Molecular Building Blocks Flashcards

Question

sphingolipids

Answer 1

formed from sphingosine (serine and palmitate) instead of glycerol. Sphingomyelin contains a phosphorylcholine group attached to ceramide and is a component of cell membranes and myelin sheath

Answer 2

synthesised from 20 C atoms. Acids with. 3,4,5 double bonds. biological functions

Answer 3

comprise of glycerol with one or more fatty acids attached through ester linkages. Can be phosphorylated to form phosphoacylglycerols

Answer 4

Micelle, Bilayer formed due to hydrophilic and hydrophobic interactions

Answer 5

atoms → molecules → macromolecules (large molecules formed from simple molecules)

Answer 6

universal solvent (can dissolve ionic and polar substances) • polarity leading to hydrogen bonding • liquid range from O to 100 °C • max density at 4°C • Doesn’t interact with non-polar substances, lipids, aromatic groups - hydrophobic compounds Hydrogen bonding = electronegative atoms such as O or N can attract hydrogen atoms from other molecules. Partial sharing of this proton leads to a mutual attraction between the 2 atoms. Weak on its own but strong in a collective

Answer 7

long unbranched polysaccharide radiating from a core protein, form (along with proteins such as collagen) the extracellular matrix cells exist on

Answer 8

contain an amino group instead of a hydroxyl group on one of the carbons (usually carbon 2)- often acetylated to form an N-acetylated sugar eg. Glucosamine

Answer 9

Heparin, chondroitin sulphate

Answer 10

Asymmetric so have a chiral carbon- enantiomers. Monosaccharide stereoisomers are designated D or L based on position of hydroxyl group farthest from the carbonyl carbon that matches D or L glyceraldehyde

Answer 11

straight chain C compounds (mostly 16-20) with a methyl group and a carbonyl group • Melting point decreases with degree of unsaturation (fluidity) as kink prevents surface contact but increases with chain length • Double bonds are commonly cis (hydrogen on same side of double bond) and spaced at 3 C intervals- trans fatty acids produced by hydrogenation of polyunsaturated fatty acids in vegetable oils

Answer 12

18:1, (triangle)9 18 = number of carbon atoms 1 = number of double bonds Triangle = the position of double bond (between 9th and 10th Carbon) Also classified by distance of double band closest to w end eg W-6 20:4 (triangle) 5,8,11,14

Answer 13

compounds that contain an amino group and a carboxylic acid group • 20 naturally occurring • Most naturally occur in L form • Charge determined by R group, amino group and carboxylic acid group which changes with pH • Polarity determined by R group • At different pH, carboxylic and amino group (and some R groups) are ionised • peptide bond formed by condensation reaction- cleaved by proteolysis enzymes (proteases or peptidases) • very stable • flexibility around C not involved in bond so multiple conformations

Answer 14

React intramolecularly to form a zwitterion = both positively and negatively charged. Because of this, there are strong intermolecular forces of attraction between amino acids so they are soluble crystalline solids

Answer 15

Act as protein buffers as can accept or donate H+

Answer 16

very stable • Cleaved by proteolysis enzymes- proteases or peptidases • Partial double bond • Flexibility around C atom not involved in bond, allows multiple confirmations • Usually one preferred native conformation, determined mainly by side chain and sequence of amino acids

Answer 17

simplest amino acid R group= -H

Answer 18

amine group and amino acids added to carboxylate group (N-terminus)

Answer 19

compact, roughly spherical in shape+ soluble in water Non-polar R groups orientate inwards with polar R groups on outside so are soluble and easily transported and involved in metabolic reactions. Specific shapes and some are conjugated (prosthetic group) e.g. Enzymes + immunoglobulins

Answer 20

long strands of polypeptide chains that have hydrogen bond cross-linkages Insoluble due to large number of hydrophobic R groups Repetitive sequence of amino acids eg. Collagen

Answer 21

X-ray diffraction of protein crystals

Answer 22

• backbone (line of peptide bonds- no side groups) • cartoon (shows fundamental secondary structures)

Answer 23

structural protein that's a component of connective tissue in vertebrates (eg tendons, skin, cartilage) Insoluble fibrous protein formed from 3 polypeptide chains closely held together by hydrogen bonds to form a triple helix- tropocol lagen Each chain is a helix shape and every 3rd amino acid is glycine and chains are offset allowing then to be arranged closely together Covalent bonds form cross-linkages between R groups in parallel to form fibrils Fibrils are positioned so that they are staggered to form fibres that are lined up with forces they are withstanding.

