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TCD from molecules to cells 2nd year > Genetics > Flashcards

Genetics Flashcards

1
Q

Frederick Griffiths transformation experiment

A

Transforming principle

Two kinds of streptococcus pneumonia

Type IIR is nonvirulent. Injection of mice with this will not kill them. No bacteria are recovered

Injection of a type IIIS will kill the mice. These bacteria are recovered

When Griffith injected a heat-killed version of the virulent bacteria, the mice survived.

However, when Griffith injected the mice with heat-killed virulent bacteria, as well as living nonvirulent bacteria, the mouse died and virulent bacteria was recovered.

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2
Q

What was the hypotheses from Griffiths experiment?

A

That the transforming agent was an IIS protein, However this was wrong and it was actually DNA

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3
Q

Transformation principle

A

DNA is transformed from one individual to another in genetic transformation

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4
Q

Oswald T Avery’s Transformation experiment

A

He took a mixture of DNA and RNA from smooth bacteria, (The virulent kind)
Treated them with RNase,which kills RNA - He then mixed them up with the non-virulent kind, which resulted in virulent smooth colony bacteria The transformants were still produced

However, when he treated them with DNase, no transformants were produced.

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5
Q

What did Averys experiment prove

A

That DNA is the transforming factor

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6
Q

Alfred Hershey and Martha Chase

A

Bacteriophage - little viruses that infect bacteria. Bind to the bacteria and inject material that will infect the bacteria and generate new phage
After they inject this material - phage ghosts (dead phage) fall off and vanish into the medium
Labelled proteins specifically in the phage with radioactive sulfur, which is not found in amino acids.
All of the radioactivity was found in phage ghosts

The other phage set was labeled with radioactive phosphorus.

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7
Q

DNA strand directionality

A

5’phosphate end
3’ hydroxyl end
The strands are in anti-parallel orientation

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8
Q

How was the 3D structure of DNA determined

A

Using X-Ray crystallography

DNA can be precipitated from solution by the addition of ethanol

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9
Q

The meselson-stahl experiment

A
  1. Grow E. coli cells for many generations in N-15 medium
  2. Centrifuge culture to obtain the cells (their DNA is N-15)
  3. Transfer the N-15 cells into fresh N-14 growth medium
    The cells will replicate their DNA every 20 min Newly
    made DNA will be labelled with N-14
    N-14
  4. Extract DNA and investigate its banding
    position on a CsCl gradient
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10
Q

Testing models of DNA replication by isotopic labeling of DNA

A

Extract DNA labeled with N-15

Centrifugation of DNA in a CsCi gradient
This heavy Nitrogen, hence the bands will be near the bottom.

Heavy high concentration of cesium chloride is at the bottom and a light concentration is at the top.
DNA bands at the density that corresponds to this gradient.
In the first replication you get a hybrid density, in the middle - indicating its a semi conservative mechanism
In this second replication you get a light band, as two of the daughter strands are not made up of N14.

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11
Q

phosphodiester bonds

A

a chemical bond joining successive sugar molecules in a polynucleotide.

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12
Q

How is DNA replicated

A

DNA replicated via template- directed base pairing of nucleotides and step wise creation of phosphodiester bonds.
The nucleotide comes in, a phosphorylation happens
ATP , and an ADP is released.

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13
Q

Replication fork

A

The replication fork is a structure that forms within the long helical DNA during DNA replication. It is created by helicases, which break the hydrogen bonds holding the two DNA strands together in the helix.

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14
Q

Topoisomerases

A

Positive supercoiling - overwinding. If you have a tightly wound segment of DNA and trying to pull it apart. You have increased tension. Topoisomerases fix this.

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15
Q

DNA polymerase

A

Adds nucleotides to a pre-existing primer strand
(primer extension)

can only add nucleotides to the 3’ end
hence DNA synthesis occurs exclusively in a 5’ to 3’ direction

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16
Q

DNA synthesis directionality

A

Adds nucleotides to a pre-existing primer strand
(primer extension)

can only add nucleotides to the 3’ end
hence DNA synthesis occurs exclusively in a 5’ to 3’ direction

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17
Q

DNA replication is semi discontinuous

A

The Leading strand copies continuously i.e. a single long
molecule
– The Lagging strand copies in segments (= Okazaki
fragments) which must eventually be joined up!

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18
Q

Replication as a process

A 
. Double-stranded DNA must first unwind
3. A new strand is formed by pairing
complementary bases with an old strand
2. The junction of the unwound molecules
is called a Replication Fork

Each replicated DNA molecule has
one old and one new DNA strand

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19
Q

The enzymatic activities of DNAP 1

A

5’ to 3’ DNA polymerizing activity (forward synthesis)
• 3’ to 5’ exonuclease activity (backwards, base error
correcting)
• 5’ to 3’ exonuclease activity (forward, removal of
RNA primers)

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20
Q

The 5 different DNAPs in E coli

A

Five different DNAPs have been discovered in E. coli:
• DNAP I: functions in replication & repair
• DNAP II: functions in repair (proven in 1999)
• DNAP III: the principal DNA replication enzyme
• DNAP IV: functions in repair (discovered in 1999)
• DNAP V: functions in repair (discovered in 1999)

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21
Q

How does the synthesis of a new strand get started?

