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Medicine Year 1 Semester 1 > FUN Quiz 2 > Flashcards

FUN Quiz 2 Flashcards

1
Q

What type of replication is DNA replication?

A

Semi-conservative

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2
Q

What are the requirements for DNA Replication?

A

DNA Replication has the following requirements:

  1. Single-Stranded Template
  2. Deoxyribonucleotide Triphosphates (dNTPs)
  3. Replisome: Nucleoprotein complex that co-ordinates the replication activities. Numerous enzymes and proteins
  4. A Primer with a free 3’ end hydroxyl group
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3
Q

True or False?

Eukaryotes have multiple points of replication

A

True

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4
Q

Where or when is the replisome assembled

A

Origin of Replication

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5
Q

DNA Helicase

A

Separates DNA strands in an ATP-dependent process

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6
Q

Single-Strand Binding Proteins (SSB)

A

Binds to prevent strands from reassociating

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7
Q

Topoisomerase

A

Regulates twisting of the DNA Supercoiling.

It has nuclease and Ligase Activity

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8
Q

Nuclease

A

Cut DNA

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9
Q

Ligase

A

Bind DNA (backbone)

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10
Q

Describe the process of initiation in DNA Replication

A

Initiation: Involves the separation of two complimentary strands and occurs at the origin of replication which are specific points where DNA replication begins. Eukaryotes have several points of replication. From the origin, two replication forks move outwards in opposite directions. The replisome is assembled at the origin.

  1. DNA Helicase: Separates the DNA strands in an ATP-dependent process
  2. Single-Strand Binding Proteins (SSB): Binds to prevent the strands from re-associating
  3. Topoisomerase: Regulate twisting of the DNA supercooling. It has nuclease (cut DNA) and ligase activity (binding
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11
Q

RNA Polymerase

A

Makes RNA Primer

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12
Q

What is the Significance of RNA Polymerase in DNA Replication

A

DNA polymerase can only extend a chain and cannot synthesize DNA from scratch. After replication, the RNA primer is removed and the gap is filled

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13
Q

What are the general functions and roles of DNA Polymerase in DNA Replication

A

• DNA Polymerase:

  1. Uses single strand DNA as template
  2. Reads its template 3’ to 5’
  3. Makes new DNA from 5’ to 3’
  4. Aligns and adds nucleotides along the template strand with the correct pairing
  5. Catalyzes the formation of the phosphodiester bond
  6. Has Proofreading activity prevent errors. DNA polymerase’s active site can bind all 4 dNTP types. Catalysis occurs only when the correct one is bound. It also has proofreading with 3’-5’ exonuclease activity which removes nucleotides at the 3’ end of the new strand that are mismatched
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14
Q

What is PCNA?

A
  1. PCNA is a protein that acts as a sliding clamp which encircles the DNA template and keeps DNA polymerase closely associated to template allowing it to speed up the process
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15
Q

Do both strands in DNA Replication have the same process of Elongation? Explain

A

DNA replication occurs in two directions but DNA Polymerase can only work in one direction. Because of this, there is a leading and a lagging strand. The leading strand is replicated continuously whereas the lagging strand is replicated discontinuously. In the lagging strand, DNA is replicated into pieces of 1000bp Okazaki Fragments. This means that there are several primers, and DNA nuclease removes the RNA primers replacing it with the correct nucleotides (DNA pol), DNA ligase connects the backbone

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16
Q

What are the different types of DNA Polymerase

A

Alpha, beta, epsilon, delta, and gamma

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17
Q

Function of Alpha DNA pol

A
  1. Alpha a-DNA Pol is involved in initiating replication and associates tightly with primase to make the complex. Replicates DNA by extending primer 5’-3’. It has no Exonuclease activity
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18
Q

Functions of epsilon and delta DNA polymerases. What are the differences between them

A
  1. Epsilon e-DNA pol and Delta d-DNA pol: Not associated with primase, replicates DNA, highly processive, and has 3’-5’ exonuclease activity. eDNA pol works with the leading strand synthesis and d-DNA pol works with the lagging strand synthesis
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19
Q

Function of Beta DNA polymerase

A

Involved in DNA Repair

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20
Q

Function of Gamma DNA polymerase

A

Replicates mitochondrial DNA

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21
Q

Function of RNAse H1

A

Removes most of the RNA leaving one 5’ ribonucleotide adjacent to the DNA

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22
Q

Does DNA pol A have endonuclease activity?

