Exam 1 Flashcards

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1
Q

What did Rosalind Franklin discover?

A

Using X-ray diffraction on DNA the double helix structure was discovered

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2
Q

Structure of DNA

A

Each nucleotide is composed of a sugar, a phosphate, and a base (guanine, cytosine, adenine, thymine or uracil).

Nucleotides are linked by covalent phosphodiester bond between sugar and phosphate
(sugar–phosphate backbone)

DNA: two polynucleotide chains held together by hydrogen bonds between the paired bases.
(A = T) (G ≡ C)

Run antiparallel (opposite chemical polarities: 5’ of sugar and 3’ of sugar give rise to polarity)

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3
Q

Human genome

A

full set of human chromosomes

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4
Q

Human karyotype

A

ordered display of full set of 46 chromosomes

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5
Q

Which DNA sequence elements are necessary for the duplication of a chromosome and then segregation at mitosis?

A

Replication origins
Centromere
2 telomeres (marking ends of each chromosome)

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6
Q

Overview of the cell cycle

A

Interphase: DNA replication begins at replication origin giving duplicated chromosomes
M phase: centromeres attach duplicated chromosomes to mitotic spindle = one copy distributed to each daughter cell

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7
Q

What is the nucleolus

A

a well-organized region in the interphase nucleus where parts of different chromosomes that carry genes that encode ribosomal RNAs come together

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8
Q

What is the function of the nucleolus

A

rRNAs are synthesized + combine with proteins to form ribosomes

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9
Q

Structure of the interphase nucleus

A

Interphase chromatin:
Heterochromatin: contains few genes that don’t usually get expressed due to its highly condensed form
Concentrated around centromeres + telomeres
Appear as dense regions of chromatin
Euchromatin: actively transcribed + not as condensed

Nucleolus contains the genes for rRNAs

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10
Q

What is the nucleosome

A

first and most fundamental structural unit of chromatin packing

convert DNA molecules in interphase nucleus into a chromatin fiber (clusters of closely packed nucleosomes)

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11
Q

Features of experimentally unpacked chromatin fiber

A

looks like beads on a string
string - the DNA
bead - a nucleosome core particle (DNA wound around a core of histone proteins)
Exposed DNA between core particles - linker DNA

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12
Q

Process of experimentally unpacking chromatin fiber

(structure of chromatin fiber / nucleosome core)

A

Nucleosome core particle can be released from chromatin by digestion of the linker DNA with a nuclease, which cleaves the exposed linker DNA but not the DNA wound tightly around the nucleosome core.

When the DNA around each isolated nucleosome core particle is released, its length is found to be 147 nucleotide pairs; this DNA wraps around the histone octamer (2 sets of 2 histones)

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13
Q

Levels that give rise to highly condensed mitotic chromosome

A

short region of DNA doube helix > beads on a string chromatin fiber > further nucleosome packaging via H1 (pulls adjacent nucleosomes together) > chromatin fiber folded into loops via nonhistone chromosomal proteins = interphase chromosome > more packing to form mitotic chromosome

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14
Q

How do chromatin-remodeling complexes work?

A

Reposition the DNA wrapped around nucleosomes

Use energy from ATP hydrolysis to loosen the nucleosomal DNA + push it along the histone octamer

Either exposing or hiding a sequence of DNA controlling its availability to other DNA-binding proteins

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14
Q

What aids changes in nucleosome structure to allow access to DNA

A

chromatin-remodeling complexes
histone-modifying enzymes

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15
Q

How do histone-modifying enzymes work?

A

All histones in the core have tails that extend from each nucleosome core particle

Each histone can be modified by the covalent attachment of diff groups to the tails (e.g addition/removal of acetyl, phosphate, methyl etc)

These modifications generally serve as docking sites on histone tails for nonhistone chromosomal proteins which then have different functional outcomes (e.g gene silencing, gene expression, heterochromatin formation, etc)

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16
Q

How does heterochromatin form and spread?

A

heterochromatin-specific histone tail modifications attract heterochromatin-specific proteins that reproduce the same histone modification on neighbouring nucleosomes
Heterochromatin spreads until it encounters a barrier DNA sequence that blocks further propagation into regions of euchromatin

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16
Q

Study that demonstrated DNA is the genetic material

A

Avery, MacLeod, McCarty
Prepared an extract from S strain + fractioned it into classes of molecules (RNA, protein, DNA, lipid, carbohydrate)
Tested the molecules ability to transform R-strain cells
Only the DNA was able to transform R into S strain

Concluded: Molecule that carries heritable ‘transforming principle’ is DNA

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16
Q

Outline of study regarding pathogenic and harmless bacteria

A

Study by Griffith:
S strain of S.pneumoniae = disease causing
R strain = harmless strain

Injected S strain = mouse dies of infection
Injected R strain = mouse lives
Injected heat-killed S strain = mouse lives
Injected mix of heat-killed S and live R = mouse dies of infection + living S strain is recovered

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17
Q

X-chromosome inactivation

A

males (with one X chromosome) females (with two X chromosomes but one inactivated)

Females have two X chromosomes (one from the mother and one from the father).
Early in development, randomly one of these X chromosomes becomes inactivated and condensed into heterochromatin.

