ANSWERS BOOKS/Group 4 - Sciences... · 2020. 11. 10. · Answers 373 Chapter 1 1 Themean is...
Transcript of ANSWERS BOOKS/Group 4 - Sciences... · 2020. 11. 10. · Answers 373 Chapter 1 1 Themean is...
ANSWERS
Answers
373
Chapter 1
1 Themeanisdeterminedbyaddingthedatatoobtain27.Themeanisnowfoundbydividingbythenumberofitemsofdata(10)toobtainameanof2.7.
Directentryofthedataintoagraphiccalculator(inthiscaseaTexasTI-83)givesanimmediateanswer.
First use theSTAT modeand enter thedata as a list.
Mean
Standard deviation
First use theSTAT modeand enter thedata as a list.
Mean
Standard deviation
To calculate the standard deviation manually usingtheformula
x x–( )2∑
n 1–-------------------------- ,
wecoulduseatable.
Number of eggs x – _ x ( x –
_ x ) 2
5 2.3 5.293 0.3 0.095 2.3 5.293 0.3 0.094 1.3 1.692 -0.7 0.490 -2.7 7.292 -0.7 0.491 -1.7 2.892 -0.7 0.49
Thesumofthethirdcolumngives
x x–( )2∑ =24.1
Thesamplestandarddeviation
= x x–( )2∑n 1–
--------------------------24.1
10 1–--------------- 1.6363917= =
2 (i) ThepopulationmeansfordrugsAandBarethesame.
(ii) The population means for drugs A and B aredifferent.
(iii)3.899(iv) 2.10(v) Greater(vi) Yes
Chapter 2
1B,2C,3D,4A,5C,6B,7B,8C,9A,10B,11A,12B,13D,14A,15B,16A,17B,18A
19 The correct order is: atom, DNA double helix(thickness, length is much more), thickness ofmembrane, organelle, prokaryotic cell, eukaryoticcell
20 (a) Thechildhasthelargersurfacearea.(b) dog:SA/V=0.13/2=0.065 child:SA/V=0.9/24=0.0375 thedoghasahigherSA/Vratiothanthechild(c) IftheSA/Vratioislarger,theorganismislikelyto
loosemoreheatsobasedonthis,thedogwouldneedmorefoodperkgbodyweightthanthechild.However,asthechildislikelytobegrowingandthe dog is fully grown, the child is likely to eatmore.Thelevelofactivitymayalsobedifferentwhichwouldaffecttheamountoffoodneeded.
21 (a) RefertoFigure215.(b) RefertoFigure222.(c)
organelle structure functionrER small structures,
consisting of 2 subunits, each made of RNA and protein
proteins for use outside the cell are produced here
lysosome membrane surrounding (hydrolytic) enzymes
intracellular digestion and autolysis
Golgi apparatus stack of flattened, membrane bound sacs, forming an extensive network in the cell
intracellular transport, processing and packaging
mitochondrion double membrane, inner membrane folded into cristae, surrounding matrix.
involved in the release of energy from organic molecules
nucleus contains DNA in linear chromosomes, surrounded by nuclear envelope
controls the activity of the cell by transcribing certain genes and not others (see 3.5)
374
Answers
22 (a) RefertoFigure233 Themembraneismadeofaphopholipidbilayer.
This consists of a number of phosholipidmolecule. Each phospholipid molecule is madeof a polar/hydrophilic phosphate, attachedto a central glycerol molecule which also has2 nonpolar/hydrophobic lipid tails. In thephospholipid bilayer, the phosphate moleculesarefoundoneithersideofthelayerandthelipidtailsarefacingeachotherinthecentre.
In between the phospholipid molecules, thereare cholesterol molecules and proteins. Integralproteins are proteins which are mostly foundin between the phospholipid molecules of themembrane and peripheral proteins are mostlyfound outside the phospholipid bilayer butinteractingwiththephosphateheads.
(b) the middle of the phospholipid bilayer ishydrophobic/non-polar which means that thepartoftheproteinthatisfoundinthisareaalsoneedstobenon-polar.If itwerepolar, itwouldmove through the phospholipid bilayer to findotherpolarmoleculestomakehydrogenbonds.Thiswouldobviouslydisruptthestructureofthemembrane.
23
creation of two genetically identical nuclei
mitosis telophase
biochemical reactions interphase
separation of sister chromatids mitosis anaphase
DNA replication interphase S-stage
chromosomes moving to the equator
mitosis metaphase
protein synthesis interphase
24 Asexualreproduction,cloning,growth,repair.Mitosis,withmeiosis,isalsopartoftheprocessofproducinggametes.
25 Volumeis lengthxwidthxheightsoasthecellgetsbigger,volumeincreasestothepower3.Surfaceareaislengthxwidthsosurfaceareaincrasestothepower2.
Rateofuseofresources(food,oxygen)andtherateofheatandwasteproduction(urea,carbondioxide)arelinkedtothevolumeofthecell.Rateofexchange(ofe.g.oxygen)islinkedtothesurfacearea.Sotheneedfore.g.oxygenwillincreasefasterthantheabilitytotakeitup.
26 realsizexmagnification=measuredsize realsize=measuredsize/magnificatio 8x10-3/400 magnification20x10-6
theactualsizeofthenucleusis20micrometres.
Chapter 3
1D,2C,3B,4A,5D,6A,7D,8A,9A,10A
11 Thermal properties: water needs a lot of energy towarmupandalotofenergytogofromsolidtoliquidorliquidtogas.Thisisbecauseofthestrongcohesionforcesbetweenmolecules.
Watermoleculeshaveanegativeandapositiveside.This means that they are polar. There are hydrogenbonds(attractionforces)betweenthenegativesideofonemoleculeandthepositivesideofanotherwhichcausesthecohesionforcesbetweenwatermolecules.
Water is a good solvent for polar molecules. Polarmolecules can interact with water molecules in thesameway thatotherwatermoleculesdo.Non-polarmoleculesdonot interactwithwatermolecules andthereforedonotdissolvewell.
12 (a) carbohydrates:carbon,hydrogen,oxygen(b) lipids:carbon,hydrogen,oxygen(c) proteins:carbon,hydrogen,oxygen,nitrogen
13 (a) carbohydrates:e.g.glucose-see Figure 313(b) lipids:glycerolandfattyacids-see Figure 317(c) proteins:aminoacids-see Figure 319
14 Thecondensationreactionbetweenmonosaccharidesis between two OH groups, forming water. ThisleavesoneOinplacewhichformsanoxygenbridgebetween the monosaccharides. The condensationreactionbetweentwoaminoacidsisbetweenNH2ofoneaminoacidandCOOHoftheotheraminoacid.TheHoftheNH2groupcombineswiththeOHoftheCOOHgroup,formingwaterandapeptidebond.
15 (a) Enzymesareproteins.Increasingthetemperaturechanges the three dimensional shape of anyprotein, so also that of the active site of theenzyme.Iftheactivesitechangesshape,thenthesubstratewillnolongerfitandtheenzymedoesnotworkanymore.
(b) Increasing substrate concentration will makeitmore likelythattheactivesiteofanyenzymehasa substrateattached.Therefore thespeedofthe reaction will increase untill all active sitesare filled with substrate at any time. Further
Answers
375
increasing the substrate concentration will notincreasetherateofreaction.
(c) Denaturationoccurswhentheshapeoftheactivesite of the enzyme changes. The substrate nolongerfitsandtherateof thereactionbecomesverylow.
16 (a) covalent bond caused by a condensationreaction
(b) covalent bond caused by a condensationreaction
(c) hydrogen bonding caused by positive-negativeattraction forces between either oxygen ornitrogeninonebaseandhydrogenintheother.
(d) covalentbondsarestrongerbecausetheyinvolvesharingelectrons.
(e) becauseAandTbothformtwoHydrogenbondsandCandGbothform3Hydrogenbonds.Also,A and G are purines (have two rings) while CandTarepyrimidines(haveonering)andthereisroomforexactlythreeringsbetweenthesidesoftheDNAladder.
17 Similarities: Both are nucleic acids, made of nucleotides.
Nucleotidescontainapentosesugar,aphophateandanorganicbase.
Differences: DNAhasde-oxyribose,RNAhasribose. DNAhasthymine,RNAhasuracil DNAhastwostrands,RNAhasonestrand DNAisusuallyintheshapeofadoublehelix,RNAis
not.