Answer 24

Many hydrogen bonds within triple helix results in great tensile strength - can withstand large pulling forces Staggered ends of fibrils provide strength Insoluble as too long

Answer 25

Globular protein with a quaternary structure 4 polypeptide chains (2 alpha-globins and 2 beta-globins) Each subunit has a prosthetic haem group- porphyria ring which holds an iron atom Held together by disulfide bonds and arranged so hydrophobic R groups face inwards (preserving 3D shape) and hydrophilic R groups face outwards (preserving solubility) Prosthetic haem group = contains Fe2+ ion that can reversible combine with O2 Each haemoglobin molecule has 4 haem groups so can carry 4 oxygen molecules Role of Hb- transport oxygen around body in red blood cells (erthrocytes)

Answer 26

• temperature • H+ • CO2 concentration

Answer 27

occurs when the microcirculation is obstructed by sickled RBCs, causing ischemic injury to the organ supplied and resultant pain. Endothelial damage causes multicellular (platelets, white cells) aggregates which occlude the capillary- cause inflammation

Answer 28

1. Gradient of curve initially shallow = the shape of Hb means it is difficult for the first O2 molecule to bind to a subunit as they are closely united. So at low ppO2, little oxygen binds to Hb 2. The binding of O2 changes the tertiary structure of Hb, causing it to change shape and create another binding site. So it takes a smaller increase in ppO2 for the second O2 to bind - positive cooperativity (ensures rapid intake of O2 in lungs) 3. Gradient reduces and curve plateaus = 4th O2 molecule is hard to bind as less empty sites to bind to as molecule becomes saturated Where ppO2 is low (e.g. Respiring tissue) Hb has a low affinity for oxygen which means it unloads O2 rather than loads it - % saturation is low Where ppO2 is high (e.g. Lungs) Hb has a high affinity for oxygen so readily loads O2 - % saturation is high

Answer 29

Hb has a reduced affinity for O2 in the presence of CO2. The greater the concentration of CO2, the more readily the Hb releases its oxygen - the curve shifts right At the gas exchange surface - CO2 is constantly being removed and so the pH is slightly raised due to the low pCO2. The higher pH changes the shape of Hb into one that enables it to readily load oxygen. The shape also increases Hb affinity for oxygen so it is not released while being transported in the blood to tissues At the tissues- CO2 is produced by respiring cells and so the blood pH is lowered. The lower pH changes the shape of Hb so it has a lower affinity for oxygen. Hb releases its oxygen into respiring tissues High rate of respiration→ more CO2 produced → lower pH→ greater change in Hb shape → more readily unloads oxygen→ more oxygen available for respiration

Answer 30

Autosomal recessive genetic disorder characterised by the formation of hard, sticky, sickle-cell shaped red blood cells in contrast to the biconcave-shaped red blood cells found in ‘normal’ individuals The gene defect is a single nucleotide polymorphism (GAG to GTG) of the beta-globin gene which results in glutamic acid being substituted by valine at position 6. Haemoglobin S with this mutation is referred to as HbS. This is normally. Benign mutation, causing no apparent effects on the structure of haemoglobin in normal oxygen concentrations. But at low ppO2 eg hypoxia, HbS polymerises and forms fibrous precipitates because the deoxy form of haemoglobin exposes a hydrophobic patch on the protein between the E and F helices. The amino acid change causes polymerization of this mutant hemoglobin (hemoglobin S [HbS]) to form fibers upon deoxygenation in the tissues, the root cause of the pathology of the disease. The fibers make the red blood cells less flexible and distort the shape of the cells, a process typically referred to as sickling.

Answer 31

• hydroxyurea- stimulates bone marrow to make foetal haemoglobin • Embolising drugs • Plasma exchange to reduce number of sickle cells

Answer 32

Early presentations can be pain in the hands and feet with severe pain in bones such as femur, humerus, ribs and pelvis occurring in older children. These are due to vaso-occlusive events in the small vessels. Repeated events in the bones can lead to chronic infarcts.