A

No lol

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23
Q

Flap Endonuclease (FEN1)

A

FEN performs the endonuclease activity removing the last 5’ primer and up to 15 BP from the DNA synthesized by Pol a

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24
Q

When is DNA replication normally terminated

A

When each fork collides with the fork from the adjacent replicon

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25
Q

Briefly list the steps of DNA replication

A

Full Overview of DNA Replication: Helicase => SSB => Topoisomerase => RNA Polymerase => DNA pol a => DNA pol e & d with PCNA => Proofreading with DNA pol => Ligase => RNAse H1 => FEN1 => DNA pol e & d fill in gaps

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26
Q

Explain the process of primer removal in DNA Replication

A

Primer Removal: Involves 2 enzymes: Removes 5’ ribonucleotide. Pol a lacks proofreading so it cannot detect the error in the RNA primer.

  1. RNAse H1: Removes most of the RNA leaving one 5’ ribonucleotide adjacent to the DNA.
  2. Flap Endonuclease 1 (FEN1): FEN1 performs this endonuclease activity removing the lat 5’ primer and up to 15 BP from the DNA synthesized by Pol a
  3. DNA Pol d/e fills gaps
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27
Q

LEARNING OUTCOME: Describe and Explain the full process of DNA replication (Initiation, Elongation, Primer Removal, and Termination)

A

Initiation: Involves the separation of two complimentary strands and occurs at the origin of replication which are specific points where DNA replication begins. Eukaryotes have several points of replication. From the origin, two replication forks move outwards in opposite directions. The replisome is assembled at the origin.

  1. DNA Helicase: Separates the DNA strands in an ATP-dependent process
  2. Single-Strand Binding Proteins (SSB): Binds to prevent the strands from re-associating
  3. Topoisomerase: Regulate twisting of the DNA supercooling. It has nuclease (cut DNA) and ligase activity (binding)

Elongation:
• RNA Polymerase makes an RNA primer. This happens because DNA polymerase can only extend a chain and cannot synthesize DNA from scratch. After Replication, the RNA primer is removed and the gap is filled
• DNA Polymerase:
1. Uses single strand DNA as template
2. Reads its template 3’ to 5’
3. Makes new DNA from 5’ to 3’
4. Aligns and adds nucleotides along the template strand with the correct pairing
5. Catalyzes the formation of the phosphodiester bond
6. Has Proofreading activity prevent errors. DNA polymerase’s active site can bind all 4 dNTP types. Catalysis occurs only when the correct one is bound. It also has proofreading with 3’-5’ exonuclease activity which removes nucleotides at the 3’ end of the new strand that are mismatched.
7. PCNA is a protein that acts as a sliding clamp which encircles the DNA template and keeps DNA polymerase closely associated to template allowing it to speed up the process.
DNA replication occurs in two directions but DNA Polymerase can only work in one direction. Because of this, there is a leading and a lagging strand. The leading strand is replicated continuously whereas the lagging strand is replicated discontinuously. In the lagging strand, DNA is replicated into pieces of 1000bp Okazaki Fragments. This means that there are several primers, and DNA nuclease removes the RNA primers replacing it with the correct nucleotides (DNA pol), DNA ligase connects the backbone.
There are 3 main enzymes involved in replication
1. Alpha a-DNA Pol is involved in initiating replication and associates tightly with primase to make the complex. Replicates DNA by extending primer 5’-3’. It has no Exonuclease activity.
2. Epsilon e-DNA pol and Delta d-DNA pol: Not associated with primase, replicates DNA, highly processive, and has 3’-5’ exonuclease activity. eDNA pol works with the leading strand synthesis and d-DNA pol works with the lagging strand synthesis

Primer Removal: Involves 2 enzymes: Removes 5’ ribonucleotide. Pol a lacks proofreading so it cannot detect the error in the RNA primer.

  1. RNAse H1: Removes most of the RNA leaving one 5’ ribonucleotide adjacent to the DNA.
  2. Flap Endonuclease 1 (FEN1): FEN1 performs this endonuclease activity removing the lat 5’ primer and up to 15 BP from the DNA synthesized by Pol a
  3. DNA Pol d/e fills gaps

Termination: DNA Replication proceeds until each replication fork collides with fork from adjacent replicon.

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28
Q

Telomeres

A

Telomeres are at the 3’ end of each chromosome and it is made up of 1000s of tandem repeats

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29
Q

Telomerase

A

Telomeres are synthesized and maintained by Telomerase

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30
Q

What are the components of telomerase

A

Ribonucleoprotein made up of RNA and Protein

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31
Q

LEARNING OUTCOME: Outline the Role of Telomerase:

A

No need to include everything below as long as you include the main ideas

Termination: DNA Replication proceeds until each replication fork collides with fork from adjacent replicon. Continuous synthesis on leading strand can proceed to very tip of template but the lagging strand will have some missing nucleotides at the beginning when the primer is removed. As this is the lagging strand, the primer is removed but no OH group is available for DNA polymerase to add nucleotides. In normal removal of the primer, DNA pol would use the preceding fragment to fill in the gap but at the beginning there are no preceding fragments. This causes DNA to be shorter after every replication. These ends are called Telomeres.

