MGD Flashcards

Question

Name 5 important uses of proteins

Answer 1

``` Catalysts Transporters Structural support Immune protection Ion channels ```

Answer 2

As zwitterions

Answer 3

Bond formed in a condensation reaction between two amino acids C(=O)--N(-H)

Answer 4

Planar- all elements lie in one plane Rigid (double bond characteristics)- no rotation about double bond; delocalised electrons = shorter and more stable bond Trans orientation - of carboxyl oxygen and amide hydrogen

Answer 5

Aliphatic/ Aromatic Polar/ non polar Charge of R groups

Answer 6

``` Glycine Alanine Valine Leucine Isoleucine Methionine Proline Phenylalanine Tryptophan ```

Answer 7

``` Serine Threonine Tyrosine Cysteine Glutamine Asparagine ```

Answer 8

Histidine, lysine, arginine

Answer 9

Glutamate, aspartate

Answer 10

Proline and glycine

Answer 11

Alanine and leucine

Answer 12

Very small R group

Answer 13

Peptide bond is in a cyclic arrangement and so rotation of bonds either side of the peptide bond is IMPOSSIBLE So groups can't move to form bonds which would form the helical structure

Answer 14

The amino acid sequence of a protein.

Answer 15

Stretches of the polypeptide chain that form a-helices and b-sheets Bonds on either side of the peptide bond can rotate freely. When these angles remain the same throughout a segment of polypeptide the protein adopts a regular secondary structure

Answer 16

The full 3D structure of the protein. Involves the folding up of the secondary structures. Improper folding (Amyloidoses) may cause disease. Most proteins fold spontaneously, but some require the help of molecular chaperones.

Answer 17

Interaction between and arrangement of different polypeptide chains (subunits) within the same protein. The polypeptide chains may be identical (homomeric) or different (heteromeric).

Answer 18

3.6 amino acids per turn 0.54 nm pitch Right handed helix C=O group of one amino acid is H bound to N--H group of a residue 4 amino acids away H bonds are roughly parallel to axis of helix R groups found on outside of alpha helix

Answer 19

``` Extended conformation 0.35nm between adjacent amino acids R groups alternate Can form anti parallel, parallel or mixed sheets Multiple inter strand H bonds ```

Answer 20

Spontaneously or with the help of chaperones

Answer 21

Improper folding of polypeptide sequence (tertiary structure of protein) which may cause disease

Answer 22

Misfolded protein Highly ordered B sheet H bonds between B sheet cause protein aggregation Interchain assembly of B strands are stabilised by hydrophobic interactions between aromatic amino acids (aromatic amino acids block the interaction of amino acid side chains- and stops the aggregations of fibrils)

Answer 23

One repeating secondary structure Little/ no tertiary structure Long strands/ sheets Usually insoluble

Answer 24

Several types of secondary structure Complex tertiary structure Compact Usually soluble

Answer 25

Structure Support Protection

Answer 26

Enzymes | Regulatory proteins

Answer 27

Covalent (peptide)

Answer 28

Hydrogen bonds

Answer 29

``` Covalent (disulphide) Hydrophobic interactions Ionic interactions Hydrogen bonds VDW ```

Answer 30

Complex of protoporphyrin IX and Fe2+ | Fe2+ bound to 4 nitrogens (ring structure), and bound to histidine residue on globin chain

Answer 31

Heart and skeletal muscle

Answer 32

Acts as an oxygen reservoir in heart and skeletal muscle Oxygen carrier Increases rate of transport of oxygen in muscle cell

Answer 33

1 polypeptide of 153 amino acids Compact 75% a- helical His93 residue is in the 8th a- helix covalently bound to Fe2+ Hydrophobic centre and hydrophilic surface

Answer 34

Hyperbolic dependence Fe 2+ in deoxygenated myoglobin is slightly below the plane of the protoporphyrin ring Oxygen binding causes the movement of Fe2+ into the plane of the ring Causing the movement of Histidine F8 and a change in the protein conformation

Answer 35

Rectangular hyperbole

Answer 36

Exclusively in RBCs

Answer 37

``` 2 a (141 aa) and 2 b (146 aa) subunits = heterotetramer Each polypeptide is associated with haem Non covalent interactions hold the 4 chains together ``` Carry oxygen from lungs to tissue

Answer 38

Through crystallisation and a series of X ray shots

Answer 39

Mb had a very high affinity for oxygen and so will only release oxygen when pO2 is very low Hb however can exist in 2 states- a low affinity T state and a high affinity R state. The transition between these two states gives Hb its sigmoidal binding curve - Hb's affinity for oxygen increases as more oxygen binds- cooperativity

Answer 40

Tense state Low affinity of oxygen Right T state Stabilised by low pH, high co2 and high BPG When positive histidine residue is attracted to the negative aspartate residue

Answer 41

Relaxed High affinity for oxygen Left More stable Stabilised by higher pH low co2 and low BPG When histidine residue moves down- so there is no more interaction between histidine and aspartate

Answer 42

Binding of one molecule promotes the binding of another molecule

Answer 43

Decreases the affinity of Hb for O2 1 binds per tetramer Favouring the T state Curve shifts to the right

Answer 44

BPG conc increases at high altitudes which lowers Hb's affinity for oxygen - promoting the release of oxygen into tissues R converts to T state

Answer 45

Large amounts of metabolism produces large amounts of BPG which lowers Hb's affinity for oxygen- so O2 is released more readily in areas performing large amounts of metabolism