Once an X chromosome is inactivated in a cell, that same X stays inactivated in all cells that come from that original cell.

The inactivated X chromosome is seen as a Barr body (a small mass of chromatin in the nucleus).

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18
Q

What is the biological significance of X-chromosome inactivation in females?

A

dosage compensation: Men and women now have similar levels of X-chromosome-related gene expression

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19
Q

Consequences of incomplete x-chromosome inactivation in females?

A

more expression of certain genes, especially at the tip of the chromosomes. Tip of chromosomes contain certain genes of the immune system, this is one of the ideas why females are better in combatting infections diseases such as Covid 19. The price they pay is: more autoimmune diseases (about 4x more) in women.

Genes at the tip of X, that are not silenced, even in female cells with inactivation of one X chromosome.
= female cells have the capacity to make more IFNs = protect them against developing severe Covid-19 symptoms

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20
Q

What enables a cell to replicate its genes

A

The ability of each strand of DNA to act as a template for producing a complementary strand

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21
Q

What is the first step of DNA replication

A

Parent DNA separation
- initiator proteins (helicase) bind to replication origins
- locally pull apart the two strands of the double helix (breaking the hydrogen bonds between base pairs)

Form replication forks at each replication origin

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21
Q

What is proofreading in DNA replication?

A

Done by DNA polymerase (has a P site for polymerization activity and E site for proofreading/editing)
If incorrect base pair then it clips off the mispaired nucleotide and tries again

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22
Q

Replication forks at a replication origin

A
  • two replication forks form at a replication origin
  • move away in opposite directions (bidirectional)
  • movement driven by the replication machine (opening up 2 strands + copying DNA) which involves helicase, DNA polymerase, and single-strand DNA-binding proteins (prevents strands from reforming + keeps them elongated)
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23
Q

Which parent strand is the lagging and leading strand? and in which direction is new DNA synthesized?

A

New DNA is synthesized in a 5’ to 3’ end
Remember strands run antiparallel

Leading strand:
continuous replication as DNA is synthesized towards helicase

Lagging strand: noncontinuous replication, okazaki fragments, moves back along the template strand towards the fork to synthesize the next fragment

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24
Q

How is replication started

A

Primase synthesizes the RNA primer which makes a short length of RNA giving a 3’ starting point
For leading RNA primer is only needed to start replication @ replication origin
For lagging new primers repeatedly needed

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25
Q

What is the role of the sliding clamp in DNA replication

A

keeps DNA polymerase attached to the template, allowing it to move along without falling off

On lagging: clamp detatches each time the polymerase completes an okazaki fragment and reattaches each time to start a new one

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26
Q

What is the role of the clamp loader in DNA replication

A

uses energy of ATM hydrolysis to lock the sliding clamp onto DNA

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27
Q

How are replication proteins arranged when a replication fork is moving

A

Lagginf strand has been folded to bring its DNA polymerase in contact with the leading strand DNA polymerase + bring the 3’ end of each completed okazaki fragment close to the start site

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28
Q

What is the role of the DNA ligase in DNA replication

A

uses energy of ATP hydrolysis to join okazaki fragments

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28
Q

What is the role of DNA topoisomerase in DNA replication

A

produces transient nicks in the DNA backbone to relieve the tension built up by unwinding the DNA ahead of helicase

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29
Q

How are okazaki fragments joined together

A

nuclease degrades RNA primer > DNA polymerase replaces RNA primer with DNA > DNA ligase joints the 5’ phosphate end of a DNA fragment to adjacent 3’ hydroxyl end of the next

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30
Q

How is DNA replication finished off and what are the issues encountered?

A

Leading strand is replicated all the way to the chromosome tip

Lagging strand ends cannot be completed because once the final RNA primer has been removed there is no mechanism for replacing it with DNA (chromosome ends would keep shrinking with each cell division)

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31
Q

How is the finishing up dilemma of the lagging strand dealt with?

A

Template strand is extended beyond the DNA that is to be copied via telomerase (adds to the telomere repeat sequences at the 3ʹ end of the template strand) which then allows the newly synthesized lagging strand to be lengthened by DNA polymerase

Telomerase carries its own short piece of RNA template that is complementary to the DNA repeat sequence

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31
Q

What happens when damage occurs to DNA during replication
+ examples

A

Chemical modifications of nucleotides, if left unrepaired, produce mutations.