Telomeres: telomeres are at the 3’ end of each chromosome and it is made up of 1000s of tandem repeats (TTAGGG in humans). Telomeres DNA is synthesized and maintained by Telomerase (Ribonucleoprotein make up of RNA + Protein). Telomerase’s RNA acts as a template for synthesis of DNA adding tandem repeats to the 3’ end. Telomerase activity is normal in gamete production and germ line cells. During development, as cells divide and differentiate, Telomerase function declines
When the telomeres become too short, it causes the cell to enter G0 or go for apoptosis or can develop into a cancer where there is uncontrollable division and hence shortening affecting coding sequences

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32
Q

Does the proof-reading activity of all endonuclease activity eliminate errors that could arise?

A

No lmao

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33
Q

What factors may damage DNA?

A

• Radiation: UV light - Fuse adjacent pyrimidines
• High Energy Radiation: Double stranded breaks
• Chemicals: Nitrous Acid - Deaminates amines turning C into U binding with A and A into Hypoxanthine binding with C
Most of this damage is repaired by the cell

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34
Q

What are the 5 mechanisms of DNA repair? List the names

A 
Mismatch Repair
Base Excision Repair
Nucleotide Excision Repair
Non-Homologous End-Joining (NHEJ)
Recombination or Homologous Repair
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35
Q

Describe the mechanism of Mismatch repair

A

Mismatch Repair (MMR): Occurs shortly after replication and replaces mismatched bases or loops. First MUT S detects which is the Parental and which is the daughter strand through methylation (parental strand is methylated), then MUT L and PCNA form the Tetrameric Complex and cut the strand recruiting Exonuclease 1 and DNA polymerase. Exonuclease removes bases around the mutation and DNA polymerase rebuilds the copy strand. DNA Ligase then joins the backbone. Defects in human MMR causes a high chance of cancer, specifically HNPC (Hereditary Non Polyposis Cancer). This cancer affects MLH1 and MLH2 genes which produce MutL proteins

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36
Q

Describe the mechanism of Base Excision Repair

A

Base Excision Repair: Replaces bases lost though chemical processes such as Depurination (purines) or Deamination (Pyrimidines). DNA Glycosylase identifies and removes damaged base leaving an Apurinic or Apyrimidinic Site where an OH is now present. AP Endonuclease identifies the site and cuts the backbone and Exonuclease removes the sugar and several adjacent bases. DNA pol and Ligase then repair the issue

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37
Q

Describe the Process of Nucleotide Excision Repair

A

Nucleotide Excision Repair: Responds to helix distortion caused by Pyrimidine Dimers. Replaces regions of damaged DNA. It provides defense against 2 notable carcinogens being Tobacco smoke and Sunlight. This process involves 16 proteins and mutations affecting these proteins are identified in two disorders being Cockayne Syndrome where the patient experiences premature aging, extreme sensitivity to light, and shortened lifespan as well as Xeroderma Pigmentosum which is a rare human skin disease that causes extreme sensitivity to light and skin cancer with frequent tumors as well as shortened lifespan

38
Q

What are the mechanisms of repair of double stranded breaks

A

Non-homologous End-joining (NHEJ)

Recombination or Homologous Repair

39
Q

Describe the mechanism of Non-homologous End-Joining

A

Non-Homologous End-Joining (NHEJ): The Ku Protein is a broken DNA sensor which recognizes these breaks. This protein holds both strands of broken DNA together leaving the ends accessible to Nucleases, Polymerases, and Ligases. These broken strands are not at the same level. As this DNA is susceptible to error while being repaired, it can cause deletions as there will be a hanging portion being single-stranded which can cause cancer and immunodeficiency syndromes. The most ideal situation would be the ligase

40
Q

Describe the Process of Recombination or Homologous Repair

A

Recombination or Homologous Repair: Non Mutagenic - Uses Enzymes and proteins that perform genetic recombination between homologous chromosomes during meiosis. Uses DNA sequence information in homologous chromosome to correct the break. During S phase, sister crhomatids are physically close, providing a homology donor for repair. In humans, defects in proteins BRCA1 and BRCA2 causes the increase in breast, ovarian, prostate, and pancreatic cancers. Basically uses homologous chromosome’s DNA to repair issues

41
Q

What does the number of mRNA determine

A

The amount of protein to be produced

42
Q

What is the function of mRNA

A

They are a transcript of a protein-coding gene. It is transported out of the nucleus to be translated for protein synthesis.