Answer 33

Liver function: acid phosphatase/ alanine aminotransferase Prostate cancer: alkaline phosphorylase Pancreas function: amylase Renal function: angiotensin converting enzyme Liver function: aspartame aminotransferase Myocardial infatuation: creatinine kinase

Answer 34

the loss of tertiary (and/or secondary) structure within a protein, which can be caused by heat, acid, or other agents that interfere with hydrogen bonding and usually causes a decrease in solubility (precipitation).

Answer 35

Biological catalysts = speed up rate of reaction by providing an alternative reaction pathway with a lower activation energy without being used up or changed Globular proteins with a specific active site where specific substrates bind, forming an E/S complex • intracellular enzymes= produced and function inside cell • extra cellular enzymes = secreted by cells + catalyse reactions outside cells e.g. Bacteria release proteases to digest protein so can absorb amino acids for growth

Answer 36

fitting into active site puts stresses + strains on bonds in substrate so bonds more likely to break

Answer 37

being attached to the enzyme holds then closer together, reducing any repulsion so they can bond more easily

Answer 38

Both enzymes and substrates are rigid structures that lock into each other exactly. Comments on specificity but not how activation energy is lowered

Answer 39

the enzyme's active site can change shape slightly as the specific substrate molecule enters the enzyme (conformational changes) so they are complementary. The substrate molecule temporarily bonds with amino acids in the active site and, as it changes shape, it places stresses and strains on bonds in the substrate making it more likely to react. The temporary E/S bonds then break + products expelled as enzyme returns to its original shape If active site altered by temperature, pH or mutation in base sequence, E/S complexes can not form

Answer 40

Can be regulated by altering concentration of substrates, products, inhibitors or activators as well as phosphorylation of the enzyme

Answer 41

enzymes that have a different structure and sequence but catalyse the same reaction

Answer 42

cannot in themselves catalyse a reaction but can help enzymes to do so- bind with enzyme protein molecules to form active enzymes

Answer 43

• very close proximity of the antibody CDR regions and the antigen surface • Intimate contact allows the combination of relatively weak interactions to produce a strong binding surface • CDR loops have a sequence of amino acids that complement the surface of the antigen Hydrogen bonds, charge, bulk (CDR can form a pocket to interact with and bury the bulky group), CDRs can form a compatible hydrophobic surface which affect recognition

Answer 44

portion of antigen bound

Answer 45

Proteins (immunoglobulins) with specific binding sites complementary to a specific antigen synthesised by B cells (plasma cells) Composed of 4 polypeptide chains: 2 long heavy chains and 2 short light chains- joined by disulphide bonds Each antibody has a specific binding site that fits precisely onto a specific antigen to form an antigen-antibody complex- Van der Waals forces hold together (instantaneous dipole)

Answer 46

primary sequence forms a specific tertiary structure, resulting in a specific binding site

Answer 47

gives flexibility which allows the antigen-binding site to be placed at different angles when binding to antigens, so allows it to bind to 2 antigens at once

Answer 48

• agglutination of cells - clumping then together, making it easier for phagocytes to locate and engulf them • opsonisation - serve as markers that stimulate phagocytosis • can act as anti-toxins, neutralising them • can attach to flagella of bacteria, making them less active so phagocytosis easier • can combine with viruses and toxins to block them from entering or damaging cells • complement activation - can create holes in cell wall, causing them to burst (lysis) when water moves in by osmosis - work inconjunction with other proteins

Answer 49

• agglutination of cells - clumping then together, making it easier for phagocytes to locate and engulf them • opsonisation - serve as markers that stimulate phagocytosis • can act as anti-toxins, neutralising them • can attach to flagella of bacteria, making them less active so phagocytosis easier • can combine with viruses and toxins to block them from entering or damaging cells • complement activation - can create holes in cell wall, causing them to burst (lysis) when water moves in by osmosis - work inconjunction with other proteins

