2. Cells Flashcards

Question

Describe the general structure of prokaryotic cells

Answer 1

Always present: - Cell surface membrane - Cell wall: contains murien, a glycoprotien - Cytoplasm - Small ribosomes - Circular DNA: free in cytoplasm, not associated with proteins) Sometimes present: - Capsule (on the outside) - Plasmitd: small rings of DNA - Flagella

Answer 2

Eukaryotic cell: - Has membrane-bound organelles eg. mitochondria, endoplasmic reticulum - Has a nucleus Containing DNA - DNA is long & linear & associated with histone proteins - Larger (80S) ribosomes (in cytoplasm) - Cell wall only in plants, algae and fungi Containing cellulose or chitin - Plasmids / capsule never present (sometimes flagella) - Larger overall size Prokarytoic cell: - No membrane-bound organelles eg. no mitochondria, endoplasmic reticulum - No nucleus DNA is is free in cytoplasm - DNA is short & circular & not associated with proteins - Smaller (70S) ribosomes - Cell wall in all prokaryotic cells Containing murein, a glycoprotein - Plasmids, flagella and a capsule sometimes present - Much smaller overall size

Answer 3

● Acellular - not made of cells, no cell membrane / cytoplasm / organelles ● Non-living - have no metabolism, cannot independently move / respire / replicate / excrete

Answer 4

1. Nucleic acids surrounded by a capsid (protein coat) 2. Attachment proteins allow attachment to specific host cells 3. No cytoplasm, ribosomes, cell wall, cell-surface membrane etc. 4. Some also surrounded by a lipid envelope eg. HIV

Answer 5

● Magnification = number of times greater image is than size of the real (actual) object ○ Magnification = size of image / size of real object ● Resolution = minimum distance apart 2 objects can be to be distinguished as separate objects

Answer 6

Optical microscope: - Light focused using glass lenses - Light passes through specimen, different structures absorb different amounts & wavelengths - Generates a 2D image of a cross-section - Low resolution due to long wavelength of light - Can’t see internal structure of organelles or ribosomes - Specimen = thin - Low magnification (x 1500) - Can view living organisms - Simple preparation - Can show colour Transmission electron microscope (TEM): - Electrons focused using electromagnets - Electrons pass through specimen, denser parts absorb more and appear darker - Generates a 2D image of a cross-section - Very high resolution due to short wavelength of electrons - Can see internal structures of organelles and ribosomes - Specimen = very thin - High magnification (x 1,000,000) - Can only view dead / dehydrated specimens as uses a vacuum - Complex preparation so artefacts often present - Does not show colour Scanning electron microscope (SEM): - Electrons focused using electromagnets - Electrons deflected / bounce off specimen surface - Generates a 3D image of surface - High resolution due to short wavelength of electrons - Can’t see internal structures - Specimen does not need to be thin - High magnification (x 1,000,000) - Can only view dead / dehydrated specimens as uses a vacuum - Complex preparation so artefacts often present - Does not show colour

Answer 7

● Scientists prepared specimens in different ways ● If an object was seen with one technique but not another, it was more likely to be an artefact than an organelle

Answer 8

1 Note formula / rearrange if necessary (I = AM) 2 Convert units if necessary - image and actual size must be in same unit 3 Calculate answer and check units required or if standard form etc. is required

Answer 9

1. Line up (scale of) eyepiece graticule with (scale of) stage micrometre 2. Calibrate eyepiece graticule - use stage micrometre to calculate size of divisions on eyepiece graticule 3. Take micrometre away and use graticule to measure how many divisions make up the object 4. Calculate size of object by multiplying number of divisions by size of division 5. Recalibrate eyepiece graticule at different magnifications