Answer 46

``` High carbon dioxide and H+ (highly metabolic tissues) Decreases the affinity of Hb for oxygen Curve shifts to the right R to T state H+ protonates His residue Carbon dioxide binds at N terminus ```

Answer 47

Binds to Hb 250x more readily than oxygen Decreases affinity of Hb for oxygen Fatal when COHb > 50%

Answer 48

Binding of CO acts to increase the affinity of unaffected sub units for oxygen

Answer 49

HbF = 2alpha and 2gamma HbA = 2alpha and 2beta HbF has a higher affinity for oxygen than HbA Ensures oxygen can be obtained from mother

Answer 50

Glutamate to valine

Answer 51

Glutamate is a polar negative amino acid Valine is a non polar neutral amino acid In the T state= Formation of hydrophobic pocket as a result of polymerisation of haemoglobin molecules due to valine residues This results in a distortion of the RBCs to a sickle shape Sickle Hb = 6-8g.dl-1 Normal Hb = 14g.dl-1

Answer 52

Decreased or absent alpha chains Beta chains can form stable tetramer a with a higher affinity Caused by 2 genes on chromosome 16 - varying levels of severity Onset before birth

Answer 53

Decreased or absent beta chains Lack of beta chains causes alpha chains to precipitate out of solution or form tetramers with gamma chains Chromosome 11- major and minor Symptoms appear after birth

Answer 54

Enzymes work by lowering the activation energy needed for a reaction to occur. Binding of substrate to a distinct part of the enzyme, the active site, increases the local concentration of reactants and also stabilises the formation of the high energy transition state

Answer 55

Increases the number of molecules with energy greater than the activation energy and hence rate of reacton

Answer 56

Increases the chance of molecular collisions and hence the rate of reaction

Answer 57

Cleft or crevice in the globular structure of an enzyme where the substrate weakly binds to form the enzyme substrate complex It is formed by only a few amino acids from different parts of the primary sequence It excludes water - preventing hydrolysis from occurring and interference of reaction

Answer 58

No they just increase the rate of attainment of equilibrium

Answer 59

Substrate has a complementary shape to the active site and binds to it without changing its shape

Answer 60

Active site only forms a complementary shape after binding to the substrate as the active site slightly changes shape to fit around the substrate

Answer 61

Vo= Vmax [S] / km + [S]

Answer 62

Rectangular hyperbolic | Simple Enzymes which can only exist in one conformation

Answer 63

The maximal rate of reaction when all enzyme active sites are saturated with substrate

Answer 64

Substrate concentration that gives half the maximal velocity

Answer 65

Gives a measure of the affinity of an enzyme for its substrate

Answer 66

High affinity of enzyme for substrate

Answer 67

Low affinity of enzyme for substrate

Answer 68

X intercept = -1/km | Y intercept = 1/vmax

Answer 69

Bind covalently to the active site

Answer 70

Binds at the active site temporarily Affect km Does not affect vmax

Answer 71

Binds at a secondary site, alters enzyme conformation Affects vmax Does not affect km

Answer 72

Changing substrate and product concentration | Changing the conformation of the enzyme- allosteric, covalent modification, proteolytic cleavage

Answer 73

Change in rate of protein synthesis | Change in rate of protein degradation

Answer 74

Availability of substrate affects rate of enzyme activity

Answer 75

Accumulation of product inhibits the forwards reaction

Answer 76

Accumulation of G6P in step 1 of glycolysis inhibits hexokinase activity

Answer 77

With allosteric affectors- which bind to a site other that the active site, and change the enzyme conformation, changing the activity of the enzyme by stabilising the R(highaff) or T(lowaff)state of the enzyme Enzymes have more than one sub unit and exist in 2 states (high affinity R state and low affinity T state) Sigmoidal relationship between [S] and rate (so km and vmax do not apply) The binding of substrate to one subunit makes subsequent binding to other subunits progressively easier- positive cooperativity These allosteric effectors may either inhibit or activate an enzyme

Answer 78

Bind to site other than active site Alter the conformation of the enzyme Increase the proportion of the R state enzymes

Answer 79

Bind to site other than active site Alter the conformation of the enzyme Increase the proportion of the T state enzymes

Answer 80

Phospho fructokinase | Step 3 of gycolysis key regulatory step

Answer 81

AMP, fructose 2 6 BISPHOSPHATE

Answer 82

ATP citrate H+

Answer 83

Many different types of group can be attached covalently to proteins (in this case enzymes) via amino acids. Most importantly, phosphate groups can be added (phosphorylation) [Ser,Thr OH]. The attachment of a phosphate group is carried out by kinase enzymes and their removal by phosphatase enzymes (regulated by hormonal levels)

Answer 84

Phosphatase

Answer 85

When enzymes activate other enzymes, the number of affected molecules increases in the enzyme cascade. Kinases transfer the phosphate group from ATP to the –OH group of Ser, Thr, Tyr Phosphatases remove phosphate groups through hydrolytic activity. Phosphate groups are bulky, charged groups that can significantly affect enzyme conformation and substrate binding. The addition and removal of these groups therefore regulates enzyme activity.

Answer 86

Enzyme secreted as an inactive protein precursor (zymogen) and cleaved by proteases to the active enzyme.

Answer 87

Inactive precursor of an enzyme

Answer 88

Blood clotting cascade

Answer 89

Inactive: Pepsinogen Active: Pepsin Activated by pH

Answer 90

Inactive: Trypsinogen Active: Trypsin Activated by enteropeptidase

Answer 91

Inactive: chymotrypsinogen Active: chymotrypsin Activated by trypsin

Answer 92

Inactive: procarboxypeptidase Active: carboxypeptidase Activated by trypsin

Answer 93

Inactive: proelastase Active: elastase Activates bye trypsin

Answer 94

Example of proteolytic cleavage Coagulation (intrinsic and extrinsic pathway) + common pathway Formation of a fibrin clot through a series of proteolytic cleavages

Answer 95

Endothelium lining tear Endothelial cells produce von willebrands factor VW factor activates platelets and stimulates their adhesion to exposed laminin and collagen Activates factor 12 Activates factor 11 Activates factor 9 Activates factor 10 with help of factor 8 Common pathway

Answer 96

``` Endothelium lining tear Endothelial cells produce thromboplastin (tissue factor) Thromboplastin activates factor 7 Activates factor 10 Common pathway ```

Answer 97

Activated Factor 10 activates prothrombin to thrombin with the help of factor 5 Thrombin activates fibrinogen to fibrin Thrombin also activates factor 13 Factor 13 helps fibrin become stabilised fibrin= fibrin clot

Answer 98

By the removal of activated proteins (dilution of factors by bloodstream to liver) Proteolytic digestion (protein c activated by thrombin binding to thrombodulin- digests factors 5 and 8 Binding of inhibitor molecules (antithrombin 3 acts on unbound thrombin - enhanced by heparin binding) Fibrinolysis (tPa and streptokinase activates plasminogen to plasmin; plasmin activates breakdown of fibrin clot to fibrin fragments)

Answer 99

Fibrinolysis- tPa and streptokinase activates plasminogen to plasmin; plasmin activates breakdown of fibrin clot to fibrin fragments

Answer 100

Vitamin K is required to produce inactive factors in the liver Forms gla residues that targets factors to the calcium in the membranes using carboxylase enzyme

Answer 101

Calcium accumulates in the endothelial lining Positive charge attracts the negatively charged gla residues on the inactivated factors bringing them to the site of damage Calcium required to assist with the activation of factors 2,7,9,10 (same as those that require vit K in their synthesis)

Answer 102

Synthesised by vit k on factors Negatively charged Attracted to calcium in endothelial lining- activated

Answer 103

Inhibits the vitamin K dependent factors in the intrinsic and extrinsic pathways 2thrombin, 7, 9, 10 Prevents the formation of gla residues on the factors So factors not targeted to endothelial lining and calcium So factors not activated

Answer 104

Used to activated conversion of plasminogen to plasmin | Hence breakdown of a fibrin clot by plasmin action

Answer 105

Activation of thrombin promotes further activation (of factors 8,5,13 assistant factors)

Answer 106

Because only a small amount of initial factor is required to produce a big response of clotting

Answer 107

Phosphate Pentose sugar Base

Answer 108

Pentose sugar and base

Answer 109

Can cross cell membranes more easily than nucleotides (without the large and negative phosphate)

Answer 110

OH at carbon 2 | Found in RNA

Answer 111

H at carbon 2 | Found in DNA

Answer 112

Double ring structured bases | A, G

Answer 113

Single ring structured bases | C, U, T

Answer 114

Purine with a pyrimidine A with T (2 H bonds) C with G (3 H bonds)

Answer 115

Single stranded A C G U Forms a stem loop structure- hydrogen bonds are formed e tween anti parallel complementary sequences on the same strand of RNA (single strand loops back on itself do that one side will run anti parallel and H bonds will form between complementary bases)

Answer 116

Double stranded helix A C T G Two polynucleotides are completely complementary and anti parallel 10 BP per turn 3.4nm per turn (between turns) 0.34 nm between base pairs Stabilised by van der waals forces above and below the ring Have major (exposed bases) and minor grooves (which are not at 180 degrees from one another)

Answer 117

Both DNA and RNA are labelled 5’ to 3’ (5’ starts with phosphate) Top strand is 5’ to 3’ Various bases are given letters, eg A, T, G, C, U Duplex structure includes the complimentary antiparallel strand Hydrogen bonds are denoted by dotted lines

Answer 118

Nucleotides are covalently linked via phosphodiester bonds. Each single-strand nucleic acid chain has a polarity. Two distinct ends – 5’ end with free phosphate and a 3’ end with free –OH

Answer 119

Nucleosomes – DNA is wound (~ twice) round histone core, which is charged Each nucleosomes is coiled to form solenoid structures Solenoid structures are further condensed into chromatids Shcjhebihcbuiwdbciubwdoucbouwbecuwed chromo abno

Answer 120

Condensed DNA which is not being expressed / transcribed/ cannot replicate Appears darker (Mitosis chromosomes- ie are involved in cell division)

Answer 121

Uncondensed DNA which is being expressed / transcribed/ can replicate Appears lighter (Interphase chromosomes)

Answer 122

23 chromosomes

Answer 123

2 identical diploid cells

Answer 124

4 non identical haploid cells

Answer 125

Interphase - G1- cell content replication S- DNA replication G2- cell check and quick repair before division Mitosis- cell division

Answer 126

At the end of G1 | At the end of G2

Answer 127

Important for regulation | In cancer cell cycle no longer functions so there is continuous growth of cells

Answer 128

Comes off of G1 | Importance icuhbefhu bweifcjniedcedcjinedc?

