2 Flashcards

Question

anaoblic

Answer 1

building up

Answer 2

diffusion of free water molecules across a partially permeable membrane ↳ from high conc -area to low cone . area movent of water against solute conc gradient- from a solution of low conc of solute to a solution of a high conc of solute

Answer 3

moves ions / molecules across plasma membrane against concentration gradient -molecule attaches to carrier protein , & protein changes shape , moving the moleculeacross the membrane & releasing on other side -ATP provides source of energy to go against concentration gradient ↳ ATP produced by respiration , immediate source of energy in cell, binds to carrier proteins- atp hydrolysed releasing phosphate group which releases energy as phosphate bonds with water-changes shape of carrier protein ↳hydrolysis of ATP releases energy

Answer 4

substances taken into the cell by a creation of a vesicle from the cell surface mebrane- part of the cell membrane engulfs substances some molecules are too large to be taken into cells via carrier - proteins , lipids , & some carbohydrates- taken in by endocytosis

Answer 5

if the mucus layer is too runny, the presence of excess water is detected by the epithelial cell membranes that line the airways. carrier proteins in the babsal membranes of the epithelial cells actively pump na+ out of the cell, conc of na+ in cell falls setting up a conc grdirjy across apical membrane causing na+ to diffuse from mucus down the conc gradient into epithelial cell by facillitated diffusion through epithelial na+ channels in apical membrane raised conc of na+ in tissue fluid on the basal mebrane- creates a potential diff between tissue fluid an mucus on the apical membrane- tissue fluid contains more pos charged ions than mucus, electrical gradient cuases cl- to diffuse down electrical gradient out of the mucus into tissue fluid via gap of neighbouring epithelial cells. increases na+ and cl- conc in tissue fluid, water drawn out of epithelial cells by osmosis across basal membrane into tissue fluid due to high slat conc in tissue fluid , water loss increases the overall solute conc in cell-higher than in mucus, water drawn out of the mucus by osmosis across apical mebrane

Answer 6

cl- transpported across basal mebrane into cells by active transport, creates conc gradient in apical mebrane with higher cl- in cell then in mucus, causes CFTR protein chanelle to open allowing cl- to diffuse out of cell and into mucus, when open CFTR blocks the epithelial sodium channel in apical membrane, build up of cl- creates an electrical gradient between mucus and tissue fluid, na+ diffuse out of tissue fluid and move down electrical gradient passing between cells into mucus, elevated salt conc in mucus draws water out of epithelial cells by osmosis until solutions are isotonic

Answer 7

CFTR protein not functioning or missing , when too little water in mucus and is sticky-cl- cannot be secreted across apical membrane, epithelial sdiem ion channels are not blocked and allow na+ into epithelial cells, raised level of na+ in cell drawns cl- and water out of mucus, causes mucus too be sticky due to absence of water

Answer 8

globular proteins that act as biological catalysts for metabolic reactions precise 3Dshape including specific active site

Answer 9

active site of enzymes certian shape for a specifc substrate, form temporray bonds with amino acids of active site to produce enzyme-substrate complex- allowing reaction to take place- product released leaving enzyme unchanged

Answer 10

flexible active site- when substrate enters active site the enzyme molecule changes shape slightkly, fitting more closely around substrate- only specifically shaped substrates will induce the correct change in shape

Answer 11

breaking down

Answer 12

synthesis: enzyme forces molecules together and reduces repulsion so molecules can bond more easily Breakdown : fitting in an active site strains bonds in substrate , breaking it down

Answer 13

too large to move across membrane no carrier/channel proteins for molecule to move through are polar and are repelled by hydrophobic fatty acid tail

Answer 14

CF cuases devlopment of thick sticky mucus which blocks pancreatic duct preventing panncreas enzymes from reaching small intestine. This reduces enzymes digesting food in intestine fundmenatlly reducing abosrbtion of products of digestion into the blood. This reduced amino acid/ mineral/ vitmins slows growth rate- enzymes and supplemnts would increase growth rate as help aid digestion with absence of digestive enzymes from pancreas

Answer 15

cell recognition, receptors, antigens

Answer 16

different primary structure results in a different sequence of amino acids, formation of teritarty structure from interaction of R groups so change in R groups changes tertiary structure. This could change shape of active site preventing substrate from binding reducing purpose as catalyst

