3. Molecular and Medical Genetics Flashcards

Question

What is the charge of the phosphate-sugar backbone in DNA?

Answer 1

* DNA -\> Deoxyribose * RNA -\> Ribose

Answer 2

Carbon-1 is the carbon with the base attached to it.

Answer 3

On carbon-2: * Deoxyribose has a H * Ribose has an OH

Answer 4

On carbon-1, the carbon on the opposite side of the O to the "tail" of the sugar.

Answer 5

Thymine in DNA is replaced by uracil in RNA. T -\> U

Answer 6

They are split into two categories: * Purines -\> 2 rings * Pyrimidines -\> 1 ring

Answer 7

Nitrogenous bases with 2 rings. They include: * Adenine * Guanine

Answer 8

Nitrogenous bases with 1 ring. They include: * Cytosine * Thymine * Uracil

Answer 9

* Nucleoside = Base + Sugar * Nucleotide = Base + Sugar + P

Answer 10

It replaces the H from the OH on carbon-5 (on the tail), so that the P is surrounded by 4 O atoms.

Answer 11

* Base = Adenine * Nucleoside = Adenosine * Nucleotide = e.g. AMP or dCTP

Answer 12

A 'd' is added to the start of the shorthand name. For example, deoxyadenosine triphosphate is dATP.

Answer 13

Deoxyribonucleoside triphosphates (dNTPs) -\> The two phosphates that are not in the phosphodiester bond are cleaved off.

Answer 14

3^I and 5^I

Answer 15

DNA (or RNA) polymerases catalyse this: * OH on the 3-prime carbon carries out a nucleophilic attack on the phosphate closest to the sugar of a different nucleotide (remember: there are 3 phosphates attached) * The other 2 phosphates are lost as a molecule of pyrophosphate * A phosphodiester bond has been formed, which looks like this: C-O-P-O-C

Answer 16

Oligonucleotides, then polynucleotides.

Answer 17

* 3-prime -\> OH * 5-prime -\> Phosphate

Answer 18

5-prime to 3-prime direction, because polymerase only works in 1 direction.

Answer 19

It is negative, which can create dipoles.

Answer 20

* A and T -\> 2 bonds * G and C -\> 3 bonds

Answer 21

They are anti-parallel.

Answer 22

Act as templates for transcription and replication.

Answer 23

The double helix has major and minor grooves, which are important for molecules, such as transcription factors, to associate with the DNA.

Answer 24

The mass of genetic material composed of DNA and proteins that condense to form chromosomes during eukaryotic cell division.

Answer 25

DNA -\> Chromatin -\> Territories

Answer 26

Nucleosome

Answer 27

It is the fundamental unit of DNA packing in a chromosome. It consists of: * A section of DNA (about 160 nucleotides) * Wrapped twice around 8 histone proteins

Answer 28

2 times: H2A, H2B, H3, H4

Answer 29

* Sections of 160 nucleotides are wrapped twice around a histone * 8 histones are in a nucleosome * Nucleosomes are joined by linker DNA * Association of non-histone proteins creates chromatin * 3D architecture -\> Caused by organisation into complex dynamic structures (such as chromatin loops that rae formed by ring-shaped proteins such as cohesin

Answer 30

PRIMARY * DNA double helix -\> DNA sequence, methylation, genes SECONDARY * Nucleosomes -\> Nucleosome positioning, epigenetic modification, Chromatin, Accessibility 3D STRUCTURE 1. Chromatin loops -\> Promoter enhancer interactions 2. TADs (Topologically Associated Domains) 3. A/B compartments -\> Chromatin scale: A = Active, B = Inactive 4. Chromosome territory -\> Nuclear space occupied 5. Nucleus -\> Relative positioning of chromosome territories

Answer 31

Semi-conservative

Answer 32

The two antiparallel strands must be unwound and each may be used as a template for the synthesis of new complementary strands by dNTPs (nucleoside triphosphates) and polymerases.

Answer 33

At origins of replication.

Answer 34

DNA helicases

Answer 35

* Pre-replication complex * In the gap phase of the cell cycle (G₁)

Answer 36

RNA primers are synthesised by primase on both strands.

Answer 37

* Single strand specific binding protein: SSB * For example, Replication protein A (RPA)

Answer 38

* In the gap phase of the cell cycle, a pre-replication complex is formed at the origins of replication. * DNA helicase moves to origins of replication, breaks open the hydrogen bonds and opens up the double helix. This process requires energy from ATP. * DNA polymerase requires an RNA primer in order to begin replication, which is a strand of RNA about 10-60 nucleotides long. * A primer is required on both strands of the DNA, so each strand can act as a template. * Replication protein A (RPA) binds single-stranded DNA and prevents reannealing.