A

DNA polymerase always requires a primer to start synthesis of a new strand

the primer must be base paired with the template strand

however when double-stranded DNA unwinds for copying, there is no primer

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22
Q

Where does the initiation primer and Okazaki primers come form

A

It was observed that the initiation of DNA replication was very sensitive to an inhibitor of RNA polymerase. (Rifampicin)

However, the continuation of DNA replication is not sensitive to this inhibitor.
Hence it can be concluded that since RNA polymerase is required to begin DNA replication, that RNA is also required to begin DNA replication.

Okazaki fragments can incorporate H - Uracil, which only RNA molecules can do
If Okazaki fragments are treated with alkali they get shorter. Only RNA is degraded by alkali.

The conclusion from this is that Okazaki fragments get started by a short RNA primer.
These RNA primers are made by a special RNA polymerase called RNA primase.

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23
Q

How are okazaki fragment primers and leading strand primers made?

A

The RNA primer that primes synthesis of the leading strand is made by the main e coli RNA polymerase , RNA polymerase 2

Okazaki fragments get started by a short RNA primer. These primers are made by a special RNA polymerase called RNA primase.

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24
Q

How do the okazaki fragments become linked together?

A

The physiological function of Pol I is mainly to support repair of damaged DNA, but it also contributes to connecting Okazaki fragments by deleting RNA primers and replacing the ribonucleotides with DNA.

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25
Q

Helicase role in replication

A

Unwinds DNA

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26
Q

Role of RNA primase in DNA replication

A

Okazaki fragments are initiated by short RNA primers, these primers are made by special DNA polymerases called primases

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27
Q

Role of DNAP 3 in replication

A

Adds nucleotides to the 3’ end of the RNA primer

i.e its role is primer extension

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28
Q

Joining up of Okazaki fragments

A

DNA pol III while making new Okazaki DNA (red) approaches
the next RNA primer at
(2) DNA pol I takes over, removing the RNA primer and replacing it
with new DNA - this is called “Nick translation”

DNA ligase seals the nicks between the pieces of DNA

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29
Q

What is an origin of replication and how many do e coli have

A

An origin of replication is a sequence of DNA at which replication is initiated on a chromosome, plasmid or virus. For small DNAs, including bacterial plasmids and small viruses, a single origin is sufficient.

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30
Q

What is bidirectional replication

A

Bidirectional replication is a method of DNA replication found in organisms from each of the main kingdoms. Bidirectional replication involves replicating DNA in two directions at the same time resulting in a leading strand, where replication occurs more rapidly, and a lagging strand with slower replication

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31
Q

Why do origins with AT rich elements attract the helicase?

A

AT has a much weaker base pairing and is easier to pull apart

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32
Q

The end replication problem

A

he end replication problem causes a progressive shortening of telomeric DNA at each round of DNA replication, thus telomeres eventually lose their protective capacity. This phenomenon is counteracted by the recruitment and the activation at telomeres of the specialized reverse transcriptase telomerase -
teolmerase is an enzyme responsible for the maintenence of the length of telomeres.

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33
Q

telomeres

A

Telomeres are the protective caps on the ends of the strands of DNA called chromosomes, which house our genomes. (only in eukaryotes, prokaryotes have circular chromosomes)

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34
Q

Oligonucleotides

A

Oligonucleotides are used as probes for detecting specific sequences that are complementary to the oligonucleotides. When a certain sequence needs to be detected, a complementary oligonucleotide is synthesized in the laboratory.

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35
Q

PCR process

A

Separate the DNA strands by heating to 98 degrees celsius
Reduce the temperature to allow oligo primers to anneal (to base pair with the respective bases)
Reduce the temperature, so as not to kill the enzyme. Then add the polymerases. dNTPs and DNA polymerase.

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36
Q

Threshold of detection and PCR

A

Signal after 20 cycles - high infection of viruses within the person
30 - dubious whether you even have it or not.

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37
Q

Taq DNA polymerase

A

PCR uses a heat stable DNA polymerase from the thermophilic bacterium
this enables automation and means we do not have to cool the dna every time we want to add the enzymes

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38
Q

RT PCR covid diagnostics

A

In a real-time RT-PCR machine, the sample is tested using fluorescent light because each nucleotide that builds the reverse-transcribed DNA has a fluorescent dye.

If the virus is present in the sample, it fluoresces. The more DNA present, the more it will glow. If there is no virus in the sample, it will not fluoresce, because no fluorescent chain was built.

As the DNA accumulates over time, the machine measures how much fluorescence is in each sample and displays it on a screen. This data helps calculate whether the positive result is strong or weak.