Answer 17

protein which reduces activation energy of biological reactions

Answer 18

both are chains of amino acids joined by peptide bonds both contain bonds such as - hydrogen, ionic, disulphide globular proteins have hydrophilic groups on the outside whereas fibrous have hydrophobic groups on the outside globular folded into a compact shape whereas fibrous have long chains

Answer 19

CFTR protein allows chloide ions to leave cell and move into mucus. This electornic gradient cuases sodium ions to move into mucus. This increase of solute concentration in mucus causes water to move out of cell and into mucus through osmoisis

Answer 20

Rate if diffusion proportional to SA-alveoli have larger SA Rate of diffusion proportional to differences in concentration-breathing maintains a difference in gas conc Rate of diffusion proportional to Diff in conc-blood flow maintains a diff in gas conc Rate of diffusion inversely proportional to diffusion distance-walls of alveoli and capillaries once cell thick Diffusion distance is reduced is due to flattened cells forming alveoli and capillary walls Rate of diffusion proportional to diffusion constant- cell elevator relatively permeable to non polar gas molecules Alveoli covered in capillaries causing a short diffusion distance

Answer 21

Fluid refers to movement of the phospholipids in the membrane and mosaic refers to random association of proteins within membrane

Answer 22

Barrier to some solutes but not water Enables a conc gradient of solutes/water

Answer 23

1. i ethanol causes the membrane to be disrupted 2. due to phospholipids dissolve in ethanol ; 3. membrane proteins denatured by ethanol 4. disruption of the vacuole membrane 5. i{betalain / pigment} can escape from the {cell /vacuole /eq } when the membrane is disrupted ;

Answer 24

1. beetroot cells may have been damaged when cutting / eq ; 2. idea that beetroot pieces not rinsed before being placed in ethanol solution ; 3. idea that colorimeter was not {calibrated / zeroed / eq}(properly) ; 4. idea that test 2 is done some time after test 1 OR beetroot left in solution longer than 20 minutes in test 2 ; 5. idea that different parts of the beetroot may have different pigment concentrations ; 6. smaller volume of ethanol used in test 2 ;

Answer 25

appropriate standardisation of source of beetroot tissue ; standardisation of size of beetroot pieces / eq ; need for {washing / rinsing / eq} {beetroot / eq}(thoroughly) ; use of waterbath (to maintain / change temperature) ; reference to repeats at each temperature / replicates / eq ; more of a range of temperatures {below 20 / above 90 0C / smallerntervals /eq reference to one other suitable variable e.g. time beetroot pieces left between cutting and use ; reference to {calibration / zeroing / eq} of colorimeter ; {size / mass / surface area / volume / shape} of beetroot ; beetroot storage conditions {sa / type / species / eq} beetroot ; 5. {age of beetroot / storage time} ; ( cubation) time / eq ; {vol e / concentration / eq} of {water /solution}(added to beetroot) ; pH

Answer 26

1. Choose a beetroot sample and using a boarer take out samples of betroot, cut into equal sized samples 2. reference to {washing / soaking} {beetroot/ eq} (thoroughly)- to remove excess dye- night before 3. reference to waterbath (to maintain /change temperature) 4. reference to {range / at least 5] {temperatures / alcohol concentrations} ; 5. appropriate controlled variable named e.g. length of time in water bath, size of beetroot, if repeated same age beetroot, volume of water added to beetroot, distilled water indication of what is being used to judge permeability colour of solution, -absorbance, transmission ;7. description of how permeability can be assessed e.g. use of colorimeter, standard- calibrate colormitre with putting water in cuvette. blue/green filtre- reference to {calibration / zeroing / eq} of colorimeter 8. reference to repeats / replicates ; as temp increases absorbance decreases

Answer 27

sequence of baseson a DNA molecule that codes for a sequence of amino acids in a polypeptide chain

Answer 28

deoxyribonucleic acid

Answer 29

deoxyribose, a phosphate group, orgainc base containing nitrogen

Answer 30

condenstaion reaction between sugar of one nucleotide and the phosphate of another- forms a phosphodiester bond

Answer 31

sugar is ribose same sugar and phosphate group - ↳ different bases (a, c, U,g) transfer genetic information single stranded

Answer 32

non overlapping- each triplet code is adjacent, no bases is used for more than one amino acid degenerate- several triplets can code for the same amino acid

Answer 33

2 antiparallel polynucleotide strands twisted around each other in a x2 helix- sugar and phosphates form 2 sugar-phosphate backbones. Bases of the 2 strands are joined together by hydrogen bonds- complementary bases ( A+T and C+G)