Answer 39

* Human -\> Multiple, which allows DNA to be transcribed at multiple points simultaneously. * Bacterial -\> Just one.

Answer 40

* Leading * Lagging

Answer 41

2 (one is created by DNA being synthesised on each strand in antiparallel directions)

Answer 42

* As the DNA polymerases travel in antiparallel directions on each of the two strands, a fork is created in each direction. * This means that behind the leading strand there is a gap where only the opposite strand’s leading strand is forming. * This gap is filled by lagging strands that are continuously synthesised starting with new RNA primers.

Answer 43

Short sequences of DNA nucleotides which are synthesized discontinuously and later linked together to create the lagging strand during DNA replication.

Answer 44

Unwind the DNA, using energy provided by hydrolysis of ATP.

Answer 45

* Replication protein A * Binds to single-stranded DNA and prevents refolding

Answer 46

Reduce the torsional strain on DNA caused by unwinding.

Answer 47

* DNA polymerase α (a.k.a. Primase) -\> Synthesises the RNA primer * DNA polymerase δ -\> Synthesises the DNA strand

Answer 48

* Clamp * Tethers DNA polymerase δ to DNA, displacing the primase. * Enables polymerase δ to be highly processive and synthesis can occur in long segments.

Answer 49

Loads the PCNA “clamp” onto DNA.

Answer 50

Degrades and removes the RNA primer.

Answer 51

* When the lagging strand reaches the leading strand the DNA polymerase is displaced with the DNA helicase. * Okazaki fragments are joined when the DNA polymerase and helicase push aside the primer at the end of the next Okazaki strand, and FEN1 cuts the DNA just past the primer. * The remaining DNA is synthesised as the primer is removed and then the fragments are joined by DNA ligase.

Answer 52

Removes RNA flap when Okazaki fragments are completed.

Answer 53

Fuses the completed Okazaki fragments together in the lagging strand.

Answer 54

* Prevent shortening of chromosomes during proliferation * Enable correcting matching of the two DNA strands -\> Stop end to end fusions of the DNA strands

Answer 55

* They get shorter with each replication, until cell apoptosis happens when they are too short * This is because there is no template for primase available to synthesis the next RNA primer for the last Okazaki fragment

Answer 56

Extends telomeres, so they are not shortened too quickly.

Answer 57

It has an RNA component.

Answer 58

* It is a reverse transcriptase (TERT) * It uses an RNA template to synthesise the complementary DNA strand, which extends the telomeres

Answer 59

Overactivity of telomerase can lead to cell immortality in cancer cells.

Answer 60

* 3' to 5' exonucleolytic proofreading by the DNA polymerase * Non-polymerase mismatch repair

Answer 61

1 in 10⁵ nucleotides

Answer 62

3' to 5' exonuclease activity

Answer 63

3' to 5' (this is opposite to the direction of DNA synthesis)

Answer 64

* If a wrong nucleotide is selected and inserted, base-pairing will be disrupted, causing a shift from the polymerase activity to the exo-activity. * The 3’ to 5’ exoribonuclease activity enables the polymerase to remove the ‘wrong’ nucleotide. * Polymerisation then resumes.

Answer 65

1 in 10²nucleotides

Answer 66

* They are mechanisms that recognise the newly-synthesised strand after replication (the mechanism for this in most species is not fully understood) and excise the section with the mismatch, before synthesising DNA to fill the gap. * Error rate: 1 in 10²nucleotides

Answer 67

* Histones are removed when the replication fork moves forward. * In the daughter strands, one half of each histone is recycled from the parental, the other is newly synthesised.

Answer 68

* Spontaneous * Changes in chemical properties of the nucleic acids * Problems during replication or cell division * Induced * Radiation * Chemicals

Answer 69

* Deamination * Depurination

Answer 70

* It is removal of an amino group from a nucleotide * This can cause the type of nucleotide to be changed (e.g. from C to U) * The result is that the nucleotide sequence is altered in ONE of the daughter molecules in DNA replication

Answer 71

Cytosine can be converted to uracil.

Answer 72

* It is removal of the purine base from a nucleotide * This leaves just the sugar-phosphate backbone in that part of the DNA strand, which is effectively a deletion * The result is that the nucleotide sequence is altered by frame shift in ONE of the daughter molecules in DNA replication

Answer 73

* Base mismatch (e.g. an A pairing with a C) * Polymerase slippage

Answer 74

A repeat region can be extended.

