BIO

Question 1

Inorganic ions

Answer 1

ca2+ transmission of nerve impulses+release of insulin from pancre,cofactor
na+ generate nerve impulses+muscle contraction
k+ generate nerve impulses+muscle contr
H+ affects ph
NH4+ source of nitrogen+absorbed from soil
NO3- from soil source of nitrogen
HCO3- buffer maintain ph of blood
cl- chloride shift to maintain ph co factor for amylase
PO43- photos resp make nucleotides
oh- affects ph

Question 2

Translation

Answer 2

mRNA attaches to ribosome, trna brings amino acid, trna with anticodon complementary base pairs to mrna and next trna moves onto next codon, ribosomal rna catalyses peptide bond between the two amino acids on trna joins em first trna moves away leaving aa behin,stopcodonppchainmovesoff

Question 3

Drugs and metabolic poisons

Answer 3

eg antiviral drugs-reverse transcriptase inhibitors viral dna from replicating
penicillin inhibits transpeptidase which forms proteins in bacterial cell wall - weakens cell wall-bacteria cant regulate osmotic pressure-cellbursts

poisons-cyanide irreversible inhibitor of cytochrome c oxidase - stops resp
malonate inhibits succinate dehydrogenase, arsenic inhibits pyruvate dehydr

end product inhibition regulates pathway controls amount of end product cos end product inhibits an enzyme earlier on in pathway which made it

Question 4

Meiosis

Answer 4

Gametes for reproduction, full number of chromosomes to start then half the normal number - haploid cells
Crossing over- prophase 1 chromos of homologous pairs come together and cross over so 1 chromos from each homolog pair ends up in each cell - diff chromatid diff alleles - genetic variation in offspring
Independent Ass - each homolog pair made up 1 chromos from dad and 1 from mum-wen line up in metaphase 1 and separated in anaphase 1 random which chromos ends up in which daughter cell so 4 daughter cells made have diff combo of maternal paternal chromosomes-shufflin-gevariati

Question 5

Meiosis

Answer 5

Homolog pair - 1 chromo from mum 1 from dad - same size-same genes but can have diff alleles
Anaphase1-spindsctract separate homolog pairs-1chromos goes to each end
Meiosis II two daughter cells undergo pmat II, anaphase II pairs of sister chromatids separated each new daughter cell gets one chromatid from each chromos producing 4 genetically different haploid daughter cells

Question 6

Transp in Plants 1

Answer 6

multicellular org Exchanging substances by direct diffusion too slow to meet metabolic demands so needtranspsystemstomovesubances to and fromcells
Xylem transports water and mineral ions up from roots to leaves, phloem transp sugars up and down-xylem is green cross in middle alwayson inside

Root-xylem centre surrounded by phloem to provide support
Stem-xylem+phloem outside -scaffolding less bending
Leaf-x+p make up network of veins to support thin leaves
Xylem vessels long tube like cells joined end to end, no end walls-water passes thru easy, dead cells no cytoplasm,walls thickened w lignin prov supp

Sieve tube-living cells to transport solutes,joined end to end-tube,sieve parts have holes for solutes to pass thru
Companion cells-lack nucleus and other orgs in sieve tube so companion cells carry out functions for them and sieve cells like nrg for active transp

Question 7

Transp in Plants 2

Answer 7

Water enters thru root hair cells thru root cortex incl endodermis to reach xylem, drawn into roots via osmosis down water pot grad, soil has high wp leaves have low
Symplast pathway-living parts-cytoplasm-connected thru plasmodesmata-channels in cell wall
Apoplast-non-living-cell walls-water diffuses thru carrying solutes from high hydrostatic pressure to low, when water gets to endodermis casparian strip blocks it so takes symplast-goes thru cell membrane-partiallyperm

Xylem vessels transport water all around and at leaves water moves into cells by apoplast and evaporates from cell walls to spaces between cells-stomata open water diffuses out intoo air-transpiration
Cohesion+tension against force of grav, water evaporates from leaves,creates tension sucks water into leaf, water=cohesive so other water molecules follow too-column of water in xylem moves up, adhesion-water molecules attracted to walls of xylem vessels so rise up

Question 8

Transp in Plants 3

Answer 8

Transpiration rate:
Light-lighter the faster as stomato open so co2 diffuses in for photos
Temp-higher the faster-water molec more ke so evaporate from leaf fast-waterpotgrad increases so diffuses out fast
Humidity-lower the faster-if air dry water pot grad betw leafnair high
Wind-windier the faster-watermolc blown away around stom-waterpotgrad+

Xerophytes- marram grass sunken stomata in pits-sheltered fromwind-slows transp, hairs trap moist air around stom-reduces watpot grad red transp, hot/windy they roll their leaves to trap moist air slows down transp, cacti+marram hve thick wproof waxy layer to reduce water loss by evap, cacti have spines to reduce surface area for water loss instead of leaves and close stom at hottest times when transp rate highest

Hydrophytes-water lilies- air spaces inleaves helps plant float stores oxygen for resp, air spaces in roots and stem so o2 moves down to parts underwater, stomata only on upper surface of floating leaves-max gasexch, flexible leaves+stem-suppo by wateraroundthem-preventsdamagby currents

Question 9

Transp in Plants 4

Answer 9

Translocation moves dissolved substances iesugars - assimilates. happens in phloem requires nrg source to sink - source-substance made-sink where used up. Sugars transported as sucrose cos sucrose is soluble and metabolically inactive so not used up in transp. Enzymes maintain conc grad-change dissolved subs in sink-always low conc.

Mass flow-active transp actively loads solutes into sieve tubes at source, lowers wp-water enters by osmosis from xylem and companion cells-high pressure inside sieve tubes at source, at sink solutes removed, increase wp water leaves by osmosis-lowers pressure-presu grad pushes solutes from s-s

Substances enter compan cells by active loading and to sieve tubes against conc grad-conc of sucrose higher in companion cells than surrounding tissue cells, higher in sieve tube cells than companion - active transp used-in companion cell atp active transp h+ ions out of cell into surrounding tissue cells-conc grad-h+ ion binds to co transport protein in comp cell memb and reenters-sucrose molecu binds to co transp too- movement of h+ ion moves sucrose molec into cell against conc grad. Sucrose transp out same process.

Question 10

Factors affecting rate of enzyme catalysed reactions

Answer 10

As temp increases, molecules vibrate internal bonds ionic,h bonds, disulphide bridges break, changes tert struc, changes shape of active site, no more e-s complex formed- enzyme denatured
pH-optimum, increase or decrease from it lowers rate cos oh/h messes w hydrogen ionic bonds changes tert struc,shape of active site,no e-s, denature

Question 11

Factors affecting membrane permeability

Answer 11

Temp - freezing - carrier proteins+channel proteins lose shape,open cell memb damaged by ice crystals, membrane v permeable, cold-low ke phospholipids close together, memb stiff-least perm, warm-phospholips moving-more space betw more per, hot(40+)-carrier proteins and protein channels denature-opens-bilayer melts v perm

Solvent type- some solvents dissolve in phospholipds in cell memb ie ethanol- loses struc-more perm

Question 12

Defence against pathogens

Answer 12

Skin-physical barrier blocks pathogens entering, mucous membranes protect body openings exposed to env+secrete mucus to trap pathogens, blood clot-plug wounds to prevent pathogen entry and blood loss, inflammation-cells damaged release histamines, vasodilation for blood flow-redness, wbcs arrive to fight pathog, makes capillary walls permeable to wbcs, expulsive reflexes-cough/sneeze removes pathogens from nose,lungs, wound repair-reforms protective barrier of skin

Plants- physical-waxy cuticle-physical barrier to prevent pathogen enter and stop water collecting on leaves stops pathogens transferring in water, cell walls physical barrier, callose deposition between cell wall and plasma membrane in pathogen invasion to make it harder for pathogens to enter and at plasmodesmata limit spread of viruses, chemical-antimicrobial chemicals to kill pathogens/toxic to insects to prevent pathogen transferring thru vector

Question 13

Immune response

Answer 13

Phagocyte detects foreign antigen on pathogen and engulfs it, opsonins attach too to aid phagocytosis, phagosome formed,lysosome contains dygestive enz to break pathog down, phagocyte presents pathogens antigens on its surface to activate other cell signalling molec like t lymphocyte-its receptors bind to complementary antigens on apcs which activates tlympho-clonal selection, clonal expansion-tlymp divides to produce clones of itself

T regulatory cells-suppress immune response to stop immune system mistakenly attackng own body cells, t lymp activate b lymp which have antibodies that binds to comple antigens, t help release interleukins activates b lymp-clonal selec, b lymph divides by mitosis into clones-plasma cells+memory cells-clonal expan plasma cells secrete loads of antibodies form antig-antib comp