Answer 34

A and G have two ring structure whereas C and T have one ring structure- pair to make 3 rings A and T form 2 hydrogen bonds and G and C for 3

Answer 35

-found in cytoplasm, folded to specific pattern & held by H bonds -amino acid binding site w/3 base sequence at other end -anti-codon held by H bonds -carries amino acids to ribosomes during translation -A,C,G,U -ribose -1 strand

Answer 36

DNA helicase uzips strand of DNA hydrogen bonds between complemtary bases break- 2 polynucleotide strands seperate free RNA nucleotides complemntary bases, pair to the exposed bases on the template strand by forming hydrogen bonds RNA polymerase forms sugar phosphate bonds betweeb RNA nucleotides mRNA detaches from template strand DNA strand rejoins by complementary base pairing pre-mRNA strand includes exons (sections that code for proteins) and introns (non coding) introns removed by enzymes and exons are spliced together mRNA leaves nuclear pore and goes to ribosome

Answer 37

mRNA strand enter the ribosme passing through a codon at a time- each codon coding for a specific amino acid as complementary to an anticodn on tRNA tRNA also move to ribosome- each tRNA has an anticodon and a specific amino acid attached specific tRNA antidcodon attaches to mRNA codon in ribosome, carrying a specific amino acid this process continues as mRNA moves through ribosome- AA join by peptide bonds continues until a stop codon is reached

Answer 38

insertions and deletions of a number of bases where the number is not divisable by 3 causes a 'frame shift'- all subsquent triplets from that point onwards are affected meaning different amino acids could be coded for substitution could lead to a different amino acid being coded for

Answer 39

conservative- replication results in 2 DNA molecules, one made of og strand and one new semi conservative- 2 DNA moelcules made up of one og strand and one new fragmentary- 2 DNA molecules made up of strands of fragments of og and new - mixture One sample of bacteria was grown in a N 14 nitrogen broth and the another in a heavy nitrogen broth N 15, each one taking in nitrogen during replication to make new nucleotides nitrogen gradually became part of bacteria's DNA 2 samples were taken from each tube & spun in a centrifuge heavy nitrogen DNA settled at the bottom & light nitrogen DNA settled at the top heavy nitrogen placed in light nitrogen broth and allowed to undergo one round of replication before removing a sample and spinning it in a centrifuge-meant all new nucleotides incorprated into replication of DNA were light whilst original was heavy DNA settled in the middle of the centrifuged tube , showing that it was between heavy and light one strand is heavy and the other is light, proving semi -conservative replication .

Answer 40

DNA helicase seperates H bonds in a DNA strand free DNA nucleotides bind to complementary bases DNA polymerase joins adjacent nucleotides to form a new strand of DNA results in 2 DNA molecules, each with 1 new synthesied strand of DNA- one template strand and one new strand- semi conservative replication

Answer 41

subsititution mutation in the gene that codes for 1 of the polypeptide chains in the haemglobin- A instead of T, mrna contains GUA instead of GAA- as a resultthe protein produced contains a non polar valine instead of polar glutamic acid- haemglobin less soluble and when o2 levels are low the molecule forms long fibres that stick together inside thr RBC distorting its shape - sickle shape

Answer 42

small sample of placenta tissue is removed either through wall of abdomen or through the vag carried out earlier in preg than amnios- 8-12 weeks 1-2% chance of causing miscarriage allows decision about abortion to occur earlier , reducing physical trauma Initial results with major issues available after a few days , complete set available after 2 weeks

Answer 43

characteristic controlled by 1 gene

Answer 44

carried out at 15-20 weeks of pregnancy involves inserting a needle into the abdomen- amniotic fluid to collect fetal cells that have fallen off the placenta and fetus- risk of infection from needle 1% risk of causing miscarriage results available after 2-3 weeks , but expedited test for most common disorders can be done a ready after 3- 4 days

Answer 45

DNA fragments in mothers blood plasma - 10-20% from the embryo cell free fetal DNA becomes detectable 4-5 weeks- collected 7-9 weeks into preg used in screening for a limited number of single gene disorders

Answer 46

PGD carried out on IVF embryos , where they are screened for genetic disorders before implantation in uterus reduces chance of having baby with a disorder as only embryos without disorder are implanted before implantation prevents ethical issues regarding abortion IVF expensive and stressful procedure- low success rate - 30% for live births