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39
Q

Amplification of alleles based on length polymorphisms

A

We design a forward and reverse primer to target the gene of interest, and then we distinguish them based on size using gel electrophoresis.

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40
Q

Multiplex genotyping

A

The genotyping procedure included a one-round multiplex polymerase chain reaction (PCR) with simultaneous incorporation of a fluorescent label into the PCR product and subsequent hybridization on a biochip with immobilized probes.

We use fluorescently labelled primers to amplify alleles, and then we separate them by size on the same gel.

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41
Q

DNA sequencing

A

Determining the base sequence of a DNA molecule

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42
Q

Sanger sequencing

A

Sanger sequencing results in the formation of extension products of various lengths terminated with dideoxynucleotides at the 3′ end. The extension products are then separated by Capillary Electrophoresis or CE.

The absence of the 3’ -OH group in a dideoxynucleotide prevents further elongation of the chain and terminates chain synthesis.

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43
Q

General principle of sequencing by synthesis

A

Use a radiolabeled primer, and add dNTPSs and DNA polymerase to extend the primer.
the site to which the primer anneals defines the starting point.
The newly synthesised strands are separated by size using gel electrophoresis through a polyacrylamide gel.

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44
Q

Fluorescently labelled dideoxynucleotides

A

Fluorescently labelled dideoxynucleotides make sequencing more efficient.
Each dideoxynucleotide is tagged with a unique colored fluorophore.
Combine the 4 reactions and separate them on a capillary

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45
Q

Next generation sequencing

A 
Very fast, parallel short reads of
Millions of (35-400bp) sequences reads.
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46
Q

Shotgun sequencing

A

Sequence millions of small random fragments (10X coverage).

Computers assemble the sequence

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47
Q

Where are proteins made

A

in the cytoplasm

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48
Q

How does DNA in the nucleus specify the production of proteins in the cytoplasm?

A

DNA is located in the nucleus (in chromosomes) But
proteins are made in the cytoplasm
2. RNA is synthesised in the nucleus
3. RNA migrates to the cytoplasm where proteins are
synthesized
4. In viral infections: RNA synthesis precedes new viral
particle formation.

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49
Q

Structure of RNA

A

Structure of RNA
• A polymer with a ribose-phosphate backbone
• Ribonucleic acid (as against Deoxy-Ribonucleic acid)
• Bases (4 kinds):
Purines: Adenine, Guanine
Pyrimidines: Uracil, Cytosine
• Usually a single-stranded molecule (unlike DNA) - it
could form double helices, but the complementary
strand is not usually present in cells.

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50
Q

The key biochemical differences

between RNA and DNA

A
  1. The sugar: Ribose in RNA vs 2-deoxyribose in DNA
  2. Base difference: Uracil in RNA vs Thymine in DNA
    - but uracil can pair to adenin like thymine
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51
Q

Different types of RNA polymerases

A 
in Prokaryotes:
a single RNA polymerase (RNAP)
in Eukaryotes:
RNAP I - makes rRNA
RNAP II - makes mRNA
RNAP III - makes tRNA
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52
Q

Promoters

A

(Prokaryotes) DNA sequences that guide RNAP to the beginning of a gene to start transcription

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53
Q

Terminators

A

(Prokaryotes) DNA sequences that specify the termination of RNA
synthesis (STOP) and release of RNAP from the DNA.

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54
Q

Nomenclature

A

Transcription initiation site is called +1
no zero positon
upstream and downstream
template strand = coding strand

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55
Q

What does a promoter look like and how does RNAP recognise one

A

Prokaryotic promoters contain 2 highly conserved regions of base pairs, and RNAP binds a similar region in the promoters of many different genes.

(TATA box)

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56
Q

Roles of RNA polymerase

A

Roles of RNA polymerase - scans the DNA to identify a promoter.
Initiates transcription of mRNA.
Elongates the mRNA chain.
Terminates transcription
Interacts with regulatory proteins, activators and repressors.

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57
Q

Promoter structure in Prokaryotes

A

Promoters typically consist of a 40 bp region
located upstream of the transcription start site
• Promoters contain 2 conserved sequence elements:
– The “-35 region”, with consensus TTGACA
– The “-10 region”, with consensus TATAAT

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58
Q

RNA polymerase structure in prokaryotes

A

two alpha subunits, two beta subunits

A sigma subunit, which is required for specific recognition of promoters

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59
Q

Binding of polymerase to template DNA

A
  1. RNA polymerase binds non-specifically to DNA
    and scans, looking for a promoter.
  2. Sigma subunit recognizes promoter sequence.
  3. RNA polymerase and promoter form a “closed
    promoter complex” (DNA not yet unwound).
  4. RNA polymerase then unwinds about 12 base
    pairs to form “open promoter complex”.
    In the open promoter complex it no longer requires the sigma subunit. It gets kicked off, you now have polymerase with an open bit of DNA.
    Starts transcribing from it and the RNA gets kicked off the other side.
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60
Q

Termination of transcription in e coli

A

Simple - No protein factor required. Hairpin structure forms near the end of the mRNA, causing the RNAP to dissociate.
2. Complex, rho dependent. A protein factor called rho is required.