Deamination of cytosine produces uracil.
During replication base pairs the U with A (when it should be G and C which correctly forms on the other template strand)

Depurination, if uncorrected, can lead to the loss of a nucleotide pair.
When the replication machinery encounters a missing purine on the template strand, it can skip to the next complete nucleotide
Producing a daughter DNA molecule that is missing one nucleotide pair

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32
Q

Cancer incidence with age

A

Because cells are continually experiencing accidental changes to their DNA—which accumulate and are passed on to progeny cells when the mutated cells divide—the chance that a cell will become cancerous increases greatly with age.

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33
Q

How does mismatch repair work

A

Mismatch repair eliminates replication errors and restores the original DNA sequence.

Repair machinery must replace the incorrect nucleotide on the newly synthesized strand, using the original parent strand as its template.
Eliminates the error, and allows the original sequence to be copied during subsequent rounds of replication.

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34
Q

Difference between RNA and DNA

A

RNA contains the sugar ribose
DNA has deoxyribose sugar
RNA contains the base uracil, which differs from thymine
RNA is single stranded

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35
Q

Process of transcription

A

Opening small portion of DNA to expose bases
One strand serves as the template
RNA transcript produces RNA chain 5’ to 3’ direction
RNA polymerase covalently links incoming ribosenucleoside triphosphates to growing RNA chain, and unwinds DNA helix ahead, and catalyses formation of phosphodiester bond between nucleotides

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36
Q

DNA strands in transcription

A

Transcription of a gene produces an RNA complementary to one strand of DNA

The template strand (DNA) is used to guide the synthesis of the RNA molecule.
The nontemplate strand (DNA) called the coding strand because its sequence is equivalent to the RNA product

Which DNA strand serves as the template varies

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37
Q

What signals bacterial RNA polymerase where to start and stop

A

Bacterial RNA polymerase contains a subunit called sigma factor that recognizes the promoter of a gene

Once transcription starts, sigma factor is released, + polymerase moves forward, synthesizing the RNA

Elongation continues until the polymerase encounters a sequence in the gene called the terminator

After transcribing this sequence into RNA, enzyme halts + releases both the DNA template and newly made RNA

Polymerase then reassociates with a free sigma factor and searches for another promoter to begin the process again

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37
Q

What determines which DNA strand is transcribed

A

The polarity of the promoter orients the polymerase

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38
Q

How is transcription initiated in eukaryotes?

A

Eukaryotic RNA polymerase II requires a set of general transcription factors

Eukaryotic promoters contain a DNA sequence called the TATA box which is recognized by a subunit of the general transcription factor TFIID, called the TATA-binding protein (TBP)

The binding of TFIID enables the adjacent binding of TFIIB.

The rest of the general transcription factors, as well as the RNA polymerase itself, then assemble at the promoter.

TFIIH pries apart the double helix at the transcription start point, using the energy of ATP hydrolysis = exposes the template strand of the gene

TFIIH phosphorylates RNA polymerase II, releasing the polymerase from most of the general transcription factors, so it can begin transcription

Once the polymerase moves away from the promoter, most of the general transcription factors are released from the DNA; the exception is TFIID, which remains bound through multiple rounds of transcription initiation

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39
Q

What are exons and introns

A

exons: coding sequences/regions on a gene that become expressed

introns: noncoding sequences that interrupt exons

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40
Q

What is RNA splicing

A

introns are removed from freshly synthesized RNA and exons are stitched together

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41
Q

What is alternative splicing

A

Some pre-mRNAs undergo alternative RNA splicing to produce different mRNAs and proteins from the same gene

Exons can be skipped over by the spliceosome
Skipping occurs when the splicing signals at the 5ʹ end of one intron are paired up with the branch-point and 3ʹ end of a different intron.

Order of exons cannot be rearranged

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41
Q

How is splicing carried out

A

via snRNPs (U1, U2, and U6)

U1 recognizes the 5’ slice site and U2 recognizes the lariat banch-point site through complementary base-pairing

U6 rechecks the the 5’ splice site by displacing U1 + base-pairing with this intron sequence

Formation of spliceosome active site

Splicing reactions occur

Spliceosome deposits exon junction complex on the mRNA to mark splice site as successfully completed

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42
Q

What are the 3 mRNA modifications that occur

A

Capping: modifies 5’ end of RNA by attaching G bearing a methyl group
Advantage of CAP - stabilizes RNA which helps transport the mRNA from the nucleus to the cytosol

Splicing

Polyadenylation : 3’ end of RNA is trimmed by an enzyme, 2nd enzyme adds a series of repeated A nucleotides (poly-A tail)

PolyA tail differentiates mRNA from RNA
PolyT tail can help isolate mRNA

43
Q

Where do the following processes take place in the cell?
Transcription
Translation
RNA splicing
Polyadenylation
RNA capping

A

Transcription - in the nucleus
Translation - on the ribosome (on both free ribosomes and boundER)
RNA splicing - in the nucleus before the RNA migrates to the cytoplasm
Polyadenylation - in the nucleus
RNA capping - in the nucleus

44
Q

How does a cell know mRNA is ready for export out the nucleus to the cytosol

A

RNA-binding proteins signals that a completed mRNA is ready for export to the cytosol (e.g exon junction complex)

A nuclear transport receptor associates w/ the mRNA + guides it through the nuclear pore

45
Q

How many codons represent amino acids

A

most amino acids are represented by more than one codon (but some are represented by 1)

46
Q

Which codons do not specify for an amino acid?