18 (a) light intensity, colour of the light, amount ofcarbondioxideavailable,temperature.
(b) oxygen,ATP(andNADPH)
19 (a) ATP,NADHandpyruvate(b) ATP,CO2,ethanol,lacticacid,heat
20 It isdonebyattaching theenzyme lactase toa largemoleculeandthenbringingitintocontactwithmilk.Anylactosepresentwillbebrokendownintoglucoseandgalactosebythelactase.
Chapter 4
1A,2B,3D,4D,5B,6A,7C,8B,9D,10D,
11 RefertoFigure413 Theresultsoftheself-fertilisationoftheF1willbe: genotypes:25%TT,50%Tt,25%tt phenotypes:75%tall,25%short
12 w+=wildtype w=black AxBgivesonlywildtypecoloursoatleastoneparent
mustbew+w+ AxCgivessomeblack(ww)sobothAandCmust
haveawalleleandbew+w BxCgivesonlywildtypecoloursoatleastoneparent
mustbew+w+ Therefore Aisw+w Bisw+w+ Cisw+w
13 (a) ExpA:Yellowisdominantoverwhite ExpB:Shortisdominantoverlong(b)
phenotypes possible genotypesyellow YY or Yy
white yy
short SS or Ss
long ss
(c) sincetheF1containssomelongplantswithwhiteflowers, theirgenotype isyyss.Thegenotypeofthe parent plant with white flowers and longstems is also yyss. So the yellow flowered shortstemmedparentmustbeheterozygousYySs.
Punnett square yyss genotype parent
ys gametes
YySs YS YySs
yellow short
genotype offspring
phenotype offspring
Ys Yyss
yellow long
genotype offspring
phenotype offspring
yS yySs
white short
genotype offspring
phenotype offspring
ys yyss
white long
genotype offspring
phenotype offspring
genotype parent
gametes
There is 25% chance of each of the types ofoffspring.Theresultsfoundmatchthisratio
14 If a mouse has two c alleles, it will not be able toproducepigment,soitwillbealbino.Itwillstillhavetheallelesforwildtypecolourorblackcolourbuttheywillnotbeseenbecausethemousewillbealbino.Ifamouseiswild-typecolourorblackcolour,ithasatleastonealleleformakingpigmnt(C).
376
Answers
phenotypes possible genotypes
albino ccAA or ccAa or ccaa
wild type CCAA or CCAa or CcAA or CcAa
black CCaa or Ccaa
(a)
Punnett square ccAA(albino)
genotype parent
cA gametes
CCaa(black)
Ca CcAawild type
genotype offspringphenotype offspring
genotype parent gametes
(b)
Punnett square
CcAa genotype parent
CA Ca cA ca gametes
CcAa CA CCAAwild type
CCAawild type
CcAAwild type
CcAawild type
genotype offspringphenotype offspring
Ca CCAawild type
CCaablack
CcAawild type
Ccaablack
genotype offspringphenotype offspring
cA CcAAwild type
CcAawild type
ccAAalbino
ccAaalbino
genotype offspringphenotype offspring
ca CcAawild type
Ccaablack
ccAaalbino
ccaaalbino
genotype offspringphenotype offspring
genotype parent
gametes
Theratioofthephenotypesoftheoffspringwillbe
wildtype: 9 black: 3 albino: 4
15 (a) Refer to Figure 414. BabyhasphenotypeBsocanhavegenotypeIBIB
orIBi. MotheristypeAsocanbegenotypeIAIAorIAi. Father’sfatheristypeAsocanbegenotypeIAIA
orIAi. Father’smother is typeB so canhavegenotype
IBIBorIBi. BabydoesnothaveanalleleIAsohemusthave
receivedalleleIfromhismother. BabyistypeBsohemustbeIBi,ashehasreceived
Ifromhismother.
(b) InordertobeIBi,hemusthavereceivedIBfromhisfatherwhomusthavebeenIBi,havingreceivedIfromhisfather(thepaternalgrandfatherofthebaby who was type A) and IB from his mother(thepaternalgrandmotherwhowastypeB).
PaternalgrandfathermusthavebeenIAi. PaternalgrandmotherwaseitherIBIBorIBi.
16
possible phenotypes genotypes
red polled CR CR PP or CR CR Pp
red horned CR CR pp
roan polled CR CW PP or CR CW Pp
roan horned CR CW pp
white polled CW CW PP or CW CW Pp
white horned CW CW pp
(a) P: redpolled x whitehorned (phenotypes)
CRCRPP x CWCWpp (genotypes)
CRP x CWp (gametes)
F1 CRCWPp (genotype) roanpolled(phenotype)
(b) APunnettsquarecanbeusedtohelpworkouttheF2generation
Punnett square
CR CW Pp genotype F1
CR P CR p CW P CW p gametes
CR CW Pp
CR P CR CR PPred polled
CR CR Ppred polled
CR CW PProan
CR CW Pproan polled
genotype F2phenotype F2
CR p CR CR Ppred polled
CR CR ppred horned
CR CW Pproan polled
CR CW pproan horned
genotype F2phenotype F2
CW P CR CW PPRoanpolled
CR CW Pproan polled
CW CW PPwhite polled
CW CW Ppwhite polled
genotype F2phenotype F2
CW p CR CW Pproan polled
CR CW pproan horned
CW CW Ppwhite polled
CW CW ppwhite horned
genotype F2phenotype F2
genotype F1
gametes
F2: redpolled(3)+redhorned(1)+roanpolled(6)+roanhorned(2)+whitepolled(3)+whitehorned(1)
Answers
377
17
phenotypes possible genotypesmale normal XHY
male haemophiliac XhY
female normal XHXH
female carrier XHXh
female haemophiliac XhXh (usually fatal)
(a) MothermustbecarrierXHXh.FatherisnormalXHY. Since Mohammed is male, he cannot beacarriersohehasnochanceofpassingonthehaemophiliaalleletohischildren.Hecouldhavechildrenwithhaemophiliaifhiswifeisacarrier.
(b) Latifa inherits a normal allele from her fatherXH.Shehas50%chanceofinheritinganormalallelefromhermotherXHbutalso50%chanceofinheritingthehaemophilaallelefromhermotherXh.Soshehasa50%chanceofbeingacarrier.Ifsheisacarrier,shewillpassherhaemophiliaallele on to 50% of her children. That meansthat each of her sons will have a 50% of beinga haemophiliac and each of her daughters willhavea50%chanceofbeingacarrier..SoifLatifaisacarrier,25%ofherchildrenare likely tobehaemophiliacs.
The above is based on the assumption that Latifa’s husband is not a haemophiliac.
18 Similarities:• in both divisions, chromosomes move to the
equator• meiosisIIissimilartomitosis• bothinvolvespindleformation• in both chromosomes/homologous pairs move
tothepoles• bothincreasethenumberofcells
Differences
mitosis meiosispurpose growth, repair production of gametesnumber of divisions
one two
number of cells produced
two four
possible in haploid/diploid nuclei
diploid (or more) nuclei
nuclei produced as parent haploid (reduction division)
chromosome movement
chromosomes line up at the equator
meiosis I: homologous pairs line up at the equator
crossing over no possiblechiasmata no possible
19 Justmatchtheshapes.
Chapter 5
1B,2D,3A,4C,5A,6D,7A,8C,9A,10A,11C,12C,13C,14B,15A,16D,
17 Therearemanypossibilitiesherebutoneexampleisgivenbelow:
1. a. insect has wings . . go to 2 b. insect does not have wings . go to 6
2. a. wings are shorter than body . go to 3 b. wings are as long as body .. go to 4
3. a. body is gray .. .. E b. body is striped . . F
4. a. antennae are short . .. G b. antennae are long . . go to 5
5. a. antennae are feathered . H b. antennae are not feathered . I
6. a. insect has tails . . go to 7 b. insect does not have tails .. go to 8
7. a. insect has two tails . .. A b. insect has three tails . .. B
8. a. insect has antennae . .. C b. insect does not have antennae . D
18 (a) Species:agroupoforganismsthatcaninterbreedandproducefertileoffspring.
(b) Ahabitatdescribesjusttheenvironmentinwhicha species normally lives while an ecosystemdescribesthecommunityanditsenvironment.