Answer 129

DNA polymerase

Answer 130

``` Primate Helicase DNA polymerase- required activated precursors dNTPs and ATP since every addition to chain requires hydrolysis of ATP Ligase DNA nucleotides ```

Answer 131

DNA helicase unravels the DNA double helix Primase binds to the origin of replication recruiting DNA polymerase to the site (DNA polymerase can only add bases in a 5' to 3' direction)

Answer 132

Leading strand is replicated in a 5’ to 3’ direction continuous, as normal Lagging strand is replicated discontinuously in Okazaki fragments (in 5'-3' direction too) [Since DNA polymerase can only extend strands in the 3' direction only one of the two new strands can be synthesised in a continuous manner in the same direction in which the replication fork is moving]

Answer 133

Replication forks move from the ends of the DNA strands towards each other and meet in the middle [Lead strands move towards lagging strands and vice versa] Okazaki fragments are then joined by DNA ligase from OH group to Phosphate group covalently

Answer 134

Semi conservative in that after replication each chromosome now consists of one strand of original DNA and one new strand

Answer 135

Growth Repair Maintenance

Answer 136

``` Prophase Prometaphase Metaphase Anaphase Telophase Cytokinesis ```

Answer 137

Nuclear membrane disappears Chromosomes condense (heterochromatin) Spindle fibres appear

Answer 138

Spindle fibres attach to the centromeres of chromosomes | Chromosomes continue to condense

Answer 139

Chromosomes line up randomly in single file along the metaphase plate (centre of cell)

Answer 140

Spindle fibres contract Centromeres divide Sister chromatids move too opposite poles

Answer 141

Nuclear membrane reforms Chromosomes decondense Spindle fibres disappear

Answer 142

Cytoplasm divides | Parent cell becomes 2 daughter cells with identical genetic info

Answer 143

Maintains diploid chromosome number in zygote

Answer 144

Prophase - cytokinesis 1 Prophase - cytokinesis 2 (Effectively two rounds of mitosis)

Answer 145

Meiosis produces 4 non identical gametes, whereas e mitosis produces 2 identical daughter cells In meiosis there are two cell divisions whereas there is only one in mitosis In meiosis, homologous pairs are first divided and then sister chromatids, whereas in mitosis sister chromatids are divided straight away

Answer 146

Independent assortment of chromosomes at meiosis | Crossing over

Answer 147

Independent assortment is the random distribution of maternal and paternal chromosomes into gametes during meiosis.

Answer 148

In bivalent/tetrad forms of chromosomes- bits of DNA swap from one homologous chromosome to another- increasing genetic variation

Answer 149

Genetic make up of an organism either as a whole or for a specific genetic locus

Answer 150

All observable characteristics of an individual or the observable trait as the result of the genetic make up of the one or more specific genetic loci

Answer 151

``` Radiation Mutagens Chemicals that affect cell growth Diet Lifestyle ```

Answer 152

Where two alleles of a gene have equal dominance and both affect the phenotype equally Blood type

Answer 153

When more than one gene is involved in the expression of a genotype- means that parents with different affected alleles can have unaffected children Albinism

Answer 154

Genes close to one another on the same chromosome are said to be linked Genes on different chromosomes or far away from one another on the same chromosome are said to be not linked Linked genes do not show independent assortment at meiosis (as they co segregate and are inherited together)

Answer 155

Linked genes

Answer 156

The likelihood of crossing over occurring between two linked genes

Answer 157

High likelihood of crossing over occurring between two genes Genes are unlinked RF=50% because when material splits during meiosis the 2 genes will either end up on the same chromatid or will separate (2 options with equal chance)

Answer 158

Low likelihood of crossing over occurring between two genes | Genes are linked

Answer 159

Total number of confirmed recombinant progeny/ total number of confirmed progeny Confirmed- we are certain of their genotype

Answer 160

Gene - A unit of heredity; a length of DNA on a chromosome that contains the code for a protein

Answer 161

Allele - An alternative form of a gene; each individual has two alleles for every gene, which can either be the same or different.

Answer 162

Dominance - A phenotypic trait is dominant when it occurs in both homo and heterozygotes.

Answer 163

Recessive – A phenotypic trait is recessive when it occurs only in homozygotes

Answer 164

When the gene in question is located on an autosome

Answer 165

When the gene in question is located on a sex chromosome

Answer 166

Two alleles are different

Answer 167

Two alleles are the same

Answer 168

When there is only one allele instead of two for a gene in a diploid organism Normal- sex chromosomes Abnormal- missing autosomal chromosomes

Answer 169

When individuals have a heterozygous or homozygous dominant (fatal) genotype for a disease on their autosomes Males ~ females Cannot skip a generation Affected individuals have a 75% chance of having affected offspring Hunting tons Marfan's

Answer 170

When individuals have a homozygous recessive genotype for a disease on their autosomes Heterozygotes are carriers Males ~ females 2 homozygous recessive individuals will only have affected children 2 carriers will have 25% chance of having affected offspring (parents don't need to be affected for child to be) Can skip generations Cystic fibrosis, albinism

Answer 171

When individuals have a homozygous recessive genotype (women) or a hemizygous recessive genotype (men) on their sex chromosomes Males > females Heterozygous female has a 50% chance of having affected sons Affected males cannot give their allele to their sons (only give Y chromosome) Can skip generations Girls can inherit defective allele from mum and dad, boys can only inherited defective alleles from mother Haemophilia A Duchenne muscular dystrophy

Answer 172

Complex of rRNA and many proteins that act as a mini protein factory Makes proteins in process of translation by reading RNA template

Answer 173

2% of all RNA 1000s of different kinds but few copies of each Single strand with no stem loops, have codons Made by RNA polymerase II

Answer 174

15% of all RNA 100s of different kinds, many copies of each Single strand with stem loops, have anticodons Made by RNA polymerase III

Answer 175

80% of all RNA 4 kinds in eukaryotes but many copies of each Associated with proteins to form ribosomal subunits Made by RNA polymerase I

Answer 176

``` Bacteria have: Simpler promoter regions Different transcription factors Coupled transcription and translation No PTM so mRNA is short lived 70S ribosomes Different transcription factors and initiation mechanisms ```

Answer 177

Transcription factor binds to Promoter sequence TATA box 30 bases upstream of origin of transcription Recruits RNA polymerase II - separates DNA to allow RNA nucleotides to bind and forms phosphodiester bonds

Answer 178

RNA polymerase II binds to template strand (3’-5’) and makes pre mRNA in 5’-3’ direction (copies coding strand)