Answer 10

1. Homogenise tissue / use a blender ● Disrupts the cell membrane, breaking open cells to release contents / organelles 2. Place in a cold, isotonic, buffered solution ● Cold to reduce enzyme activity ○ So organelles not broken down / damaged ● Isotonic so water doesn’t move in or out of organelles by osmosis ○ So they don’t burst ● Buffered to keep pH constant ○ So enzymes don’t denature 3. Filter homogenate ● Remove large, unwanted debris eg. whole cells, connective tissue 4. Ultracentrifugation - separates organelles in order of density / mass ● Centrifuge homogenate in a tube at a low speed ● Remove pellet of heaviest organelle and respin supernatant at a higher speed ● Repeat at increasing speeds until separated out, each time the pellet is made of lighter organelles (nuclei → chloroplasts / mitochondria → lysosomes → ER → ribosomes)

Answer 11

Stage 1 Interphase ● DNA replicates semi-conservatively (S phase) ○ Leading to 2 chromatids (identical copies) joined at a centromere ● Number of organelles & volume of cytoplasm increases, protein synthesis (G1 / G2) Stage 2 Mitosis ● Nucleus divides ● To produce 2 nuclei with identical copies of DNA produced by parent cell Stage 3 Cytokinesis ● Cytoplasm and cell membrane (normally) divide ● To form 2 new genetically identical daughter cells

Answer 12

Stage 1 Prophase ● Chromosomes condense, becoming shorter / thicker so visible ○ Appear as 2 sister chromatids joined by a centromere ● Nuclear envelope breaks down ● Centrioles move to opposite poles forming spindle network ● Spindle fibres start to attach to chromosomes by their centromeres Stage 2 Metaphase ● Spindle fibres attach to chromosomes by their centromeres ● Chromosomes align along equator Stage 3 Anaphase ● Spindle fibres shorten / contract ● Centromere divides ● Pulling chromatids (from each pair) to opposite poles of cell Stage 4 Telophase ● Chromosomes uncoil, becoming longer / thinner ● Nuclear envelopes reform = 2 nuclei ● Spindle fibres / centrioles break down

Answer 13

● Within multicellular organisms, not all cells retain the ability to divide (eg. neurons) ● Only cells that do retain this ability go through a cell cycle

Answer 14

Parent cell divides to produce 2 genetically identical daughter cells for: ● Growth of multicellular organisms by increasing cell number ● Replacing cells to repair damaged tissues ● Asexual reproduction

Answer 15

Mitosis is a controlled process so: ● Mutations in DNA / genes controlling mitosis can lead to uncontrolled cell division ● Tumour formed if this results in mass of abnormal cells ○ Malignant tumour = cancerous, can spread (metastasis) ○ Benign tumour = non-cancerous

Answer 16

● Some disrupt spindle fibre activity / formation ○ So chromosomes can’t attach to spindle by their centromere ○ So chromatids can’t be separated to opposite poles (no anaphase) ○ So prevents / slows mitosis ● Some prevent DNA replication during interphase ○ So can’t make 2 copies of each chromosome (chromatids) ○ So prevents / slows mitosis ## Footnote (These are more effective against cancer cells due to uncontrolled cell division, but also disrupts cell cycle of rapidly dividing healthy cells)

Answer 17

Binary fission: 1. Replication of circular DNA 2. Replication of plasmids 3. Division of cytoplasm to produce 2 daughter cells ○ Single copy of circular DNA ○ Variable number of copies of plasmids

Answer 18

1. Attachment proteins attach to complementary receptors on host cell 2. Inject viral nucleic acid (DNA/RNA) into host cell 3. Infected host cell replicates virus particles: a. Nucleic acid replicated b. Cell produces viral protein / capsid / enzymes c. Virus assembled then released ## Footnote Being non-living, viruses do not undergo cell division

Answer 19

Preparation of stained squashes of cells from plant root tips; set-up and use of an optical microscope to identify the stages of mitosis in these stained squashes and calculation of a mitotic index. Students should measure the apparent size of cells in the root tip and calculate their actual size using the formula: actual size = size of image / magnification