Answer 75

* Exposure to UV can cause dimerisation of thymine. * Two thymines are covalently linked -\> Affects the structure of the DNA -\> Affects replication.

Answer 76

* Direct repair * Excision repair * Mismatch repair * Nonhomologous end-joining

Answer 77

* Single nucleotides that have been damaged by transfer of methyl/ethyl group onto the base (e.g. by alkylating agents) are repaired this way * This is done by transferring the alkyl group onto the enzyme

Answer 78

* Used to repair cytosine nucleotides that have been deaminated to give uracil * This involves single nucleotide excision repair DNA glycosylases that cut the nucleotide out and resynthesise the correct DNA

Answer 79

* The mismatched nucleotide is cut out and the correct DNA is resynthesised * Thymidine dimers require excision of a larger area

Answer 80

* They are methods of repairing DNA molecules that have a double-strand break in them * Non-homologous involves just joining the strands back up, while homologous reocmbination involve creating sort of sticky ends (CHECK THIS)

Answer 81

A mutation that alters a codon so that it is recognised by a different tRNA and consequently a different amino-acid is introduced into the polypeptide chain.

Answer 82

* A change that introduces a premature stop codon in the mRNA. * This results in the production of a shortened protein that might be nonfunctional or displays an altered function or regulation

Answer 83

* Synonymous * Silent -\> No effect on protein * Non-synonymous * Missense -\> Altered amino acid * Nonsense -\> Stop codon produced * Splicing -\> Various

Answer 84

* Multiple of 3 -\> Various effects * Not a multiple of 3 * Frame shift -\> Gain/Loss of function, Premature termination * Large deletion * Partial or whole gene deletion -\> Loss of function/expression

Answer 85

* Multiple of 3 -\> Various effects * Not a multiple of 3 * Frame shift -\> Gain/Loss of function, Premature termination * Large deletion * Partial or whole gene duplication -\> May affect dosage * Trinucleotide * Dynamic mutation -\> Altered function, stability

Answer 86

When an organism has two copies of the same allele for a given gene.

Answer 87

When an organism has two different alleles for a given gene.

Answer 88

Phenotype = Genotype + Environment

Answer 89

* Nuclear genome * Mitochondrial genome

Answer 90

Nuclear: * Linear chromosomes * Maternal and paternal inheritance Mitochondrial: * Circular chromosomes * Maternal inheritance only

Answer 91

* Nuclear genome features mostly poorly conserved sequences * Mitochondrial genome features mostly highly conserved sequences

Answer 92

37 genes: * 13 encode polypeptides * 24 encode non-coding RNAs (22 t-RNAs and 2 rRNAs)

Answer 93

Nuclear -\> The polypeptides are then imported into the mitochondria.

Answer 94

Transcription

Answer 95

RNA polymerase

Answer 96

* Template strand -\> The one that free nucleotides bind to and is transcribed * Coding strand -\> The one that is not transcribed, but the base order matches the RNA strand

Answer 97

Thymine (T) is replaced by uracil (U)

Answer 98

Yes, the RNA polymerase can use both the top and bottom strand as the template strand.

Answer 99

* DNA strand reading -\> 3' to 5' direction * RNA strand synthesis -\> 5' to 3' direction

Answer 100

No, unlike DNA polymerases they do not.

Answer 101

mRNA, snRNA, miRNA

Answer 102

tRNA, some snRNA

Answer 103

Code for proteins

Answer 104

Form the core of ribosomes and catalyse the formation of proteins.

Answer 105

Regulate gene expression.

Answer 106

Involved in converting from mRNA to amino acid chain during protein synthesis.

Answer 107

Used in: * RNA splicing * Telomere maintenance * Other processes

Answer 108

They bind to promoter sequences on the DNA.

Answer 109

Prokaryotes: * Sigma factors * Different numbers of sigma factors exist in different species and each type guides the polymerase to the promoter region for a different gene Eukaryotes: * General transcription factors (GTFs). * Different combinations of transcription factor attract each of the three types of eukaryotic RNA polymerase by binding to the core promoter first

Answer 110

Upstream of the gene

Answer 111

General transcription factors (GTFs)

Answer 112

* Pribnow box at the -10 region, which is a TATAAT sequence where the DNA strands first separate (partly due to the small number of hydrogen bonds between A and T bases). * There is a second consensus sequence at the -35 region.