A

Stop codons that act as termination sites, signaling the end of the protein-coding sequence in an mRNA

46
Q

Why can some tRNAs tolerate mismatching at the 3rd nucleotide and still produce the correct amino acids?

A

codons for the same amino acid tend to contain the same nucleotides at the first and second positions and vary at the third position

47
Q

What is the intiation codon

A

AUG, signaling the start of a protein-coding message, and specifies the amino acid methionine

48
Q

Structure of tRNA molecules

A

They are molecular adaptors

Anticodon loop at the base of the tRNA contains the sequence of three nucleotides that base-pairs with the corresponding amino acid codon

Attached amino acid is at the 3’ end of the tRNA (at the top of the molecule)

49
Q

Process of translation

A

Initiation:
- small ribosomal subunit binds to mRNA near 5’ cap with translation intiation factors + scans for start codon
- special initiator tRNA w/ methionine binds to start codon
- large ribosomal subunit joins forming complete ribosome w/ 3 sites A, P, E

elongation
- ribosome reads mRNA codons one at a time

1) a charged tRNA carrying the next amino acid to be added binds to vacant A site on the ribosome by forming base pairs with the mRNA codon that is exposed there

2) the carboxyl end of the polypeptide chain is uncoupled from the tRNA at the P site and joined by a peptide bond to the free amino group of the amino acid linked to the tRNA at the A site.

3) a shift of the large subunit relative to the small subunit moves the two bound tRNAs into the E and P sites of the large subunit.

4) the small subunit moves exactly three nucleotides along the mRNA molecule, bringing it back to its original position relative to the large subunit.

Ejects the spent tRNA in E site and resets the ribosome with an empty A site so that the next charged tRNA molecule can bind

Stopping
- when a stop codon is encountered this signals the ribosome to stop
- release factors bind to the stop codon in A site
- completed polypeptide is released
- ribosome dissociates into its 2 separate subunits

50
Q

what makes the translation so fast and efficient

A

proteins are synthesized on polyribosomes on one mRNA

51
Q

How are defective proteins degraded

A

proteins marked by a polyubiquitin chain interact with the polyubiquitin-binding site on the stopper of a proteasome
Stopper unfolds the target protein and threads it into the proteasome’s central cylinder, which is lined with proteases that chop the protein to pieces

51
Q

What are the functions of the large and small subunit of the ribosome in translation

A

Large - catalyses the formation of peptide bonds that covalently link amino acids into a polypeptide chain

Small - matches the tRNA to the codons

52
Q

How is complexity and differentiation achieved?

A

All cells have the same DNA
Different cell types express different sets of proteins
Cells respond to external signals

52
Q

How is gene expression controlled in eukaryotic cells

A

main site of control is step 1

1) controlling when and how often a given gene is transcribed from DNA into RNA
2) controlling how an RNA transcript is spliced or processed
3) selecting which mRNAs are exported from the nucleus to the cytosol
4) regulating how quickly certain mRNA molecules are degraded
5) selecting which mRNAs are translated into protein by ribosomes
6) regulating how quickly specific proteins are destroyed

53
Q

Example of gene expression control in bacteria

A

Balance of glucose and lactose determine Transcription of lac operon

When lactose is absent:
Lac repressor binds to the Lac operator and shuts off expression of the operon (cluster of genes that are transcribed)

Addition of lactose increases the intracellular concentration of a related compound, allolactose
Allolactose binds to the Lac repressor, causing it to undergo a conformational change that releases its grip on the operator DNA

When glucose is absent:
cyclic AMP is produced by the cell, and CAP binds to DNA
For the operon to be transcribed, glucose must be absent (allowing the CAP activator to bind) and lactose must be present (releasing the Lac repressor).
LacZ, the first gene of the operon, encodes the enzyme β-galactosidase, which breaks down lactose to galactose and glucose

54
Q

How can chromatin-modifying proteins help initiate gene transcription?

A

Recruitment of histone acetyltransferases promotes the attachment of acetyl groups to specific histones. These acetyl groups serve as binding sites for proteins that promote transcription

Recruitment of chromatin-remodeling complexes render the DNA packaged in nucleosomes more accessible to other proteins in the cell, including those required for transcription initiation

54
Q

How does eukaryotic gene expression control occur?

A

Gene activation can occur at a distance

An activator protein bound to a distant enhancer (on DNA) attracts RNA polymerase and the general transcription factors (transcription initiation complex) to the promoter

DNA loops to allow contact between the activator and the transcription initiation complex bound to the promoter.