Agglutins- antibody has two binding sites so binds to two pathogens, glues pathogen together with diff antigens then engulfed all at once by phagocyte, antitoxins bind to toxin and neutralise them

Primary resp-slow-not many blymphocytes,need enough antibodies, symptoms, t and b lymph produce memory cells that record specific antigen and specific antibodies needed
Secondary resp- fast-clonal selection faster memory b lymph divide into plasma cells for right antibody, memory lymp divide into right tlymph to kill antigcell, no symptoms

Question 14

Immunity and Vaccines

Answer 14

Active-IS make own antibodies, natural-immune after disease, artificial-immune after vaccine,memcellsproduced
Passive-given antibodies made by diff org, natural-baby immune from antibodies from mother-placentabremilk, artificial-injected antibodies from someone else

Autoimmune disease-immune system attacks normal body tissue- arthritis- immune system attacks lining of joints-inflammation+pain, lupus- is attacks cells in connective tissue-inflammation+damages tissues

Vaccines contain antigens cause body to make memory cells wout pathogen-no symptoms immune, herd immunity- community vaccinated-disease bcoms rare-prevents epidemics, vaccination causes immunisation, routine vaccine-mmr against measles, meningitis c bacteria vaccine
Influenza vaccine changes every year, antigens change, new strains, memory cells dont recognise it so diff vaccine made

drugs made from natural compounds from plants animals micro org, others exist that treat incurable, protect by maintaining biodiv,
species used in traditional med already have medicinal propertieis+ tested for side effectno need to find new plants and develop new, antimicrobial compounds already active

Personalised medicine- if uno patients genome, adapt drug by genome sequencing-maybe less side effects-tailored to individal dna
Synthetic biology- designing creating new biological molecules

Antibiotics can inhibit/kill growth of bacteria, inhibitribosomes and cell wall formation so cell dies/bursts, penicillin fungus penicillium but bacteria develop abr thru mutations and reproduce, allele for abr passed on-cos increased use
ie bact: Clostrodum difficile- infects digestive system, mrsa- wound infections

Question 15

Studying biodiv

Answer 15

Species- group or similar orgs able to reproduce give fertile offspring,-biodiv-variety of living orgs in area
Samping- choose area, count individuals of species, plants use quadrat,flying insect sweep net, ground insec pitfall trap, aquatic use net, repeat, use same sampling technique and estimate total number of indivs/diff species, Non random-systematic-sample at fixed intervals, opportunistic- sample chosen by researcher-bias,stratified-diff areas identified and sample in proportion to their part of habitat as a whole.

Species richness-no diff species in area, species eveness- relative abundance of each species in area(noindivseach)
D=1-(E(n/n^2))- closer to 1, more diverse

Question 16

Genetic div

Answer 16

If pop low genetic div, not able to adapt to change in env, isolated po like those bred in captivity-pedigree, breeding prog in zoo managed to max div, polymorphism-locus that has two/more alleles, proportion=number poly loci/total loci

Human pop growth-destruction of habitat, separation of pop cant interbreed, unsustainable fishing/hunting, pollution, monoculture in agricul- cultivating only 1type crop more land needed to feed inreases habitat destruction, contious monoculture-soil deple-reduced plant species survi-decreases yield, increa fertiliser-expen, climate change- co2 burnt greenhouse gas, weather changes-changes abund+distrib of species-extinction-less gen biodiv

Ecological-species part of ecos provide habitat/niche for other org part of food chain, species r genetic resource+source of med, keystone species: plays vital role in maintaining ecos if v take em out ecos collapses-conserve
Economic-resources for non medical reasons egwood, ecotourism, reduce soil deplet as result of continous monocult, aesthetic-pleasure, need to protect landscapes

Question 17

Conservation

Answer 17

In situ-on site conservation of species in nat habitat eg wildlife, habitat+species saved, hunting hard to control tho, controls pop of invasive species, supplementary feeding resting boxes, legal protection
Ex situ- off site conservation outside nat habit eg zoo,botanic gardens-health monitored, protection from hunting, less comp in species, selected mating more biodiv but expensive in long term, some dont breed in captivity less genetic biodiv

Convention on International Trade in Endangered Species- stops trade of endangered species+wild plants, not allowed to kill and raise awareness thru education
Rio convention on Biological diversity- international strategies in using resources sustainably-everyones responsibility, gives guidance to gov, cooperate onbiodivissu
Countryside Stewardship Scheme- Financial incentives to look after env, preservation/rest of habitats for conservation

Question 18

Classification

Answer 18

Taxonomy- classifying species based on observable characteristics, taxonomic hierarchy DKPCOFGS
5 kingdoms-prokaryotae-bacteria- prokaryotic,unicellular, protoctista-algae-eukaryotic single cell, fungi- yeast- eukaryotic chitin cell wall saprotrophic, plantae- eukaryotic cell wall cellulose, autotrophicmake own food, animalia- euk, no cell wall, heterophobic-consume plants,animals

Phylogeny- evolutionary relationship, first branch point is common ancestor, phylogentics- species is smallest group sharing common ancestor, closely related species diverged away more recently

Prokaryotae reclassifed into two domains from family, evidence showed large diff in archae and bact, molecular- rna polymerase diff in each, cell memb-bonds of lipids diff

Observable features not enough, other evidence molecular-similarities in proteins+dna, seq of dna base,seq of amino acids in proteins-cytochrome C more sim seq more closely related comparitve anatomy- structure and function of body plan sim, behavioural- sim of beh

Question 19

Variation and Evolution

Answer 19

Continous- range varies, ie height polymorphic, discont- distinct categ, controlled by single gene
Intraspecific within species, Inter-between species
Genetic factors- diff speces diff genes, same species have same genes but diff versions-alleles make up genotype, diff leads to variation in phenotype, inherited, ENV- diff in env change phenotype, both
s= squareroot all E(x-x*)2/n-1

Adaptations- increase chance of surv and reprod cos of evolution by natural selc, best adapted surv
Behavioural- ways it acts, physiological-processes inside, anatomical- structure, diff taxonomic group organisms have similar features even tho not closely related-evolved in similar env and sim ecological niches- convergent evo ie marsupial kangaroos- short gestation period, dont develop placenta born early into mothers pouch, milk and develop, placental opp, born once fully develop, arent closely related but share anatomical features

Natural selection- variation in phenotypes, selection pressures ie disease struggle for survivval, indivs w better adapations more likely to survive+reprod and pass onto offspring, over time this advantageous allele freq incr leads to speciation, fossils in chrono order so graduachangobserv

Insects evolve resistance to pesticides- variation-genetic mutations create allele for resistance and survive and reprod and its passed on-pesticide resistance- crops ruined more variety of pesticide used time and expen-insects harder to control spread diseases
Drug resistance- protoctists causing malaria resistant to drugs, other pathogens evolved resistance to drugs, ABR

Question 20

Cellular control

Answer 20

Cellular control - Mutations are changes in the sequence of nucleotides in DNA molecules.

Insertion/deletion mutations where one or more nucleotide pairs are inserted or deleted from the sequence. This type of mutation alters the sequence of nucleotides after the insertion/deletion point known as a frameshift.
Point mutation/substitution occurs where one base pair is replaced by another.

Mutations can either have neutral effects where the mutation causes no change to the organism, for example in a case where the mutation occurs in a non-coding region of DNA or is a silent mutation, as described above. A mutation can also be neutral when a change in tertiary structure of the protein has no effect on the organism. Some mutations are beneficial. Harmful mutations include a mutation in the CFTR protein which causes cystic fibrosis. Whether a mutation proves to be beneficial or detrimental to an organism will depend on the environment of the organism.

Controlling gene expression - Gene expression can be controlled the transcriptional, post-transcriptional and post-translational levels.
Transcriptional control is the lac operon, which is a length of DNA composed of structural genes and control sites which controls the expression of betagalactosidase responsible for hydrolysis of lactose in E.coli. The operon consists of a promoter region which is the binding site for RNA polymerase to initiate transcription, operator region where the inhibitor binds and structural genes which give rise to 3 products, beta galactosidase, lactose permease and another enzyme. The inhibitor protein is coded for by a regulator gene, located outside the operon.. In a case where the concentration of glucose is high and the concentration of lactose is low, the transcription of the structural genes is inhibited due to binding of the repressor to the operator region. However, in a case where the concentration of glucose is low and concentration of lactose is high, lactose binds to the repressor thus causing the shape of its DNA binding site to change, therefore making it ineffective. This means that it can no longer bind to the operator region therefore RNA polymerase is able to bind to the promotor region and transcription of the structural genes takes place.

Gene expression can also be controlled by transcription factors which have the ability to switch genes on and off. They do so through interaction with the promoter sequence of DNA to either initiate or inhibit transcription.