Answer 113

* Eukaryote promoters are more complex and include core, proximal and distal promoters. * Eukaryotes are also distinct in their downstream promoter regions (past the 3-prime end of the TSS).

Answer 114

* Core -\> Required for transcription and contain the TSS (transcription start site) * Proximal and distal -\> Regulatory role

Answer 115

Note the core, proximal and distal promoters.

Answer 116

They all start with TF. e.g. TFIIA

Answer 117

It is the point just past the promoter region at which transcription of a gene starts.

Answer 118

* General transcription factors (GTFs) -\> Bind to the core promoter and are required for transcription since they recruit the RNA polymerase * Activators and co-activators -\> Bind to other regions on the DNA to regulate transcription of specific genes

Answer 119

* RNA polymerase moves along the template DNA strand in the 3-prime to 5-prime direction (polymerase can only synthesise DNA in the 5-prime to 3-prime direction). * The polymerase separates the two DNA strands, forming an open complex where free RNA nucleotides can join to the template strand by complementary base pairing. This region is called the transcription bubble. * Phosphodiester bonds are synthesised between the NTPs * Once synthesised, the growing RNA chain does not remain associated with the DNA strand (as in DNA replication), but is removed through an RNA exit channel.

Answer 120

* Phosphodiester bonds are synthesised between the NTPs when the -OH on the 3-prime end of the previous NTP carries out a nucleophilic attack on the first phosphate (alpha phosphate) of the next NTP. * This releases a pyrophosphate molecule.

Answer 121

Prokaryotes: * Genes are set by default to “on”, so that without any regulatory molecules there will be some degree of transcription. * Therefore, the most basic form is negative control. * Prokaryotes feature both positive and negative control in structures called operons, which are sets of (usually functionally related) genes that are under the control of a single operator within a regulatory region. * The operon contains the genes to be transcribed, the operator and the promoter (where the RNA polymerase binds). * Negative control is due to repressors binding to the operator, while positive control is due to activators binding to a different site within the regulatory region, which increases the affinity of polymerase for the promoter. Eukaryotes: * Genes are by default set to “off” and require a set of transcription factors in order to initiate transcription. * This gives the opportunity for control pathways that affect the general transcription factors. * Aside from this, eukaryotic gene expression can be influenced across a larger distance, namely by enhancer and silencer regions that activate and repress transcription from regions outside of the promoter. * In eukaryotes, another level of gene regulation is added by the presence of histones, which can be enzymatically modified by acetylation or methylation.

Answer 122

* Operons are sets of (usually functionally related) genes that are under the control of a single operator within a regulatory region. * Found in PROKARYOTES. * The operon contains: * Genes to be transcribed * Operator * Promoter (where the RNA polymerase binds). * Negative control is due to repressors binding to the operator, while positive control is due to activators binding to a different site within the regulatory region, which increases the affinity of polymerase for the promoter.

Answer 123

Folding of the DNA in, for example, loops, so that transcription factors bound at the distal regulatory regions can interact with the general transcription factors and RNA polymerase.

Answer 124

* Heterochromatin -\> Tightly wrapped DNA regions containing inactive genes * Euchromatin -\> Loosely wrapped DNA regions containing active genes

Answer 125

* Acetylation * Methylation

Answer 126

* Acetylation -\> Increases transcription * Methylation -\> Decreases transcription These are general effects, but they can vary.

Answer 127

* It decreases the rate of transcription. * This is because it impedes the binding of transcriptional proteins to the gene and methylated DNA may be bound by transcription repressor proteins.

Answer 128

* Nucleosomes are tightly packed and associated with additional heterochromatin proteins that bind to the specifically modified histone tails of nucleosomes in those regions. * This means the DNA in these regions is inaccessible and prevents the expression of genes that are located within these regions.

Answer 129

* It is DNA that remains in a condensed state throughout the cell cycle. * It contains repeat sequences which are not transcribed and plays a role in chromosome structure * Found in: Centromere + Telomeres + Microsatellites throughout the chromosome

Answer 130

* Nucleosomes are more loosely packed. * This exposes regions that are accessible to regulatory proteins. * Level of accessibility of DNA can still vary greatly -\> Modifications of the histone tails also regulate accessibility at this scale.

Answer 131

* The largest subunit of RNA polymerase II has a characteristic structure at its C-terminal domain (CTD) * The CTD consists of 52 repeats of a seven aminoacids YSPTSPS. * Phosphorylation of the serine residues in the CTD is critical for the transition to elongation.