Answer 47

carrier testing can be offered to individuals with a family history of genetic disorders -to see if they carry an allele causing a disorder find a partner can be done to couples before getting pregnant to see if child is at risk of developing a disorder -↳ may help with decision making regarding children & pre-natal testing ethical issues finding out you are a carrier may cause emotional stress or affect ability to find a partner tests aren't always 100% accurate , may give false result other abnormalities may be found , causing further stress

Answer 48

increases risk of miscarriage false results may lead incorrect info abortion of fetus with disorder may be seen as unethical

Answer 49

Bronchodilator-inhaled, relaxes chest muscles and opens airways Antibiotics-kill or prevent growth of bacteria DNAase- wbc in the lungs release DNA which increases stickiness Steroids- reduce lung inflammation Medication to increase functioning of CFTR protein

Answer 50

Diet Physiotherapy Digestive enzyme supplements Heart and lung transplant

Answer 51

substitution only affects one codon many not effect sequence of amnio acids as substitution many code for same amino acid due to degenarte nature of genetic code

Answer 52

antibiotics can bind to ribosmome and alter its shape. mrna is prevented from binding meaning translation cannot occur so protein is not synthesised

Answer 53

different sequence of amino acids a different amino acid will have a differnt R group tertiary strcuture will change as it is dependent on interactions of R groups such as hydrogen bond and ionic bonds , - R group may form a different bond then intented differnce in structure may effect its functioning

Answer 54

primary structure detrmines folding into terirtary structure as R groups on amino acids interact this forms a globular structure where hydrophobic R groups are located in the centre of protein and hydrophllic are located on theoutside water forms hydrogen bonds with hydrophllic groups

Answer 55

frame shift- triplet code altered meaning every codon could be effected and code for a different amino acid this could effect position of start and stop codon could result in a different sequene of amino acids

Answer 56

caused by a falty allele that is only expressed in the homozygous condition

Answer 57

alternative form of a gene found at the same locus

Answer 58

more kinetic energy available, more frequent collisions between substrate and enzymes, more enzyme substrate complexes formed,

Answer 59

protein which reduces activation energy of biological reactions

Answer 60

change in triplet code, changes codons on mrna, results in a different amino acid in primary structure

Answer 61

when offspring is heterozygous for two alleles, but both are expressed in codominance

Answer 62

Insertion or section mutation causes frame shift effect where all triplets are effected which could cause gene to code for a different amino acid chain that does not fold properly as folding of the protein is dependent on R groups of amino acids. Could also cause no start codon or ribosome binding site

Answer 63

1. tRNA is folded (and mRNA is {straight /unfolded}) / eq ; 2. tRNA has hydrogen bonds (holding thestructure together) (but the mRNA doesnot / eq) ; 3. tRNA is a fixed {size/ length} (but mRNA {is not / lengthdepends on size ofgene}) / eq ; 4. tRNA has an anticodon (but mRNA has codons) ; 5. tRNA has an amino

Answer 64

a copy of a section of a DNA strand that codes for a sequence of amino acids

Answer 65

as the enzyme concentration increase the rate of reaction increases- enzyme lowers activation energy, higher concentration of enzyme means that more active sites are available ; more chance of a collision between enzyme / active site and substrate. substrate is a limiting factor so line pleatus when all substrates have been in am enzyme substarte complex

Answer 66

Prepare a series of trypsin dilutions, ranging from 0.1% trypsin concentration to 0.5% using serial dilutions Pipette 2 cm3 of milk solution into a cuvette. Calibrate the colorimeter to 1.05% Pipette 2 cm3 of the trypsin solution into the cuvette. Mix thoroughly and immediately put this cuvette into the colorimeter and start the stop clock. Measure absorbance at 30 second intervals for 5 minutes. Discard the content of the cuvette and rinse with distilled water. Repeat steps 2 to 5 of the experiment using a range of different enzyme concentrations, ensuring that other conditions are unchanged.

Answer 67

to ensure that substrate is not a limiting factor as when reaction proceeds substrate concentration decreases as substrate gets used up in enzyme-substrate complexes. Substrate conc should be constant in each test.