Gene expression is controlled at post-transcriptional level by editing of the primary mRNA transcript, during which the non-coding regions called introns are removed, thus creating a mature transcript consisting only of protein-producing regions known as exons. Gene expression can be controlled at the post-translational level. For example, proteins such as adrenaline can be activated with the help of cyclic AMP. This occurs when adrenaline binds to a complementary receptor, which activates the enzyme adenylate cyclase which converts ATP to cyclic AMP which starts a cascade of enzyme reactions within the cell, thus activating the protein.

Homeobox genes are involved in controlling the development of body plan of organisms thus aiding the development of a zygote to a complete organism. They code for transcription factors that bind to DNA to regulate transcription by switching genes on and off when they are required at particular stages of development, for instance during limb formation in humans.

Apoptosis is a form of programmed cell death which can act as a mechanism to control the development of body plans. It is a means of controlling the number of cells and ensuring that it remains constant to prevent cancer. During the process, enzymes break down the cytoskeleton of the cell, DNA and proteins. As the contents of the cell are broken down, the cell begins to shrink and break up. Subsequently, the cell fragments are engulfed by phagocytes and destroyed. Mitosis involved in development of body plans too, one cell differentiate to 2daughter cells differentiation to create body parts, apoptosis removes unwanted, internal stiumulus-dna damage, external stim-stresss-lacknutr,patho

Question 21

Patterns of Inheritance

Answer 21

Genotype- allele organism has, dominant- single allele copy present for chracteristic Bb or BB, recessive two copies bb
Co dominant- eg I^A I^B blood group AB I^O recces,1:2:1
Monogenic- RR x rr=Rr - 3dom:1rec, Dihybrid RRYY x rryy= RrYy 9domboth:3dom1strecc2nd:3recc1stdom2nd:1reccbth

Dihybrid epistatis gene-allele of one gene masks expression of allele of others - YYRR x yyrr= YyRr
Dominant epistasis gives a ratio of 12:3:1.

Sex linkage – expression of an allele dependent on the gender of the individual as the gene
is located on a sex chromosome, for instance, males are more likely to inherit an Xchromosome linked condition because they only have a single copy of the X chromosome. An example of sex linkage is haemophilia which is a recessive condition (hh).
Autosomal linkage – genes which are located on the same chromosome (which is not a sex
chromosome) and tend to be expressed together in the offspring

Question 22

Evolution 2

Answer 22

Hardy Weinberg:
p+q=1 p-dom, q=recc
p^2(AA)+2pq(Aa)+q^2(aa)=1
non carriers=p^2 carriers=2pq cystic fib=q^2toq

Genetic Drift - change in allele freq due to random chance cos not all indivs mayb reprod- small pop more affected
Genetic bottleneck- few individuals survive, decrease in gendiv lower alleles in surviving pop-small pop more affec, same geographical loc
Founder effect- migration-new colony formed by small pop by founders w few alleles lower gendiv, diff geog loc

Speciation - form new species when pop reproductive isolated-no gene flow, allopatric-geographically isolated-reprod isolated-no gene flow, sympatric-not geog isolated but reprod isolated
Variation exists due to mutations diff selection pressure, directional selection of diff phenotypes differnetial reprod succ change in allele freq for many gen

Artifical selection- humans interbreed organisms with useful traits to improve usefulness ie dairy cow- select female cow w desired trait,select male w mother w desired trait, cross breed, select offspring w desired, repeat over generations.
adv-high efficiency, bigger phenotypes, high freq of destraits
disadv-low gendiv high genetic disorders cos inbreeding causes reccessive alleles to be expressed, ethical issues

Question 23

Manipulating genomes

Answer 23

Dna profiling-identify indivs w their repeitive non coding base sequences - non coding dna dont code for protein dont affect phenotype, no of bases each repeat varies, forensic use- sample of dna from crime, amplify w pcr, label w marker genes, gel electrophoresis, medical-test for combo of alleles and diagnose genetic disease computational bio-using computers to make models of biological system+bioinformatics to study dises+mutations-epidemiol

Sequencing projects- human genome project-entire genome, only sequence short fragments at once, split genome into small sections to sequence then putbacktog
Chain termination-high throughput pyrosequencing cheap fast, prok-gene-protein few regulatory genes, little non coding dna, euk-lots noncod, reg genes turn other genes on or off

Isolating target genes- restriction enzymes-dna contains pallindromic sites can b read for and back, restriction enzymes cut target gene out at specific restriction sites, restriction enzymes leaves dna w sticky ends-unpairedbases
Reverse transcriptase-transcription backwordsmrnatocdna

Insert target gene into vector- vector moves dna from one place to another, use same restriction enzyme to cut plasmid open, sticky ends complementary, dna ligase reforms phosphodiester bonds, forms recombinant dna-more than one source, insert vector into bacteria- transformed organism contains recombinant dna, electroporation to increase permeability of bacteria cell wall so takes up recombinant plasmid

Question 24

Manipulating genomes 2

Answer 24

Marker genes-paired w target genes to check if vector inserted prop, transformed bacteria flouresce under UV, easily identified, select+culture transformed bact

PCR-amplifies dna, heat to 95 degrees to break h bonds and make dna single stranded, cool to 50degrees to allow primers to bind by complementary base pairing making dna double stranded so dna polymerase can bind, heat to 70degrees to allow dna polymerase to add complementary nucleotides form phosphodieter bonds, repeat todoubledna

Gene therapy- changing faulty alleles that cause disease - huntingdons dominant, suffer is heterozygous, already has functional allele so need to silence dominant alle use vector to add dna fragment to dominant allele so reccessive allele expressed instead, cystic fib rec, suffer homozygous, dom allele expressed, but allele may insert into wrong locus, silence wrong genes, over express
Germ line gene therapy changes alleles of games so offspring inherit, somatic gene therapy changes alleles of body cells offspring doesnt inherit

GM organisms:
Agriculture- express protein from bacteria protein toxic to pests- use less chemical pesticide-more efficient food chain, but monoculture of crop-low gdiv disease, have to buy seeds yearly, lower biodiv, gene from corn-rice to express vitamin a prevents blindness caused by vita deficiency
Industry-make enzymes low energy and cost, fast, transformed pathogens to treat disease-less suffering but could mutate and infect humans+b used in war
Medicines pharming- transform bact to express proteins, mammals to produce useful products in milk- make human proteins cheap+easy than to make proteins synthetically but maybe unexcpected problems ie cancer, and animals as commodities

Question 25

Cloning and biotech 1

Answer 25

Natural plant cloning: Vegetative propagation - asexual reproduction by plants, new plants grow from parts of parent plant and genetically identical.

Bulbs - divide,form new plants eg onions
Tubers- sprout eyes, form new plant eg potato
Rhizomes- horizontal stem below ground sprout to form new plants eg ginger
Runners- horizontal stems above ground sprout to form new plants egstrawb
Suckers- if plant damaged, new shoot forms from base of stem eg elm

Artificial plant cloning - offspring genetically identical, few clones at once
Cuttings- remove shoottip, root tip/leaf, clean cut thru xylem+phloem, apply rooting powder containing auxin, transfer to soil, control conditions/env,roots+shoots develop
Tissue culture - many clones at once from 1 parent plant. take cells from parent plant, sterilise to kill microorg eg bleach, use aseptic technique, put on growth medium containing glucose, add growth hormones eg high auxins forms mass of unspecialised cells called callus, subdivide to make many cells, change plant horm ratio so roots+shoots start to form, transf to soil.

Animal cloning -
Enucleation and somatic cell nuclear transfer- enucleate egg cell, remove nucleus from somatic body cell, inject into enucleated egg cell, electric shock-electrofused, embryo grown invitro, until it splits then insert into womb of surrogate mom-embryo genetically identical to somatic nucleus donor
Artificial embryo twinning - egg fertilised in vitro, zygote divides to form an embryo, unspecialised early embryo cells separated, each cell develops to new embryo grown in vitro until embryo splits then inserted into womb of surrogate mother, embryo genetically identical to each other.
Natural-fertilised egg-embryo-splits in two-identical twins

Question 26

Cloning and Biotech 2

Answer 26

Advantages of artificial plant cloning include the fact that a large number of plants can be produced easily and independently of the season or weather. Disadvantages include the lack of variation as the plants are genetically identical, meaning that they wouldn’t respond well to changes in conditions or attack of pathogen. Moreover, it is harder to grow plants this way than it is to sow seeds.

Advantages of artificial animal cloning include the fact that animals such as cows which are of benefit to humans can be cloned quickly. Moreover, artificial cloning can be used to preserve an endangered species. Disadvantages include the fact the lack of genetic variation, uncertainty whether cloned animal will be of good health in the long term and concerns about the welfare of animals.