Answer 20

1. Cut a thin slice of root tip (5mm from end) using scalpel and mount onto a slide 2. Soak root tip in hydrochloric acid then rinse 3. Stain for DNA (eg. with toluidine blue) 4. Lower coverslip using a mounted needle at 45 o without trapping air bubbles 5. Squash by firmly pressing down on glass slip but do not push sideways

Answer 21

● Where dividing cells are found / mitosis occurs

Answer 22

● To distinguish chromosomes ● Chromosomes not visible without stain

Answer 23

● (Spreads out cells) to create a single layer of cells ● So light passes through to make chromosomes visible

Answer 24

● Avoid rolling cells together / breaking chromosomes

Answer 25

● Separate cells / cell walls ● To allow stain to diffuse into cells ● To allow cells to be more easily squashed ● To stop mitosis

Answer 26

1 Clip slide onto stage and turn on light 2 Select lowest power objective lens (usually x 4) 3 a. Use coarse focusing dial to move stage close to lens b. Turn coarse focusing dial to move stage away from lens until image comes into focus 4 Adjust fine focusing dial to get clear image 5 Swap to higher power objective lens, then refocus

Answer 27

Stage, Appearance, Explanation Prophase ● Chromosomes visible / distinct → because condensing ● But randomly arranged → because no spindle activity / not attached to spindle fibre Metaphase ● Chromosomes lined up on equator → because attaching to spindle Anaphase ● Chromatids (in two groups) at poles of spindle ● Chromatids V shaped → because being pulled apart at their centromeres by spindle fibres Telophase ● Chromosomes in two sets, one at each pole ## Footnote In interphase (not mitosis), chromosomes aren’t visible but nuclei are. In mitosis, chromosomes are visible.

Answer 28

● Proportion of cells undergoing mitosis (with visible chromosomes) ● Mitotic index = number of cells undergoing mitosis / total number of cells in sample

Answer 29

● Count cells in mitosis in field of view ● Count only whole cells / only cells on top and right edges → standardise counting ● Divide this by total number of cells in field of view ● Repeat with many / at least 5 fields of view selected randomly → representative sample ● Calculate a reliable mean

Answer 30

1. Identify proportion of cells in named phase at any one time ○ Number of cells in that phase / total number of cells observed 2. Multiply by length of cell cycle

Answer 31

● Molecules free to move laterally in phospholipid bilayer ● Many components - phospholipids, proteins, glycoproteins and glycolipids ## Footnote ● Many components - phospholipids, proteins, glycoproteins and glycolipids The basic structure of all cell membranes (cell-surface membranes & membranes around eukaryotic organelles) is the same

Answer 32

● Phospholipids form a bilayer - fatty acid tails face inwards, phosphate heads face outwards ● Proteins ○ Intrinsic / integral proteins span bilayer eg. channel and carrier proteins ○ Extrinsic / peripheral proteins on surface of membrane ● Glycolipids (lipids with polysaccharide chains attached) found on exterior surface ● Glycoproteins (proteins with polysaccharide chains attached) found on exterior surface ● Cholesterol (sometimes present) bonds to phospholipid hydrophobic fatty acid tails

Answer 33

● Bilayer, with water present on either side ● Hydrophobic fatty acid tails repelled from water so point away from water / to interior ● Hydrophilic phosphate heads attracted to water so point to water

Answer 34

● Restricts movement of other molecules making up membrane ● So decreases fluidity (and permeability) / increases rigidity

Answer 35

● Phospholipid bilayer is fluid → membrane can bend for vesicle formation / phagocytosis ● Glycoproteins / glycolipids act as receptors / antigens → involved in cell signalling / recognition

Answer 36

● Lipid-soluble (non-polar) or very small substances eg. O2 , steroid hormones ● Move from an area of higher concentration to an area of lower conc., down a conc. gradient ● Across phospholipid bilayer ● Passive - doesn’t require energy from ATP / respiration (only kinetic energy of substances)