Answer 132

TFIIH (A transcription factor)

Answer 133

No, that is right-hand side content.

Answer 134

In eukaryotes, exons are parts of genes that are expressed, while introns are also transcribed, but they are they spliced out during pre-mRNA processing.

Answer 135

* Capping the 5' end * Splicing of introns * 3' end formation

Answer 136

During -\> It is a co-transcriptional process.

Answer 137

* When phosphorylated, the CTD (C-terminal domain) of RNA polymerase II serves as a landing pad for the 3 types of processing factor * This means that pre-mRNA processing can occur alongside transcription

Answer 138

Capping of the 5' end of the mRNA (the 5' end is the end that is produced first)

Answer 139

* A GMP nucleotide is added to the 5' end (in an unusual 5' to 5' linkage) * The G is then methylated * A cap binding complex (CBC) then binds to the GMP nucleotide

Answer 140

The cap: * Protects the mRNA from degradation by 5' exoribonucleases. * Facilitates translation initiation.

Answer 141

Trans-esterification

Answer 142

Cis-elements (near the splice sites) define the starts and ends of DNA.

Answer 143

Small nuclear RNA

Answer 144

* A large and complex molecular machine found primarily within the nucleus of eukaryotic cells. * It is responsible for splicing.

Answer 145

* 5 RNA components -\> U1, U2, U3, U4 and U5 snRNAs * These are bound to specific proteins Together, each U snRNA and bound proteins are referred to as snRNPs.

Answer 146

* Small nuclear ribonucleic RNA proteins * These are the protein-RNA complexes that form the spliceosome required for splicing.

Answer 147

Cis-elements in the pre-mRNA.

Answer 148

It creates an mRNA molecule with a continued Open Reading Frame (ORF).

Answer 149

It allows for alternative splicing.

Answer 150

* Allows permutation of exons, enabling cells to potentially generate several proteins from a single gene * Plays an important role in tissue specific gene expression

Answer 151

Proteins that associate with certain sequences on the pre-mRNA can strengthen or weaken splice sites and so regulate exclusion or inclusion of exons.

Answer 152

* The association of enhancer proteins with the pre-mRNA makes it more likely that the splicing factors recognise the 3’ and 5’ splice sites that flank an exon. * This increases the likelihood of it being kept in the mRNA.

Answer 153

* Negative factors binding to specific sequences can prevent recognition of both the 3’ and 5’ splice sites that flank an exon and splicing will skip the exon. * This results in potential exclusion of the exon.

Answer 154

3' end processing

Answer 155

* Cleavage * Polyadenylation

Answer 156

It has two parts: * The major signal for both cleavage and polyadenylation is the hexamer AATAAA (AAUAAA) which is located about 30 bases upstream of the cleavage and polyadenylation site. * A second signal, the downstream sequence element (DSE) is GT (GU) rich and is found after the cleavage site.

Answer 157

AAUAAA (In DNA it is: AATAAA)

Answer 158

* Downstream sequence element * It is GU rich (In DNA, it is GT rich)

Answer 159

* The cleavage and polyadenylation machinery associates with the AAUAAA and DSE sequences on the pre-mRNA. * Cleavage occurs and the 3’end of the mRNA is then polyadenylated by poly(A) polymerase PAP.

Answer 160

Poly(A) polymerase (a.k.a. PAP)

Answer 161

The poly(A) tail is covered by poly(A) binding proteins (PABP).

Answer 162

Polyadenylation creates a uniform 3’-end to all protein encoding mRNAs which is critical for nuclear cytoplasmic export, stability and translatability of the mRNA.

Answer 163

Mutations that affect the splicing pattern and/or splicing efficiency of a particular exon are common.

Answer 164

* Mutations affecting splice sites * Mutations affecting splicing factors * Mutations affecting poly(A) signals

Answer 165

* Can lead to reduced splicing out of particular introns -\> Leads to faulty mRNAs * Mutations affecting 3’ splice sites can result in exon skipping

Answer 166

* Mutations affecting poly(A) signals can reduce or increase cleavage and polyadenylation. * This affects the total output of mRNA.

Answer 167

Codon (3 bases)

Answer 168

One trinucleotide codon codes for an amino acid.

Answer 169

Methionine -\> This is because the start codon (AUG) codes for methionine.

Answer 170

* UAA * UAG * UGA

Answer 171

They are recognised by proteins instead of tRNA molecules, so this forces the termination of translation.

Answer 172

The section of mRNA from the start codon that includes all of the codons that code for amino acids.