The use of microorganisms
Microorganisms are used in biotechnological processes various reasons:
easy to grow as they grow rapidly, grow well at low temperatures and are
not climate dependent. Apart from this, they can be grown on unwanted waste of no use to humans, genetically engineered to produce desired products, which often are purer than those produced in chemical processes
Microorganisms are used in processes such as brewing, baking, cheese making, yoghurt production, penicillin production, insulin production
However, less publically accepted cos food grown on waste products+contaminated easy cos ideal cond for all microorganisms.

The growth curve of a microorganism in a closed culture has various distinct features:
The first phase of microorganism growth is the lag phase where microorganisms are adjusting to the environment before starting to reproduce, thus meaning during the lag phase the population remains constant.
The next part of the growth curve is the log phase where the population size grows exponentially meaning that every round of division doubles the population size, so long as the dividing organism has a sufficient amount of nutrients.
The stationary phase is where the population size reaches its maximum due to
decreasing nutrient levels and build of up toxic substances. Cellprod=cell death
The stationary phase is followed by a decline phase where lack of nutrients and increase in toxic products causes death of organisms.

Culturing microorganisms
Batch culture, the fermentation is carried out in a closed fermenter. The microorganisms and nutrients are added and then left to grow for a particular period of time. No further nutrients are added, and products are removed at the end of the period.
Continuous culture takes places in an open fermenter, where nutrients are continuously added and products are removed at a steady rate. Even though the batch culture is easier to set up and maintain than the continuous culture, the growth rate isn’t as fast.

However, in the case of contamination of batch culture, only a single batch is lost whereas in the case of continuous culture, it can lead to huge amount of product lost. To maximise the yield of product, the temperature needs to be maintained at the optimum with a sufficient nutrient supply and the aerobic conditions to prevent the formation of undesired products through anaerobic respiration. The pH needs to be kept constant to ensure that the enzyme activity is not altered.
It’s important for the microorganisms to be manipulated under aseptic conditions, where unwanted organisms are absent. In the case where unwanted organisms are present, the medium is said to be contaminated. This is undesired as contaminants compete with the culture for nutrients and space, thus reducing the product yield. Some contaminants might produce toxic chemicals, thus destroying the culture microorganisms and the products.

Methods of enzyme immobilisation include: enzymes attached to inert materials making them stable and easy to remove from product
Adsorption where enzyme bind to a support through hydrophobic and ionic
interactions
Covalent bonding where enzymes covalently bind to a support with the help of a cross linking agent
Entrapment – enzymes are trapped in a semi-permeable material such as gel beads which allows the passage of substrate and product only
Membrane separation – partially permeable membrane serves to separate the enzymes from the substrate

Examples of immobilised enzymes in biotechnology include: glucose isomerase for conversion of glucose to fructose, penicillin acyclase for the formation of semi-synthetic, penicillins, lactase for the hydrolysis of lactose to glucose and galactose, aminoacyclase for production of pure samples of L-amino acids, glucoamylase for the conversion of dextrins to glucose, nitrilase for the conversion of acrylonitrile to acrylamide for use in the plastics
industry.

Using immobilised enzymes can be advantageous due to the product not being contaminated with enzyme therefore removing the need of filtering/purification costs. The enzymes are also less susceptible to the effect of temperature and ph-more stable. Can be reused saving money.

Question 27

Ecos

Answer 27

efficiency=biomass transfer/intake x 100
productivity/received x 100

An ecosystem includes all the organisms living in a particular area known as the community as well as all the non-living elements of that particular environment.
The distribution and abundance of organisms in a habitat is controlled by both biotic (living) factors e.g. predators and pathogens and abiotic (non-living factors) such as light levels and temperature. Each species has a particular role in its habitat called its niche which consists of its biotic and abiotic interactions with the environment.

Succession is the change of one community of organisms into the other.

Primary succession occurs when area previously devoid of life is colonised by communities of organisms for instance after the eruption of a volcano which lead to formation of a rock surface. The area is first colonised by the pioneer species such as lichens which are adapted to survive in such harsh conditions. As organisms die, they are decomposed by microorganisms thus adding
humus, this in turn leads to formation of soil which makes the environment more suitable for more complex organisms. Over time, the soil becomes richer in minerals thus enabling larger plants such as shrubs to survive. Eventually, a climax community is established which is the final seral stage of succession, a self-sustaining, largest complex stable community of organisms.

Secondary succession occurs in a previously colonised area in which an existing community has been cleared. This type of succession can occur after events such as forest fires. As a soil layer is already present, succession begins at a later stage.

As succession goes on ecos more complex, new specieis move in with existing, species diversity increases, biomass increases cuz plants more large,dense.

Question 28

ecos2

Answer 28

Biomass
In any ecosystem, plants synthesise organic compounds from either atmospheric or aquatic carbon dioxide. Most of the sugars synthesized by plants are used by the plant as respiratory substrates whereas the remaining sugars are used for synthesis of biological molecules which form the biomass of plants. The biomass can be measured in terms of mass of carbon or dry mass of tissue per given area per given time. The chemical energy stored in dry biomass can be estimated using calorimetry.
Net primary productivity (NPP) – the rate at which energy is transferred into the
organic molecules that make up new plant biomass, that is the chemical energy store in plant biomass after respiratory losses to the environment have been taken into account.
Gross primary productivity (GPP) – the rate at which energy is incorporated into
organic molecule in the plants in photosynthesis, that is the chemical energy store in plant biomass, in a given area or volume, in a given time
Therefore: NPP = GPP – Respiratory loss
The net primary production is available for plant growth and reproduction as well as to other trophic levels in the ecosystem such as decomposers and herbivores
The net production of consumers (N) such as animals can be calculated by:
N= I –(F+R) where I represents the chemical energy store in ingested food, F represents the chemical energy lost to the environment in faeces and urine and R represents the respiratory losses to the environment.

Sampling
Abundance of organisms can be measured with the use of:
Line transect- where a line is placed down across the habitat and species in contact with the line are recorded
Quadrat – a square frame of a given size, randomly placed in the area being sampled, species inside the quadrat are identified and counted to determine the abundance
Belt transect – 2 tape measures are laid out and samples are taken between the two at set intervals along the tapes

Question 29

ecos 3

Answer 29

Nitrogen cycle- Living organisms need nitrogen for protein and nucleic acid synthesis, and it is cycled through an ecosystem between the biotic and abiotic components, and bacteria are involved in the processes of ammonification, nitrogen fixation, nitrification, and denitrification. Fixed nitrogen in the form of ammonia (NH3) are fixed by the bacteria Azotobacter living freely in the soil and Rhizobacteria, found in the root nodules of leguminous plants; leghaemoglobin absorbs oxygen and provides the anaerobic conditions under which nitrogen reductase is synthesised. Ammonia in the soil is converted to nitrites (NO2-) and then nitrates (NO3-) through nitrification by oxidation in well-aerated soils by nitrifying bacteria, such as Nitrosomonas, a chemoautotroph which obtains energy from the oxidation of ammonia into nitrites, and Nitrobacter which oxidises nitrites to nitrates. As these nitrogen-containing molecules are highly soluble, they are absorbed by plant roots are transferred to organic nitrogen in animals through feeding, and these are cycled back to the soil through ammonification when excretory matter and dead organisms are decomposed by saprotrophs, converting nitrogen-containing molecules into ammonium compounds. Denitrification occurs in anaerobic soils, where denitrifying bacteria use nitrates as a source of oxygen for respiration and produce nitrogen gas (N2) and nitrous oxide (N2O), releasing it into the air.

Carbon cycle- Carbon is transferred from abiotic and biotic components of the cycle, driven by the processes of respiration and photosynthesis with carbon dioxide being the main vehicle for cycling of carbon. Autotrophs fix carbon in biotic organisms through photosynthesis, and it is transferred through ecosystems by heterotrophs, and animals, plants and microorganisms all respire to release carbon dioxide back into the atmosphere. Microorganisms cycle carbon from both hetero and autotrophic in the decomposition of dead organisms and waste. Some of this decayed matter is trapped in rocks and over millions of years forms carbon-based compounds that we use as fossil fuels. The increased use of fossil fuels has caused an imbalance in the carbon cycle, increasing atmospheric levels of carbon dioxide.
Aquatic plants use dissolved carbonates in water for photosynthesis. When carbon dioxide dissolves in water it reacts to form carbonic acid. Carbon can also enter rivers and lakes from the weathering of limestone and chalk in the form of hydrogen carbonate.

Question 30

Pop and Sust

Answer 30

The limiting factors which determine the carrying capacity, that is the maximum population size that can be supported by the environment include: food, water, light, oxygen, nesting sites, shelter, parasites and predators. Competition between organisms takes place in a case where a particular resource is limited supply.
Predator-prey relationships give rise to a fluctuating population size of predators and prey. This is because, as the predator population grows, more prey is consumed meaning that prey population decreases to a point where there isn’t a sufficient amount of prey to feed the predator population. As a result of that, there are fewer predators thus increasing the chance of survival of prey, thus causing the prey population to increase in size. This in turn means that the predator population can grow again, thus restarting the cycle.