Answer 37

● Restricts movement of water soluble (polar) & larger substances eg. Na + / glucose ● Due to hydrophobic fatty acid tails in interior of bilayer

Answer 38

● Water-soluble / polar / charged (or slightly larger) substances eg. glucose, amino acids ● Move down a concentration gradient ● Through specific channel / carrier proteins ● Passive - doesn’t require energy from ATP / respiration (only kinetic energy of substances)

Answer 39

● Shape / charge of protein determines which substances move ● Channel proteins facilitate diffusion of water-soluble substances ○ Hydrophilic pore filled with water ○ May be gated - can open / close ● Carrier proteins facilitate diffusion of (slightly larger) substances ○ Complementary substance attaches to binding site ○ Protein changes shape to transport substance

Answer 40

● Water diffuses / moves ● From an area of high to low water potential (ψ) / down a water potential gradient ● Through a partially permeable membrane ● Passive - doesn’t require energy from ATP / respiration (only kinetic energy of substances) ## Footnote Water potential is a measure of how likely water molecules are to move out of a solution - pure (distilled) water has the maximum possible ψ (0 kPA), increasing solute concentration decreases ψ

Answer 41

● Substances move from area of lower to higher concentration / against a concentration gradient ● Requiring hydrolysis of ATP and specific carrier proteins

Answer 42

1. Complementary substance binds to specific carrier protein 2. ATP binds, hydrolysed into ADP + Pi, releasing energy 3. Carrier protein changes shape, releasing substance on side of higher concentration 4. Pi released → protein returns to original shape

Answer 43

● Two different substances bind to and move simultaneously via a co-transporter protein (type of carrier protein) ● Movement of one substance against its concentration gradient is often coupled with the movement of another down its concentration gradient

Answer 44

Absorption of sodium ions and glucose (or amino acids) by cells lining the mammalian ileum: 1 ● Na + actively transported from epithelial cells to blood (by Na + /K + pump) ● Establishing a conc. gradient of Na + (higher in lumen than epithelial cell) 2 ● Na + enters epithelial cell down its concentration gradient with glucose against its concentration gradient ● Via a co-transporter protein 3 ● Glucose moves down a conc. gradient into blood via facilitated diffusion ## Footnote The movement of sodium can be considered indirect / secondary active transport, as it is reliant on a concentration gradient established by active transport

Answer 45

● Increasing surface area of membrane increases rate of movement ● Increasing number of channel / carrier proteins increases rate of facilitated diffusion / active transport ● Increasing concentration gradient increases rate of simple diffusion ● Increasing concentration gradient increases rate of facilitated diffusion ○ Until number of channel / carrier proteins becomes a limiting factor as all in use / saturated ● Increasing water potential gradient increases rate of osmosis

Answer 46

● Cell membrane folded eg. microvilli in ileum → increase in surface area ● More protein channels / carriers → for facilitated diffusion (or active transport - carrier proteins only) ● Large number of mitochondria → make more ATP by aerobic respiration for active transport

Answer 47

Production of a dilution series of a solute to produce a calibration curve with which to identify the water potential of plant tissue.

Answer 48

You can rearrange and use the formula: C1 x V1 = C2 x V2 with V2 = V1+ volume of distilled water, or: 1. Calculate dilution factor = desired concentration (C2) / stock concentration (C1) 2. Calculate volume of stock solution (V1) = dilution factor x final desired volume (V2) 3. Calculate volume of distilled water = final desired volume (V2) - volume of stock solution (V1)