In the case shown here, a large protein complex called Mediator serves as a go-between. The broken stretch of DNA signifies that the segment of DNA between the enhancer and the start of transcription varies in length. The TATA box is a DNA recognition sequence for the first general transcription factor that binds to the promoter.

55
Q

What is the difference between general transcription factors and transcription regulators?

A

GTFs are universal and essential for the transcription of all genes transcribed by RNA polymerase II

Transcription regulators are specific to different genes. They are proteins that bind to regulatory DNA sequences (which are located at various positions relative to the promoter)
Transcription regulators control whether a particular gene is transcribed and at what rate.
Their binding sites and roles vary depending on the gene being transcribed.

56
Q

How can a single transcription regulator coordinate expression of many different genes

A

Transcription regulator binds to the regulatory DNA sequence (promoters/enhancers)
If multiple genes share the same or similar regulatory sequences, the same transcription regulator can influence their expression simultaneously.

57
Q

How can differentiated cells de-differentiate into induced pluripotent stem cells (iPS)

A

A set of 3 genes are introduced into fibroblast nucleus
These genes are artificially expressed (each of which encode a transcription regulator) which reprograms a fibroblast into an iPS cell
Such iPS cells can proliferate indefinitely in culture and can be stimulated by appropriate extracellular signal molecules to differentiate into almost any cell type in the body

57
Q

Who was Dolly

A

First cloned mammal from an adult mammary gland derived somatic cell

Evidence that somatic cells can be reprogrammed

58
Q

How can a cell reinforce its cell identity?

A

In vertebrate cells, DNA methylation occurs on selected cytosine bases that fall next to a guanine
This modification blocks gene transcription
DNA methylation patterns are passed onto progeny cells by an enzyme that copies this pattern on the parent DNA strand to the daughter DNA strand as it is synthesized

58
Q

Apart from methylation what is another type of regulation of gene expression that can be inherited?

A

Histone modifications

When a cell replicated its DNA, each daughter double helix recieves half of its parent’s histone proteins (containing modifications)
Enzymes responsible for the modifications can bind to the parental histone modifications and can confer the same modifications to the new histones nearby

59
Q

What is the function of miRNAs

A

Control gene expression by base-pairing with specific mRNAs and reducing their stability and their translation into protein

60
Q

How does miRNA destruct target mRNA?

A

Precursor miRNA transcript processed to into a mature, single-stranded miRNA.

This miRNA assembles with a set of proteins into a complex called RISC, which then searches for mRNAs that have a nucleotide sequence complementary to its bound miRNA.

Depending on how extensive the region of complementarity is, the target mRNA is either rapidly degraded by a nuclease within the RISC or transferred to an area of the cytoplasm where other nucleases destroy it

61
Q

Principle of western blotting

A

Detects a specific protein in a blood or tissue sample

62
Q

Procedure of western blotting

A

Preparation of proteins:
Treat with SDS which denatures proteins, unfolding them and negatively charging them (makes proteins similar in charge)

Proteins loaded onto polyacrylamide gel: travel through the gel from -ve to +ve anode (as they are -vely charged) = separates the proteins by size

Staining polyacrylamide gel = creates visible bands where proteins are

Use marker proteins as a reference to determine molecular weight of the proteins in the sample (but not sure which exact protein the band is representing)

Gel placed on top of nitrocellulose membrane, sandwiched between filter paper soaked in buffer

Transfer proteins from gel onto membrane: ^ fixed between cathode and anode where electric current pulls the -vely charged proteins towards the +ve anode

Blocking: Incubate membrane w/ milk or BSA which prevents antibodies from non-specifically binding to membrane or other proteins

Incubate w/ primary antibody which binds selectively to protein of interest

Wash

Incubate w/ secondary antibody specific to primary antibody

Wash

Detection: secondary antibodies are conjugated to an enzyme (horseradish peroxidase)

Enzyme produces chemiluminescence signal that visualizes the protein band

62
Q

Principle of immunohistochemistry

A

Staining that allows you to see where proteins are localized in a cell

63
Q

When is western blotting used

A

Separate + identify a wide range of proteins from a complex sample (e.g if you want to see if a protein has been expressed in a cell)

Highly specific, can detect proteins that are present at relatively low samples = investigate the purity of a sample (based on relative abundance)

Disease identification: confirmatory test after a positive ELISA test has been produced to ensure the ELISA test did not produce a false-positive

Detecting post-translational modifications: (a synthesized protein may undergo modification e.g phosphorylation or acetylation) indicates whether the modification has occurred by using antibodies specific to the modification

64
Q

Immunohistochemistry procedure

A

Tissue preparation: tissue sample fixed with formalin > embedded in paraffin wax > cut into thin slices placed on glass slides

Antigen retrieval: unmask epitope of the antigens for antibody binding (using heat for example)