Answer 50

pregnancy testing, cancer treatment, diagnosing diseases

Answer 51

B cells, dendritic cells, macrophages

Answer 52

Cross-linking of collagen fibres through the formation of strong collagen cross-links between lysine residues

Answer 53

RNA polymerase binds to the promoter sequence and unwinds part of a DNA molecule by breaking the hydrogen bonds between complementary base pairs, exposing the gene to be transcribed. Free activated RNA nucleotides pair up via hydrogen bonds with their complementary bases on the template antisense strand of the 'unzipped' DNA molecule. RNA polymerase then joins the RNA nucleotides together by condensation reactions, forming phosphodiester bonds in the 5’ to 3’ direction. Terminators signal RNA transcription is complete so the hydrogen bonds break between the DNA and mRNA molecules and the double-stranded DNA reforms. The pre-mRNA is then spliced - introns are removed and exons are joined together, resulting in mature mRNA which exits nucleus via a pore

Answer 54

during and after transcription, introns in the precursor (pre) mRNA are excised, and the noncontiguous coding exons are spliced together to form a shorter mature mRNA before its transportation to the ribosomes. The boundary between the exons and introns consist of a 5' donor GT dinucleotide and 3' acceptor AG dinucleotide. These, along with surrounding short splicing consensus sequences, another intronic sequence (branch site), small nuclear RNA (snRNA) molecules and associated proteins are necessary for splicing

Answer 55

the 5' cap is thought to facilitate transport of the mRNA to the cytoplasm and attachment to the ribosomes, as well as protect the RNA transcript from degradation by endogenous cellular exonucleases. After 20-30 nucleotides have been transcribed, the nascent mRNA is modified by the addition of a guanine nucleotide to the 5' end of the molecule by an usual 5' to 5' triphosphate linkage. A methyltransferase enzyme then methylates the N7 position of the guanine, giving the final 5' cap

Answer 56

transcription continues until specific nucleotide sequences are transcribed that cause the mRNA to be cleaved and RNA polymerase II to be released from the DNA template. Approx. 200 adenylate residues are added to the mRNA (poly(A) tail), which facilitates nuclear export and translation

Answer 57

The regulatory elements in the promoter region are cis-acting- they only affect the expression of the adjacent gene on the same DNA duplex, whereas the transcription factors are trans-acting, acting on both copies of a gene on each chromosome being synthesized from genes that are located at a distance.

Answer 58

proteins which bind to DNA (the promoter sequence) and increase/decrease the transcription of genes Enter nucleus from cytoplasm and are activated through a signalling pathway

Answer 59

1. The transcription factor binds to the promoter sequence of DNA (a section of DNA upstream of the coding region) near to the gene to be activated 2. RNA polymerase binds to the DNA-transcription factor complex and is activated, which causes it to move away from the complex along the gene 3. RNA polymerase transcribes the strand of DNA until it reaches a terminator sequence- gene is expressed If not expressed, the transcription factor inhibits RNA polymerase from binding A transcription complex forms around the TATA box 5’ of first exon Helix opens, DNA strand separation RNA Pol II builds mRNA

Answer 60

• activation of depressors (inhibitors of RNA polymerase binding) • Each step of RNA transcription or processing finds no longer actively produced transcription an processing proteins • Complexes do not form anymore for lack of phosphorylation • Enzymes no longer activities • RNA stability

Answer 61

After leaving the nucleus, the mRNA molecule attaches to a ribosome. A tRNA molecule (carrying a specific amino acid), with an anticodon complementary to the first mRNA codon, attaches itself to the mRNA by complementary base pairing. First amino acid is always methionine, codon = AUG on mRNA. Second tRNA molecule attaches itself to the next mRNA codon in the same way. The two specific amino acids attached to the tRNA molecules are joined by a condensation reaction forming a peptide bond using ATP. The first tRNA molecule moves away, leaving its amino acid behind. The ribosome moves along the mRNA in the 5’ to 3' direction. A third tRNA molecule binds to the next mRNA codon, bringing a specific amino acid, which binds to the first two amino acids and the second tRNA moves away. This continues producing a polypeptide chain until there is a 'stop' codon (UAG, UAA, UGA) on the mRNA molecule. The polypeptide chain moves away from the ribosome