Answer 49

Step, Control variables 1. Create a series of dilutions using a 1 mol dm-3 sucrose solution (0.0, 0.2, 0.4, 0.6, 0.8, 1.0 mol dm-3 ) ● Volume of solution, eg. 20 cm3 2. Use scalpel / cork borer to cut potato into identical cylinders ● Size, shape and surface area of plant tissue ● Source of plant tissue ie variety or age 3. Blot dry with a paper towel and measure / record initial mass of each piece ● Blot dry to remove excess water before weighing 4. Immerse one chip in each solution and leave for a set time (20-30 mins) in a water bath at 30 oC ● Length of time in solution ● Temperature ● Regularly stir / shake to ensure all surfaces exposed 5. Blot dry with a paper towel and measure / record final mass of each piece ● Blot dry to remove excess water before weighing Repeat (3 or more times) at each concentration 6. Calculate % change in mass = (final - initial mass)/ initial mass 7. Plot a graph with concentration on x axis and percentage change in mass on y axis (calibration curve) ○ Must show positive and negative regions 8. Identify concentration where line of best fit intercepts x axis (0% change) ○ Water potential of sucrose solution = water potential of potato cells 9. Use a table in a textbook to find the water potential of that solution

Answer 50

Increase in mass ● Water moved into cells by osmosis ● As water potential of solution higher than inside cells Decrease in mass ● Water moved out of cells by osmosis ● As water potential of solution lower than inside cells No change ● No net gain/loss of water by osmosis ● As water potential of solution = water potential of cells

Answer 51

Investigation into the effect of a named variable on the permeability of cell-surface membranes.

Answer 52

1. Cut equal sized / identical cubes of plant tissue (eg. beetroot) of same age / type using a scalpel 2. Rinse to remove pigment released during cutting or blot on paper towel 3. Add same number of cubes to 5 different test tubes containing same volume of water (eg. 5 cm3 ) 4. Place each test tube in a water bath at a different temperature (eg. 10, 20, 30, 40, 50 oC) 5. Leave for same length of time (eg. 20 minutes) 6. Remove plant tissue and measure pigment release by measuring intensity of colour or concentration of surrounding solution semi-quantitatively or quantitatively

Answer 53

Semi- quantitative ● Use a known concentration of extract and distilled water to prepare a dilution series ● Compare results with these ‘colour standards’ to estimate concentration Quantitative ● Measure absorbance (of light) of known concentrations using a colorimeter ● Draw a calibration curve → plot a graph of absorbance (y) against concentration of extract (x) and draw a line / curve of best fit ● Read off sample absorbance value on curve to find associated concentration

Answer 54

● More permeable / damaged ● As more pigment leaks out making surrounding solution more concentrated (darker)

Answer 55

● As temperature increases, cell membrane permeability increases ○ Phospholipids gain kinetic energy so fluidity increases ○ Transport proteins denature at high temperatures as hydrogen bonds break, changing their tertiary structure ● At very low temperatures, cell membrane permeability increases ○ Ice crystals can form which pierce the cell membrane and increase permeability

Answer 56

● High or low pH increases cell membrane permeability ○ Transport proteins denature as H / ionic bonds break, changing tertiary structure

Answer 57

● As concentration increases, cell membrane permeability increases ● Ethanol (a lipid-soluble solvent) may dissolve phospholipid bilayer (creating gaps)

Answer 58

● Foreign molecule / protein / glycoprotein / glycolipid ● That stimulates an immune response leading to production of antibody

Answer 59

● Each type of cell has specific molecules on its surface (cell-surface membrane / cell wall) that identify it ● Often proteins → have a specific tertiary structure (or glycoproteins / glycolipids)

Answer 60

1. Pathogens (disease causing microorganisms) eg. viruses, fungi, bacteria 2. Cells from other organisms of the same species (eg. organ transplants) 3. Abnormal body cells eg. tumour cells or virus-infected cells 4. Toxins (poisons) released by some bacteria