Answer 179

RNA polymerase II recognises a stop codon and falls off

Answer 180

Post transcriptional modification 5’ capping- methylated guanine added to 5’ end 3’ Polyadenylation- lots of A’s added to 3’ end PREVENT DEGRADATION Splicing- introns removed by endo/exonucleases from pre mRNA = mRNA (only contains exons)

Answer 181

40S ribosome subunit with methionyl tRNA binds to AUG start codon on mRNA.. Binding causes 60S subunit to combine with the 40S subunit initiating translation to occur Can be free in the cytoplasm or bound to ER

Answer 182

Met tRNA occupies the P site of the ribosome Another anticodon is recognised, so next amino acyl tRNA occupies the A site Peptidyl tranferases catalyse the formation of peptide bond between the amino acids in the P and A site tRNA in P site becomes uncharged and so is released- both amino acids now occupy the A site Ribosome shifts one to the right- moving the dipeptide into the P site and freeing up that A site for the addition of another amino acyl tRNA Continues

Answer 183

Stop codon is recognised on mRNA No tRNA with complementary anticodons exist so polypeptide is hydrolysed from tRNA using H20 to release it into the cytoplasm

Answer 184

There is more than one triplet code coding for one amino acid

Answer 185

Membrane or secretory pathway

Answer 186

Cytosol or post translational import into the cell

Answer 187

Via co translational insertion

Answer 188

Signal, receptor, translocational machinery, Energy

Answer 189

Ribosome begins to synthesise a polypeptide chain from mRNA The polypeptide chain will have a hydrophobic signal at it's N terminus This signal is recognised by a signal recognition particle which binds to the signal, stopping translation temporarily, and the SRP directs the ribosome to the ER At the ER the SRP binds to a receptor which temporarily fixes the ribosome to the ER SRP dissociates and this restarts translation The polypeptide sequence enters the ER via a membrane pore, whilst being synthesised If protein is for secretory pathway signal peptide is cleaved within the ER If protein is for the membrane it will have another signal peptide which anchors it in the membrane and continues translation on the cytosolic side of ER

Answer 190

Multi domain riboprotein which mediates a 3 way association with SRP receptor in the ER, the ribosome and signal peptide

Answer 191

Hydrophobic N terminus signal sequence 13-36 amino acids long 1+ positively charged residues & 10-15 hydrophobic residues at N terminus A few hydrophilic residues within C terminal region

Answer 192

SRP, SRP receptor, protein translocator

Answer 193

Cleaved by signal peptidase

Answer 194

Yes- hydrolysis of GTP by SRP

Answer 195

NLS= nuclear localising signal Basic (Arg and lys residues) May be multipartite Various positions, must be on the surface of the folded protein

Answer 196

Folded (can pass through the large pores in the double membrane of the nucleus)

Answer 197

Importin - recognises NLS and mediates transport to nucleus | RanGTP displaces importin in nucleus and drives out export cargo

Answer 198

Signal is retained - to facilitate the reimporting of proteins when nucleus reforms after cell division

Answer 199

Yes - the hydrolysis of GTP

Answer 200

Amphipathic signal for the initial targeting to matrix (may be extra signals to other final destinations) at the N terminus

Answer 201

Held partially unfolded by chaperones (mitochondrial import stimulation factor)

Answer 202

TOM (Translocase of the outer membrane)- recognise proteins on outer membrane and form import channels TIM (Translocase of inner membrane)- proteins transported across the inner mitochondrial membrane Chaperones of the HSP70 family- assist in folding

Answer 203

Cleaved by mitochondrial processing peptidase

Answer 204

Yes- ATP hydrolysis by mitochondrial HSP70, drives translocation into the lumen and keeps some proteins unfolded prior to delivery. Electrical potential across IMM assists translocation

Answer 205

Post translational addition of mannose 6 phosphate to N linked oligosaccharides in the Golgi ( using N-acetylglucosamine phosphotransferase and N-acetylglucosamine phosphoglycosidase) Proteins destined for lysosomes are targeted for the addition of M6P groups by the presence of a signal patch, a sequence of several amino acids from different parts of the amino acid sequence

Answer 206

Folded (delivered via vesicles)

Answer 207

M6P receptor in the trans Golgi

Answer 208

Phosphate group is removed from the M6P group on the protein by a phosphatase- to ensure that the protein does not return to the Golgi with the receptor

Answer 209

Yes- phospho transferase | UTPNAG + mannose --> UMP + NAG + mannose-6-P

Answer 210

``` KDEL signal (Lys-Asp-Glu-Leu) At the C terminus ```

Answer 211

Folded (transported as a vesicle)

Answer 212

KDEL receptor in the cis Golgi - protein binds at low pH

Answer 213

Not directly- involves binding and release dependent on pH Golgi- low pH ER- neutral pH But formation of budding vesicle requires GTP hydrolysis

Answer 214

``` Insertion of proteins into membranes Specific proteolytic cleavage N linked glycosylation Formation of disulphide bonds Proper folding of proteins Assembly of multi sub unit proteins Hydroxylation of proline and lysine residues ```

Answer 215

Signal proteolytic cleavage Disulphide bond formation N linked glycosylation

Answer 216

O linked glycosylation Trimming and modification of N linked oligosaccharides (ie. addition of M6P for targeting to lysosomes) Further proteolytic cleavage of some proteins- e.g activating an enzyme

Answer 217

Signal sequences are cleaved off (removal of pre sequence) | Signal peptidase

Answer 218

In collagen synthesis when 150 N terminal aa and 250 C terminal aa (pro peptide sequences which don't form the triple helix) are cleaved off (Removal of pro sequence)

Answer 219

N linked glycosylation Oligosaccharides are built up on a dolichol phosphate carrier molecule sitting in the membrane Oligosaccharides is then transferred to the amide group of asparagine Oligosaccharide protein transferase

Answer 220

O linked glycosylation Attachment of oligosaccharides to the hydroxyl group of serine and threonine Glycosyl transferase builds up sugar chain from substrates (Important in proteoglycans)

Answer 221

They increase the stability of the protein in environments that can be harsher than those inside the cell Protein disulphide isomerase

Answer 222

Protein disulphide isomerase | Signal peptidase

Answer 223

Binding immunological protein | Calnexin and calreticullin

Answer 224

As a result of a mutation, causing the protein to be incorrectly associated with other proteins

Answer 225

Binds to exposed amino acid sequences that would normally be buried in the interior of the folded protein Retain folded protein in the ER Act as sensors to monitor the extent of misfolding Mediate increased transcription of chaperones and reduction in translation

Answer 226

Bind to oligosaccharides on incompletely folded proteins Retain folded protein in the ER Act as sensors to monitor the extent of misfolding Mediate increased transcription of chaperones and reduction in translation

Answer 227

Protein may be returned to the cytosol for degradation | Protein may accumulate to toxic levels in the ER resulting in disease

Answer 228

Continuous process Proteins packaged into vesicles and released continuously by exocytosis. E.g. Serum albumin, collagen

Answer 229

Proteins released in response to a signal e.g. hormone | Proteins packaged into vesicles but not released until stimulus received E.g. insulin

Answer 230

The basic unit of Collagen is Tropocollagen Primary sequence is (Glycine-X-Y)n X,Y= proline (helix breaker-prevents individual chains from forming helices), hydroxyproline (hydroxylated proline using prolyl hydroxylase, increases amount of interchain H bonds), lysine and hyroxylysine 3 tropocollagen alpha chains arranged in a triple helix structure (Left-handed triple helix – Non-extensible/compressible, high tensile strength) o Prolyl Hydroxylase requires Vitamin C and Fe2+ ions for activity. - Scurvy is due to low Vitamin C, therefore weak tropocollagen triple helices.