There are two types of competition between organisms; interspecific competition which is the competition between individuals of different species and intraspecific competition between individuals of the same species.
Conservation serves to maintain or increase the biodiversity within a particular habitat by allowing a sustainable use of the resources whereas preservation serves to maintain the biodiversity levels and the habitat intact by minimising the effects of human activities on the particular habitat.

There are many economic, social and ethical reasons for conservation. For instance, many species provide a source of food and medicine, and are important for processes such as pollination of crops as well as for maintaining a good quality of water as well as tourism. Social reasons include conservation for aesthetic reasons and recreation whereas ethical reasons include the right to survive+ we have moral resp to conserve for future generati

Question 31

Pop and Sust 2

Answer 31

The Masai Mara region (Kenya)-The Masai Mara reserve has consolidated land to create conservancies to conserve biodiversity and generate tourist income. Conservancies are paid payment for wildlife conservation (PWC) by tourism operators according to the amount of land set aside for conservation. This means that there is constraints on how they use the land, such as local landowners needing to move livestock off the land during tourist season, and being unable to live on land in the conservancies.

The Terai region (Nepal)- Marshy grassland, savannah and forest is home to endangered species such as the Bengal tiger and great one-horned rhinoceros. The local people are heavily reliant upon the forest, which is under pressure from increased agriculture and grazing. The WWF and Nepalese government used the Terai-arc landscape initiative to introduce community forestry initiatives that give local people rights to exploit the forest in return for adopting responsibility to look after it. These community forest user groups help create forest corridors between national parks for dispersal and survival of tigers, as well as counteracting poachers and illegal fellers in exchange for diversified on and off-farm activity, built entrepreneurial skills, and stimulated small credit and marketing schemes. The scheme introduced biogas plants and wood-efficient stoves to reduce demand for firewood. Also done is constructing waterholes, monitoring endangered species, and eradicating invasive species. The data from southern Nepal suggests that tiger numbers are increasing, and the groups have earned hundreds of thousands of dollars from tourism activities in the buffer zone which protect the edges of the national parks in the Terai region.

Peat bogs: home to unique species, fragile ecosystems slow to recover from damage, grazing animals damage ecosystem, conservation programmes and charities reward farmers for using peatbogs sustainably.

Question 32

pop sust 3

Answer 32

Environmentally-sensitive ecosystems
The Galápagos Islands- Since Charles Darwin’s discoveries made the islands famous they have been severely affected by human activities, for example:
fishing and whaling, which have upset the marine ecosystem
the introduction of new species, such as:
1. Goats, which compete with native species for the vegetation
2. Dogs and cats, which chase and eat the native species
3. Rats and mice, which damage the eggs of native species
4. Plants such as elephant grass, which compete with native vegetation
tourism
disturbance due to scientific research and collection of samples
increasing population, which requires more land for housing and agriculture, produces sewage and waste, and uses more water and energy supplies
These effects must be managed to enable survival of the indigenous species and maintain biodiversity. The strategies include: searching boats and tourists for foreign species, using natural predators to reduce pest populations,
education and fostering a culture of conservation, captive-breeding programmes for species such as tortoises, management of the Galápagos Marine Reserve with cooperation from local stakeholders

Antarctica-The antarctic ecosystem is under pressure from over fishing — particularly of krill which is a keystone species. Exploitation of krill is controlled by catch size — once the catch drops to a certain level the boats must move on. Whales are protected by marine reserves where fishing is restricted. Albatrosses and petrels are protected by fishing at night and only during non-breeding seasons so they are less likely to be trapped in nets. Whales are protected by Sanctuaries such as the Southern Ocean Whale sanctuary (established in 1994) within which it is illegal to hunt and kill whales, and sanctuaries are monitored to ensure rules are obeyed.

Snowdonia- national park with a wide diversity of rare plants and animals. It attracts walkers and climbers. Trampling damage is reduced by providing well-maintained footpaths. Draining moorland and planting conifers have reduced the value of the land as a diverse plant habitat and nesting area for rare birds. It also reduces water quality in local rivers and increases the risk of flooding. These problems have been improved by using temporary dams such as hay bales and cut branches in the drainage ditches. Farmers are also encouraged to reduce sheep grazing on the mountain itself to avoid creating bare ground open to erosion. Farmers are being encouraged to plant hedges and take action to conserve ancient woodland.

The Lake District- The landscape is largely a result of man’s activity causing deflected succession and continued management is essential.
Farmers are offered financial incentives to reduce the use of chemicals, plant hedges and care for the diverse habitats including hay meadows, heather moorland, wetland, limestone pavement and woodland.
Management techniques include:planting native broadleaf species that support a wider diversity of insects and birds, rotational timber harvesting to create a mosaic of different-aged trees, removal of invasive species, such as rhododendron and laurel which reduce growth of other plants legal protection of certain habitats such as limestone pavement, controlled burning of heathland to encourage new growth reintroduction of waterlogged areas by artificially raising the water table

Question 33

pop and sust 4

Answer 33

managing ecosystems
The management of an ecosystem can provide resources in a sustainable way, maintaining Biodiversity without losing economic benefits. For example:
Coppicing – cutting down of tree stems close to the ground to encourage growth of many narrow stems which can be harvested, effective as stumps have good root systems.
Selective felling – the harvesting/removal of largest, mature trees, as well as diseased ones to enable other trees to grow, trees which are removed are replaced with seeds
Pollarding is a form of coppicing where the trees are cut higher up, to prevent deer from eating the new shoots
Efficient production where most of the tree is used to minimise wastage
Rotational coppicing is good for Biodiversity as it prevents the trees from growing so big that the light is blocked from reaching the forest floor, meaning that a larger variety of species are able to live there.

Fishing quotas-limit fish caught prevents species extinction and protects fish while theyre spawning, protect breeding pop, regulate mesh size of nets to reduce bycatch so only fish of certain size caught, protect important areas by stopping fishing their

Question 34

Specialised cells

Answer 34

Alzhiemers-nerve cells die-memory loss-grow them using stem cells, parkisons-uncontrollable tremors-nerve cells hve dopamine producing cells controlling movement, stem cells regenerate movement in dopamine prod cells

Alveoli air sacs in bronchioles lots+lots of folded membs+capillaries increase sa+ air enters trachea splits into bronchitwobronchus branches off into bronchioles, conc grad ventilation brings air w high o2 replaces air w low, circulation repl blood low in o2 w blood high in o2-maintains conc grad, alveolis+capillaries thin epithelial cells-short diff path-fast diff

Ciliated epithelial cells - cilia waft mucus along cell hair like, squamous epithelial-thin layer of cells short diff pathway fast rate of diff, neutrophil wbcs-type of phagocyte-connectedones engulf pathog and digest/hydrolyse with lysosomal enzymes, flexible shape to engulf pathog, sperm cells-flagellum allows sperm to swim to egg, acrosome has lots of mitoch-atp for swim, hydrolytic enzymes to hydrolyse jelly layer surrounding egg

Palisade mesophyll-lots of chloroplasts for photos, thin walls sh diff dist, root hair cells-absorb water+minerals, large sa for max absorp, lots of mitoch-atp-active transp-thin cell wall, guard cell thick inner wall, thin outer force guard cells to bend outwards to open stom for gas exchange for photos, at day take up water turgid, night when plant cell flaccid, stom close prevents water loss

Tissues- xylem provide structural support for stem-transp wat+min up stem, dead xylem vessels+living xylem parenchyma cells, phloem- transp sugards up and down step-sieve tube cells+companion cells, cartilage- strong connective tissue in ears nose and wind pipe-formed when chondroblasts secrete matrix which they bcom trapped inside, animal: ciliated epithelium+squamous-thinsingle layer of cilia-lining inside tissues, muscle: cardiac,smooth-lining inside stomach wall, skeletal-for movement

Question 35

Photos

Answer 35

stroma has enzymes grana has pigments
LDR- uses light, thylakoid membrane, light energy absorbed by photosynthpigmens(primary r reactioncentres(rc) electron excited, accessory make up light harv syst+transf light nrg to rc) protein+pigment=photosystem) converted to chemical, adds phosphorylates adp to atp nadp to reduced nadp-nadph, splits water into protons, electrons+oxygen -photolysis, atp transf nrg, nadph transfers hydrogen, h20 oxidised to o2, lir-calvin cycle, stroma, atp+nadph from ldr for nrg and hydrogen to make glucose from co2.

ETC-chain of proteins which excited e-s flow thru, light nrg absorbed by ps2 excites es in chlorphyll, es move to higher nrg lev and move along to ps1, photolysis of water makes protons,e-s, o2, es lose nrg as they move along-used to transp proteins into thylakoid via proton pump so thylakoid has higher conc than stroma using atp synthase combining adp and pi-atp, light nrg absorbed by ps1 electrons excited higher nrg level, transf to nadp w h+fromstrom-nadph
Cyclic photophosphorylat only makes atp, only uses psi, electrons recycled passed back to ps1.