19 (a) Carbondioxide levels increase from 315ppm in 1960 to 368 ppm in 2000. Each year,carbondioxide levelsfluctuatearound themeanvalue.Thefluctuationisabout4ppmaboveandbelowthemeanvalue.
(b) Suggestreasonsforthesechanges.Theoverallincreasefrom1960to2000canberelatedtotheincreaseduseoffossilfuelwhichproducescarbondioxidewhenburned.Deforestation has reduced the number ofplantsavailabletoreducecarbondioxidelevelsbyphotosynthesis.The yearly fluctuations are caused by theseasons: in summer, plants photosynthesizemore which reduced carbon dioxide levelswhiletheyincreaseinwinterwhenplantshave
•
•
•
378
Answers
dropped their leaves and reduced/stoppedphotosyntesis.
20 (a) Methaneandnitrousoxided(b) EnergyfromthesunreachesEarthaslightrays
withashortwavelength.70%of thiswillwarmuptheEarthsurface,bechangedintoInfraRedradiation (with a longer wavelength) and re-radiateintospace.GreenhousegasesabsorbsomeofthisIRradiationwhichcausesanincrease inthetemperatureoftheatmosphere.
21 (a) The precautionary principle comes from theconcept of do no harm. Normally, if there is acertain change planned, e.g. building a nuclearpower station at a new site, those against thechange (e.g.people living in thearea)willhavetoprovethatitwillbeharmful.Thisissometimesdifficulttodowhichwouldmakethechangegoahead.
If the precautionary principle is applied, thenthosewhowant thechange (e.g. thosebuildingthenuclearpowerstation)willhavetoprovethatitwillnotdoharm.Thismayalsobedifficultbutfailuretodosowillstopthechangefromtakingplace.
(b) When it isnotcertain towhatdegreea speciesis endangered, the precautionary principle willdictatethattheyareplacedonthelevelofthemostendangeredspecies,givingitthehighestlevelofprotection.Ifitturnsoutthatthespeciesisindeedendangered,thenitisplacedcorrectly.Shoulditbe less endangered than originally thought, thehigher level of protection has not caused harmto the species. However, if the species were tobe placed incorrectly and too low, then it willbegiven lessprotectionandthespeciesmaybeseriously reduced in numbers. This would do aconsiderableamountofdamagetothespecies.
22 Itshouldlooksomethinglikethis:
23 Sexual reproduction increases variation because thegenetic material of each individual comes from aspermandaneggcell.
Mendelsfirst law(lawofsegregation)says thateachtraithastwoallelesbuttheyareseperatedduringthe
formationofgametes so thatagametehasonlyonealleleforeachgene.
Mendel’ssecondlaw(lawofindependentassortment)says that the combination of alleles from differentgenesisrandomandthatthereforethegametesofoneorganismarenotallthesame.
As a result, the combination of male and femalegametes(whichissexualreproduction)willleadtoanumberofdifferentindividuals.Thisshouldbeknowsbecauseitisclearthatsiblingsfromonesetofparentsarenotidenticaltoeachother.
In addition, it is possible that during Prophase I ofMeiosis, an exchange of genetic material betweennon-sister chromatids occus. This is called crossingoverandwillleadtomorenewcombinationsofgenes,causingfurthervariation.
24 ThePepperedMothrestsontreetrunkscoveredinoff-white lichens.Themoth is thesamecolourandwillspread its wings to blend in with their background.Occasionally a black one appears which is quicklyseenandeatenbybirds.
As a result of the industrial revolution, trees werecovered in soot. Lichens died and the tree trunksbecameblack.Thewhitemothnowwereeatenmoreoften and the few black ones had the advantage. Inonlyafewyears,theblackmothsweremorecommonthanthewhiteones.
25 (a) Homo sapiens.Noticethatthenameshouldbeinintalics or underlined (when writing by hand).Thefirst letterof thegenusname iscapitalised,everythingelseinsmallprint.
(b) Kingdom, phylum, class, order, family, genus,species.
26 Similarities:bothareautotrophs/producershavechlorophyllforphotosyntesishavetrueroots,stems,leavesproduceseeds
Differences
coniferophyta angiosperma
have needles have leaves
no flowers flowers (although small in wind pollinated plants)
seeds in cones seeds in ovaries which become fruits
27 (a) allexceptporifera(b) allexceptporifera(c) platyhelminthes, annelida, mollusca and
arthropoda.(d) porifera,cnidaria
••••
Answers
379
Chapter 6
1C,2B,3C,4A,5B,6B,7C,8A,9D,10A,11B,12D,13B,14B,15A,16A,17B,
18 (a) RefertoFigure603(b)
structure function
villus increase surface area which increases absorption
epithelial cells these cells are permeable to nutrients which need to be absorbed
microvilli increase surface area which increases absorption
capillary bed near surface of villus
removes absorbed nutrients and maintains concentration gradient
proximity to surface reduces distance and speeds up absorption
lacteal part of the lymphatic system which absorbs and tranports lipids
19 (a) Thebloodpressureinarteriescanbehighandathick wall is needed to prevent the artery frombursting.
(b) Becausealotoftheplasmahasleftthecapillaryto become tissue fluid and the remaining fluidis more viscous. This is an advantage becausethereismoretimetoexchangematerialswiththetissue
(c) Topreventbackflowofblood.Duetothehigherpressureandspeedofthebloodinarteries,thereisnoriskofbackflowtherebutthelowpressurein veins makes it likely to flow in the wrongdirection, e.g in the legs when we are standingup.
20 • nutrientsoxygenhormonesantibodiesheatwaste:ureaandcarbondioxide
21 (a) antigenenterstheorganism(b) lymphocyte that produces the correct antibody
willrecognisetheantigen(c) lymphocytewillformaclone(d) allclonedcellsproducethesameantibody(e) antibodywillmakeantigenharmless
•••••
22 HIV reduces the number of lymphocytes.Lymphocytes make antibodies so HIV reducesantibodyproduction.
23 (a) RefertoFigures620and621.(b)
structure functionlarge surface area a lot of gas exchange can take place at
the same time
thin short diffusion distance
moist gases can only cross a membrane when dissolved
good blood supply maintains concentration gradient
(c) An alvelous privides a large surface area fordiffusionofgasesintoandoutoftheblood.
(d) Tomaintainthehigherconcentrationofoxygenandthelowerconcentrationofcarbondioxideinthelungssothatoxygenwilldiffusefromtheairin the lungs into the blood and carbondioxidefromthebloodintotheairinthelungs.
24 Both refer to a difference in charge across themembrane of a neuron. A resting potential is fairlystableandusuallyaround-70mV.Anactionpotentialis a change in the potential across the membrane.The electric potential will change from -70 mV to+30mVandbackagain.Anactionpotentialiswhatismeasuredasanimpulse isconductedthroughtheneuron.
25 (a) Heat centre in the hypothalamus which sensesthe temperature and compares it to the desiredvalue.Skinarterioles,musclesthatflattenorraiseuphairsandsweatglandsarealleffectors.
(b) • vasodilationsweatingIncreasedmetabolism’fluffing’ofhairorfeathersthicklayerofbrownfatorofblubberspecialstructurehair
26 Both estrogen and progesterone inhibit secretionofFSHwhich isnecessary for thedevelopmentof afollicleandhenceaneggcell.IntheabsenceofFSH,nofolliclewillripenandnoeggcellcanmaturesonofertilisationcantakeplace.
27 • dryskinisdifficulttopenetratebecauseitismadeoftoughcellsandthereareonlyfewgapsnormalbacteriagrowingontheskinwillmakeitmoredifficultforpathenogenicbacteriatogrowpHofskinisslightlyacidicandmanypathogenscannotgrowthere
•••••
•
•
380
Answers
mucus is found where there is not enoughprotectione.g.airpassagesstickymucustrapsbacteriaandstopsthemfromspreading
Chapter 7
1A,2C,3A,4D,5B,6D,7D,8B,9D,10A,
11 RefertoFigure703andmentionthesepointsDNAismadeoftwostrandsstrandsareantiparallelstrands are kept together by hydrogen bondsbetweentheorganicbasesbuildingblocksofDNAarenucleotidesnucleotides are linked together by covalentbondscovalent bonds are created by condensationreactionsnucleotideconsistofadeoxyribose,aphosphateandanorganicbasebaseisatachedtoC1,phosphatetoC5organicbasecanbeAdenine,Thymine,Cytosine,GuanineabasepairbetweenthestrandsisalwaysA-TorC-Gladderistwistedapprox10nucleotidesperfullturn
12 (a) helicase unwinds the twisted ladder and breaksthehydrogenbondsbetween the strandswhichseparatesthestrands
(b) RNAprimasewillformcovalentbondsbetweentheRNAnucleotidestoformtheRNAprimer
(c) DNA polymerase III will form covalent bondsbetween the DNA nucleotides and the growingstrand.Itworksina5’to3’direction.