Blocking background: reduce non-specific antibody binding, tissue sections are treated with a blocking solution (e.g., normal serum, BSA, or casein)

Antigen detection + visualization:

Fluorescence-based detection method: the antibody is conjugated to a fluorophore (a molecule that fluoresces under light of a specific wavelength)
Visible in fluorescence microscopy

Colorimetric method: the antibody is conjugated to an enzyme (e.g. horseradish peroxidase) which catalyzes a reaction that yields a precipitating colored product after incubation with a buffer
Visible in light microscopy

64
Q

Controls of immunohistochemistry

A

Positive control: The tissue sample or cell type used should be known to express the protein (antigen) of interest that the primary antibody targets

Negative control: The tissue sample or cell type should not express the protein of interest

65
Q

What is double staining in immunohistochemistry + what considerations should be taken

A

Detection of two different antigenic epitopes on one slide/tissue

Considerations:
Antibody Selection: Choose primary antibodies from different host species or isotypes to minimize crossreactivity

Pretreatment: Test each antibody separately before performing double labeling. If both antibodies survive the “double pretreatments”, immunohistochemistry double staining can be performed

Secondary Antibodies: Select secondary antibodies with different fluorophores or chromogens to distinguish between the two antigens

Antigenic epitopes are in different cell types or in different cellular compartments (e.g. nuclei and cytoplasm)

66
Q

Procedure for double staining immunohistochemistry

A

Procedure:
Tissue Preparation: Fix and process tissue samples according to standard IHC protocols.

Primary Antibody Incubation: Incubate tissue sections with both primary antibodies (antigen-specific) simultaneously, using optimized pretreatment conditions for each antibody.

Secondary Antibody Incubation: Incubate tissue sections with secondary antibodies (antibody-specific) conjugated to different fluorophores or chromogens.

Detection: Visualize the labelled antigens using fluorescence microscopy or chromogenic substrates.

67
Q

What is gold immunohistochemistry

A

Uses gold particles conjugated to secondary antibodies

Visible with electron-microscopy = high-resolution detection = allows precise localization of antigens at the ultrastructural level

68
Q

Flow cytometry principle

A

Analyzes characteristics of individual cells or particles in a population:

sorts cells into size + cellular complexity

Based on fluorescence

Analyse cell cycles or cell viability

69
Q

Procedure for flow cytometry

A

Identification:
Tag cells w/ antibodies
Proteins on outside - mix w/ antibodies
Proteins on inside - internal staining
Addition of fluorophores (directly linked to antibody + emits color)

Cells/particles are suspended in fluid
Sample is passed through a narrow stream, where the cells line up in single file

Cells injected into a stream of sheath fluid which makes them move at a constant rate (low conc of cells + travel as slow as possible = best results)

As cells pass through laser beam light is scattered + detected

Detectors capture:
Forward scatter (FSC): Measures cell size (larger cell = scatter more light forward)
Side scatter (SSC): Indicates cell complexity or granularity (cells w/ more internal structures = scatter more light sideways

70
Q

Data interpretation for flow cytometry

A

Histogram
Positive control:
Expected level of fluorescence from cells known to express the target antigen
Clear separation from the negative control suggests effective binding of the antibody

Negative control:
Baseline fluorescence levels of cells that should not express the target antigen
Ideally tightly grouped near the origin (low fluorescence)
Broad or shifted peak = issues e.g high autofluorescence or nonspecific binding

71
Q

Usage of flow cytometry

A

Determine cell cycle phases
Study cell proliferation
Apoptosis detection
Detecting chemical + physical differences of cells

71
Q

Direct ELISA principle

A

The detection antibody directly binds to the antigen
No amplification caused by the secondary antibody = less sensitive but less prone to error
Intensity of the color change (absorbance @ specific wavelength using a spectrophotometer) = amount of antigen present in sample

72
Q

Direct ELISA procedure

A

Antigen coated well
Wash well from excess antigens
Add blocking buffer
Rewash
Add enzyme-conjugated primary detection antibody
Wash to remove any unbound antibodies
Horseradish peroxidase added to induce color change

73
Q

Advantages + disadvantages of direct ELISA

A

Advantages
Reduced cross-reactivity (direct binding, specific results)
Rapid procedure

Disadvantages
High cost
Tedious (each primary antibody has to be labelled separately
Less sensitive

74
Q

Advantages + disadvantages of indirect ELISA

A

Advantages
Higher sensitivity than direct ELISA
More economical

Disadvantages
Longer process than direct ELISA
Risk of cross-reactivity (multiple binding sites could cause unwanted proteins to bind + give false positives)

74
Q

Indirect ELISA procedure

A

After primary antibody is added and well is washed
Complementary secondary enzyme-conjugated antibody is added
Wash well

75
Q

Sandwich ELISA procedure

A

Immobilize capturing antibody
Add sample
If antigen is present, will bind to capture antibody
Enzyme-conjugated detection antibody added
Wash
Add substrate