Answer 231

Prolyl hydroxylase assists in the hydroxylation of proline residues in collagen Requires vitamin C and Fe2+ for normal activity

Answer 232

Cleavage of signal peptide Hydroxylation of proline and lysine residues Addition of N linked oligosaccharides and galactose to hydroxylysine residues Disulphide bond formation at C terminal propeptide sequence to hold alpha chains in place for helix formation Formation of triple helical pro collagen from N terminal to C terminal Addition of O linked oligosaccharides Transport vesicle to cell membrane Exocytosis at membrane into EC space Removal of the N and C terminal propeptide sequences Lateral association of collagen molecule Covalent cross linking of collagen molecule Aggregations of fibrils

Answer 233

Contains A and B peptide joined via 2 disulphide bonds

Answer 234

C peptide is a good marker for measuring the levels of endogenous insulin being produced in diabetic patients Since insulin and C peptide are produced in the same amounts

Answer 235

PreProInsulin – Contains Signal Sequence, A, B and C peptides. Signal Sequence is cleaved by signal peptidase and 2 disulphide bonds are formed between A and B proteins ProInsulin – Contains A, B and C peptides Endopeptidases cleave C peptide

Answer 236

Determines the sequence of nucleotide bases in a DNA fragment

Answer 237

SANGER method · Uses modified ‘terminator’ nucleotides (ddNTP which lacks 3’ OH and thus stops elongation) · 4 test tubes are set up each containing a single stranded DNA fragment (with an unknown sequence), free nucleotide bases, ‘terminator’ nucleotide bases (ddNTP, where N is A, T, C, G), a radioactively or fluorescently labelled primer, DNA polymerase · Depending on which ddNTP is used and where the ddNTP binds to the DNA template (in 5’- 3’ direction), the synthesis of the new strand will stop at different places and thus varying lengths of DNA are synthesised · All the fragments of new DNA in each of the test tubes will end with a nucleotide that has the same base · DNA Gel electrophoresis: Samples are then loaded onto a polyacrylamide gel and the DNA strands separate according to their length · Fragments can be seen on the gel because of their labelled primers MODERN method Fluorescent dideoxyDNA sequencing - the samples of DNA are run on one lane · Each different ddNTP is labelled a different colour · Cheaper and faster methods

Answer 238

· Determine the length and sequence of DNA · Forensics · Cloning

Answer 239

Bacterial enzymes that are able to recognise specific DNA sequences and then cut the double stranded DNA at that specific site

Answer 240

· Specific restriction endonucleases recognise and cut specific DNA sequences by breaking the phosphodiester bonds · Restriction endonucleases cut at palindromic sites · Produce staggered ends- ‘sticky ends’ between which H bonds can form, but not phosphodiester bonds (can only be formed in the presence of DNA ligase) · Restriction endonucleases also methylate the sticky ends of the DNA fragment to prevent other DNA fragments from rebinding

Answer 241

Locate specific sequences of DNA | Comparing number and location of restriction sites on two DNA strands- wild type and mutant type

Answer 242

Use of restriction endonucleases and them gel electrophoresis

Answer 243

· Investigate the size of the DNA fragments (identify deletions) · Investigate mutations (sickle cell disease) · Investigate DNA variation (DNA fingerprinting) · Gene cloning (IN VITRO= PCR)

Answer 244

To amplify the quantity of target DNA Cam be done in vivo or in vitro

Answer 245

· Using plasmids o Small circular DNA Can be transferred to other bacteria o Can contain antibiotic genes · Plasmid is cut using restriction endonucleases- gene is added to create a Recombinant DNA molecule · Recombinant DNA is introduced into bacteria in transformation · Antibacterial genes are used to positively select for bacteria that have taken up any plasmid (e.g. will now contain gene for ampicillin resistance so will survive when ampicillin is added), and bacteria which have taken up the recombinant DNA (e.g. recombinant DNA would have been added into the tetracycline gene; so replica plating occurs, and then tetracycline is added; cells containing recombinant DNA will die) · Bacteria containing recombinant DNA are placed in an environment to multiply

Answer 246

Separates strands of DNA according to size

Answer 247

· Gel: Agerose · Buffer: Allows charge on DNA samples across the gel · Power supply: generates charge difference across the gel · Stain/ detection: Ethidium bromide · DNA fragments of known size are used as a reference · DNA molecules are loaded into wells on a gel made of agerose (which has small holes in it and thus acts like a sieve) · An electric field is supplied across the gel · The negatively charged DNA (due to negative phosphate group) moves towards the anode (from negative electrode to positive electrode) · Shorter DNA fragments travel further along the gel as longer DNA fragments get trapped in the small holes of the agerose gel more easily · DNA fragments are stained with ethidium bromide (fluorescent dye) so that they can be seen on the gel

Answer 248

· Size of DNA fragments | · Used with most other methods of analysis when identifying a gene

Answer 249

· Separation of DNA fragments: DNA fragments, primers and DNA polymerase are placed in a vessel in the thermocycler; temperature is increased to 95 degrees causing the two strands of DNA to separate · Addition (annealing) of primer: mixture is cooled to about 55 degrees (ranges from 55- 65; 65 is most common for G and C bases due to there being 3 H bonds being formed as opposed to 2 H bonds with A and T) causing the forwards and reverse primers to anneal to the complementary bases at the ends of the DNA fragment; Primers provide the starting sequence for DNA polymerase to begin DNA copying because DNA polymerase can only work in the 5’ to 3’ direction and attach at the end of an existing chain; Primers also prevent renaturation · Synthesis of DNA: temperature is increased to 72 degrees (optimum temperature for DNA polymerase to add complementary nucleotides along each of the separated DNA strands); It begins at the primer on both strands and adds nucleotides in sequence until it reaches the end of the chain.

Answer 250

Reverse Transcriptase-PCR Form of PCR where mRNA is converted into cDNA using reverse transcriptase Poly A tail on mRNA acts as a primer, enabling reverse transcriptase to convert the mRNA into cDNA The cDNA is then amplified in PCR

Answer 251

· Amplify specific DNA fragments · Genetic tests -with the addition of a restriction endonuclease, we can identify conditions where a restriction site is created (such as Junctional Epidermolysis Bullosa) and where a restriction site is destroyed (Tay Sachs Disease) · Single base mutations- Tay Sachs, Sickle Cell disease · Investigating small deletions or insertions (Cystic Fibrosis) · Investigating variation, genetic relationships (DNA profiling)

Answer 252

RT PCR- · Gene expression (if gene is being expressed then mRNA will be present) · Produces DNA for PCR · Can be used for genes from bacteria/viruses

Answer 253

Investigation of one gene on a DNA strand using a DNA probe

Answer 254

A technique where DNA separated by electrophoresis is transferred to a membrane filter and is detected by the hybridisation of a labelled probe

Answer 255

· Extraction- DNA is extracted from samples and increased using PCR · Digestion- DNA is cut into fragments using restriction endonucleases · Separation- DNA fragments are separated according to size using gel electrophoresis (agerose gel); DNA fragments are immersed in alkali which causes the DNA to denature into single strands · Blotting- DNA fragments are transferred from gel to nylon membrane by southern blotting- wodge of paper towels placed on top which assists the DNA being drawn up into the nylon; DNA fragments are fixed to the nylon with UV light · Hybridisation- DNA probes are added to label the fragments by binding to specific fragments (complementary nucleotide base sequence to the gene we are interested in potentially finding) · Development- membrane with labelled DNA fragments, is placed onto an X ray film; development of film shows dark bands where radioactive DNA probes have bound