(d) DNA polymerase I removes the RNA primers(mostly from Okazaki fragments) and replaceRNAnucleotideswithDNAnucleotides
(e) DNA ligase will attach the DNA fragments thatweretheOkazakifragmentstoformacompletestrand
13 Bothneedenergyandworkinthesamedirection(5’to3’)
transcription translationDNA to RNA RNA to protein
more limited number of mRNA is made each time.
many protein molecules are made almost at the sme time
enzymes needed enzymes and ribosomes and tRNA needed
in nucleus in cytoplasm
•
•
•••
••
•
•
••
•
••
14 (a) mRNA is an RNA copy of the nonsense strandoftheDNA.ThesequenceofthemRNAcodonsdetermines the sequence of the amino acids inthepolypeptide.
(b) ribosomes are needed for the tRNA anticodonto bind to the mRNA codon so that the aminoacidofthetRNAcanbeattachedtothegrowingpolypeptidechain.
(c) a codon is a sequence of 3 organic bases thatcodesforaspecificaminoacid;ananticodonisasequenceofthreeorganicbases(foundontRNA)complementarytothecodon.
(d) each tRNA has a specific anticodon. Related tothis anticodon, tRNA also carries a particularaminoacidsothatthesameaminoacidisalwaysfoundwiththesameanticodon.tRNAtakestheaminoacidtotheribosomeandallowsittoattachtothegrowingpolypeptidechain.
15 • primarystructure:covalentbondssecondarystructure:hydrogenbondstertiary structure: hydrophobic interactions,disulfidebonds,hydrogenbondsquaternarystructure:hydrogenbonds,positive/negative attraction forces, hydrophobic forces,disulfidebridges
16 • enzymese.g.maltasewhichbreaksdownmaltoseintoglucoseandfructosehormonese.g.insulindefensee.g.antibodies/immunoglobinsstrucruree.g.spindlefibreincelldivisiontransporte.g.hemoglobinforoxygentransport
17 • theyareusedtokeeptheproteininitsplace:polaraminoacidsinteractwithwateroneithersideofthe membrane, non-polar amino acids interactwiththenon-polar lipidbilayerinthecentreofthemembraneahollowtubecanbecreated,usingpolaraminoacidsat theends, interactingwith thewateroneither side of the membrane. In the centre, theinsideofthecylinderismadeofnon-polaracids,creating a hydrophilic channel through themembrane.polarandnon-polaraminoacidscanhelpcreatetheshapeoftheactivesiteofanenzymeandhelpprovidetheforcesthatkeepthesubstrateintheactivesiteandmakesurethatothersubstratedonotfitwellanddonotstaythere.
18 (a) attheactivesite(b) anywhere on the enzyme as long as it is away
fromtheactivesite(c) competitiveinhibitor:prontosil non-competitiveinhibitor:cyanide
••
•
••••
•
•
Answers
381
19 • allosteryisnon-competitiveinhibitionendproductistheinhibitorendproductwillattachtotheenzymeofanreactionearlierinthemetabolicreactionsataplacewhichisNOTtheactivesitewillchangetheshapeoftheactivesiteenzymenolongerworksnomoreproductproducedwhenproductrunsoutthen no more product to act as allostericinhibitorsoreactiomproceedsnewproductformed.
20 • ifreactionisA+B→CAandBmayneedtocollidewithahighamountofenergyinordertohavereactiontakeplaceonlyfewparticleshavethisenergyincreasingtemperaturewillincreasethenumberofpartcipleswithsufficientenergybut this could denature proteins in a livingsystemwhensubstratesmeetattheactivesitethe conditions are such that it is easier for thereactiontotakeplacelessactivationenergyisneededmoreparticlescanreactatthesametimerateofreactionincreases.
Chapter 8
1D,2A,3D,4A,5D,6D,7D,8D,
9 (a) inthecytoplasm(b) inthematrixofthemitochondria(c) on the cristae (inner membranes of the
mitochondria)
10 RefertoFigure804
11 (a) Thesite for theKrebscycle is thematrixof themitochondria. It is a watery fluid contaning allthe enzymes and compounds needed for theKrebscycletoproceed.
(b) The site for the electron transport chain isthe cristae (folded inner membrane) of themitochondrion. It has a large surface area sothere isa lotofroomformanyATPsynthetasemolecules. It is also impermeable to protons(exceptwherethereareATPasemolecules).
•••••••••
••
•
•••
•
••
•••
(c) Thenitwouldnotbepossibletobuildupahigherconcentration of protons and chemiosmosiswould not work. As a result, no ATP would beproduced.
12 RefertoFigure809.
13 RefertoFigure810. The light dependent stage produces ATP and
NADPHwhichareneededtodrive theCalvincycleinthelightindependentstage.Thelightindependentstage fixes carbon dioxide to produce a triose. ThisprocessrequiresATPandNADPH,madeinthelightdependentstage:ATPisneededtophosphorylateGPand to change TP into RuBP. NADPH is needed toreduceglyceratediphosphateintoanthertriose.
14 (a) theyprovidetheenergyandthereducingpowertodrivetheCalvincycleinthelightindependentstage.
(b) itisasimplemethodofproducingATP(c) non-cyclic photophosphorylation produces
NADPH which is needed to drive the lightindependent stage. The advantage of the lightindependentstageisthatitmakescarbohydrateswhichcanbeusedforlongtermenergystoragebutalsoasbuildingmaterialsfornewcellssothattheplantcangrowandreproduce.
(d) intheCalvincycle,3moleculesofcarbondioxidearecombinedtoformatriosemolecule;2triosemoleculeswillbecombinedtoformglucose.
15 The grana thylakoid provide a large surface forabsorbinglightthemembraneofthegranaallowstheelectron carriers needed for photophosphorylationtobekept intherightorder,helpingthereactiontoproceed.Thevolumeinsidethegranaissmallsothatarelativelysmallnumberofprotonswillcausealargedifferenceinconcentrationacrossthemembranethestroma is awateryfluidwhichcontainsall enzymesandintermediatesneededfortheCalvincycle.
16 Similarities:bothusechemiosmosisbothinvolvetransportofelectronsandprotonsboth use ATP synthetase to produce ATP fromADPandPiboth take place across a membrane which isimpermeabletoprotonsboth buld up a proton gradient across themembranebothtransportprotonsintoasmallspace
•••
•
•
•
382
Answers
Differences:
chloroplasts mitochondria
site membrane of granum membrane of crista
organelle chloroplast mitochondrion
source of electrons water (photolysis) NADH + H+
ultipmate electron acceptor
NADP oxygen
protons are pumped into
lumen of the grana intermembrane space
protons diffuse via ATPase into
stroma matrix
Chapter 9
1C,2A,3C,4D,5A,6D,7A,
8 (a) RefertoFigure903(b) upper epidermis reduceswaterloss preventgasexchange allowslighttopass secretescuticle barrieragainstinfection palisade layer photosynthesis spongy layer allowsrapiddiffusionofoxygen
andcarbondioxidethroughairspaces
photosynthesis lower epidermis stomataopen/closetoallowgas
exchangebutreducewaterloss secretecuticle barrieragainstinfection vascular tissue xylem transports water and
mineralfromrootstoleaves phloem transports carbohy-
dratesfromsourcetosink(c) upper epidermis cells close together - no space
betweenforpathogenstoenterorwatertoevaporate
transparent(nochloroplasts)-lightcanpasseasily
palisade layer tighly packed, narrow end totop - maximum exposure tolight
spongy layer airspacesallowrapiddiffusionofrespiratorygases
chloroplastsforphotosynthesis lower epidermis cells close together - no space
betweenforpathogenstoenterorwatertoevaporate
nochloroplasts stomata allows rapid gas
exchange
vascular tissue xylem dead cells - no crosswalls - facilitates transport ofwaterandminerals
phloem livingcells-activetransportofcarbohydrates
9
monocotyledenous plants
dicotyledenous plants
veins in leaf parallel reticulate (net-like)
distribution of vascular tissue
scattered in a ring
number of cotyledons in seed
one two
floral organs multiples of three four or five
roots unbranched branched
examples grass, onion, lily and tulip
daisy, oak tree and rose
10 • auxinisgroupofplanthormonesauxinisproducedbytheapicalbudofaplantauxinistransporteddownthestemasneededauxinstimulatesgrowthbypromotingcelldivisionandcellstretchingauxin accumulates on the shaded side of thestemwhichmakestheshadedsidegrowfastersotheplantgrowstowardsthelightthisincreasestheamountoflightontheleaveswhichincreasesphotosynthesis
11 Wateristakenupbyosmosisbecausetheconcentrationof dissolved particles in the root hair cell is greaterthan that outside. The plant roots take up mineralsby active transport this increases the concentrationofdissolvedparticlesintheroothaircells.Roothairsincreasesurfacearearootsneedalargesurfaceareatoallowalotofwateruptakeatthesametimewateristakenfromrootsbyxylem.Transpirationpullmoveswaterupthexylem.Cohesionistheforceneededfortranspiration pull, adhesion enables the water creepupthesidesofacapillarytubeandsupportstheflowofwaterintheplant.