61
Q

transcription

A

how mRNA is made from a DNA template

62
Q

3 different RNA polymerases

A

in eukaryotes - RNAP 2 - makes ribosomal mRNA
RNAP 1 - makes ribosomal RNA
RNAP 3 - makes transfer RNA

In prokaryotes; single RNA polymerase, which transcribes all of them

63
Q

transcription factors

A

repressors or activators

64
Q

lac operon in e coli

A

In the absence of lactose, the lac repressor binds the operator, and transcription is blocked

in the absence of lactose, the lac repressor is released from the operator and transcription proceeds at a slow rate

65
Q

operator

A

An operator is a genetic sequence that allows proteins responsible for transcription to attach to the DNA sequence. The gene, or genes, which get transcribed when the operator is bound are known as the operon

66
Q

Transcriptomics

A

Allow us to analyze gene sequences and expression levels of all genes in an organism
In many cases, simultaneously, this allows us to study how this varies; in populations, cells types, in health and disease and in many other situations.

67
Q

Genetic totipotence of somatic cells

A

Every cell has in it the same DNA that another cell has
Shown by john gurdon
He took an adult frog and cultured its skin cells, sucked the nucleus
Unfertilised egg, destroyed nucleus
Injected the original nucleus
This nucleus allowed the egg to develop

68
Q

Core promoter

A

This is where RNA polymerase associates to start transcription

Access to the core promoter is regulated by transcription factors and chromatin state

69
Q

Chromatin state

A

in eukaryotes, DNA is wrapped around histones to give chromatin. Promoters have many control sites for transcription factors

Chromatin controls access of all proteins - transcription factors, activators ect, to the promoter

70
Q

Typical control of a eukaryotic transcription via poll 2

A

Access of all of the molecules to the promoter is controlled by upstream sequences that contain regulatory proteins.
Binding of these proteins allows for loosening of the chromatin, so that RNA polymerase can bind to the promoter sequence.
Proximal control elements - GTFs - general transcription factors control the loosening of chromatin so RNA polymerase can bind to the promoter sequence.

Regulatory proteins - enhancer binding proteins/silencer binding proteins. These proteins influence the state of chromatin, and allow access of general transcription factors to bind to the operator so that RNAP can bind to the promoter and begin transcription.

71
Q

Major differences between eukaryotic and prokaryotic transcription

A

In eukaryotes - a nucleus surrounded by a double membrane

DNA is wrapped around histones to give chromatin
Promoters have many control sites for transcription factors

RNAP has 12 protein subunits and can associate with many different regulatory proteins.

72
Q

The proximal core promoter in eukaryotes

A

GTFs (General transcription factors) substitute for the sigma factor.

Distal control elements

enhancer and silencer elements controlled by transcriptional activators/repressors and influence GTF and Pol 2 recruitment

Regulatory proteins - enhancer binding proteins/silencer binding proteins. These proteins influence the state of chromatin, and allow access of general transcription factors to bind to the operator so that RNAP can bind to the promoter and begin transcription.

73
Q

pribnow box

A

The bacterial homolog of the TATA box is called the Pribnow box which has a shorter consensus sequence.

74
Q

Transcription initiation in Eukaryotes

A

TATA binding protein binds to the TATA box in the genes promoter region

General transcription factors are recruited to form the pre-initiation complex - that includes RNA polymerase
(GTFs are a complex version of sima factor)

RNAP is activated by phosphorylation of its C terminal tail and then begins transcription

75
Q

TATA binding protein

A

The TATA-binding protein (TBP) is a general transcription factor that binds specifically to a DNA sequence called the TATA box.

76
Q

Pre initiation complex assembly

A

binding of TATA binding protein and transcription factor IID

Assembly of a Pre- Initiation complex (PIC) around RNA polymerase (RNAP)

Transcription bubble opens up

77
Q

transcription bubble

A

A molecular structure formed during DNA transcription when a limited portion of the DNA double strand is unwound.

78
Q

How is RNAP activated

A

RNAP is activated by phosphorylation of its C - terminal tail and then begins transcription which leaves many of the general transcription factors behind

79
Q

Processing steps required to produce mature mRNA in eukaryotes

A

The mRNA precursor has EXONS and INTRONS
INtrons are none coding segments of DNA, these are spliced out and removed in the nucleus
A 5’ cap and a 3’ polyA tail are added to the mRNA

80
Q

Alternative splicing

A

yields different mRNAs and proteins isoforms

81
Q

Transcriptome

A

a collection of all the gene readouts present in a cell

82
Q

DNA hybridization and melting

A

Two highly negatively-charged polymers are held together by base pairing

A Velcro-type reaction, many low-energy linkages multiply their effects to hold DNA together • DNA can be melted to single strands when temperature is high • Short DNA duplexes are particularly unstable

83
Q

DNA hybridization

A

Hybridization is the process of combining two complementary single-stranded DNA or RNA molecules and allowing them to form a single double-stranded molecule …

84
Q

Northern blotting

A

Can be used to detect specific mRNAs
It is a laboratory technique used to detect a specific RNA sequence in a blood or tissue sample.
The sample RNA molecules are separated by size using gel electrophoresis. The RNA fragments are transferred out of the gel to the surface of a nylon membrane, via electroblotting. The membrane is exposed to a DNA probe labeled with a radioactive or chemical tag. If the probe binds to the membrane, then the complementary RNA sequence is present in the sample.