Answer 256

A technique where RNA, separated by electrophoresis is transferred to a membrane filter and is detected by the hybridisation of a labelled probe

Answer 257

· Investigating gene structure- large deletions/ duplications · Investigating gene expansions and triplet repeats- (E.g. Fragile X syndrome, Huntington’s) · Investigating mutations in genetic tests- using allele specific probes (E.g with Sickle cell disease AàT) · To investigate variation and genetic relationships- DNA fingerprinting

Answer 258

Investigation of 1000’s of genes simultaneously on a DNA strand using a DNA probe (can also be used with RNA- reverse transcriptase= cDNA)

Answer 259

· DNA fragments organised into arrays are hybridised to labelled DNA from 2 different sources · 2 different sources are labelled either red (e.g cancerous) or green (e.g healthy) · Red and green fluorescence can be studied to work out the cancerous: healthy ratio for each cell (red: green) 1. An array of DNA probes covering the entire genome is applied to the surface of a solid matrix 2. Patient DNA and normal control DNA are each labelled with different coloured fluorescent tags e.g. Patient’s DNA labelled red and normal DNA labelled green 3. Equal amounts of the labelled DNA are then hybridised to the probe array and the hybridisation signals are detected and compared. 4. For probes where the signal of the normal DNA exceeds that of the patient’s DNA, the patient has a deletion of the chromosomal region from which that probe was derived. (If Green > Red on the probe array solid matrix

Answer 260

· Investigate deletions/ duplications · Investigate conditional gene expression- (Comparing cancerous and normal genes; Comparing patient and normal genes) · Used to screen for sub microscopic chromosomal deletions for which location (locus) cannot be deduced from the patients phenotype

Answer 261

Looks collectively at all chromosomes

Answer 262

A karyotype is a picture of the full set of stained metaphase chromosomes of an individual organised in pairs according to chromosome number

Answer 263

· Chromosome 1 is the largest chromosome and chromosome 22 is the smallest · Chromosomes are grouped and numbered according to their size and position of their centromere

Answer 264

· Investigating chromosome deletions/ duplications · Reasons for karyotyping include: Constitutional (congenital) abnormalities (prenatal screening, birth defects, abnormal sexual development, infertility, recurrent foetal loss) AND Acquired abnormalities (Leukaemia and other related disorders)

Answer 265

Used to identify the presence and location of a region of DNA or RNA within morphologically preserved chromosome preparations, fixed cells or tissue sections

Answer 266

· Fluorescent probes for a specific gene (or several genes) can be used or probes for specific DNA stretches on chromosomes such as telomeres or centromeres · For chromosome investigation chromosome painting is often used whereby each chromosome is visualised using a different coloured fluorescent probe ``` Denature the chromosomes Denature the probe Hybridization Fluorescence staining Examine slides or store in the dark ```

Answer 267

· WITHIN cells · Locating a specific gene on a chromosome · Identifies chromosome abnormalities · Degree of sequence identity can be determined

Answer 268

Separates protein fragments on the basis of molecular weight (size), shape or charge

Answer 269

Sodium dodecyl sulphate polyacrylamide gel electrophoresis (SDS PAGE)- Separation on basis of size/ molecular weight Isoelectric Focusing- Separation on basis of Isoelectric Point/ Charge 2D polyacrylamide gel electrophoresis (2D-PAGE)- Separation of proteins with complex structure

Answer 270

Protein molecules are denatured by B-ME (breaks disulphide bonds)and SDS (eliminates secondary/ tertiary structure) [One molecule of SDS binds for every 2 amino acids] · The bound SDS has a large negative charge which masks the intrinsic charge of the protein · Therefore protein fragments are separated due to molecular weight/ size

Answer 271

containing a pH gradient · Protein will migrate in the electric field until it reaches a pH that matches it’s pI (Isoelectric point- protein has no overall charge at this point)

Answer 272

· Combination of SDS PAGE and Isoelectric focusing | · Allows separation of proteins that have identical pI values but different molecular weights

Answer 273

Where amino acid deletions have occurred | Nonsense mutations have resulted in a PTC and a truncated protein

Answer 274

· Gel: Polyacrylamide · Buffer: maintains charge on protein samples · Power supply: generates charge difference across the gel · Stain/ detection: coomassie blue

Answer 275

a technique where a protein, separated by electrophoresis is transferred to a membrane filter and is detected by the hybridisation of a labelled probe

Answer 276

Determines whether a particular protein is present in a sample Measures the concentration of proteins in solution (a complex mixture e.g. serum) by analysing the binding of antibodies

Answer 277

· Coating- Polythyrene plate is treated with a solution of either antigen or antibody · Blocking- an unrelated protein based solution is used to cover all unbound sites on the plate · Detection- Enzyme conjugated antibody or antigen binds specifically to the target antigen or antibody · Read Results- Substrate is added and the signal (fluorescence) produced by the enzyme substrate reaction is measured

Answer 278

Antibodies used in ELISA’s can either be monoclonal or polyclonal

Answer 279

Antibodies derived from unique antibody producing cells called hybridomas and capable of specific binding to a single unique epitope

Answer 280

A pool of antibodies purified from animal sera that are capable of binding to multiple epitope

Answer 281

· Elisa tests are used to detect proteins (substances with antigenic properties)- hormones, bacterial antigens and antibodies

Answer 282

Measures the activity of an enzyme | Gives an indication of its presence at normal levels

Answer 283

· Measures the production of product/ disappearance of substrate · Performed at optimal pH, temperature and ionic strength ( with ions and cofactors) · Performed with a high concentration of substrate so that the enzyme works optimally · Continuous (where the assay gives a continuous reading of activity)- E.g. spectrophotometry or chemiluminescence · Discontinuous (where samples are taken, the reaction stopped and then the concentration of substrates/products determined)- E.g. radioactivity or chromatography

Answer 284

Important CLINICALLY- | The activity of various enzymes in serum is an indicator of tissue damage

Answer 285

Liver damage

Answer 286

Pancreatitis

Answer 287

Liver damage due to alcohol

Answer 288

Bone dosorder

Answer 289

Prostate cancer

Answer 290

Decreased in liver disease

Answer 291

Myocardial infarction

Answer 292

Mutation – ‘a change in a nucleic acid sequence, which can be the addition of one or more (or many) nucleotides [insertion], the removal of one or more (or many) nucleotides [deletions], or the rearrangement of several (or many) nucleotides’

Answer 293

A single base substitution

Answer 294

Substitution of a purine for another purine | Substitution of a pyrimidine for another pyrimidine

Answer 295

Substitution of a pyrimidine to a purine or vice versa

Answer 296

A mutation that does not alter the amino acid specified

Answer 297

A mutation that changes the amino acid specified to a stop codon Results in a premature termination codon (PTC) mRNAs containing PTCs are degraded by nonsense mediated decay (NMD) and little or no protein is produced

Answer 298

A mutation that replaces one amino acid with another

Answer 299

Addition or subtraction of nucleotides not in multiples of 3

Answer 300

Deletion of nucleotides not in multiples of 3 Insertion of nucleotides not in multiples of 3 Intron splice site mutation - since exon after mutated site is deleted- if its not a multiple of 3 = frame shift

Answer 301

Mutation to a start codon which results in the protein no longer being transcribed or translated as it is not recognised

Answer 302

Mutation to nucleotides in the promoter sequence where transcription factors bind Activates / deactivates the promoter region Alters transcription and gene expression