12 Phloemtransportssugarsandaminoacidsbypressureflow hypothesis from source (area of production/storage)e.g.leavesorrootstosink(areaofuse/storage)e.g.fruits/growingpointsorroots.Socarbohydrates/amino acids may go up or down stem at differenttimes. Phloem is made of living cells and transportinvolvesactivetransporti.e.energyisneeded,energyissuppliedbycompanioncells.
•••••
••••
Answers
383
13 PleaserefertoFigure925.
14 Water is absorbed, gibberelins are made in thecotyledons,amylase ismade,starchbrokendowntomaltose which is used to release energy and enablegrowth.
15 (a) oxygen,water,suitabletemperature(b) oxygen:respirationrequiresenergyandoxygenis
neededtoreleaseenergy water:seedsswellupaftertakingupwaterwhich
burststhetestaandthepresenceofwateractivatesenzymeswhichhydrolyseslargemolecules
suitable temperature: germination requiresenzymes which need to have a temperaturesuitableforthemtowork
16 • there are long day plants, e.g.carnations andclover, and short day plants, e.g. coffee andstrawberrieslong day plants flower in summer, short dayplantsinautumn(andspring)long day plants only flower when nights areshortshort day plants only flower when nights arelongphytochromeregulatesthefloweringofplantstwostates:PrandPfrPr becomes Pfr quickly when exposed to redlightPfrbecomesPrslowlyindarknesslongdayplantsneedPfrtoflowershortdayplantsonlyflowerwhenPfrisabsent
Chapter 10
1B,2D,3C,4C,5C,6A,7B,8A
9 RefertoFigure1001Interphase: DNAreplication.Prophase I: Chromosomescondense. Nucleolusbecomesinvisible. Spindleformation. Synapsis: homologous chromosomes
sidebyside. (thepair isnowcalledabivalent, the
crossoverpointsarecalledchiasmata). *Nuclearmembranedisappears(some-
timesconsideredasearlymetaphase).Metaphase I: Bivalentsmovetotheequator.Anaphase I: Homologous pairs split up, one
chromosomeofeachpairgoestoeachpole.
•
•
•
•••
•••
Telophase I: Chromosomesarriveatpoles. *Spindledisappears.Prophase II: Newspindle is formedatrightangles
tothepreviousspindle.Metaphase I: Chromosomesmovetotheequator.Anaphase II: Chromosomes separate, chromatids
movetooppositepoles.Telophase II: Chromosomeshavearrivedatpoles. Spindledisappears. Nuclearmembranereappears. Nucleolusbecomesvisible. Chromosomesbecomechromatin.
10 (a) Mendel’s second law: law of independentassortment.
Anyoneofapairofcharacteristicsmaycombinewitheitheroneofanotherpair.
(b) Mendel’s second law does not apply to linkedgenes. They inherit together except for thosecaseswherecrossingoveroccurs.
11 (a) wildtypecolour,wildtypewings(b) Allpossiblephenotypesandgenotypesare: r+isredeyes,riswhiteeyes w+isnormalwings,wisvestigialwings
possible phenotypes corresponding genotypes
red normal r+r+w+w+, r+r w+w+, r+r+w+w, r+r w+w
red vestigial r+r+ww, r+r ww
white normal rr w+w+, rr w+w
white vestigial rr ww
Possiblegenotypesoftheparents:
parent phenotype possible genotypes
1 white eyes, vestigial wings
rr ww
2 red eyes, normal wings
r+r+w+w+, r+r w+w+, r+r+w+w, r+r w+w
(c)
P : white, vestigial x red, normal phenotype rr ww r+r+w+w+ genotype rw r+w+ gametes
F1 : +r w+w genotype red, normal phenotype
r+w+ r+w rw+ rw possible gametes
r
384
Answers
PUNNET SQUARE
Red, normal
r+w+ r+w rw+ rw
Red, normal r+w+ r+r+ w+w+
red, normalr+r+ w+wred, normal
r+r w+w+
red, normalr+r w+wred, normal
r+w r+r+ w+wred, normal
r+r+ wwred, vestigial
r+r w+wred, normal
r+r wwred, vestigial
rw+ r+r w+w+
red, normalr+r w+wred, normal
rr w+w+
white, normalrr w+wwhite, normal
rw r+r w+wred, normal
r+r wwred, vestigial
rr w+wwhite, normal
rr wwwhite, vestigial
(d) genotypes of the F2: r+r+w+w+: 1 r+rw+w+: 2 rrw+w+: 1 r+r+w+w: 2 r+rw+w: 4 rrw+w: 2 r+r+ww: 1 r+rww: 2 rrww: 1
phenotypes of the F2: red,normal: 9 red,vestigial: 3 white,normal: 3 white,vestigial: 1
12 Since haemophilia is a sex-linked trait, boys eitherdisplaythediseaseordonotpossesstheallele.Girlsmaybecarriers,i.e.appearhealthybutbeabletopassonthetrait.
EdwardVIIwasQueenVictoria’soldestson.Hedidnotdisplayhemophiliasodidnotpossessthetraitsocouldnotpassitontohischildren.SeveralofVictoria’sdaughtersprobablywerecarriersandpassedthetraiton to their offspring. Males would display the trait,e.g.Alexis,sonofNicolasIIofRussiaandAlexandra(granddaughterofQueenVictoria).Daughterscouldbecarriers.
13 (a)
possible phenotypes corresponding genotypes
grey, straight g+g+s+s+, g+g s+s+, g+g+s+s, g+g s+s
grey, curly g+g+ss, g+g ss
black, straight gg s+s+, gg s+s
black, curly gg ss
Aheterozygousgrey,straightwingedflyhasgenotypeg+gs+s.
Ablack,curlywingedflyhasgenotypeggss.
P : grey, straight black, curly phenotypeg+g s+s x gg ss genotypeg+s+, g+s, gs+, gs gs possible gametes
F1 g+g s+s g+g ss gg s+s gg ss genotypesgrey straight grey curly black straight black curly phenotypes
(b) Theexpectedratiosforthephenotypeswouldbe grey straight: grey curly: black straight: black
curly=1:1:1:1(c) As the ratios found show a much higher
occurrenceoftheparentalphenotypes,itsuggeststhattheLawofindependentassortmentdidnotapply here, which suggests that the traits arelinked.
14 (a) C=coloured,c=albino;G=grey,g=black
possible phenotypes corresponding genotypes
coloured grey CCGG, CCGg, CcGG, CcGg
coloured black CCgg, Ccgg
albino ccGG, ccGg, ccgg
Parent1:homozygousrecessivealbino:ccggParent2:homozygousgrey:CCGG
P: albino x grey phenotypeccgg CCGG genotypecg CG gametes
F1: CcGg genotype grey phenotype CG Cg cG cg gametes
AllF1havethephenotypegreyandthegenotypeCcGg.