You can the estimate levels of DNA complementary to the mRNAs (cDNA)
DNA is easier to isolate and work with, as it is more stable than RNA.
Reverse transcriptase makes DNA from an RNA template.
To do this you can make an oligo dT primer, which is specific to the polyA tail, as almost all mRNAs have one. We then add reverse transcriptase, which will transcribe a complementary DNA strand to the mRNA strand. Add RNAse H, which degrades the mRNA. If you then add the appropriate polymerases, you can make a double strand DNA.
This is called complementary DNA, or cDNA.

85
Q

Measuring the abundance of mRNA using amplification of cDNA

A

Reverse Transcription PCR

This quantifies mRNA indirectly via PCR amplification of cDNA copies

86
Q

gene

A

A transcriptional unit
encoding an RNA or protein, where
the gene product has some
biochemical or cellular function.

87
Q

•Gene products can function either as RNA molecules or proteins

A

All genes encode RNA molecules, some of which (i.e. messenger RNAs,
mRNAs) are translated into proteins. There is
no doubt that the protein-coding fraction of
our genomes is a major part of what makes
cells different from each other and is also
largely responsible for the different activity
states that cells can achieve. However, it is
very important to remember that many genes
encode RNAs that do not become translated
into protein and these RNAs possess
enzymatic or regulatory function as RNA
molecules. The key non-translated RNAs are
ribosomal RNA (rRNA), transfer RNA
(tRNA) and regulatory RNAs such as snRNA (small nuclear RNA),
microRNA (miRNA) and long non-coding RNA (lncRNA) that influence the
expression and translation of the protein-coding genes into proteins

88
Q

None coding parts of genes - eukaryotes versus prokaryotes

A

•Prokaryotic genomes are relatively compact and gene dense but the
genomes of higher organisms contain a huge proportion of non-coding
sequence that has gene regulatory function
In higher eukaryotic cells (such as mammals), only a tiny fraction of the DNA
is actually transcribed (2%) and translated into protein. Of the approximately
3 million genes that the human
genome could potentially
contain (if all of the DNA coded
for proteins and the average
gene was 1000 base pairs long), it
turns out that there are only
25,000 or so protein-coding
genes. The rest of the genome
consists of non-coding sequence
that regulates the expression of
the coding fraction of the
genome. Thus, the regulation of
gene expression is a key source
of complexity in higher
organisms and is more complex
and sophisticated than the gene
expression controls that exist in prokaryotes.

89
Q

How do higher eukaryotes get more from their genes when they dont necessarily have more protein coding genes?

A

•Higher eukaryotes don’t have considerably more protein-coding genes
but can get more from their genes through alternative splicing of mRNAs
Bacteria (prokaryotes) have approximately 4,500 genes, yeasts (lower
eukaryotes) approximately 6,500, fruit flys about 14,000 genes and mammals
only 25,000 or so. Thus, although we don’t have considerably more genes
than the much simpler fruit fly,
how we regulate the expression of
these genes is under more
sophisticated control. In addition
to greater levels of control over
the expression of their genes, as
we shall see, higher eukaryotes
also extensively splice the
mRNAs of their protein coding
genes to make different variations
of the same protein, or proteins
with very distinct function, rather
like using the same set of building blocks to make different creations by
using combinations of different blocks within the set.

90
Q

Differential gene expression

A

Differential gene expression is achieved through environmental
or developmental cues that activate expression of genes in a celltype specific manner.
As outlined above, gene expression is regulated by a variety of mechanisms
that permit cells to express only subsets of their genes, at specific times (i.e.
in response to developmental or
environmental cues), or specific
places (cells in different parts of
an organism ‘know’ where they
are and express the appropriate
cohort of genes for that particular
location. Differential gene
expression is controlled through
activating signaling pathways that
activate the correct transcription
factors (TFs), leading to further cascades of gene expression as many TFs also
induce the expression of other TFs.