43
Q

Function of tRNA

A

Central to protein synthesis as adaptors between mRNA and amino acids

44
Q

rRNA function

A

Forms the basic structure of the ribosome and catalyzes protein synthesis

45
Q

Discuss the Promotor region and it’s significance to transcription

A

The Promotor Region: There are distinct start & stop signals within the DNA sequence of a gene. Transcription begins at the Transcription Start Site (TSS) which is found at the end of the promotor region. Transcription ends at the Terminator Site. Promoters are regions of DNA upstream of the TSS which lead to the initiation of transcription. Each gene has their own unique promotor. Enhancers are recognized by proteins that will aid transcription and hence Transcription Factors. The Basal Promotor is bound by a key transcriptional enzyme RNA Polymerase II and basal transcription factors that are required for RNA pol II.

46
Q

What is the function of Basal Promotors

A

The Basal Promotor is bound by a key transcriptional enzyme RNA Polymerase II and basal transcription factors that are required for RNA pol II. These Basal Promotors are
• essential for the transcription of all genes.
• Functions to recruit RNA Pol II
• Binds basal/general transcription factors first and once they are bound, RNA pol II will bind.
• Example includes the TATA Box

47
Q

Describe the process of Transcription in Eukaryotes: Initiation, elongation, termination.

A

Transcription is typically catalyzed by RNA Polymerase II. RNA Pol II transcribes just one of the two strands. It reads the template strand (3’-5’) and making an RNA copy 5’-3’. RNA Pol II will used NTPs when creating mRNA

Initiation: RNA Pol II binds to DNA and unwinds the promotor segment

Elongation: RNA pol II moves along the template synthesizing mRNA until reaching the terminator region.

Termination:
• Transcription continues beyond the protein-coding region: 3’UTR (untranslated region)
• Most Eukaryotic mRNA precursors have motif AAUAAA within the transcribed sequence which is a Polyadenylation Signal Sequence.
• Recruits endonuclease enzyme that cleaves mRNA (around 20 bases downstream beyond the sequence

48
Q

Describe the 5’ Cap and it’s function

A

5’ Capping: Protection of mRNA from degradation by Exonuclease, promotes nuclear export, and aids recognition by translational machinery

49
Q

Describe the function of the 3’ Post-transcriptional modification

A

3’ Poly A Tail: After the Polyadenylation Signal has been recognized and endonuclease performs its activity, Poly A polymerase adds adenine residues to the cleaved 3’end

50
Q

Discuss mRNA stability and longevity

A

MRNA Stability: mRNA’s from different genes have their longevity encoded within them which is determined by the 3’UTR. The 5’ CAP protects against degradation and hence Exonuclease will degrade from the Poly A tail. This tail is gradually shortened over the lifetime of the mRNA and hence the longer it is, the longer the mRNA will last

51
Q

Discuss mRNA splicing

A

Splicing: Genes transcribed to produce mRNA will have introns and exons. Exons only encode the protein coding region of a gene. Small Nuclear RNAs snRNA is involved in transcript processing. The non-coding intrinsic sequences are not present in mature RNA; They are spliced or cut out and exons are joined together. Removed introns are transported into the cytoplasm. Splicing is carried out by a molecular RNA and protein rich Spliceosome. SnRNA combine with a number of proteins to form Small Nuclear RiboNucleoProteins (snRNPs) which facilitate splicing by recognizing and interacting with specific sequences at each end of the intron, cutting and rejoining the RNA sequence. The process of splicing involves:

  1. Spliceosome formation (Several snRNPs)
  2. The Spliceosome attacks the 5’ intron breaking the RNA
  3. The 5’ end of the intron becomes attached to the A nucleotide forming a loop or RNA.
  4. The free 3’ end of one exon attacks the 5’ end of the other causing them to bond covalently and releasing the intro
52
Q

Discuss Differential splicing or Exon Shuffling

A

Differential Splicing: Also known as Exon Shuffling, There is an unusually small number of genes but so many possible expressions. Exons will be shuffled in different orders. Each order Codes for a different protein and allows differentiation within cells. It is needed to turn genes on and off to generate specificity

53
Q

What are the two major types of genes that are relavent to transcription and cell function

A

Consecutive Genes and Inducible genes

54
Q

What are consecutive genes

A

Consecutive Genes: Some genes are essential and necessary for life, and hence they need to be continually expressed ex enzymes involved in metabolism and DNA repair. Level of gene expression determined by how many binding sites and how much transcription factors are around

55
Q

What are Inducible Genes?