Answer 62

incorporation of amino acids into a polypeptide chain requires amino acids to be covalently bound by reacting with ATP to the specific tRNA molecule by the activity of the enzyme aminoacyl tRNA synthetase. • the ribosome, with its associated rRNAs, moves along the mRNA, the amino acids linking up by the formation of peptide bonds through the action of the enzyme peptidyl transferase to form a polypeptide chain

Answer 63

in vitro method of amplifying DNA 1. Add to the thermocycler: piece of DNA to be copied, primers (a short section of single-stranded DNA that is complementary to the DNA interested in. 2 different primers are needed as the sequences at the ends of the target DNA are different. They keep the 2 DNA strands separate and create a double-stranded section for Taq DNA polymerase to bind to), nucleotides, Taq DNA polymerase (thermostable as from bacteria found in a hot spring - thermophilic) and buffer solution 2. Denaturation - heat up to 95°C to break hydrogen bonds between 2 DNA strands 3. Annealing - cool to 55°C to anneal the primers 4. Extension - heat to 72°C and Taq DNA polymerase will attach to the DNA strand and synthesise complementary DNA to produce new identical double-stranded DNA molecules (stops when all DNA nucleotides used up)

Answer 64

95 —> 55 —> 72

Answer 65

technique to look at gene expression by detection of RNA

Answer 66

piece of DNA to be copied, primers (a short section of single-stranded DNA that is complementary to the DNA interested in. 2 different primers are needed as the sequences at the ends of the target DNA are different. They keep the 2 DNA strands separate and create a double-stranded section for Taq DNA polymerase to bind to), nucleotides, Taq DNA polymerase (thermostable as from bacteria found in a hot spring - thermophilic) and buffer solution

Answer 67

1. A virus uses attachment proteins on its surface to bind to complementary receptors on the surface of a host cell 2. The virus injects its DNA/RNA into the host and it is replicated by the cell 3. The host cell uses its nucleic acids and ribosomes to produce new viral proteins (i.e. capsid) 4. The new viral particles are released either when the host cell bursts or viral particles leave individually through the host cell membrane via budding - forming the lipid envelope Attachment →penetration →reverse transcriptase →integration →translation →assembly →budding →release

Answer 68

1. Following infection, HIV enters the bloodstream and circulates around the body. A protein on HIV readily binds to a CD4 receptor on a helper T cell 2. The capsid fuses with the cell surface membrane and so the RNA and enzymes of HIV enter the helper T cell 3. The reverse transcriptase converts the viral RNA into DNA which is inserted into the cell's DNA by DNA intergrase 4. The HIV DNA in the nucleus creates mRNA which exists through a nuclear pore and synthesises new HIV particles at a ribosome 5. HIV particles 'bud' from the helper T cell eventually killing it

Answer 69

• vaccines • production of therapeutic proteins eg insulin and growth hormone - Recombinant DNA corresponding to the A-chain of human insulin was prepared and inserted into plasmids that were used to transform Escherichia coli cells. The bacteria then synthesized the insulin chain, which was purified. A similar process was used to obtain B-chains. The A- and B-chains were then mixed and allowed to fold and form disulfide bonds, producing active insulin molecules. • Production of complex human proteins eg Factor VIII • production of siRNA to reduce expression of a particular gene • production of monoclonal antibodies used in targeted drug treatment, diagnosis • Gene therapy: Ex vivo- adult stem cells removed, virus vector altered to contain desired gene, patent's cells altered after exposure to virus then grown in the lab, transgenic cells injected into patient In vivo- liposome or adenovirus inject DNA into cell's DNA e.g. Severe combined immunodeficiency (SCID)- created by ex vivo somatic gene therapy as a virus transfers a normal allele for ADA into T-lymphocytes • Screening patients: Genetic screening can help identify individuals who are carrying an allele at a gene locus for a particular disorder e.g. CF or breast cancer • DNA fingerprinting • Transgenic animals • Proteomics is the study of proteins expressed by a cell. Differences in protein expression between normal and cancer cells can be used to identify potential targets for future therapy. CNS

Answer 70

adult stem cells removed, virus vector altered to contain desired gene, patent's cells altered after exposure to virus then grown in the lab, transgenic cells injected into patient

Answer 71

liposome or adenovirus inject DNA into cell's DNA e.g. Severe combined immunodeficiency (SCID)- created by ex vivo somatic gene therapy as a virus transfers a normal allele for ADA into T-lymphocytes

Answer 72

eg insulin and growth hormone - Recombinant DNA corresponding to the A-chain of human insulin was prepared and inserted into plasmids that were used to transform Escherichia coli cells. The bacteria then synthesized the insulin chain, which was purified. A similar process was used to obtain B-chains. The A- and B-chains were then mixed and allowed to fold and form disulfide bonds, producing active insulin molecules.