91
Q

Role of transcription factors in enabling differential gene expression

A

•Transcription factors play a key role in
enabling differential gene expression
There is a vast array of transcription factors
that bind to DNA in a sequence-specific
manner and permit the correct genes to be
expressed in the correct cells. Switching on the
correct transcription factors is typically
achieved through external signals (either from
the environment or other cells), that activate
the specific transcription factors that enable the expression of specific subsets
of genes that are required for a particular process (e.g. cell division, cell
differentiation) or cell function (secretion of insulin, secretion of antibodies,
phagocytosis). Thus, transcription factors play a
critical role in the activation of gene expression in a cell-type specific manner

92
Q

Initiation of developmental gene
expression in the Drosophila embryo through
spatial signaling from the mother

A

Fruit fly Drosophila melanogaster
During early development of a fly embryo , spatial signals are provided to the developing embryo by cells of the mother - called nurse cells - that instruct cells within the embryo to express genes involved in making the different parts of the fly along the anterior to the posterior.

The nurse cells transfer mRNAs largely encoding 2 proteins - Bicoid and Nanos into the anterior end of the embryo.

93
Q

bicoid

A

Bicoid is a transcription factor
that, upon introduction into the anterior part of the developing embryo, then
forms a gradient running from head to tail of the developing embryo and
this transcription factor gradient then induces the expression of a variety of
other transcription factors (Hunchback, kruppel, Knirps, Giant), depending
on the concentration of Bicoid protein present in the cells throughout the
embryo.

Thus, the spatial signal of Bicoid mRNA deposition at the anterior pole of the embryo, upon translation into Bicoid protein, is then rapidly translated
into the differential
expression of several
additional
transcription factors
that Bicoid directs the
expression of
(Hunchback, Kruppel,
Giant, Knirps) running
from the Anterior (left
on the figure above) to
the posterior (right on
the figure above) pole
of the embryo.
The different combinations of
Bicoid, Hunchback, Kruppel,
Giant, and Knirps running from
head to tail of the embryo then
switches on a battery of
additional developmental genes
that 'tells' each cell within the
embryo where they are and what fly parts they should specify by turning on additional genes, called Hox genes, that in turn switch on the correct genes to make a leg, a wing, an antenna, a
sex bristle, etc.
94
Q

Activation of inflammatory genes within the immune system

due to the detection of pathogen components by Toll-like receptors.

A 
Many immune cells express receptors for conserved components of infectious
agents, called pathogen-associated molecular patterns (PAMPs). PAMP
receptors enable cells of the
immune system to detect the
presence of pathogens in the body.
There are a number of different
flavors of PAMP receptors and one
well known class of these receptors
is called Toll-like receptors. Upon
binding of a PAMP to one of the
Toll-like receptors capable of
detecting these foreign substances,
the receptor becomes activated and
this results in activation of a
transcription factor called
NFkappaB that switches on a
diverse array of genes (numbering
in the hundreds) that are involved
in coordinating the process of
inflammation, which helps the
body fight infection. Thus, the
external cue (the PAMP), promotes
expression of a new set of genes on
demand, by activating the
NFkappaB transcription factor, that
directs expression of a battery of
immune defence genes that are
required to fight the infection.
There are many PAMP
receptors in higher eukaryotes,
called Pattern recognition
receptors, and they all function
similarly, turning on a battery of
genes that are involved in fighting infection.
95
Q

PAMPS

A

These receptors recognize conserved molecular structures known as pathogen- or damage-associated molecular patterns (PAMPs and DAMPs) that are found in microbes such as bacteria, viruses, parasites or fungi.

96
Q

The dominant mode of inheritance

A

The affected individual usually has at least one affected parent. The disease may be transmitted by either sex. Usually 50% of offspring have the disease

97
Q

Recessive - eg cystic fibrosis

A

Affected people are usually born to unaffected parents who are both carriers
children have a 25% change of being infected

98
Q

X linked

A

usually males are affected
Usually males are affected. The mother is an unaffected
carrier. 50% of male children are affected and there is no male-male transmission.

99
Q

crossing over

A 
Before the chromosomes
disjoin, chromatids
exchange parts by
crossing-over, or
recombination. This
ensures as great a mixing
of the parental genetic
material as possible
100
Q

meiosis

A

Somatic cells have 46 chromosomes, or more correctly 23 pairs, or homologues, one member of
each pair being inherited from either parent. This means that humans are diploid – they have two
versions of each gene (much more later). The formation of germ cells (spermatozoa and egg
cells), requires a reduction division, or meiosis. Homologous chromosomes pair up at metaphase
1, and each segregates into a different germ cell. The result, germ cells have 23 chromosomes and
not 23 pairs of chromosomes (a haploid genome). When sperm and egg cells fuse, 23 pairs of
chromosomes are generated again (a diploid genome).

101
Q

metacentric

A

the centromere divides the chromosome into equal halves

102
Q

sub-metacentric

A

centromere divides the chromosomes (p arm shorter than q arm)

103
Q

acrocentric

A

no short arm

104
Q

Y chromosome

A

the tiny Y chromosome carries few genes, but one is the SRY, or sex determining gene

105
Q

What does the SRY gene encode

A

A transcription factor that controls the expression of testis determining genes

106
Q

methods of identifyfing chromosomes

A

staining with dyes that produce banding patterns
These chromosomes are
stained with Giemsa (G-banded). Another form of staining is Qbanding.