Answer 303

By doing exon counts with MLPA

Answer 304

Results in a delayed stop codon | Elongated non functional protein formed

Answer 305

DNA replication- tautomeric shift and slippage

Answer 306

``` Spontaneous cause of mutation Altered base pairing due to a shift of a proton Bases act as other bases C--> A T--> G ```

Answer 307

Spontaneous cause of mutation Newly synthesised strand loops out = extra base in new strand Template strand loops out = loss of base in new strand

Answer 308

Mutagens- nitrous acid, ethyl methane sulphonate, iQ/ ethidium bromide Ionising radiation - UV

Answer 309

C --> U binds with A A --> H binds with C G --> X binds with C

Answer 310

Removal of purine rings | Any base can pair with a purinic site

Answer 311

Disrupts DNA packing Intercalation of IQ/ EB forces bases further apart Leads to misreading by DNA polymerase = single base deletion a at GC base pairs

Answer 312

UV light photons cause adjacent TSH to base pair = dimers with one another

Answer 313

Cooked meats and cigarettes

Answer 314

UV bActivates Vitamin D in skin UV b Can cause sun burn and skin cancer UV a and b together can destroy vitamin a in the skin UV a b and c can damage collagen fibres and cause skin ageing

Answer 315

Proof reading Nucleotide mismatch repair Excision repair P53 (guardian angel protein)

Answer 316

DNA polymerase detects mispaired 3' bases in newly synthesised strand (in DNA replication)

Answer 317

Enzymes detect bases that don't base pair in new DNA strand (mismatched) Enzymes excise and replace the base pairs

Answer 318

Removal of damaged DNA by excision if bases and replacement by DNA polymerase (damaged)

Answer 319

Nucleotide excision repair- 30 bases - post UV damage/ carcinogens Base excision repair - 1 to 5 bases - oxidised, alkylated, delaminates bases and uracil in DNA

Answer 320

Guardian angel Monitors repair of damaged DNA Promotes apoptosis if damage is too severe

Answer 321

PCR amplification and southern blotting (e.g. delta FS80 CF- 3 base deletion- phenylalanine lost) PCR amplification and restriction analysis (e.g. sbs A-->T, restriction site for MstII enzyme destroyed - so one less fragment in mutated gene) SSCP Prenatal diagnosis

Answer 322

Mutation scanning Heterozygous person= mix of normal and mutant sequences DNA is heated and denatured It is rapidly cooled - forming a partly double stranded DNA molecule DNA is electrophoresed on a PA gel Detected by silver staining A single nucleotide change in a particular sequence, as seen in a double-stranded DNA, cannot be distinguished by electrophoresis, because the physical properties of the double strands are almost identical for both alleles. After denaturation, single-stranded DNA undergoes a 3-dimensional folding and may assume a unique conformational state based on its DNA sequence. The difference in shape between two single-stranded DNA strands with different sequences can cause them to migrate differently on an electrophoresis gel, even though the number of nucleotides is the same, which is, in fact, an application of SSCP.

Answer 323

Amniocentesis (taken from amniotic fluid) 15-20 weeks in ; (0.5-1% rom) ultrasound guidance Chorion virus biopsy ( trans vascular/-abdominal; foetal virus needs to be separated from maternal tissue) 10-13 weeks in ; ( 2% rom) ultrasound guidance From mothers blood (still in development stages)

Answer 324

C to T transition 2/3

Answer 325

Wild-type – ‘an individual within a population displaying a wild-type trait, which is the trait that is most common in that population’ A mutation causes a mutant phenotype, which is a phenotype that differs from the common or wild type phenotype in the population. - A mutation in a gene causes a mutant allele, which is an allele that differs from the common allele in the population (the wild type allele). - Mutations that occur in the germline have the possibility of being passed on to offspring – germline mutations.

Answer 326

Can be sporadic or heritable If DNA is damaged to the extent that apoptosis (programmed cell death promoted by p53 protein) doesn’t occur or the damage leads to uncontrolled growth then cancerous cells are produced. Tumours - Tumours are derived from individual abnormal cells. - They arise from the lack of normal growth control. - Generated by a multistep process. - Tumours are more likely to arise from cell types undergoing frequent cell division. - All of the cells in a tumour are of the same type. - The behaviour of a tumour depends on the cell type. ``` Oncogenes Genes involved in control of cell division: - They’re present in normal cells - Many different classes - May stimulate/inhibit growth ``` Tumour Suppressor Genes - Genes involved in protecting the cell against one step on the path to cancer. When the genes are normally present in cells they’re proto-oncogenes. It is after that mutation (or increased expression) that a proto-oncogene becomes oncogene. Viruses can carry copies of oncogenes, the presence of a virus means that the gene doesn’t function as normal. E.g. HPV.

Answer 327

Occurring at random or by chance, and not as a result of a genetically determined, or inherited, trait.

Answer 328

Capable of being passed from one generation to the next; hereditary

Answer 329

Centromere at top of chromosome

Answer 330

Centromere in top portion of chromosome Chromosome p arms have satellites Group D and G chromosomes

Answer 331

Centromere in top portion of chromosome

Answer 332

Centromere in middle of chromosome

Answer 333

Chromosomes are made up of chromatin. Chromatin is made up of: - DNA - Non-histone proteins - RNA - Histones (H1. H2A, H2B, H3, H4) Of the histones H2A, H2B, H3 and H4 interact directly with the DNA. H1 varies between species and H3 and H4 are highly conserved. The histones are responsible for the ‘beads on a string’ structure of chromatin forming nucleosomes. DNA in the cell is organised into tightly folded chains around histone proteins to form Chromosomes. In a human cell, there are 23 pairs of chromosomes. These replicate during the S phase of the cell cycle.

Answer 334

``` Numerical – a number of chromosomes other than 46 o Polyploidy (e.g. triploidy, tetraploidy) o Aneuploidy (an abnormal number that is not a multiple of the haploid number) ```

Answer 335

A number of chromosomes which is a multiple of the haploid number and greater than the diploid number of chromosomes

Answer 336

Poly spermy

Answer 337

Fertilisation of an egg by more than one sper

Answer 338

A number of chromosomes which is not a multiple of the haploid number of chromosomes

Answer 339

Type of aneuploidy | Monosomy is a loss of one chromosome i.e. one ‘chromosome pair’ exists as a single chromosome.

Answer 340

Type of aneuploidy | Trisomy is a gain of one chromosome i.e. one ‘chromosome pair’ exists as a triplet

Answer 341

Structural – physical changes to one or more of the chromosomes Balanced, when the change does not cause any missing or extra genetic info. Unbalanced, when the changes cause missing or extra genetic info.

Answer 342

o Deletion – loss of genetic information o Duplication – some genetic material is doubled o Inversion – no loss of genetic material, but a rearrangement of genetic materia o Ring chromosome – loss of telomeres or ends of both arms and formation of a ring o Isochromosome – creation of two non identical chromosomes, one is a combination of the two short arms, the other is a combination of the two long arms.

Answer 343

Inversion – No loss of genetic material, but a rearrangement of genetic material to a non-homologous chromosome Reciprocal translocation – no loss of genetic material, but an exchange of genetic material between two non-homologous chromosomes Robertsonian translocation – rearrangement of genetic material between two chromosomes; the q-arms (long) of two acrocentric chromosomes combine to form one ‘super-chromosome’ with the loss of both p-arms.