A

Inducible Genes: Many genes are only synthesized when required. These are also advantageous as it determines development and tissue specificity and also allows cells to respond to the environment when needed in the quantity needed. Inducible gene expression is controlled by transcription factor proteins. When the TF receives the correct signal it will immediately bind to the DNA and begin transcription. Inducible transcription factors are turned on only when needed which allows the cells and tissues to respond to environmental cues. The number of binding sites for that transcription factors determines its affect on the rate of transcription

56
Q

What are the key components of translation

A
  1. MRNA - Template, encodes information to make the protein
  2. TRNA - Carrier, each amino acid brought to mRNA can bind to a specific tRNA
  3. RRNA - Structural and functional roles in ribosome
  4. Ribosomal proteins
  5. Release/Termination Factors
57
Q

What are the main characteristics of the genetic code that define it.

A

• The Genetic Code is:
1. Specific as a specific codon always codes for the same amino acid
2. Universal as it applies to all species
3. Redundant as a given amino acid can be coded by by several different codons
4. Non-overlapping as the code is read from a fixed starting point, taken 3 at a time.
• Changes in the genetic code can cause several diseases including Huntington Disease which involved codon amplification leading to the misfolded protein accumulation as well as Sickle Cell Anemia where Glutamic acid changes to Valine by an SNP.

58
Q

How many amino acids are there

A

20

59
Q

How many nucleotides are found in DNA

A

4

60
Q

How many codon possibilities are there?

A

4^3 => 64

61
Q

What shape foes the tRNA have?

A

Clover leaf

62
Q

Explain tRNA’s role in translation as well as how the Wobble Effect influences that function

A
  • tRNA binds to the codon sequence with it’s Anticodon sequence which will have a base pair system with the codon.
  • The Wobble Effect: Since we have 50 tRNAs matching 61 codons for 20 amino acids, some tRNAs can match more than one codon creating flexible pairings. Wobble base is the last 3’ base in the codon on the mRNA which binds to the first 5’ base in anticodon
63
Q

What is the function of ribosomes as well as their structure in association to translation.

A

• Ribosomes are the sites of translation. They bring the tRNA and mRNA together to translate nucleotide sequence of mRNA into amino acid of a protein
• They are complexes of protein and rRNA
• There are 3 sites of the Ribosome involved in translation:
1. A Site: Acceptor for the Next tRNA
2. P Site: Contains the amino acid chain. The first rRNA enters through the P site (only)
3. E Site: Harbors deacetylated tRNA on the way out of the ribosome

64
Q

Explain the process of translation including all major details and events taking place until it’s termination

A

Initiation:
1. Assembly of Components:

  • Two ribosomal subunits
  • mRNA to be translated
  • tRNA specified by the first codon
  • GTP factors (provide the required energy)
  • Initiation factors (help the ribosome recognize the sequence for the start of translation)
  1. Recognition of Start Codon:
  • Assembly of components
  • Recognition of Start codon where tRNA for met will enter at the P position and bind it’s anticodon to AUG

Elongation:

  1. Next required tRNA is delivered to the A site with the help of elongation factors)
  2. Peptide bonds are formed between adjacent amino acids facilitated by the enzyme Peptidyltransferase
  3. After a bond is formed, the ribosome is translocated three nucleotides in the 3’ direction. => the growing chain is moved to the P site, Uncharged tRNA moves to E site and is released and A site free to accept next tRNA

Termination:

  • Occurs when one of 3 termination codons arrives in the A site
  • Stop codon is recognized by a release factor
  • Release factor binds to A-site which causes the newly synthesized protein to be released and the disassembly of the tRNA-ribosome-mRNA complex
65
Q

How is protein regulated in the cell . List the main ways

A

Protein level regulation allows the cell to respond to environmental stimulus quickly. Increase Transcription/translation enhances levels and degradation removes proteins when function is complete or they have become damaged. Protein regulation occurs through the following methods:

  1. Protein levels
  2. Protein location and concentration
  3. Ligand Binding
  4. Cofactor Requirements
  5. Post-translational modifications
66
Q

Discuss the process and significance of Protein turnover

A

Protein Turnover is a normal process to ensure that proteins are fit for purpose. Normally they are kept at a homeostatic balance but changes in response to environmental stimuli. Proteins with key regulatory functions are often rapidly turned over or degraded to keep them under tight regulation. Rapid turnover of proteins is necessary to allow their levels to change quickly in response to external stimuli

Normal Protein Turnover: Housekeeping proteins get damaged and need to be replaced. Misfolded proteins need to be recycled to ensure normal cell processes are conducted correctly

67
Q

How are proteins degraded?