Answer 61

1 Phagocyte attracted by chemicals / recognises (foreign) antigens on pathogen 2 Phagocyte engulfs pathogen by surrounding it with its cell membrane 3 Pathogen contained in vesicle / phagosome in cytoplasm of phagocyte 4 Lysosome fuses with phagosome and releases lysozymes (hydrolytic enzymes) 5 Lysozymes hydrolyse / digest pathogen ## Footnote Phagocytosis leads to presentation of antigens where antigens are displayed on the phagocyte cell-surface membrane, stimulating the specific immune response (cellular and humoral response)

Answer 62

T lymphocytes recognise (antigens on surface of) antigen presenting cells eg. infected cells, phagocytes presenting antigens, transplanted cells, tumour cells etc. Specific helper T cells with complementary receptors (on cell surface) bind to antigen on antigen-presenting cell → activated and divide by mitosis to form clones which stimulate: ● Cytotoxic T cells → kill infected cells / tumour cells (by producing perforin) ● Specific B cells (humoral response - see below) ● Phagocytes → engulf pathogens by phagocytosis

Answer 63

B lymphocytes can recognise free antigens eg. in blood or tissues, not just antigen presenting cells. 1. Clonal selection: ○ Specific B lymphocyte with complementary receptor (antibody on cell surface) binds to antigen ○ This is then stimulated by helper T cells (which releases cytokines) ○ So divides (rapidly) by mitosis to form clones 2. Some differentiate into B plasma cells → secrete large amounts of (monoclonal) antibody 3. Some differentiate into B memory cells → remain in blood for secondary immune response

Answer 64

● Quaternary structure proteins (4 polypeptide chains) ● Secreted by B lymphocytes eg. plasma cells in response to specific antigens ● Bind specifically to antigens forming antigen-antibody complexes

Answer 65

● Antibodies bind to antigens on pathogens forming an antigen-antibody complex ○ Specific tertiary structure so binding site / variable region binds to complementary antigen ● Each antibody binds to 2 pathogens at a time causing agglutination (clumping) of pathogens ● Antibodies attract phagocytes ● Phagocytes bind to the antibodies and phagocytose many pathogens at once

Answer 66

● Primary - first exposure to antigen ○ Antibodies produced slowly & at a lower conc. ○ Takes time for specific B plasma cells to be stimulated to produce specific antibodies ○ Memory cells produced ● Secondary - second exposure to antigen ○ Antibodies produced faster & at a higher conc. ○ B memory cells rapidly undergo mitosis to produce many plasma cells which produce specific antibodies

Answer 67

● Injection of antigens from attenuated (dead or weakened) pathogens ● Stimulating formation of memory cells

Answer 68

1. Specific B lymphocyte with complementary receptor binds to antigen 2. Specific T helper cell binds to antigen-presenting cell and stimulates B cell 3. B lymphocyte divides by mitosis to form clones 4. Some differentiate into B plasma cells which release antibodies 5. Some differentiate into B memory cells 6. On secondary exposure to antigen, B memory cells rapidly divide by mitosis to produce B plasma cells 7. These release antibodies faster and at a higher concentration

Answer 69

● Herd immunity - large proportion of population vaccinated, reducing spread of pathogen ○ Large proportion of population immune so do not become ill from infection ○ Fewer infected people to pass pathogen on / unvaccinated people less likely to come in contact with someone with disease

Answer 70

● Antigens on pathogens change shape / tertiary structure due to gene mutations (creating new strains) ● So no longer immune (from vaccine or prior infection) ○ B memory cell receptors cannot bind to / recognise changed antigen on secondary exposure ○ Specific antibodies not complementary / cannot bind to changed antigen

Answer 71

- Lipid envelope - RNA - Reverse transcripatse - Capsid - Attachment protein

Answer 72

1. HIV attachment proteins attach to receptors on helper T cell 2. Lipid envelope fuses with cell-surface membrane, releasing capsid into cell 3. Capsid uncoats, releasing RNA and reverse transcriptase 4. Reverse transcriptase converts viral RNA to DNA 5. Viral DNA inserted / incorporated into helper T cell DNA (may remain latent) 6. Viral protein / capsid / enzymes are produced a. DNA transcribed into HIV mRNA b. HIV mRNA translated into new HIV proteins 7. Virus particles assembled and released from cell (via budding)