Question 36

Photos 2

Answer 36

calv cycle:
RuBP is combined with 1c co2 in a reaction called carbon fixation
catalysed by the enzyme RUBISCO.
RuBP is converted into two 3c glycerate 3-phosphate (GP) molecules
2 atp to adp+pi, 2nadph to 2 nadp, atp and h+ from nadph nrg convert GP to 3c 2 triose phosphate (TP)
Some of TP molecules converted to useful organic substances like glucose(carbs, lipids, amino acids)
Remaining TP molecules are used to reform 5cRuBP with the help of ATP.

Three turns= six tp, 5 used to regenerate rubp, 3 turns only one tp to make hexose sugar, 6 times to make two tp to make one hexose sug, six turns needs 18atp and 12 nadph from ldr

Question 37

Photos 3

Answer 37

Light intensity- a to b only limited by light intensity,rate photos increase, at saturation point increasing makes no diff levels off-another limiting factor, higher it is=more nrg only certain wavelengths used, red and blue absorbed by pigments, green light reflected, nadp atp prod in ldr slow in low light-short supply-conversition of gp to tp, rubp slow, gp rise, tp rubp fall

Temp- 25deg, lower-enzymes involved inactive, higher-enzymes catalysing calvin cycle denature reaction slower cos enzyme slower so rubp gp tp lev fall+ stomato close-less co2 in for photos, thylakoid memb damaged less sites for electron transfer-rate of ldr lowers, membranes around chloroplasts damaged-reduces enzymes released into cell for calv cyc reduces rate of lid, chlorophyll damaged-less pigment to absorb photosboth level off when light intensity no longer limiting, 25degrees levells of at higher point so temp mustve been limiting factor at 15deg,

CO2 conc- higher rate of photos but higher than 0.4% stomata close, level off when light intensity no longer limiting, higher conc levels of higher than lower so co2 conc limiting for lower cant b temp cos both temp same. at low conc rubp to gp slow cos less co2 to combine w, rubp rise but gp tp fall

Question 38

resp1

Answer 38

Glycolysis- occurs in the cytoplasm which can take place in aerobic or anaerobic conditions. Glucose is broken down to two molecules of pyruvate.
The link reaction- occurs in the matrix of the mitochondria. Pyruvate is dehydrogenated and decarboxylated and converted to acetate.
Krebs cycle- occurs in the matrix of the mitochondria. Acetate is dehydrogenated and decarboxylated.
Oxidative phosphorylation- occurs on the folded inner membrane (cristae) of mitochondria. This is where ADP is phosphorylated to ATP.

One molecule of ATP is hydrolysed and the phosphate group produced is used to convert glucose into glucose-6-phosphate. The energy from the hydrolysed ATP molecule activates the hexose sugar and prevents it from being transported out of the cell.
Glucose 6-phosphate is rearranged, using the enzyme isomerase, into fructose 6-phosphate.
Phosphorylation via hydrolysis of a molecule of ATP occurs again forming hexose 1,6-bisphosphate.
The hexose 1,6-bisphosphate splits into two molecules of triose phosphate.
Each triose phosphate is dehydrogenated, removing hydrogen atoms using dehydrogenase enzymes.
The coenzyme NAD accepts the hydrogen atoms and becomes reduced NAD.
Two molecules of ATP are formed, via substrate-level phosphorylation (the formation of ATP directly during glycolysis and the Krebs cycle only).
The triose phosphate molecules are converted to pyruvate, which is actively transported to the mitochondrial matrix. In the process, another two molecules of ADP are phosphorylated to make two molecules of ATP.

During aerobic respiration in animals, the triose phosphate molecules are converted into pyruvate. Pyruvate that is produced during glycolysis is transported across the inner and outer membrane to the mitochondrial matrix where the link reaction takes place.

The Link reaction
The pyruvate molecule is decarboxylated by the enzyme pyruvate decarboxylase, removing a carboxyl group which eventually becomes carbon dioxide.
The pyruvate molecule is also dehydrogenated by the enzyme pyruvate dehydrogenase, removing hydrogen atoms, forming acetate.
The hydrogen atoms are accepted by the coenzyme NAD, becoming reduced NAD.
The acetate combines with coenzyme A forming acetyl CoA.
2 pyruvate + 2NAD+ + 2CoA → 2CO2 + 2NADH + 2 acetyl CoA
Coenzyme A (CoA) accepts acetate to become acetyl coenzyme A. The function of CoA is to carry acetate to the Krebs cycle.

Question 39

resp2

Answer 39

Krebs cycle- once for every pyruvate molecule, goes round twice for every glucose molec, 2c acetyl coa from link reac combines w 4c oxaloacetate-6c citrate, co enzymea goes back to link reac, 6c citrate conveted to 5c molec, decarboxylation(co2)+dehydrogenation-hydrogen used to make nadh from nad, 5c molec convert to 4c, decarbox+dehydrog makes one molec fadh and two nad from nadh, fadh, atp from adp+pi thru substrate level phosphoryl - phosphate one molec to another, 6c citrate now 4c oxaloacetate

1coenzyme a reused in link, oxaloacetate regen for next krebs, 2co2 released as waste prod, 1atp used as nrg, 3nadh+1fadh to oxid phosph

Oxid phosph - h atoms released from nadh and fadh as theyre oxidsed to nad and fad, h atoms split to protons, electrons-move along etc in inner mitochondrial memb folded into cristae-sa, nrg used to pump protons from matrix to intermemb space, conc of protons higher in intermemb space-electrochemical grad- protons move down back into matrix via atp synthase causing atp to form from adp and pi-chemiosmosis, in matrix protons,es,o2 form water so oxygen is final e acceptor.
Glycolysis - 2atp=2atp
Glycolysis - 2nadh=2x2.5=5atp
Link react x 2 - 2nadh=2x2.5=5atp
Krebs x 2 - 2atp=2
Krebs x 2 - 6nadh=6x2.5=15
Krebs x 2 - 2fadh=2x1.5=3 so 32 atp total made from 1 gluc molec

Question 40

resp3

Answer 40

Anaerobic:
Lactate fermentation in mammals=lactate
Reduced NAD is oxidised to NAD and the NAD formed goes back into glycolysis.
Pyruvate accepts the hydrogen atoms and is reduced to lactate with the help of the enzyme lactate dehydrogenase.
Occurs in mammalian muscle tissue during vigorous activity.
The lactate is carried in the blood away from muscles, to the liver.

Alcoholic ferment in yeast=ethanol
Pyruvate is decarboxylated to ethanal with the help of the enzyme pyruvate decarboxylase, releasing CO2 (pyruvate has the coenzyme thiamine diphosphate),
Ethanal accepts hydrogen atoms from reduced NAD, becoming reduced itself forming ethanol, with the help of the enzyme ethanol dehydrogenase.
The NAD formed goes back into glycolysis.

ATP yield from anaerobic lower than from aerob cuz anaerobic only has 1nrg release stage=glycolysis=2atp per glucose molec, nrg releasing krebs and link need oxygen so cant happen in anaerobic

Respiratory substrate respired to work out rq
carb nrg=15.8 rq=1
lipids= 39.4 rq=0.7
protein= 17.0kjg-1 rq=9
rq= co2 released/o2 consumed
higehr than 1 rq short of o2 respiring anaerobically, plants may have low cos co2 used for photos

Question 41

rtd

Answer 41

sens - receptor via short dendrites to long dendron to cell body and short axon to cns branches, motor- short dendrites from cns to cell body, long axon to effector cells. relay- many short dendrites from sensory neurone to cell body and many short axons from cell body to motor neurones.

Question 42

chrom

Answer 42

The mobile phase may be a liquid or a gas.
The stationary phase may be a solid (as in thinlayer chromatography, TLC) or either a liquid or solid on a solid support (as in gas chromatography, GC)
A solid stationary phase separates by adsorption, A liquid stationary phase separates by relative solubility If the stationary phase was polar and the moving phase was non- polar e.g. hexane. Then non-polar compounds would pass through the column more quickly than polar compounds as they would have a greater solubility in the non-polar moving phase.