Answer 73

study of proteins expressed by a cell. Differences in protein expression between normal and cancer cells can be used to identify potential targets for future therapy. CNS

Answer 74

study of proteins expressed by a cell. Differences in protein expression between normal and cancer cells can be used to identify potential targets for future therapy. CNS

Answer 75

prevention of blood clots, autoimmune therapies, targeting drugs… 1. Monoclonal antibodies are produced that are specific to antigens on cancer cells 2. Antibodies are given to a patient and bind to receptors on cancer cells, blocking the chemical signals which stimulate uncontrolled growth e.g. Herceptin 3. OR involves attaching a radioactive or cytotoxic drug so when the antibody attaches to the cancer cell, it kills them

Answer 76

Circulating tetramers made at the beginning of infection.

Answer 77

Monomer highly specific antibodies targeting single epitopes.

Answer 78

Likely to have developed in response to parasitic threats. Implicated in allergy, particularly alongside eosinophils

Answer 79

Expressed in mucosal tissue. Forms dimers. Protects the neonatal gut (expressed in breast milk)

Answer 80

Monomers, induction of antibodies in B cells, activates basophils and mast cells

Answer 81

Sickle cell haemoglobin is caused by a mutation that replaces glutamic acid at residue 6 in beta-globin with valine leading to the formation of linear polymers of deoxygenated HbS. The valine residue on the surface binds to the free complementary O2 site, linking the 2 tetramers together. As more tetramers become linked, linear polymers form converting normally flexible red blood cells into stiff, sickle-shaped cells which plug the capillary beds. The change in a single nucleotide (GAG to GTG) in the gene that codes for the beta-globin molecule on chromosome 17, results in a change in the primary structure of the polypeptide (glutamic acid is substituted by valine). Glutamic acid has a hydrophilic negatively charged side chain whereas valine has a hydrophobic side chain. As a result in low ppO2, as the amino acid is on the surface, the beta - 6 valine becomes buried in a hydrophobic pocket on an adjacent beta-globin chain, joining the molecules together to form insoluble polymers.

Answer 82

Chromosome 17

Answer 83

GAG to GTG

Answer 84

glutamic acid is substituted by valine

Answer 85

Sickle cell anemia is a autosomal recessive condition meaning that you must inherit a mutated haemoglobin gene (haemoglobin S) from both the maternal and paternal DNA. It can also occur by compound heterogeneity when a child inherits one haemoglobin S gene and another faulty haemoglobin gene (eg beta thalasdemia, haemoglobin C/D/E). If both parents are carriers, the child has a 25% chance of homozygosity.

Answer 86

produces both types of haemoglobin so conditions worse in cold, wet environments due to vasoconstriction

Answer 87

• hydroxyurea- stimulates bone marrow to make foetal haemoglobin • Embolising drugs • Plasma exchange to reduce number of sickle cells

Answer 88

• anemia as sickle cells break apart easily and die • episodes of pain which develops when sickle-shaped red blood cells block blood flow through capillaries in chest, abdomen or joints • swelling of hands and feet caused by sickle-shaped blood cells blocking circulation in the hands and feet • frequent infections as it can damage the spleen, increasing vulnerability to infections- encapsulated bacteria eg pneumonia • delayed growth or puberty • vision problems due to a damaged retina from reduced blood supply as a result of blocked blood vessels • Pulmonary hypertension due to reduced oxygen carrying capacity of blood • Avascular necrosis eg in head of femur

Answer 89

Fibroblasts

Answer 90

Tubulin Dynein

Answer 91

HbS polymerises and crystallises