A

How Are Proteins Degraded:

  1. Lysosomal Degradation: Degrades proteins that are imported into the cell. Lysosomes are membrane-bound organelles that arise from the Endoplasmic Reticulum. They are highly acidic (proton pumps) and contain degradation enzymes that are only active at a low pH. Lysosomes will degrade any biological material that the cell deems unfit.
  2. Proteasomal Degredation: Degrades cytosolic proteins marked for destruction. Proteins can be labelled or ‘tagged” also known as Polyubiquitination which allows the Proteasome to detect it and degrade it. The Proteasome is a multi protein, barrel shaped, complex containing protease enzymes and is found in the nucleus or cytoplasm to carry out it’s function
68
Q

What are the three ways protein activity is regulated in the cell with a brief description of each

A

No need for all details

Regulating Protein Activity in the Cell: This can be done through tight regulation of protein levels, where a protein in located, its concentration and Post-Translational Regulation. This can be carried out through the following methods:
• Ligand Binding: Ligand binding changes the shape of the protein into it’s active conformation when the steroid hormone binds to their ligand. In it’s active state, it will Dimerise and enter the nucleus and binds to DNA to carry out it’s function through gene expression (inhibitory or enhance).
• Phosphorylation: The Phosphate group is added to Serine (S), Threonine (T) and Tyrosine (Y) residues in a ratio of 100:10:1 where tyrosine would be least abundant but has the most effect. This is carried out by Kinases for adding the phosphate and Phosphatase for removing it. Kinases are great drug targets. An example of this is Leukemia which is the uncontrolled growth of Myeloid cells in bone marrow which results from the Translocation of the Abl gene to the BCR gene on another chromosome.
• Cleavage: Cleavage of proteins allows for its activation. Enzymes are released as Zymogens and need to be activated to carry out it’s function. This allows for quick response as the proteins and enzymes are ready. An example of this is Apoptosis

69
Q

What are the Large scale Genetic Variations

A

Aneuploidy
Translocations/inversions
CNVs or Copy Number Variants

70
Q

Aneuploidy

A

Aneuploidy: One or more individual chromosomes in extra copy or missing such as Down syndrome (trisomy 21). Occurrence is rare and these changes causes extreme changes in gene expression which causes learning disabilities, development delay…An example of a Monosomy is the X-Turner Syndrome. XXY is Kleinefelters

71
Q

Translocations/Inversions

A

Translocations/Inversion: Exchange of DNA during meiosis but between different chromosomes. Rare as well and clinical consequence depends on whether there is a net gain or loss of DNA or not, and whether there is a disruption in a gene sequence

72
Q

CNVs (Copy Number Variants)

A

Relatively large sections of DNA duplicated or delete. This is common in everyone and they are benign but the greater the size the greater the pathogenic consequences.
Once these copies happened, having 2 genes with the same function allows them to evolve differently which is a major spark in evolution

73
Q

What are the Small Scale Genetic Variatrions

A

SNP
Micro-satellites
Insertions and Deletions

74
Q

Single Nucleotide Polymorphism (SNPs)

A

Single Nucleotide Polymorphism (SNPs): Single base-pair changes. Everyone has them (incidence). Vast majority are benign hence it won’t have a real effect but some are massive like sickle cell anemia

75
Q

Micro Satellites

A

Repeated units of 2-5bp DNA. The number of copies varies and everyone has thousands of them. Rarely causes disease as most are benign. Can be used to identify people as it is unique to each person

76
Q

Insertions and Deletions

A

Insertions & Deletions: One or two bases, duplicated or deleted. Causes frameshift. Rarely an issue as most of the DNA is non-coding. If it occurs in the exons then GG

77
Q

Why aren’t most mutations pathogenic

A

Because most of our DNA is non-coding

78
Q

Are all genes equally pathogenic

A

No kiss emmak

79
Q

What are silent mutations

A

Mutations that dont affect aa

80
Q

What are missense mutations

A

Different aa formed

81
Q

What are non-sense mutations

A

Mutations that cause the formation of a stop codon

82
Q

Describe Gel Electrophoresis

A

What Is It: A method for separation and analysis of macromolecules (DNA, RNA, and Proteins) and their fragments based on their Size and Charge

Why Is It Important: It provides a means by which to tell differences between different DNA/RNA/Protein samples

How Does it Work:
• Mixture of DNA fragments are placed within wells of the agarose gel
• A power source is places where the negative node is at the wells and the positive on the opposite side
• DNA will migrate to the opposite side as DNA is negatively charged
• Long fragments will move slower as they are heavier and smaller fragments will move faster and hence be closer to the positive pole

83
Q

Describe Western Blotting

A

What is it: Technique used to detect a specific protein in a sample

Why is it important: It can tell us if a protein is present, and also a rough idea of the quantity of the protein

How does it Work:

  1. Separate Macromolecules/fragments on a gel (using electrophoresis)
  2. Transfer/Immobilise the separated molecules onto a membrane sheet known as the blot. This is done by putting the sheet on top of the gel and adding a weight to collect the molecules
  3. Detect a specific fragment using a probe. This is done by using the target protein as an antigen that will be detected by the primary antibody. The secondary antibody, modified to have an enzyme that will produce a detection signal. This secondary antibody will only activate the enzyme when it is attached to the primary antibody
84
Q