(b&c) ThepredictedratiosofthegenotypesoftheF2 can be found using a Punnett square. TheresultswillbeCCGG: 1 CcGG: 2 ccGG: 1CCGg: 2 CcGG: 4 ccGg: 2CCgg: 1 CcGG: 2 ccgg: 1
(d&e) ThepredictedratiosofthephenotypeoftheF2canbefoundbyusingaPunnettsquare.Theyaregrey:black:albino=9:3:4
Normally, the phenotypic ratio of this kindof dihybrid cross is 9:3:3:1 but in this case the“albinogrey”and“albinoblack”arebothalbinoand therefore have the same phenotype. Hencethenumbersareadded,givingtheaboveratio.
Answers
385
Chapter 11
1D,2B,3C,4A,5B,6C,7C,8A,9B,10D,11B,12B,13A,14C,!5C,16D,17C,18D,19B,20A,21D,
22 (a) Helper T cells will form a clone and select andactivatethecorrectBcells.
(b) B cells divide to form clones which produceplasma cells and memory cells. Plasma cellsproduceantibodiesagainsttheantigendetected.Memorycellsremainpresentaftertheinfectionhaspassedinordertospeeduptheproductionofantibodiesincasetheantigenisdetectedagain.
23 (a) Benefits:totaleliminationofthedisease(e.g.smallpox)preventionofpandemicsandepidemicsdecreasehealthcarecostsprevent harmful side effects of diseases (e.g.paralysisafterpolioorproblemswitheyesightof the child after rubella infection of themother.
(b) Dangers:some vaccines contained the preservativeThimerosal which contains mercurywhich, at high levels, causes damage to thebrain, especially in babies and infants. Thepreservativewasneededtomakesurethatnoother pathogens would grow in the solutionwith the vaccine. No evidence was found ofharmfuleffectsofmercuryinvaccinationsbutasaprecaution,itisnowusedlessandless.it ispossiblethatanoverloadoftheimmunesystem, e.g. caused by vaccinations againstmany different diseases in a short time, maymake the vaccination less effective and/orcauseotherproblems. IthasbeenspeculatedthatGulfWarsyndromemayhavebeencausedby the many vaccinations to protect soldiersfromagentsofbiologicalwarfare.apossiblelinkbetweenMMRvaccinationandanincreasedchanceofautism;studiescarriedoutsincehavefailedtoconfirmthislink.
24 When a nerve impulse arrives at the muscle, thedepolarisationofthemotorendplateispassedontothesarcoplasmicreticulumwhichcausesittoreleasecalciumions(Ca2+)intothesacroplasm.Thecalciumions attach to the troponin which is attached to thetropomyosin.Thisuncovers thebindingsitesontheactin for the myosin hooks. The muscle will nowcontract.
••••
•
•
•
(a) Ca2+isreleasedfromthesarcoplasmicreticulumintothesarcoplasm.Thecalciumionsattachtothetroponinwhichisattachedtothetropomyosin.
(b) Whentroponinbindstocalciumions,itchangesitsshape.Astroponinisattachedtotropomyosin,this then makes the tropomyosin move anduncoversthebindingsitesonactinforthemyosinhooks.
(c.) Hooksonmyosinwillattachtothebindingsitesonactin.Thehookswillthenreleaseandrepeattheactionfurtherdowntheactin.Thisiscalledtheratchetmechanism.
(d) ATPishydrolysedtoADP.Inthisprocess,energyisreleasedwhichdrivestheratchetmechanism.
25 • muscles only contract when they receive animpulsefromanervenervecellscarrydepolarisationstomusclesimpulsecausesreleaseofcalciuminthemusclemusclecontractsmuscleisattachedtoatleast2bonesviatendonsbones are attached to each other via joint andligamentsleverage of bone will make the distance ofmovement of the bone more than the distancethemusclecontractedbonesmoverelativetoeachother(usingjoints)different joints allow different kinds ofmovementantagonisticmusclescauseoppositemovement
26 Similarities: All blood cells and proteins should remain in the
blood. Since glucose and proteins will not be excreted, the
amounts should remain the same. However, it ispossiblethattheconcentrationhasincreasedifwaterhasbeenremoved.
Differences:
renal artery renal veinurea more less
oxygen more less
carbon dioxide less more
salt usually more usually less
hormones more less
toxins (e.g. medicin) more less
27 (a) oogenesis is the process of forming ova (eggcells).
(b) mitosis occurs in the germination epithelium.Onecellmitoticallydividesintotwo,oneofthem
•••••
•
••
•
386
Answers
proceeds with oogenesis (the primary oocyte),the other remains as germination cell and canhaveanothermitoticdivision.
(c) meiosis reduces the number of chromosomesin the gamete. Since a gamete needs to fusewith another gamete to start a new individual,each gamete must have half the number ofchromosomessothattheneworganismhasthesamenumberaseitherparent.
(d) spermatozoaareverysmallandthe4spermatidsthatresultfromMeiosisIandIIcanallbecomespermatozoa.Ovaareverylargesincetheyneedtocontainreservefoodforthezygote.IfeachofthecellsproducedinMeiosisIandIIwouldbean ovum, then each ovum would only containone quarter of the reserve food of the originalprimaryoocyte.Asitis,twoorthreepolarbodiesare produced to reduce the amount of geneticmaterialbutalmostall thecellmaterialgoes tooneovum,givingitthebestchancetogrowouttoanewindividualifitisfertilised.
28 (a) HCGissecretedbythetrophoblasticcellsofthedevelopingembryo
(b) HCG maintains the corpus luteum whichproducesprogesterone
(c) Atalaterstage,theplacentawillstarttoproduceprogesterone so thecorpus luteumno longer isneeded.
(d) ObtainmonoclonalantibodiesagainsttheHCG(human chorionic gonadotropin). Fix them inplace on a testing stick/strip. Add urine to thetesting stick/strip. If the HCG is present in theurine(asitwillbeifthewomanispregnant),itwillattachtotheantibodies.Thetesthasbeensodesigned that this will give a colour showing apositivetest.
Chapter 12
1 • notenoughproteinsinthebloodtissuefluidisnotreturnedtothebloodfluidretentionretardationofmentalandphysicalgrowth.
2 (a) PKUisphenylketouria.(b) PKU is a genetic disorder. PKU patients lack
the gene to produce functioning phenylalaninehydroxylase.
(c) Phenylalaninehydroxylasechangesphenylalanineintotyrosine.Withoutthisenzyme,phenylalanineaccumulates and is changed into phenylpyruvicacidwhichcausesretardation.
•••
(d) PKUcausessevereproblemsunlessitisdiagnosedearlyanda specialdiet is followed.So inmanycountries all babies are screened within hoursafterbirth.
(e)Thedietfocussesonavoidingmostproteinsincelevelsofphenylalanine(anaminoacid)mustbekeptlow.Withlowlevelsofphenylalanineinthediet,theaffectedchildrenleadnormallives.
3 (a) antioxidantactivity collagensynthesis roleinimmunesystem reducechanceofformingarterialplaque(b) CurrentFDArecommendationis70mg/day.(c) megavitamin therapy of vitamin C may reduce
chanceofrespiratoryinfectionand/orcancer.(d) reboundmalnutritioniscausedbytakinglarger
dosesofe.g.vitaminCforanextendedtime.Theorganismwillexcretethesurplusandiflevelsarethenreducedbacktonormal,itmaytaketimetostoptheexcretionsothattheorganismactuallysuffersfromalackofvitaminC.
4 (a) sex (male/female); level of activity; age (whichalsorelates toactivity, sizeandgrowth); specialcircumstancese.g.diseaseorpregnancy/nursing.