Answer 73

● HIV infects and kills helper T cells (host cell) as it multiplies rapidly ○ So T helper cells can’t stimulate cytotoxic T cells, B cells and phagocytes ○ So B plasma cells can’t release as many antibodies for agglutination & destruction of pathogens ● Immune system deteriorates → more susceptible to (opportunistic) infections ● Pathogens reproduce, release toxins and damage cells

Answer 74

Viruses do not have structures / processes that antibiotics inhibit: ● Viruses do not have metabolic processes (eg. do not make protein) / ribosomes ● Viruses do not have bacterial enzymes / murein cell wall

Answer 75

● Antibody produced from genetically identical / cloned B lymphocytes / plasma cells ● So have same tertiary structure

Answer 76

● Monoclonal antibody has a specific tertiary structure / binding site / variable region ● Complementary to receptor / protein / antigen found only on a specific cell type (eg. cancer cell) ● Therapeutic drug attached to antibody ● Antibody binds to specific cell, forming antigen-antibody complex, delivering drug ## Footnote Some monoclonal antibodies are also designed to block antigens / receptors on cells

Answer 77

● Monoclonal antibody has a specific tertiary structure / binding site / variable region ● Complementary to specific receptor / protein / antigen associated with diagnosis ● Dye / stain / fluorescent marker attached to antibody ● Antibody binds to receptor / protein / antigen, forming antigen-antibody complex

Answer 78

Example method 1 (direct ELISA): 1. Attach sample with potential antigens to well 2. Add complementary monoclonal antibodies with enzymes attached → bind to antigens if present 3. Wash well → remove unbound antibodies (to prevent false positive) 4. Add substrate → enzymes create products that cause a colour change (positive result) Example method 2 (sandwich ELISA): 1. Attach specific monoclonal antibodies to well 2. Add sample with potential antigens, then wash well 3. Add complementary monoclonal antibodies with enzymes attached → bind to antigens if present 4. Wash well → remove unbound antibodies (to prevent false positive) 5. Add substrate → enzymes create products that cause a colour change (positive result)

Answer 79

Example method (indirect ELISA): 1. Attach specific antigens to well 2. Add sample with potential antibodies, wash well 3. Add complementary monoclonal antibodies with enzymes attached → bind to antibodies if present 4. Wash well → remove unbound antibodies 5. Add substrate → enzymes create products that cause a colour change (positive result)

Answer 80

● Compare to test to show only enzyme causes colour change ● Compare to test to show all unbound antibodies have been washed away

Answer 81

● Antibody with enzyme remains / not washed out ● So substrate converted into colour product

Answer 82

● Pre-clinical testing on / use of animals - potential stress / harm / mistreatment ○ But animals not killed & helps produce new drugs to reduce human suffering ● Clinical trials on humans - potential harm / side-effects ● Vaccines - may continue high risk activities and still develop / pass on pathogen ● Use of drug - potentially dangerous side effects

Answer 83

● Was the sample size large enough to be representative? ● Were participants diverse in terms of age, sex, ethnicity and health status? ● Were placebo / control groups used for comparison? ● Was the duration of the study long enough to show long-term effects? ● Was the trial double-blind (neither doctor / patient knew who was given drug or placebo) to reduce bias?

Answer 84

● What side effects were observed, and how frequently did they occur? ● Was a statistical test used to see if there was a significant difference between start & final results? ● Was the standard deviation of final results large, showing some people did not benefit? ● Did standard deviations of start & final results overlap, showing there may not be a significant difference? ● What dosage was optimum? Does increasing dose increase effectiveness enough to justify extra cost? ● Was the cost of production & distribution low enough?

2. Cells Flashcards

(110 cards)