107
Q

ideograms

A

Ideograms are diagrammatic or idealized representations of chromosomes, showing their relative size, homologous groups and cytogenetic landmarks

Chromosomal analysis of cytological preparations involves calculation of karyotypic parameters and generation of ideograms.

108
Q

why are ideograms important

A

1 in every 160 newborn children has a recognizable chromosomal ebberation and most of these are clinically significant

Cancer cells very often display chromosomal abberations
Analysis of the karyotype can yield important function
For example, most cases of chronic myeloid leukemia displays a diagnostic chromosomal translocation - small pieces of chromosome 9 and 22 exchange places.

109
Q

Multicolour spectral karyotyping

A

Using a fluorescent probe, all 23 chromosomes are revealed in different colours. In the example shown, standard karyotyping using G banding reveals a chromosomal rearrangement appearing to involve chromosomes 8 and 14 - a translocation
However, spectral karyotypoing reveals much greater complexity - rearrangements

110
Q

karyotyping

A

A test to examine chromosomes in a sample of cells

this test can help identify problems as the cause of a disorder or disease.

111
Q

G banding

A

G banding allows each chromosome to be identified by its charachteristic banding pattern
This banding pattern can distinguish chromosomal abnormalities or structural rearrangements, such as translocations, deletions, insertions and inversions

112
Q

Synonymous substitution

A

The evolutionary substitution of one base for another in an exon of a gene coding for a protein, but the produced amino acid sequence is not modified

This is possible because the genetic code is degenerate, meaning some amino acids are coded for my more than one three-base pair codon

113
Q

third base degeneracy of the protein code

A

the sequence of each gene can vary, while still encoding the same protein

114
Q

Barr body

A

a small, densely staining structure in the nuclei of females, consisting of a condensed inactive X chromosome. It is regarded as diagnostic of genetic femaleness

115
Q

Retinis pigmentosa

A

Retina has the highest oxygen consumption per unit gramme of any tissue
It is relatively immune privileged, a unique tissue that evolved for colour vision in primates
any interruptions in the proteins that process this visual acuity can have detrimental effects on vision

116
Q

Rods - draw and label diagram

A

–

117
Q

cones draw and label diagram

A

–

118
Q

cones

A

cone photoreceptors are responsible for daytime accuity, high resolution colour reception,
in the retina

119
Q

rods

A

whereas rods are important for light/dark contract - in the periphera

120
Q

retinus pigmentosa clinical manifestation

A
  • Night blindness - as they lose their rod photoreceptors
  • loss of rod photoreceptors
  • tunnel vision
  • eventually complete vision loss
121
Q

retinus pigmentosa

A

A group of rare genetic disorders that involve a breakdown and loss of cells in the retina - which is the light sensitive tissue that lines the back of the eye

122
Q

Mutated gene for retinus pigmentosa

A

Located on chromosome 3
It encodes rhodopsin, the light sensitive pigment of rod photoreceptor cells

SNP on potsition 23 of the polypeptide

Causes a proline to change to a histidine on position 23 of the polypeptide. The protein cannot fold properly, hence aggregates and results in cell apoptosis

123
Q

Mendel’s first law

A

The First Law (Law of Segregation): We have two versions (alternative forms,
or alleles) of the same gene at the same position (locus) on homologous
chromosomes. We can pass on one allele, or the other to offspring, but not
both.

124
Q

charachteristics of autosomal dominant inheritance

A

males and females pick up the abnormal gene with equal frequency, and every perosn has an affected parent

125
Q

Recessive retinus pigmentosa

A

You may have an affected person with no affected family members
Even though the disease looks similar - it is caused by a recessive gene.

126
Q

consanguinity

A

first cousins having children with each other.

This can give rise to recessive alleles. You can get inheritance of two abnormal genes very rapidly

127
Q

Consanguinity and probability

A

When one person is a direct descendant of another, the proportion of genes they have in common is (½)^n where n is the number of generational steps separating the two

So, parent-child have ½ genes in common Grandparent-grandchild have (1/2)2 = ¼ genes in common Greatgrandparent-greatgrandchild have (1/2)3 = 1/8 genes in common

This is why we get autosomal recessive diseases, we have too many genes shared between generations.

128
Q

The coefficient of relationship

A

The proportion of alleles shared by 2 people

129
Q

Point of origin in eukaryotic cells verus prokaryotic cells

A

In prokaryotic cells, there is only one point of origin, replication occurs in two opposing directions at the same time, and takes place in the cell cytoplasm. Eukaryotic cells on the other hand, have multiple points of origin, and use unidirectional replication within the nucleus of the cell.

130
Q

differences in polymerases in eukaryotic versus prokaryotic cells

A

prokaryotic cells posses one or two types of polymerases, whereas eukaryotes have four or more

131
Q

telomere replication prokaryotes versus eukaryotes

A

In addition, eukaryotes also have a distinct process for replicating the telomeres at the ends of their chromosomes. With their circular chromosomes, prokaryotes have no ends to synthesize.