Answer 173

A point mutation that produces a premature stop codon -\> Nonsense mutation

Answer 174

Missense mutation -\> Where an amino acid is replaced by another.

Answer 175

* tRNAs * Ribosomes * Translation factors * Energy (in the form of GTP)

Answer 176

Specific aminoacyl tRNA synthetases

Answer 177

* Charge t-RNA molecules with the correct amino-acids * This requires energy from ATP

Answer 178

Their action is highly specific and “proof reading” ensures accuracy of this process.

Answer 179

* RNA * Proteins

Answer 180

RNA is the catalytic component: Ribozyme

Answer 181

In the nucleolus and then they are exported.

Answer 182

* Eukaryotic -\> 80S * Prokaryotic -\> 70S

Answer 183

The tRNA recognises its corresponding codon in the mRNA via base pairing between nucleotides in the anticodon and the codon.

Answer 184

Some tRNA molecules can have more than one version to recognise more than one codon (e.g. there can be an Ala1 and Ala2 tRNA for alanine), and each of these may be able to recognise more than one codon.

Answer 185

* It is the third base in a codon * Base paring rules at the third base (wobble position) are less strict and a base can pair with more than one complementary base

Answer 186

5' to 3' (opposite direction to DNA)

Answer 187

The C terminus.

Answer 188

* E site -\> Exit site * P site -\> Peptidyl site * A site -\> Aminoacyl site

Answer 189

It goes to the aminoacyl (A) site of the 60S subunit.

Answer 190

Peptidyl transferase

Answer 191

It causes release of the peptide and dissociation of the ribosome.

Answer 192

* Poly(A) tail -\> 3' end * Cap -\> 5' end

Answer 193

* Cap binding proteins and proteins that bind the poly(A) tail interact with each other and create a circular mRNA molecule. * The small ribosomal subunit is recruited to the cap and then scans the mRNA for the AUG start codon. * Once the start codon is identified, the large ribosomal subunit joins and translation can commence.

Answer 194

* Phosphorylation of translation initiation factors * Sequences that are present in the mRNA template

Answer 195

* Untranslated regions * They are the bits of mRNA on either side of the open reading frame * They contain many regulatory sequences that regulate translation

Answer 196

* After splicing, EJC (exon junction complex) protein assemble at the junctions between exons * These serve as markers * The presence of an EJC downstream of a stop codon identifies the mRNA as faulty and induces its destruction (degradation).

Answer 197

Exon junction complex proteins -\> These are protein complexes that assemble at junction between exons after splicing of mRNA. They serve as markers.

Answer 198

Quinolones

Answer 199

* Tetracycline * Erythromycin * Streptomycin * Chloramphenicol

Answer 200

* Hybridisation techniques * Enzyme-based techniques * Antibody techniques

Answer 201

* Southern blotting * Northern blotting * Fluorescent in situ hybridisation (FISH)

Answer 202

* PCR * RT-PCR * DNA sequencing * NGS * Manipulation of DNA -\> Restriction, ligation, cloning

Answer 203

* Western blotting * Elisa * Imaging * Immunoprecipitation

Answer 204

* Double stranded molecules can be denatured (separated) by heat * Complementary sequences re-nature (re-hybridise) at lower temperature

Answer 205

* Northern blotting -\> Detection of the presence of specific **RNA** molecules in a sample * Western blotting -\> Detection of the presence of specific **proteins** in a sample * Southern blotting -\> Detection of the presence of specific **DNA** molecules in a sample

Answer 206

* Target molecules (e.g. RNA, proteins or DNA) are isolated -\> DNA is usually fragmented * Separated by size using gel electrophoresis * Blotted onto a membrane, so that probes or antibodies can access them * Probes and antibodies are chosen to be specific to a certain RNA, DNA or protein * In Northern blotting, a hybridisation probe is used, which is an RNA or DNA sequence that is complementary to the **target RNA** sequence and is labelled fluorescently or radioactively * In Western blotting, an antibody is used that is specific for the **target protein** and is labelled or can be bound to by a labelled secondary antibody * In Southern blotting, a hybridisation probe is used, which is an RNA or DNA sequence that is complementary to the **target DNA** sequence and is labelled fluorescently or radioactively * Excess antibodies or antibodies are washed away * Labels are detected -\> e.g. Using a X-ray film for radioactive probes -\> This shows whether the target molecule is present

Answer 207

The presence of a specific RNA sequence.

Answer 208

The presence of a specific DNA in a sample.

Answer 209

The presence of a specific protein in a sample.