Describe the basic principles of Nucleic Acid Hybridization

A

• Involves using a probe to attach/detect a specific nucleic acid sequence.
◦ Nucleic Acids with complimentary sequence will anneal/join together
◦ Probes will have several sequences and only the one that matches with the DNA will completely attach and form a hybridization event. This allows us to determine the sequence
◦ The ones that do not completely match will be mismatched and will not fully bind
• The longer the probe, the more specific it is to certain parts of the DNA (less likely to have duplicates

85
Q

Describe Polymerase Chain Reaction

A

What is it: A method to generate large amounts of DNA from a limited number of copies of the template. PCR allows us to amplify DNA

Why is it important: By providing us with the ability to amplify large amounts of the specific region of DNA that we want to study, it forms the basis of many tests including DNA fingerprinting, genetic diagnostics, and early detection of cancer. Used for determining small-scale changes

How Does it Work:
• PCR produces millions of copies of just the small part you are interested in. Allows an exponential increase in the number of copies of a target sequence.
• It uses several components:
◦ Thermostable DNA polymerase such as Taq DNA Polymerase (Allows the enzyme to work at high temperatures)
◦ DNA template to be copied
◦ dNTPs + Mg2+
◦ Two Oligonucleotide Primers (Nucleotide Probe from hybridization)
1. To amplify the DNA from a particular gene, its nucleotide sequence must be known. This is important to design the two synthetic nucleotide primers
2. The system is heated to high temperatures to break the bonds of the DNA molecule (95)
3. Primers are introduced to the system and the temperature is lowered to allow for hybridization (55). It has the 3’ facing inwards
4. The temperature is then raised again for the Taq Polymerase to start copying (72)
5. The DNA strands produced will be long. Each primer will go in a direction and the overlap is the desired strand. This process is then repeated with the same stands so that the second pair of oligonucleotide primers attach and hence forming the desired sequence.
6. Process is then repeated with the desired strand obtained for amplification

86
Q

Describe G-Banded Karyotyping

A

What is it: A technique to determine the number, and banding pattern of chromosomes in an individual

Why is it Important: Because it can detect aneuploidy (Number), or large chromosome rearrangements (Banding Pattern)

How does it Work:
• Carried out on any tissue
1. Cells stimulated to divide in culture
2. Dividing cells arrested in metaphase so that chromosomes are supercoiled and burst open using hypotonic solution
3. Fixation and dropping of cells onto a microscope slide
4. Giemsa Staining (Gene Banding)
5. Examine via microscope
• Produces characteristic patterns of light (Euchromatin) or GC rich and dark or AT rich (Heterochromatin) bands
• Aneuploidy: Abnormalities of chromosome number
• Translocations: Genetic material moves from one chromosome to another
• Inversion of genetic material within a chromosome
• Very Large CNVs: Deletions of amplifications of chromosome segments (greater than 10 million bases)

87
Q

Describe Florescent In-Situ Hybridization, and the Nature of Genetic Variation the Technique Can Detect

A

What is it: A method to visualize specific regions of DNA in an individual’s cell (including genes)

Why is it Important: It has greater resolution/is more sensitive than G-banded karyotyping.

How is it Done:
• Vividly paints chromosomes or portion of chromosomes with fluorescent molecules (using principle of hybridization)
• Can determine the presence and location of a region of SNA within chromosome preparations, fixed cells or tissues.
• Can detect duplications or deletions of SNA smaller than detected by G-banded proteins
1. DNA or RNA sequences from appropriate, chromosome-specific probes are first labelled with reporter molecules
2. Fluorochrome Labelling
3. Denature Probe DNA
4. Labelled DNA or RNA probe is hybridized to the metaphase chromosomes or interphase nuclei on a slide
5. Hybridize to DNA Denatured on chromosomes
6. Screen for the reporter molecules by Fluorescence Microsopy
• Basically labeling a chromosome and putting it under a microscope to be compared to a normal one. Can detect smaller changes that typical G-banded Karyotyping

Nature of Genetic Variation:

  • Gene Fusions/ Translocation
  • Aneuploidy (gain or loss)
  • Cytogenetic causes of birth defects and mental retardation.