Answer 344

The karyotype formula starts with the total number of chromosomes in the cell, followed by a comma, then the X chromosomes, then the Y chromosomes. E.g. Normal Female = 46,XX and a Normal Male is 46,XY The plus (+) or minus (-) sign and then a number indicate an extra/missing entire chromosome. A chromosome number then a p/q and then a +/- indicates an extra/missing piece E.g. 5p- means ‘missing a segment of the p-arm on chromosome 5

Answer 345

Only one Other is inactivated and condensed down to form a BARR body seen at periphery of cell nucleus

Answer 346

Trisomy 21 More common in boys ``` 3 causes: Non disjunction at meiosis : 47, XY, +21 RT 14 & 21 : 46, XY, -14, +t(14q, 21q) RT 21 & 21 : 46, XY, -21, +t(21q, 21q) [N.B DS may arise from a carrier of superchromosomes due to RT] ``` Symptoms: Mild to moderate intellectual problems Congenital heart disease Haematological malignancies are more common: acute lymphoblastic leukaemia (all) Hypothyroidism GI: lack of colon nerves = constipation Infertility: males infertile, females have reduced fertility Eye disorders: strabismus, refractive errors, cataracts Hearing disorders due to infections and malformations

Answer 347

Trisomy 18: 47, XX, +18 More common in females (80%) Modal lifespan: 5-15 days Mostly diagnosed prenatally

Answer 348

Trisomy 13: 47, XY, +13 More than 80% die before they turn 1 Symptoms: Congenital abnormalities Polydactyly

Answer 349

Monosomy X: 45, X Females ``` Symptoms: Short stature Broad chest Low hair line Low set ears Webbed neck Cardiovascular symptoms renal problems INFERTILITY No mental retardation issues ```

Answer 350

Trisomy X : 47, XXX Females 2/3 X chromosomes are inactive- BARR bodies Mostly undiscovered Symptoms: Tall stature Microcephaly Delayed motor skills, speech and learning disabilities Auditory processing defects Scoliosis (abnormal curvature of the spine to the sides) MOSTLY NORMAL FERTILITY

Answer 351

Trisomy X : 47, XXY Males After onset of puberty Symptoms: Smaller testes = reduced testosterone production Glynocomastia (increased breast tissue) Language, learning and reading impairment (treatment with hormones and surgery) Reduced facial and pubic hair INFERTILITY

Answer 352

Trisomy Y: 47, XYY Males Phenotype relatively normal ``` Symptoms: Increase growth rate from early childhood (avg 7cm taller) Normal testosterone levels NORMAL FERTILITY Slightly lower IQ levels ```

Answer 353

Specific hormone abnormality associated with chronic myolegenous leukaemia cml Reciprocal translocation between chromosome 9 & 22 Translocation brings BCR gene on 22 close to ABL1 gene Results in a protein that makes fusion protein oncogenic

Answer 354

Balanced: Duplication Unbalanced: Interstitial and terminal deletions Ring chromosome Isochromosome Balanced and unbalanced: Inversion Reciprocal translocation Robertsonian translocation

Answer 355

X AND Y chromosomes have short regions in common at the tips of their long or short arms - PSEUDO AUTOSOMAL REGIONS (par1 and par2) These regions allow the two different chromosomes to recognise one another Turned syndrome patients will be monosomic for genes in the PARs- has a large effect on phenotype

Answer 356

Duplicated copies of rRNA genes

Answer 357

Could be large enough to be seen by standard light microscopy or karyotyping More cryptic deletions may require FISH

Answer 358

Non disjunction at meiosis | Anaphase lag

Answer 359

The failure of homologous chromosomes to separate properly during Meiosis 1 OR The failure of sister chromatids to separate properly during Meiosis 2 Part of chromosome left behind is not destroyed- goes to other cell

Answer 360

When a homologous chromosome or sister chromatid is left behind in meiosis which results in such chromosomes being excluded from daughter cell Due to a defect in spindle function Part of chromosome left behind is destroyed

Answer 361

Non disjunction at meiosis Robertsonian translocation 14&21 Robertsonian translocation 21&21

Answer 362

25% Trisomy 21 25% Monosomy 21 50% normal

Answer 363

Normal with 14&21 superchromosome (45) Normal (46) Monosomy 14 (45) Trisomy 14 with 14&21 superchromosome (46) Monosomy 21(45) Trisomy 21 with 14&21 superchromosome (46)

Answer 364

Trisomy 21 with 21&21 superchromosome | Monosomy 21

Answer 365

Production of a male gamete- sperm ``` Cell growth, Meiosis 1 (homologous chromosomes split into 2 cells) Meiosis 2 (sister chromatids split into 4 cells) Cell differentiation (into sperm- tails etc) ```

Answer 366

Production of female gametes ``` Cell growth Meiosis 1 (homologous chromosomes split= 1 cell and 1 polar body) Meiosis 2 (sister chromatids of cell from meiosis 1 split= 1 ovum and 1 polar body) ``` In females, primary oocytes enter meiosis 1 before birth and remain arrested there until ovulation - so the later you give birth (maternal age effect) the longer the time available for possible damage to occur to primary oocyte in meiosis 1

Answer 367

Mosaicism- two distinct cell lines (with different DNA) derived from one fertilised egg- caused by non disjunction in one of the early mitotic divisions of the zygote Tetragametic chimerism- two distinct cell lines (with different DNA) derived from two fertilised eggs which fuse during early embryogenesis (fusion of 2 non identical twins)

Answer 368

Non disjunction- Chromosomes divide unequally over daughter cells Anaohase lag- Chromosomes left at metaphase plate during anaphase as a result of defective attachment of spindle fibres to the centromere- chromosomes are subsequently lost by degradation in the cytoplasm

Answer 369

``` Loss of a chromosome One cell (in which mutation has occurred) and all subsequent daughter cells will be mutant ```

Answer 370

Loss of chromosome Mutant gamete formed Consequences for potential offspring

Answer 371

Amplification of signals by kinase cascades allows amplification of the initial signal by several orders of magnitude within a few milliseconds

Answer 372

Epinephrine activated adenylate cyclase Adenylate cyclase activates ATP to cyclic amp Cyclic amp activates protein kinase A (Breakdown) Protein kinase A activates phosphorylase kinase Phosphorylase kinase activates glycogen phosphorylase (Synthesis) protein kinase A activates glycogen synthase

Answer 373

* Prenatal screening (with maternal age effect ESP-DOWNS) * Birth defects (mental retardation/ developmental delay- DOWNS, KLINEFELTERS) * Abnormal sexual development (KLINEFELTERS) * Infertility (KLINEFELTERS, TURNERS, DOWNS) * Recurrent fetal loss (EDWARDS AND PATAUS) * Leukaemia (ass with DOWNS cml?)

Answer 374

 In both processes a template (a code) is read to create a macromolecule built up of basic subunits (a polynucleotide or a polypeptide).  Both processes consist of three stages: initiation, elongation and termination (of which especially the first and the last stage are highly regulated).  Both processes require energy and are driven by enzymatic activity.  In both processes the template has ‘surplus code’ that can be used for regulation; in DNA promoter sequences, terminator sequences and introns; in mRNA the 5’UTR and 3’UTR.  The products of both processes undergo further modification before they are ‘fully active’; posttranscriptional modification of RNA (splicing) and posttranslational modification of proteins

Answer 375

Transcription an RNA molecule is made takes place in the nucleus a code is copied (1 unit=1 base to 1 unit= 1 base) Translation a protein is made takes places in the cytoplasm a code is translated (3 units=triplet -> 1 unit=1 amino acid)

Answer 376

mRNA template covered with very many actively translating ribosomes

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