Answer 210

S-D -\> Southern-DNA N-R -\> Northern-RNA O-O -\> Nothing W-P -\> Western-Protein

Answer 211

Fluorescent in situ hybridisation: * Uses fluorescent probes that bind to only complementary nucleic acid sequences. * Used to detect and localize the presence or absence of specific DNA sequences on chromosomes. * Fluorescence microscopy can be used to find out where the fluorescent probe is bound to the chromosomes.

Answer 212

FISH is done within the cell (in situ), while Southern blotting involves some processing of the DNA before the probes are added. [CHECK THIS]

Answer 213

* The use of FISH to colour entire choromosomes. * This can be useful in characterising chromosomal abnormalities.

Answer 214

* Where cells use small ~21 nucleotide long RNAs (e.g. miRNAs) to regulate the expression of genes. * The small RNAs are incorporated into the RNA induced silencing complex (RISC) and act as guides for RISC to interact in a sequence specific manner with the target mRNAs. * RISC can then degrade the target mRNA or inhibit its translation. This can be used to clinically knock out certain genes.

Answer 215

It is a microbial adaptive immune system that has been repurposed as gene editing system that can be used to generate or correct sequence changes in DNA.

Answer 216

CRISPR (clustered regulatory interspaced short palindromic sequences) associated Ca9 nuclease

Answer 217

* The Cas9 nuclease can be introduced into eukaryotic cells and using guide RNAs specifically directed to any location in the genome where it can cut the DNA, leading to a break. * When the DNA tries to repair itself, two things can happen: * Homologous end joining -\> This happens when a homologous sequence is provided along with the Cas9 nuclease, so that it acts as a template and therefore the DNA in the cell is edited to whatever is desired * Non-homologous end joining -\> This happens when no homologous sequence is provided, so that the DNA either repairs itslef to be like before or it repairs itself by excising bases, leading to a non-functional protein

Answer 218

Palindromic 6 base sequences

Answer 219

* It is when there is a mutation in the recognition site of restriction endonuclease. * These can be detected by addition of restriction endonucleases and then separation of the fragments by electrophoresis. * This is because a mutation causes changes in lengths of the fragments.

Answer 220

* The restriction endonucleases can be used to create fragments * These fragments can be separated by electrophoresis * Differences in their length from normal indicate mutations

Answer 221

Separation of DNA fragments by their size.

Answer 222

It can join DNA fragments that have been cut by restriction endonucleases.

Answer 223

* Forming recombinant DNA molecules, then replicating them within a host organism * Used to obtain multiple copies of a desired section of DNA within a cell

Answer 224

* Vector is selected -\> Usually a bacterial plasmid that contains an origin of replication, drug-resistance gene, and a recognition site for restriction endonucleases * Same **restriction endonuclease** used to cut target DNA and bacterial plasmid * Restriction endonucleases cut DNA at palindromic recognition sites by, leaving sticky or blunt end fragments * Recombinant DNA is formed by joining the vector DNA with the target DNA using **DNA ligase**, which forms phosphodiester bonds. * Recombinant DNA is placed in the host cells (transformation) * Methods for increasing competency include: * Placing the bacterial cells in cold calcium chloride, then exposing then to a brief heat shock at around 42°C * Electroporation -\> Exposing cells to an electric field, which temporarily forms holes in the cell membrane * Cells are placed in agar that contains an antibiotic -\> Cells which are successfully transformed contain the plasmid with the gene for antibiotic resistance, so they survive and replicate * This results in the production of millions of copies of the recombinant DNA, which may be isolated for use or transcribed by the DNA to produce a useful product

Answer 225

A small circular DNA element that can replicate independently of the chromosome. It is typically found in bacteria.

Answer 226

How willing the cell is to take up a recombinant plasmid.

Answer 227

* Placing the bacterial cells in cold calcium chloride, then exposing then to a brief heat shock at around 42°C * Electroporation -\> Exposing cells to an electric field, which temporarily forms holes in the cell membrane

Answer 228

* Antibiotic resistance gene (Ampr) -\> For selection of cells that have actually taken up the plasmid * Origin of Replication (ori) -\> For replication * Multiple cloning site (MCS) -\> Unique restriction enzyme sites necessary for insertion of DNA * Phage promoters (T7, SP6) -\> Enable in vitro transcription of the recombinant DNA fragment

Answer 229

* Polymerase chain reaction * The in vitro production of multiple copies of a selected DNA fragment, using a minimum of a very small amount of starting material