Separation by column chromatography depends
on the balance between solubility in the moving
phase and retention in the stationary phase.
In gas-liquid chromatography GC the mobile
phase is a inert gas such as nitrogen, helium,
argon.
The stationary phase is a liquid on an inert
solid.
Types of chromatography include:
thin-layer chromatography (TLC) – a plate is coated
with a solid and a solvent moves up the plate
column chromatography (CC) – a column is packed
with a solid and a solvent moves down the column
gas chromatography (GC) – a column is packed with a
solid or with a solid coated by a liquid, and a gas is
passed through the column under pressure at high
temperature.
TLC:
a) Wearing gloves, draw a pencil line 1 cm above the
bottom of a TLC plate and mark spots for each sample,
equally spaced along line. b) Use a capillary tube to add a tiny drop of each solution to a
different spot and allow the plate to air dry.
c) Add solvent to a chamber or large beaker with a lid so that
is no more than 1cm in depth
d) Place the TLC plate into the chamber, making sure that
the level of the solvent is below the pencil line. Replace
the lid to get a tight seal.
e) When the level of the solvent reaches about 1 cm from
the top of the plate, remove the plate and mark the solvent
level with a pencil. Allow the plate to dry in the fume
cupboard.
f) Place the plate under a UV lamp in order to see the spots.
Draw around them lightly in pencil.
g) Calculate the Rf values of the observed spots.
Rf value = distance moved by amino acid
distance moved by the solvent
Wear plastic gloves to prevent contamination
from the hands to the plate
pencil line –will not dissolve in the solvent
tiny drop – too big a drop will cause different spots to merge
Depth of solvent– if the solvent is too deep it
will dissolve the sample spots from the plate
Will get more accurate results if the solvent is
allowed to rise to near the top of the plate but
the Rf value can be calculated if the solvent
front does not reach the top of the plate
lid– to prevent evaporation of toxic solvent
dry in a fume cupboard as the solvent is toxic
UV lamp used if colourless

A spot of the mixture on a TLC plate is first separated with one solvent.
Then the TLC plate is rotated 90 and the plate is placed in a second solvent for a second
separation to take place - two directional

Question 43

bp

Answer 43

boilin points
(ch3oh) H-bonds / hydrogen bonding

(ch3sh) dipole-dipole forces or vdws

H-bonds are a stronger / are the strongest IMF

(ch3seh) bigger molecule / larger Mr / larger no of electrons / Se bigger atom
With stronger/more vdw forces between molecules

van der Waals’ forces between oxygen molecules;

Hydrogen bonding between methanol molecules

HF = hydrogen bonding
HCl = (permanent) dipole-dipole bonding or even van de Waals’
Hydrogen bonding stronger / is the strongest IMF

ionic nacl- high- because of giant lattice of ions with strong electrostatic forces between oppositely charged ions.

simple molecular,iodine - low- because of weak intermolecular forces between
molecules (specify type e.g van der waals/hydrogen bond)

macromolecular - high- because of many strong covalent bonds in macromolecular structure. Take a lot of energy to break the many strong bonds eg diamond, graphite very high melting points because of strong covalent forces in the giant structure lot of nrg to break the many strong covalent bonds.

metallic mg,na - high- strong electrostatic forces between positive ions and sea of delocalised electrons

VDW- induced dipole-dipole interactions. They occur between all simple
covalent molecules and the separate atoms in noble gases.
In any molecule the electrons are moving constantly and randomly. As this happens the electron density can fluctuate and parts of the molecule become more or less negative i.e. small temporary or transient dipoles form.
These instantaneous dipoles can cause dipoles to form in neighbouring molecules. These are called induced dipoles. The induced dipole is always the opposite sign to the original one. dont occur in ionic
Main factor affecting size of Van der Waals
The more electrons there are in the molecule the higher the chance that temporary dipoles will form. This makes the Van der Waals stronger between the molecules and so boiling points will be greater.

The increasing boiling points of the halogens down the group 7 series can be explained by the increasing number of electrons in the bigger molecules causing an increase in the size of the Van der Waals between the molecules. This is why I2 is a solid whereas Cl2 is a gas. The increasing boiling points of the alkane homologous series can be explained by the increasing number of electrons in the bigger molecules causing an increase in the size of the Van der Waals between molecules. The shape of the molecule can also have an effect on the size of the Van der Waals forces. Long chain alkanes have a larger surface area of contact between molecules for Van der Waals to form than compared to spherical shaped branched alkanes and so have stronger Van der Waals.

Permanent dipole-dipole forces - in addition to vdw
between polar molecules, stronger than vdw and so the compounds have higher boiling points. Polar molecules have a permanent dipole. (commonly compounds with C-Cl, C-F, C-Br H-Cl, C=O bonds)
Polar molecules are asymmetrical and have a bond where there is a significant difference in electronegativity between the atoms. Difference in electronegativity leads to bond polarity dipoles don’t cancel therefore the molecule has an overall permanent dipole and there is an attraction between delta+ on one molecule and delta− on another

Hydrogen bonding
Hydrogen bonding is stronger than the other two
types of intermolecular bonding. The anomalously high boiling points of H2O,
NH3 and HF are caused by the hydrogen bonding between the molecules
The general increase in boiling point from H2S to H2Te is caused by increasing Van der Waals forces between molecules due to an increasing number of electrons.
Alcohols, carboxylic acids, proteins, amides all can form hydrogen bonds
It occurs in compounds that have a hydrogen atom attached to one of the three most electronegative atoms of nitrogen, oxygen and fluorine, which must have an available lone pair of electrons. e.g. a –O-H -N-H F- H bond. There is a large electronegativity difference between the H and the O,N,F

Question 44

obs

Answer 44

L= FeCl3
NH3 - red brown ppt
[Fe(H20)6]3+ + 3NH3 makes [fe(h20)3(oh)3] + 3NH4+

co3^2- = red brown ppt and effervesence
2[fe(h20)6]3+ + 3CO3^2- makes 2[fe(h20)3(oh)3] + 3co2 + 3h20

Test solution made acidic w nitric acid: no change = reacts w any carbonates to prevent formation ag2co3 ppt which mask desired observations
hno3 = FeCl3 + 3HNO3 makes Fe(NO3)3 + 3HCl

1st portion no reaction with bacl2
2nd portion: with agno3: ag+(aq) + cl- (aq) makes agcl(s) - white ppt

M= FeSO4
NH3 - green ppt
[Fe(H20)6]2+ + 2NH3 makes [fe(h20)4(oh)2] + 2NH4+

co3^2- = green ppt
[Fe(h20)6]^2+ + CO3^2- makes FeCO3 + 3H20

Test solution made acidic w nitric acid: no change = reacts w any carbonates to prevent formation ag2co3 ppt which mask desired observations
hno3 = 2HNO3 + FeSO4 makes Fe(NO3)2 + H2SO4

1st portion w bacl2 = Ba^2+ + SO4^2- makes BaSO4 forms white ppt
2nd portion: no reaction w agno3

Effervescence is observed in the reaction between FeCl3 (iron(III) chloride) and sodium carbonate (Na2CO3), but not in the reaction with FeSO4 (iron(II) sulfate), due to the difference in the chemical properties of the two compounds.

When FeCl3 reacts with sodium carbonate, a double displacement reaction occurs. The iron(III) chloride reacts with sodium carbonate to form iron(III) carbonate (Fe2(CO3)3), which is insoluble in water. The insolubility of iron(III) carbonate causes it to precipitate out of the solution. Additionally, carbon dioxide gas (CO2) is released as a product of the reaction, leading to the observed effervescence.

On the other hand, when FeSO4 reacts with sodium carbonate, no effervescence is observed. This is because iron(II) sulfate forms a soluble complex with sodium carbonate, and there is no release of carbon dioxide gas.

The formation of a soluble complex in the reaction between FeSO4 and sodium carbonate prevents the release of carbon dioxide gas and thus no effervescence is observed.

Therefore, the difference in effervescence observed in the reactions with FeCl3 and FeSO4 is due to the solubility of the resulting compounds and the release or absence of carbon dioxide gas.