76
Q

Competitive ELISA procedure

A

Incubate sample with antibodies
Antibodies bind to target antigen if present
Coat ELISA well with reference antigens
Add sample to well (unbound antibodies bind to reference antigens)
Wash
Add secondary conjugated antibodies
Wash
Add substrate for color change

76
Q

Interpreting ELISA data

A

Lighter color = less free antibodies
Darker color = more free antibodies

77
Q

ELISA usage

A

Direct: analyzing the immune response to an antigen, quantitative antibody detection

Indirect: determining total antibody concentration in samples (e.g. HIV) in order to test if a patient is infected

Detecting and estimating levels of tumor markers

Mostly used with body fluids

78
Q

ELISA controls

A

Positive control:
Test functionality of assay
Produces color change showing it is functioning correctly

Negative control:
Establishes baseline responses, ensuring no false positives
It can identify non-specific binding
Doesn’t contain target antibody or antigen
If color change observed = non-specific interactions, incomplete washing steps, or cross-reactivity

79
Q

How can you study how specific cells react + cell function

A

must be done in cell culture
Cannot be studied in intact tissue
best off using epithelial cell

80
Q

Primary culture of cells

A

Primary cells grow out of tiny bits of tissue

Very tiny bits that can be scrapped off teeth can expand until millions in first passage and subsequent passages.

Passaging of cells goes through “lifting” the cells that are adhered to the tissue culture plastic.
Done by briefly adding trypsin, which will degrade some surface proteins, including the adhesion molecules by which the cells adhere to the plastic, or by adding EDTA, which also interferes with adhesion proteins.

Cells that adhere to each other
cells that contact each other will decline to devide any further.
(exception: tumor cells)

81
Q

What is the red liquid in tissue culture bottles

A

A color indicator
At lower Ph, it will turn yellow, meaning that you have to feed the cells

81
Q

What is a characteristic of early cell propagation

A

Mixed culture
First outgrowths can be quite heterogeneous

During subsequent passing cells become a bit more homogenous (losing its in vivo characteristics)

81
Q

Requirements for tissue culture medium to propagate/proliferate cells

A

vitamins, sugars, ions, essential amino acids, growth factors – serum, temperature, O2 and CO2,
+ mechanical stimulation (do not mimic real situation but there are ways to do mechanical stimulation)

82
Q

How can cells be stored in liquid nitrogen?

A

Cells are frozen in DMSO solution, avoids crystal formation via gradual freezing

82
Q

Advantages and disadvantages primary cell cultures
+ solutions

A

Advantages
Closer to reality in tissues: they may maintain characteristics of tissue they are from.

Disadvantages
Limited life span
Large individual variation

solution:
increase the number of biological replicates
use a cell line that is h-TERT immortalized: with telomerase transfection eternal life
Other possiblity: use ips cells

83
Q

What could determine the life span of cell cultures

A

cells from primary cell cultures have a short lifespan due to telomere shortening with each cell division

84
Q

What is Fibrodysplasia Ossificans Progressiva

A

Patients develop an extra skeleton
Due to mutated BMP-receptor, increased signaling
- Patients petrify
- Initiated at bruises

Difficult to get patient biopsies:
taking a biopsy can cause extra-
skeletal bone formation

85
Q

Characteristics of cell lines

A

Can be used inexhaustibly: will multiply
Often of tumor origin
Can be exchanged between laboratories
you can compare results world-wide when more labs are using the same cell line

86
Q

What is the most famous cell line

A

HeLa
Cervical cancer cell line
From 1951
Immensely propagated
Used for polio vaccine development

86
Q

What is aneuploidy + what are the implications

A

deviant chromosome numbers

Gives rise to genetic instability: after the next cell division, there are unbalanced separation of chromosomes.

87
Q

What are the limitations/differences of typical cell biological experiments compared to in vivo?

A

Cell + reagent X

How is this different from what cells experience in vivo?
cells do not grow on plastic
cells in vivo experience different O2/CO2 concentrations
influence of other cell types around
influence of substratum that conveys a characteristic stiffness
fluctuating serum constitution
flow and movement of tissu

88
Q

Features of cells

A

Cells respond to their substrate features (e.g Stiffness of artificial substrate directs differentiation) (different stiffnesses regulate differention pathway into neuronal cells, muscle cells or osteoblasts)
Different cell types influence each other

89
Q

Skin mimicking in vitro: advantages + what is missing compared to in vivo

A

Advantage:
can be used to test drugs
recuperation after scar formation/burn wounds.

Missing:
blood supply
immune cells that are normally dispersed in the dermis

90
Q

What is a tissue/body on a chip

A

a tiny, simplified model of a human organ or system using living cells, allowing researchers to study how these tissues behave under different conditions

replicate the functions of human tissues or organs on a small, controlled device

91
Q

Features of tissue/body on a chip

A

Multiple cell types can be added
Mimic of blood flow
Cells secrete their own factors that influence other cells.
Also potential for drug testing: which cell type is affected in a multi-celltype culture?