Answer 230

* DNA template is chosen * Oligonucleotide primers are designed and synthesised, so that they flank the target sequence * Reaction mixture contains DNA template, DNA polymerase, primers and free dNTPs * First, the reaction mixture is heated to about 95\*C, which separates the double-stranded DNA, so that single-strand templates are exposed * Next, the reaction mixture is cooled to about 60\*C, which allows the primers to anneal to their complementary sequences on the DNA -\> This will allow the DNA polymerase to begin synthesis and will also prevent re-annealing of the strands * Then, the reaction mixture is heated to 72\*C, which increases the activity of the DNA polymerase, allowing it to synthesis the complementary strand of DNA using the free dNTPs * This cycle of 3 steps can be repeated multiple times, to create several copies of the initial DNA template -\> This happens exponentially until the reaction substrates are depleted

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* DNA template * Oligonucleotide primers (flanking the target DNA) * DNA template * DNA polymerase * Primers * Free dNTPs

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* 95\* * 60\* * 72\*

Answer 233

Reverse transcriptase PCR (do not confuse with real time PCR): * It is like normal PCR * But first a virus-derived reverse transcriptase is used to create a complementary copy of DNA (cDNA) from an RNA template, which is then amplified using PCR * It is often used alongside real time PCR to estimate the amount of a certain RNA present in a sample (real time PCR involves the incorporation of a fluorescent dye into the DNA after each cycle, giving a quantitative result)

Answer 234

Eukaryotic mRNA as they all share a common 3’end. A short poly (T) primer ‘TTTTTTTTTTTTTTTTT’ (oligo dT) can be used to reverse transcribe all mRNA into DNA.

Answer 235

* Uses the single-stranded DNA template that is being sequenced, a DNA primer, a DNA polymerase, dNTPs and ddNTPs that are labelled fluorescently or radioactively * Mixture of the reactants is heated to separate the strands in the template DNA, then cooled to allow the primer to bind to a strand of the template * Upon heating, the polymerase synthesises the complementary strand (starting at the primer), using dNTPs. * The number of ddNTPs is relatively small compared to the dNTPs, so they are very infrequently added to the strand being synthesised. However, they lack the OH on the 3’ carbon, so they are unable to form phosphodiester bonds with the next dNTP or ddNTP. This means that when they are added to the strand, it can no longer be extended (chain termination). * This is repeated many times, and probability dictates that (with enough repeats), the chain should be terminated at every residue in the strand at least once. * All of the synthesised single strands are processed by electrophoresis, with a detector at the end that detects the labelled ddNTPs (e.g. a laser scanner that scans the fluorescent labels). The shorter strands pass through the gel faster and are detected by the detector sooner, meaning that the identity of the base at each position in the original DNA sequence can be determined.

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They do not have an OH in the 3' position, but have a H instead.

Answer 237

* The DNA is fragmented into small pieces whic can be processed in parallel * The overlapping reads can be assembled to give the whole genome * So the method has a higher throughput.

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Oxford nanopore

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* The number of unique reads that include a given nucleotide in the reconstructed sequence. * e.g. The base at position 27 may come up 20 times in DNA fragments in whole genome sequencing

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A technique used to visualize the localization of a specific protein in cells by use of a specific primary antibody that binds to it.

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Detection and localisation of a particular protein within a cell or tissue using a fluorescently-labelled antibody. It is essentially immunocytochemistry but with fluorescence.

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Enzyme-linked immunosorbent assay

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* Antibody recognising a specific antigen is immobilised on surface of the wells * Sample containing the potential antigen is added. After incubation, the wells are washed, removing any non-captured material. * Antibody recognising the specific antigen and conjugated with enzyme is added. Detection solution is added and if enzyme is present solution changes colour.

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* A technique of precipitating a protein antigen out of solution using an antibody that specifically binds to that particular protein. * This process can be used to isolate and concentrate a particular protein from a sample containing many thousands of different proteins.

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* Many protein-coding genes have more than one polyadenylation site, so a gene can code for several mRNAs that differ in their 3′ end. * Since alternative polyadenylation changes the length of the 3′ untranslated region, it can change which binding sites for microRNAs the 3′ untranslated region contains.

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* The signal hypothesis proposes that proteins destined for secretion, which involves the movement of the protein across a biological membrane, are originally manufactured with an initial sequence of amino acids that may or may not present in the mature protein. * It is to do with the movement of proteins.

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Genes for: * Histones * Ribosomal RNA These are genes that are found multiple times in DNA.

3. Molecular and Medical Genetics Flashcards

(323 cards)