
Q

How do recombination events affect gamete formation in the F1 generation?

A

Recombination events result in the formation of gametes with both original and new combinations of alleles (recombinant gametes) in the F1 generation.

276
Q

What is the outcome of fertilization in the example provided in the diagram?

A

Fertilization results in the F1 generation (AaBb), combining genetic material from both the P generation.

277
Q

What is the difference between recombinant and nonrecombinant gametes?

A

Nonrecombinant gametes contain the original combinations of alleles from the P generation, while recombinant gametes contain new combinations of alleles due to recombination.

278
Q

What is the conclusion about gametes formed through recombination?

A

Through recombination, gametes contain new combinations of alleles, leading to genetic diversity in the offspring.

279
Q

How can genes on the same chromosome be separated?

A

Genes on the same chromosome may be separated by crossing over events, also known as recombination.

280
Q

What is the significance of crossing over in the inheritance of genes?

A

Crossing over allows genes located on the same chromosome to be separated, leading to new combinations of alleles in offspring.

281
Q

What does the diagram show about the P (parental) generation?

A

In the P generation, homologous chromosomes with alleles for different genes (B, b, A, a) undergo crossing over, leading to recombination.

282
Q

What happens to the chromosomes during crossing over in the P generation?

A

During crossing over, sections of homologous chromosomes swap, leading to new combinations of alleles (e.g., B and a, b and A).

283
Q

What is formed during the gamete formation in the F1 generation?

A

During gamete formation in the F1 generation, new combinations of alleles are formed, including recombinant gametes (e.g., B and A, b and a).

284
Q

How does crossing over affect the formation of offspring in the F1 generation?

A

Crossing over results in offspring with new combinations of alleles, which increases genetic diversity.