Limitations of FISH:
1. It’s Specific to what you design the probes for
A. FISH can only detect deletions or duplications of regions specifically targeted by the probe used
2. Size of Translocation / Genetic change:
A. Small scale genetic including very small deletions are not detected / genetic change
3. It’s very labour intensive

88
Q

Describe Array CGH, and the Nature of Genetic Variation the Technique Can Detect:

A

What is it: A method for the detection of aneuploidy and chromosomal copy numbers change on a genome-wide and high resolution scale

Why is it Important: Because it provides a rapid and cheap test to detect genetic changes that result in the gain or loss of DNA. This includes extra or loss of chromosomes, CNVs (Copy number variations; deletion, duplication, insertion), and Translocation

How Does it Work:
• Can Detect aneuploidy as well as sub-microscopic rearrangement of DNA.
• Capable of whole genome wide scan
• Less labor intensive and cheap
1. Isolate Genomic DNA from patient/test sample
2. DNA Digestion (test and reference)
3. Fluorescently label patient test (red) and control a Green sample.
4. Post-hybridization washing
5. Assays scanning and Data Analysis

• The Microarray contains probes to hybridize to test and control fragments
◦ Small chromosomal segments are dotted on a piece of glass
◦ These segments or probes are designed to overlap the nuclear genome
◦ Clinical Applications of an aCHG include extra Down syndrome with extra chromosome 21
Limitations of CGH:

  • Does not detect balanced translocation
  • Distinguishing between normal variants and pathologically significant variants can sometimes be challenging
89
Q

3- Describe How the Array Hybridization Method Can be Used to Genotype SNPs

A

What is it: A massively parallel technique to genotype multiple different specific germline SNPs in a sample

Why is it Important: It can tell us the germline genotype of an Individual across hundreds of thousands SNPs. Used to identify carriers for known mutations

How does it Work:

• These arrays are used to identify carriers for single-gene disorders and to research the genetics of complex the end
• The identify carriers: known for mutations. The array must be designed to test those specific variants
• Identify generic risk variants for complex diseases
1. The Genotyping array contains oligonucleotides (100s of thousands), designed to target the SNPs of interest. Each oligonucleotide is designed to stop one base before the SNP target
2. The oligonucleotides are bound to tiny silica beads spread across the microarray
3. DNA is fragmented and added to the array, together with bases and labelled fluorescent tags (4 colors, to distinguish each base)
4. The fragments hybridize to the oligonucleotides based on Watson-crick pairing
5. A single base-pair extension is conducted (polymerase)
6. Genotype read based on color of the single added base

Types of Genetic Variation DNA Hybridization Can Detect:
• The technique can detect germline changes in genetic variation
• It cannot detect somatic changes
• Note that the array can only detect variation targeted by oligonucleotides hence it will not detect any and all variation

90
Q

Describe “Next Generation” Sequencing and the Nature of Genetic Variation the Technique Can Detect:

A

Where array hybridisation genotyping systems focus on specific SNPs and thus provide incomplete ascertainment, sequencing gives you complete ascertainment of genetic variation.

Exome sequencing vs genome sequencing -> exome sequencing is cheaper generally and is more done.

What is it: A massively parallel technique to sequence DNA. A technique to sequence DNA that is several orders of magnitude more efficient than Sanger sequencing

Why important/what does it tell us? It allows us to sequence complete exomes or complete genomes. With this information we have ultimate resolution to detect mutations.

Process for Illumina sequencing:
1)Library/sample preparation
1. DNA is fragmented and “adaptors” sequences added. This pool of DNA with adaptors attached = the sequencing “library” break up DNA
2. The library is added to the sequencing flow cell.
➔ Adaptors are oligos of known sequence, which can hybridise to the sequencing flow cell. Flow cell contains oligos that are complementary to adaptors
Take blood sample. Fragment it. Add adaptor. Similar to probe to anchor fragment to flow cell with millions of oligos and when the library is added, there is hybridisation.

2) Cluster generation PCR based
Net result: spots with many copies of the same read
Required so that the DNA to be sequenced is present at sufficient copy number to achieve the required signal.
Create double stranded DNA -> heat -> separate and repeat to make more of the DNA

3)Sequencing in parallel
​Nucleotides in the sequencing reaction are fluorescently labelled and reversibly terminated. Each cycle adds 1 base, which emits fluorescent signal – captured by camera

4) Data analysis /Alignment to reference
​Short reads of sequencing
– reconstructed
– referenced against known whole genome assemblies
– i.e. realigned and mapped against reference sequence.
Comparisons to reference identify points of difference = variation
Blue piece on diagram is the are the 200 base pair fragments.
READING SAME POINT MULTIPLE TIMES WHICH GIVES YOU CONFIDENCE OF YOUR DIAGNOSIS
There are several different methods, here we focused on Illumina.

Types of genetic variation NGS DNA sequencing can detect:
• The technique can detect germline AND somatic changes in genetic variation!
• It will detect any and all variation within the genetic region targeted by the library preparation.
• Can detect small- and large-scale changes

91
Q

Ok

A

Ok

Medicine Year 1 Semester 1 (23 decks)

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