(b) sex:malesneedmoreenergythanfemalesbecausemalestendtobelargerandfemalestendtohavemoresubcutaneousfat(lessheatlost).
level of activity: the more active a person, themoreenergyisneeded.
age:youngerissmallersmallerpeopleneedlessenergy but (fast) growing people need moreenergywhereasolderpeopleareoftenlessactivesoneedlessenergy.
special circumstances: recovering from illness make require extra
energy pregnancy/breastfeeding will require extra
energy
5 • fibreisnotdigestedcausesslightirritationofthewalloftheintestineincreasesperistalsismovesfoodfasterthroughintestinesofterstools(feces)reducedconstipationandhemaroidsmayhelpweightlossaspeoplefeelfullsooneronhighfibredietmayreducecholesterollevelsbutnotproven
••••••
•
Answers
387
6 (a)
Human milk Artificial milk
monosaccharide rich in lactose no or little lactose
protein more whey, less casein
less whey, more casein
antibodies present absent
fats rich in omega-3 fatty acids (DHA)
some brands add DHA but usually less than in human milk
enzyme contains lipase no lipase
minerals most iron absorbed most iron not absorbed
(b) advantages of breastfeeding:breastmilkcontainsantibodieswhichhelpthebabyfightinfectionsproteinsinbreastmilkareeasierdigestedthanthoseinartificialmilkcalcium and iron in breast milk are betterabsorbedbreastfeedinghelpstheuteruscontractsothatthebleedingafterdeliverywillbeshorterbreast feeding reduces the risk of breastcancerbreast feeding improves the bond betweenmotherandchildbreast feeding ischeaper-very important insomepartsoftheworldmotherusuallydoesnotovulatewhilebreastfeeding(naturalcontraception)
disadvantages of breast feedingifthemothertakesmedicationforanillness,itmayalsobefoundinbreastmilkbreast feeding requires more strength fromthebabythanbottlefeedingsoweak/illbabiesmaynotbeabletogetenoughmilktheremaybeanincreasedchanceofpostnataldepressionbutthisisnotproven
Chapter 13
1 • musclesneedenergytocontractglycogencanbebrokendowntoglucoseglucosecanbeusedtoreleaseenergy/ATPinaerobicandanaerobicrespirationaerobicrespirationreleasesmostenergy/ATPandtakesplaceinmitochondriaifsufficientoxygenisavailable
•
•
•
•
•
•
•
•
•
•
•
••••••
2 • myoglobinisproteinmadeofonechainofpolypeptideabletobindoxygenstoredinmusclesmyoglobin has greater affinity for oxygen thanhaemoglobinmakesmyoglobinstoreoxygenoxygen released from myoglobin when allavailableoxygenusedup.
3 (a) Benefitsincreased amount of oxygen can be taken tomusclesso more aerobic respiration/less anaerobicrespirationmoreenergyyieldedpermoleculeofglucosemore energy available/energy available forlongertimeless lactate produced - lactate can lead tofatigueandmusclecrampmorechancetowin
personalachievement/honour teamachievement/honour prize money / scholarship / commercial
opportunities Risks
tobedisqualilfiedand/orprosecutedknowledge that you gained an unfairadvantageanabolic steroids: liver damage, fertilityproblems,femalesbecomemoremasculinblooddoping:moreviscousblood-heartfailure,strokeimproperlystored-infectionusingdonorblood-transmissionofe.g.HIVorhepatitisEPO: more viscous blood - heart failure,stroke
4
Sprains overstretching of a ligament in a joint
Torn ligaments overstretching of a ligament causing rupture
Torn muscle most severe case of muscle strain; muscle fibres may tear
Dislocation of joints the bones in the joint are moved, relative to each other, and do not return to their original position
Intervertebral disc damage
one of the discs may may buldge out and put pressure on the nerves in the spinal cord.
••••
••
•
•
••
•
•
••
•
••••
•
388
Answers
Chapter 14
1D,2A,3D,4A,5D,6A,7D,8D,9D,10D,
11 • primarystructure:covalentbondssecondarystructure:hydrogenbondstertiary structure: hydrophobic interactions,disulfidebonds,hydrogenbondsquaternarystructure:hydrogenbonds,positive/negative attraction forces, hydrophobic forces,disulfidebridges
12 • enzymese.g.maltasewhichbreaksdownmaltoseintoglucoseandfructosehormonese.g.insulindefensee.g.antibodies/immunoglobinsstrucruree.g.spindlefibreincelldivisiontransporte.g.hemoglobinforoxygentransport
13 • theyareusedtokeeptheproteininitsplace:polaraminoacidsinteractwithwateroneithersideofthe membrane, non-polar amino acids interactwiththenon-polar lipidbilayerinthecentreofthemembraneahollowtubecanbecreated,usingpolaraminoacidsat theends, interactingwith thewateroneither side of the membrane. In the centre, theinsideofthecylinderismadeofnon-polaracids,creating a hydrophilic channel through themembrane.polarandnon-polaraminoacidscanhelpcreatetheshapeoftheactivesiteofanenzymeandhelpprovidetheforcesthatkeepthesubstrateintheactivesiteandmakesurethatothersubstratedonotfitwellanddonotstaythere.
14 (a) attheactivesite(b) anywhere on the enzyme as long as it is away
fromtheactivesite(c) competitiveinhibitor:prontosil non-competitiveinhibitor:cyanide
15 • allosteryisnon-competitiveinhibitionendproductistheinhibitorendproductwillattachtotheenzymeofanreactionearlierinthemetabolicreactionsataplacewhichisNOTtheactivesitewillchangetheshapeoftheactivesiteenzymenolongerworksnomoreproductproducedwhenproductrunsoutthennomoreproducttoactasallostericinhibitorsoreactiomproceedsnewproductformed.
••
•
••••
•
•
•••••••••••
16 • ifreactionisA+B→CAandBmayneedtocollidewithahighamountofenergyinordertohavereactiontakeplaceonlyfewparticleshavethisenergyincreasingtemperaturewillincreasethenumberofpartcipleswithsufficientenergybut this could denature proteins in a livingsystemwhensubstratesmeetattheactivesitethe conditions are such that it is easier for thereactiontotakeplacelessactivationenergyisneededmoreparticlescanreactatthesametimerateofreactionincreases.
17 (a) inthecytoplasm(b) inthematrixofthemitochondria(c) on the cristae (inner membranes of the
mitochondria)
18 RefertoFigure804
19 (a) Thesite for theKrebscycle is thematrixof themitochondria. It is a watery fluid contaning allthe enzymes and compounds needed for theKrebscycletoproceed.
(b) The site for the electron transport chain isthe cristae (folded inner membrane) of themitochondrion. It has a large surface area sothere isa lotofroomformanyATPsynthetasemolecules. It is also impermeable to protons(exceptwherethereareATPasemolecules).
(c) Thenitwouldnotbepossibletobuildupahigherconcentration of protons and chemiosmosiswould not work. As a result, no ATP would beproduced.
20 RefertoFigure809.
21 RefertoFigure810. The light dependent stage produces ATP and
NADPHwhichareneededtodrive theCalvincycleinthelightindependentstage.Thelightindependentstage fixes carbon dioxide to produce a triose. ThisprocessrequiresATPandNADPH,madeinthelightdependentstage:ATPisneededtophosphorylateGPand to change TP into RuBP. NADPH is needed toreduceglyceratediphosphateintoanthertriose.
22 (a) theyprovidetheenergyandthereducingpowertodrivetheCalvincycleinthelightindependentstage.
(b) itisasimplemethodofproducingATP
•
•••
•
••
•••
Answers
389
(c) non-cyclic photophosphorylation producesNADPH which is needed to drive the lightindependent stage. The advantage of the lightindependentstageisthatitmakescarbohydrateswhichcanbeusedforlongtermenergystoragebutalsoasbuildingmaterialsfornewcellssothattheplantcangrowandreproduce.
(d) intheCalvincycle,3moleculesofcarbondioxidearecombinedtoformatriosemolecule;2triosemoleculeswillbecombinedtoformglucose.
23 The grana thylakoid provide a large surface forabsorbinglightthemembraneofthegranaallowstheelectron carriers needed for photophosphorylationtobekept intherightorder,helpingthereactiontoproceed.Thevolumeinsidethegranaissmallsothatarelativelysmallnumberofprotonswillcausealargedifferenceinconcentrationacrossthemembranethestroma is awateryfluidwhichcontainsall enzymesandintermediatesneededfortheCalvincycle.
24 Similarities:bothusechemiosmosisbothinvolvetransportofelectronsandprotonsboth use ATP synthetase to produce ATP fromADPandPiboth take place across a membrane which isimpermeabletoprotonsboth buld up a proton gradient across themembranebothtransportprotonsintoasmallspace
Differences:
chloroplasts mitochondria
site membrane of granum membrane of crista
organelle chloroplast mitochondrion
source of electrons water (photolysis) NADH + H+
ultipmate electron acceptor
NADP oxygen
protons are pumped into
lumen of the grana intermembrane space
protons diffuse via ATPase into
stroma matrix
Chapter 15
D1 ORIGIN OF LIFE ON EARTH
1. A MillerandUreyplacedmethane,ammonia,waterand hydrogen into their apparatus to stimulatethehydrogen-richreducingatmospherebelievedtohavebeenpresentonthepre-bioticEarth.