Question 45

mcq

Answer 45

7- c
8- b
9- a
10- c
11- d
12- c
13- d
14- c
15- b
16- b
17- d
18- c
19- b
20- a
21- b
22- d
23- a
24- c

Question 46

chem

Answer 46

1.1 concentrated sulfuric acid acts as a catalyst to increase the rate of the reaction byprovides an alternative mechanism for reaction and is an exotherm reac

1.2. reflux condenser fitted to flask and mixture gently boiled over an electric heating mantle for about 30 mins allows for controlled heating and stops flammable vapour lighting

1.3 mistake 1: direction of water flow through condenser should be other way round, problem - doesnt ensure efficient cooling, condenser wnt condense reactant vapours of water and will run down only one side
mistake 2: bung sealing top of condenser, problem- build up of gas pressure could cause apparatus to explode

1.4 sodium carbonate reacts with unreacted acid and remaining catalyst still present after distillation. reaction produces co2 so pressure builds up in separating funnel so pressure of gas needs to be released at intervals

1.5 suggestion forms upper layer as it is organic and has lower density than water, lower is aqueous layer which is discarded reason cos difference in densities

1.6 it is a drying agent so that liquid appears clear when dry, doesnt react w organic liquid

1.7. mol ethanol = mass/mr = 7.9/46=0.1717391304 d=m/v dxv=m 0.790 x 10=7.9
mol ethanoic acid = 5.25/60 = 0.0875 so ethanoic acid is limiting reagent

percentage yield ethyl ethanoate= actual yield/theoretical x 100 = moles of ethanol+ethanoic acid= 0.1717391304+0.0875= 0.2592391304
1:1 ratio so mass ethyl ethanoate = 0.2592391304 x 88 = 22.81
5.47g/22.81g = 23.98 = 24%

1.8 crystaks arent completely dried

2.1 ch3ch2ch2ch2ch2-o-h
2.2 skeletal- c-c line up oh line down c c line down line up
2.3 ch3 ch2 ch2 - o - ch2-ch3
2.4 ch3 c-ch3, ch3 ch2 - o-h
2.5 acid catalysed elim, h bond arrow to c c bond, c bond arrow to o on oh2+

6.1 n=cv n=0.1 x 100/1000 = 0.01 moles of hcl
n=mass/mr = 0.6/74.1 = 0.008 moles
1:2 ratio so 0.008 x 2 = 0.016
0.016> 0.01 so ca(oh)2 is in excess

6.2 moles of ca(oh)2 = 0.6/74.1 = 0.008
[oh] x 2 = 0.016
0.016-0.01 = 6 x 10^-3
6x10^-3/1.6 total vol= 1.5+0.1=1.6
= 3.75 x 10^-3
[H+] = 6.80 x 10^-15/3.75x10^-3 = 1.81 x 10^-12
ph=-log(1.81 x 10^-12) = 11.74 to 2dp

Question 47

ther

Answer 47

NH4Cl (s) makes HCl (g) + NH3 (g)
delta s J K-1 FOR S ON OWN= +ve
change from solid reactant to gaseous products
increase in number of molecules
both will increase disorder

Example
Calculate deltas for the following reaction at 25˚C:
2Fe2O3 (s) + 3C (s) makes 4Fe (s) + 3CO2 (g)
delta s = sproducts - sreactants = (3 x 213.6 + 4 x 27.3) – (2 x 87.4 + 3 x 5.7) = + 558.1 J K-1 mol-1 =(3 sig fig.)

delta g= delta h-tdeltas

delta h kJmol-1 = sum of deltaformationh [products] - deltaformationh [reactants]

delta g=kjmol-1
delta s= kj K^-1 mol^-1
degrees to k + 273

A reaction that has increasing entropy
(+ve delta S) and is exothermic (-ve delta H)
will make delta G be negative and will
always be feasible. if delta g positive reaction not feasible

Calculating the temperature a reaction will become
feasible
N2 (g) + O2 (g) makes 2 NO(g)
deltaH = 180 kJ mol-1 delta S = 25 J K-1 mol-1 The reaction will be feasible when delta G less than or equal to 0
Make delta G = 0 in the following equation delta g= delta h-tdeltas
0 = delta h-tdeltas
So T = ∆H / ∆S
T = 180/ (25/1000) = 7200 K
The T must be >7200K which is a high temperature

Hydration enthalpies are exothermic as energy is given
out when water molecules bond to the metal ions.
The negative ions are electrostatically attracted to the delta + hydrogens on the polar water molecules and the positive
ions are electrostatically attracted to the delta - oxygen on the
polar water molecules.
The higher the charge density the greater the hydration
enthalpy (e.g. smaller ions or ions with larger charges)
as the ions attract the water molecules more strongly.
e.g. fluoride ions have more negative hydration enthalpies
than chloride ions.
Magnesium ions have a more negative hydration enthalpy
than barium ions.

delta sol H kj mol^-1 = delta H latt formation/diss + sum of delta hyd H

When an ionic substance dissolves the lattice must be
broken up. The enthalpy of lattice dissociation is equal
to the energy needed to break up the lattice (to gaseous
ions). This step is endothermic.
The size of the lattice enthalpy depends on the size and
charge on the ion. The smaller the ion and the higher its
charge, the stronger the lattice.

Generally deltaH solution is not very exo or endothermic so the hydration enthalpy is about the same as lattice
enthalpy.
In general the substance is more likely to be soluble if the delta H solution is exothermic.
If a substance is insoluble it is often because the lattice enthalpy is much larger than the hydration enthalpy and
it is not energetically favourable to break up the lattice, making delta H solution endothermic

Question 48

ener

Answer 48

Calculate the enthalpy change of combustion for the reaction where 0.650g of propan-1-ol was completely combusted and used to heat up 150g of water from 20.1 to 45.5oC
Step 1: Calculate the energy change used to heat up the water.
Q = m x c x delta t
Q = 150 x 4.18 x 25.4
Q = 15925.8 J
Step 2 : calculate the number of moles of alcohol combusted.
moles of propan-1-ol = mass/ Mr = 0.650 / 60 = 0.01083 mol
Step 3 : calculate the enthalpy change per mole which is called delta c H (the enthalpy change of combustion)
delta H = Q/ no of moles
= 15925.8/0.01083
= 1470073 J mol-1 = 1470 kJ mol-1
to 3 sf
Finally add in the sign to represent the energy change: if temp increases
the reaction is exothermic and is given a minus sign eg –1470 kJ mol-1

Question 49

chem2

Answer 49

Based on the boiling points provided, we can deduce a method to obtain a sample of ethanoic acid through a process called fractional distillation.

Fractional distillation takes advantage of the differences in boiling points of different compounds in a mixture. In this case, the boiling point of ethanoic acid (118°C) is higher than that of ethanol (78°C) and ethanal (21°C).

To obtain a sample of ethanoic acid, we can perform fractional distillation on a mixture containing ethanol, ethanal, and ethanoic acid. The process involves heating the mixture to a temperature slightly above the boiling point of ethanal (21°C) but below the boiling point of ethanoic acid (118°C).

As the mixture is heated, the lower boiling point compound, ethanal, will vaporize first. The vapor is then collected and condensed to obtain a sample of pure ethanal. Once the ethanal has been separated, the temperature can be increased to reach the boiling point of ethanol (78°C). Ethanol will vaporize next and can be collected and condensed separately.

Finally, by further increasing the temperature to the boiling point of ethanoic acid (118°C), ethanoic acid will vaporize and can be collected and condensed as a pure sample.

The separation of these compounds during fractional distillation is based on the differences in intermolecular forces. Ethanal and ethanol have weaker intermolecular forces (e.g., van der Waals forces) compared to ethanoic acid, which can form stronger hydrogen bonds. This difference in intermolecular forces contributes to the variations in boiling points and allows for their separation through fractional distillation.

Question 50

chem 3

Answer 50

The boiling points of ethanal, ethanol, and ethanoic acid can be explained by the intermolecular forces present in each compound.

Ethanal molecules are held together by relatively weak intermolecular forces, primarily van der Waals forces .These forces arise from temporary fluctuations in electron distribution, causing temporary dipoles in neighboring molecules. Since ethanal has a small molecular size and a simple molecular structure, the van der Waals forces between its molecules are relatively weak. As a result, less energy is required to overcome these forces, leading to a low boiling point.

Ethanol (C2H5OH) is a larger molecule compared to ethanal, and it has a boiling point of 78°C. Ethanol molecules can form hydrogen bonds with each other. Hydrogen bonding occurs when a hydrogen atom, covalently bonded to an electronegative atom (such as oxygen or nitrogen), interacts with a lone pair of electrons on another electronegative atom. In ethanol, the hydrogen atom attached to the oxygen atom can form hydrogen bonds with the lone pairs on neighboring ethanol molecules. Hydrogen bonds are stronger intermolecular forces compared to van der Waals forces. The presence of hydrogen bonding in ethanol requires more energy to break these intermolecular attractions, resulting in a higher boiling point compared to ethanal.

Ethanoic acid (CH3COOH) is a carboxylic acid with a boiling point of 118°C. It is a larger and more polar molecule compared to both ethanal and ethanol. Ethanoic acid molecules can form hydrogen bonds with each other, similar to ethanol. Additionally, ethanoic acid has a polar carbonyl group (C=O) and an acidic hydrogen atom (the hydrogen attached to the oxygen in the carboxyl group). The presence of these polar functional groups increases the strength of intermolecular forces in ethanoic acid. These stronger intermolecular forces require even more energy to overcome, resulting in a higher boiling point for ethanoic acid compared to both ethanal and ethanol.

In summary, the trend in boiling points of ethanal, ethanol, and ethanoic acid can be attributed to the strength of intermolecular forces. Ethanal, with the weakest intermolecular forces (van der Waals forces), has the lowest boiling point. Ethanol, with the additional presence of hydrogen bonding, has a higher boiling point. Ethanoic acid, with the strongest intermolecular forces (hydrogen bonding and dipole-dipole interactions), has the highest boiling point among the three compounds.