132
Q

Transcription in Eukaryotes

A

In eukaryotes, transcription is achieved by 3 different types of RNA polymerases
RNA pol 1 transcribed ribosomal RNA
RNA polymerase 2 transcribes RNAs that will become messenger RNAs and RNA polymerase 3 transcribes small mRNAs sucgh as transfer RNAs

RNA pol 2 transcibes protein-encoding genes
It can bind to the promoter of many genes, known as the TATA box, to inititate transcription. Together with other common motifs, these elements consititute the core promoter.

Changes in RNA 2 affinity, and therefore gene expression can be influenced by surrounding DNA sequences, enhancers.
These in turn recruit transcription factors.

133
Q

What is the purpose of crossing over/recombination

A 
Before the chromosomes
disjoin, chromatids
exchange parts by
crossing-over, or
recombination. This
ensures as great a mixing
of the parental genetic
material as possible.
134
Q

Chromosomal aberrations in leukemia

A

Cancer cells very often display chromosomal aberrations. Analysis of the karyotype can yield important information. For example, most cases of chronic myeloid leukaemia display a diagnostic chromosomal translocation – small pieces of chromosomes 9 and 22 exchange places. This is called the ‘Philadelphia’ translocation. Those few cases that do not show the translocation run a more severe course, and those that do, can be treated orally with the drug Gleevec.

135
Q

Leber congenital amaurosis (LCA)

A

A rare inherited retinal degeneration. It is the rarest form of congenital blindness, a retinopathy. The photoreceptors are nonfunctional, bur remain alive inside the retina.

The RP65 gene

136
Q

RP65 gene

A

the RP65 gene is a gene that contains instructions vital for making a protein that is essential for normal vision
It lies in a small layer in the back of the retina, called the retinal pigment epithelium

137
Q

Gene replacement therapy for Leber congenital amaurosis

A

They took viral vectors and packaged it with genes encoding an enzymatic component of the visual cycle. The RP65 gene
visions improved in patients

138
Q

Alzheimers disease signs and symptoms

A

Memory loss for recent events
Progresses into dementia, almost a total memory loss
inability to converse, loss of language
affective personality/disturbance (hostile)

139
Q

Amyloid beta

A

Amyloid beta denotes peptides of 36–43 amino acids that are the main component of the amyloid plaques found in the brains of people with Alzheimer’s disease

140
Q

Confirmation of diagnosis of Alzheimers disease

A

Neuronal plaques

Neurofibrillary tangles

141
Q

Brain atrophy

A

In brain tissue, atrophy describes a loss of neurons and the connections between them. Atrophy can be generalized, which means that all of the brain has shrunk; or it can be focal, affecting only a limited area of the brain and resulting in a decrease of the functions that area of the brain controls.

142
Q

Neurofibrillary tangles

A

Neurofibrillary tangles are abnormal accumulations of a protein called tau that collect inside neurons

143
Q

FAD versus SAD

A

AD can be divided into two subtypes: familial AD (FAD) and sporadic AD (SAD). FAD is usually early-onset AD (EOAD), generally occurring in individuals <65 years, whereas SAD is mainly late-onset AD (LOAD), affecting people >65 years (Fig. 8.2). FAD accounts for less than 5% of disease incidence and is caused by one or more autosomal dominant mutations in the gene that encodes for APP

144
Q

Locus segregation with familial alzheimers disease

A

There was a mutation in a gene called amyloid precursor protein.
Seemed to be following a dominant mode of transmission
The locus was segregating with familial alzheimers disease. It was being mapped to chromosome 21
The families with LAD dont show linkage
Sporadic likely wouldnt show linkage to chromosome 21

145
Q

Amyloid beta precurosr protein

A

People will down syndrome will likely develope amyloid plaques in their brain
Likely producing too much amyloid precursor protein

Amyloid precurosor protein is a transmembrane protein that gets cleaved and chopped into small fragments, called amyloid beta. tHAT are normally transported across blood vessels of the brain and cleared in the blood

146
Q

APP gene substitution

A

There is valine isoleucine substitution in th APP protein.
It makes the transmembrane more hydrophobic, which anchors the protein more firmly to the membrane
This could impact its processing
Because this gene/chromosome is not associated with other forms of alzheimers disease, it suggests it is highly geneticically heterogenous

This leads to an accumulation of amyloid beta and a buildup of plaques in the brain

147
Q

X linked recessive disease - haemophelia

A

Leads to genetic mutations in factor 8 and factor 9 genes that encode these proteins that are important for formation of clots
People with haemophelia dont have this ability
we can produce recombinant protein, make factor 8 and 9 prproteins in the lab and deliver them systemically in the lab.
Using viruses ect

148
Q

Human gene mapping

A

Genome mapping is used to identify and record the location of genes and the distances between genes on a chromosome.

TCD from molecules to cells 2nd year (8 decks)

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