Answer 68

both made up of nucleotides joined together by hydorgen between bases and phospodiester bonds between sugar and phosphate group tRNA is a folded single strand whilst DNA is double stranded DNA contains deoxyribose whislt trna contians just ribose DNA has thymine whlst tRNA has uracil tRNA has a (specific) amino acid binding site

Answer 69

Volume of distilled water – 10cm³ of distilled water should be used to fill the boiling tubes each time Time left in water – leave each boiling tube containing beetroot for 30 minutes Size of beetroot piece – use a ruler and knife to cut cylindrical beetroot pieces of 1cm in length Colorimeter used – same colorimeter should be used on the same blue/green setting each time, measuring percentage absorbance. Calibration with distilled water should be carried out each time Volume of beetroot solution – 2cm³ of beetroot solution should be added to a cuvette each time

Answer 70

Dilute stock solution of trypsin with distilled water to produce solutions with concentrations of 0.2%, 0.4%, 0.6% and 0.8%. 1. Make a control by adding 2cm3 of trypsin solution and 2cm3 of distilled water. Use this to set the colorimeter absorbance to zero. 2. To another cuvette, add 2cm3 of milk suspension and 2cm3 of the stock trypsin solution. Mix, place in the colorimeter and measure absorbance at 15 second intervals for 5 minutes. 3. Rinse the cuvette with distilled water

Answer 71

Use a cork borer and knife to cut 8 x 1cm lengthed cylinders of beetroot over a white tile. Place all the cut pieces in a beaker of distilled water and leave overnight to remove any dye (betalains) released when the beetroot was cut. Wash and blot dry (with filter paper or a tissue) the 8 pieces of beetroot. Fill 8 boiling tubes each with 10cm³ of distilled water and place them into 8 separate water baths of different temperatures (e.g. 0°C, 10°C, 20°C, 30°C, 40°C, 50°C, 60°C, 70°C). Once at the desired temperature, add a piece of beetroot to each boiling tube and leave for 30 minutes. Remove the beetroot pieces gently with a pair of forceps and then shake the tubes to disperse the dye. Set a colorimeter to percentage absorbance on the blue/green filter. Calibrate by filling a cuvette with distilled water first then add 2cm³ of beetroot solution from the first temperature to a new cuvette. Place this cuvette into the colorimeter to read the percentage absorbance. Repeat this for all other pieces.

Answer 72

introns are removed • exons can then be arranged in several different ways /exons can be removed • this produces different mRNA (from the same gene) • translation (of the mRNA) gives different {amino acid sequences / different polypeptides / different proteins} (1)

Answer 73

• selecting one embryo over another / loss of a potential life (1) • risk of identifying other conditions / selecting for other characteristics (1) • {costly / emotional strain } as not always accurate (1)

Answer 74

• doubling of the DNA (1) • with each original DNA strand acting as a template (onto which a new strand is formed) (1) • (so) (each double-stranded) DNA formed contains one original strand of DNA and one new strand of DNA (1)

Answer 75

Adult screening advantages and disadvantages • Identifies risk of developing a particular disease in the future so choices can be made e.g. extra screening for breast cancer or preventative mastectomy/screening and lifestyle changes for some types of CVD • Identification of carriers so choices can be made about family planning – both partners tested, risk can be identified and have prenatal screening • May not want to know if you have a high likelihood of developing a disease, if one person is tested it may give other family members information they would chose not to know, may potentially affect life insurance Prenatal screening advantages and disadvantages • Amniocentesis – prepares parents for child with disease/gives choice of abortion • Chorionic villus sampling – as amniocentesis, carried out earlier in pregnancy • Some of the conditions tested for are very unpleasant and may be life limiting • NIPD non-invasive, less traumatic procedure, no increased risk of miscarriage • PGD only implant healthy embryos, do not have to make decision about abortion • Both amniocentesis and CVS carry increased risk of miscarriage, especially CVS (although it can be carried outearlier in the pregnancy) • Can’t cure the disease, only choice is to have an abortion-not acceptable to everyone • For conditions such as CF, where there is more than one possible mutation, test is only for most common mutations so there may be false negatives • NIPD currently only available for chromosome disorders such as Down’s syndrome • PGD involves IVF, which can be emotionally traumatic and only has about 30% success rate • All pre-natal screening has a risk of false positives with abortion of a healthy fetus. • Procedures involving IVF can be regarded as unethical because many embryos are discarded • Invasive nature of some of the tests

Answer 76

joins fragments of DNA together

Answer 77

molecule that slows down or stops an enzyme's activity by binding to a site other than the active site, causing a conformational change that affects the enzyme's shape and its ability to bind the substrate effectively

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