92
Q

bone marrow on chip study: overview

A

Background: osteoblasts play a role in maintenance of the haematopoietic niche
Mimic: using mesenchymal stromal cells (2D) vs. 3D culture on discs
3D device with microfluidics.

Aim: develop an in vitro bone marrow-on-a-chip model that can sustain long-term culture of HSPCs
This model aims to replicate the human bone marrow niche, which is crucial for HSPC maintenance and differentiation

Finding: the developed 3D system, which uses a hydroxyapatite-coated scaffold + includes MSCs (cells that can differntiate into osteoblasts) and HSPCs, was able to sustain the culture of primitive HSPCs for up to 28 days

93
Q

What does figure 1 show us in the bone marrow on a chip study

A

It is a scheme for the generation of the bone marrow model

A) MSCs were seeded onto the hydroxyapatite coated ceramic scaffold and cultured in DMEM +10% FCS + 1% P/S in a static environment for 1 week.

B) HSPCs were added to the prepared ceramics. The picture shows a staining of MSCs cultured on the ceramic for 7 days

C) The entire model is transferred to a microfluidic system (MOC), where it is cultured dynamically for up to 4 weeks. This system mimics the bone marrow microenvironment, and the bone marrow model is inserted into a compartment while the other compartment serves as a medium reservoir

94
Q

What does figure 2 show us in the bone marrow on a chip study

A

Scaffold for the bone marrow model

A) A comparison between the structure of the ceramic scaffold vs the in vivo bone marrow structure.
Porous scaffold mimics the cancellous (spongy) bone architecture, which is important for creating a bone marrow-like environment

B) A schematic view of the microfluidic system (MOC) from below. The bone marrow model was positioned in the culture compartment opposing the micro pump. The black arrow indicates the direction of the medium flow

95
Q

What does figure 3 show us in the bone marrow on a chip study

A

MSCs build up a suitable environment for HSPC culture within 7 days

A) Immunofluorescent staining of MSCs cultured in the ceramic for 7 days. The image shows the presence of stem cell factor (SCF) on the surface of MSCs (red) and fibronectin (green), which is part of the extracellular matrix. This environment is crucial for the support of HSPC culture. (blue stained nuclei of MSCs showing their location

B) Expression of various bone marrow niche-related genes (e.g., SCF, fibronectin) in MSCs. The graph shows higher expression of these genes when MSCs are cultured on the 3D ceramic scaffold compared to a 2D monolayer, indicating that the 3D environment better supports niche formation

96
Q

What does figure 4 show us in the bone marrow on a chip study

A

MSCs express various bone marrow niche related genes in the ceramic

A) RNA expression levels of MSCs cultured in two different mediums (2D and 3D ceramic scaffold) over 4 weeks. MSCs in the top row medium show better survival and gene expression. Bottom row: cells didnt divide properly

B) Gene expression of bone marrow niche-related markers (e.g., VEGF, SCF, fibronectin, nestin) over 4 weeks of culture in both the ceramic scaffold and the 2D monolayer. Checked that gene expression was constant over time. The 3D scaffold consistently supports the expression of these key niche-related genes, indicating the stability of the bone marrow environment.

96
Q

What does figure 6 show us in the bone marrow on a chip study

A

HSPCs remain in their native state after 4weeks culture

A) Flow cytometry plots showing that a sig proportion of HSPCs retain their primitive state even after 4 weeks of culture within the scaffold

B-D) Graphs showing the %s of different HSPC populations over time. CD34+CD38− cells (primitive HSPCs) make up around 31.71% of the population after 4 weeks, with CD34−CD38+ cells remaining stable and CD34+CD38+ cells decreasing.

E) Shows the proportion of native HSC markers, which remain similar to the population of freshly isolated HSPCs. (Key point: percentages do not differ that much compared to the beginning WOW they can maintain this niche !!!!)

F) Results of the CFU-GEMM assay, indicating that the extracted HSPCs are still functional and can differentiate into various colonies after 4 weeks of culture

G) A table comparing the average number of colonies formed by HSPCs cultured for 4 weeks and freshly isolated HSPCs.

Conclusion: Niche was maintained i.e it is a stable environment

97
Q

Why did they study these mRNA in the bone marrow on a chip study?

A

They are known to play key role in maintaining HSPCs + supporting their function and behavior in the bone marrow

Can assess whether MSCs cultured in the 3D ceramic scaffold were successfully creating a microenvironment that mimics the bone marrow niche

98
Q

What is an issue of the bone marrow on a chip study?

A

A control of only ceramics without MSC was not used

99
Q

usage of immunohistochemistry

A

to detect specific proteins in tissue sections
visual evidence of protein presence and localization
identify tumor markers
clinical diagnostics