•••
•
•
•
2. D An endosymbiont is an organism that has amutallybeneficial symbiotic relationshipwithahostorganismwhilelivinginthehost.Eukaryotesprobably evolved when larger prokaryotesabsorbed smaller prokaryotes and formed asymbioticrelationship.
3. B Current living organisms on the Earth’s surfaceare protected from the harmful and damagingeffectsofultravioletradiationbytheozonelayerin the upper atmosphere. The ozone moleculesabsorbsolarultravioletradiation.
4. A The early atmosphere of the pre-biotic Earth isthoughttohavebeenreducinginnatureduetotheabsenceofoxygen.Itisbelievedthatoxygenonly entered the atmosphere following theappearanceofphotosyntheticprokaryotes.
5. A The first organic molecule able to replicate onthe pre-biotic Earth is currently believed bymany Molecular Biologists to have been RNA.This hypothesis has been strengthened by thediscovery of ribozymes which consist of shortlengthsofcatalyticRNA.TheribosomemayalsoemploycatalyticRNAduringproteinsynthesis.
6. C The pre-biotic atmosphere is thought to beenreducinginnatureandcharacterisedbyhydrogenand carbon rich molecules. The presence ofoxygen would have made the atmosphereoxidisinginnatureandpreventedthepre-bioticsynthesisreactions.
7. C The endosymbiotic theory suggests that theancestorsofsomeeukaryoticorganelles,notablythechloroplastandmitochondria,wereoriginallyfree-living prokaryotes that entered into asymbiotic relationship with another prokaryotetoformanancestraleukaryotecell.
8. D There is no atmosphere or liquid water on theMoonandthedailyrangeinsurfacetemperaturesistoohightosupportlife.
9. D A common feature of all living organisms,including viruses, is that they contain nucleicacids(DNAand/orRNA).
10.A Oxygenwasprobablynotpresent in theEarth’searly atmosphere. The majority of the oxygenpresentinthecurrentatmospherewasgeneratedby photosynthesis. The Earth’s early pre-bioticatmosphere probably originated from volcanicgases-freeoxygengaswasnotpresent.
390
Answers
11.B The current estimate of the Earth’s age, basedupon dating of the oldest rocks, is about 4.5billionyearsago.
12.B Claysmayhavecatalysedtheformationoforganicpolymers,suchasproteinsandnucleicacids,onthesurfaceoftheirlattices.
13.D The observation that DNA in the eukaryoticnucleus codes for many of the enzymes inmitochondria is not strong and direct evidencefor the role of endosymbiosis in the origin ofeukaryotes.(However,itcouldbearguedthatovertimemanymitochondrialgenesweretransferredtothenucleus).
14.B Proteinoid microspheres are only formedwhen amino acids are heated at relatively hightemperaturesforaprolongedperiodoftime.
15.D This is a description of the endosymbiotictheory.
16.A Prokaroytes do not possess lysosomes or Golgiapparatus.Theirdigestiveenzymesarepresentinthecytoplasm.
17.C ThetheorythatlifeonEarthwasinitiatedbythearrival of organic molecules and, perhaps evenbacteria on asteroids and meteorites, is termedpanspermia.
18 Both mitochondria and chloroplasts can arise onlyfrompre-existingmitochondriaandchloroplastsbyaprocessofbinaryfission.(Theycannotbeformedinacellthatlacksthembecausenucleargenesencodeonlysomeoftheproteinsofwhichtheyarecomposed).
Both mitochondria and chloroplasts have their owngenome and it resembles that of prokaryotes notthatofthenucleargenome:bothgenomesconsistofa single circular molecule of DNA and there are nohistonesassociatedwiththeDNA.
Both mitochondria and chloroplasts have theirown ribosomes (70S), which closely resemble thatof prokaryotes not the 80S ribosomes found in thecytoplasmofeukaryotes.
Anumberofantibioticsthatactbyblockingproteinsynthesis in bacteria also block protein synthesiswithin mitochondria and chloroplasts. They do notinterfere with protein synthesis in the cytoplasm oftheeukaryotes.
Conversely, inhibitors of protein synthesis byeukaryotic ribosomes do not have any effect onbacterial protein synthesis nor on protein synthesiswithinmitochondriaandchloroplasts.
19 MillerandUreyusedarefluxapparatustore-circulatewater vapour and a mixture of methane, ammoniaandhydrogen(believedtoreflectthecompositionoftheearlyatmosphere)throughachamberwheretheywereexposedtoacontinuoushighvoltageelectricaldischarge (spark) that stimulated lightning. AfterseveraldaysMillerandUreyanalysedthemixtureanddetectedanumberofaminoacids.Laterexperimentsinvolvinghydrogencyanideandammoniaresultedintheformationofthebaseadenine.
20 Panspermia is a theory or hypothesis, that suggeststhat the ‘seeds’ of life are prevalent throughout theUniverse and life on Earth began by such ‘seeds’landingonEarthandreplicating.Onelineofevidencecomesfromresearchthatshowstherearemanymorepotential habitats for life than Earth-like planets.There is someevidence tosuggest thatbacteriamaybeabletosurviveforverylongperiodsoftimeevenindeep space (andmay thereforebe theunderlyingmechanism behind Panspermia). Bacteria and morecomplexorganismshavebeenfoundinmoreextremeenvironments, such as black smokers or oceanicvolcanic vents. Some strains of bacteria have beenfoundlivingattemperaturesabove100°C,othersinstronglyalkalineenvironments,andothersinextremeacidic conditions. Semi-dormant bacteria have beenfoundinicecoresoveramilebeneaththeAntarctic- this lends credibility to the concept of sustainingthecomponentsof lifeonthesurfaceof icycomets.ThepresenceofpastliquidwaterontheplanetMars,suggestedbyriver-likeformations,wasconfirmedbytheMarsExplorationRovermissions.PossiblewateroceansonEuropa,oneofSaturn’smoons,andperhapsothermoonsintheSolarsystem.
21 Ribose sugars (found in RNA) are much easier tosynthesis under simulated pre-biotic conditionsthanaredeoxyribosesugars(foundinDNA).Singlestranded RNA, unlike single stranded DNA, canform, via hydrogen bonding a variety of complexthree-dimensionalconfigurations.ItissuspectedthatribosomalRNAplaysanactiveroleinthesynthesisofproteins in ribosomes. Several RNA sequences havebeendemonstratedtobecatalytic, forexample,self-splicingintrons–so-calledribozymes.
D2 SPECIES AND SPECIATION
1. C Thefinches,althoughoriginatingfromthesameancestral species, evolved separately due to thedifferenthabitatscolonised.Aprocessofadaptiveradiationoccurredresultingintheformationof
Answers
391
thethirteennewspecieseachadaptedtoauniquehabitatontheisland.
2. D The relatively sudden appearance of a numberofnewspecies,withoutintermediates,istermedpunctuatedequilibrium.
3. A Natural selection is slowest when migration isabsent (nonewvariation is introduced into thepopulation);whentheselectionpressureislowtherateofevolutionwillbelowandwhenvariationduetogenemutationislow,consequentlytherewillbelittlegeneticvariationfornaturalselectiontoacton.
4. B Alltheallelespresentinthegametesofasexuallyreproducing population are known as thepopulation’sgenepool.
5. B BirdsAandBsuccessfullyinterbreedandproduceviableoffspring.ThisistheBiologicaldefinitionofaspecies(thatreproducessexually)
6. C The term ‘Gradualism’ in Darwin’s theory ofevolution refers to the emergence of complexBiological adaptations, for example, the eye,evolvedinmanysmallincrementalsteps.
7. D Members of the same species do vary inphenotype, frequently, two or more varieties ormorphscanbeidentified.
8. B Reproductive barriers prevent different speciesfrom successfully interbreeding. The variousreproductive barriers which isolate species canbeclassifiedaseitherprezygoticorpostzygotic.Prezygotic barriers are mechanisms whichprevent the fusion of the ovum and the spermsothatnozygotecanform.Postzygoticbarriersare mechanisms that prevent a zygote fromdevelopingintoafertileadultoffspring.
9. A The periodic events in which extinction ratesincreasedramaticallyaretermedmassextinctionevents.Theymaybecausedbyvolcaniceruptions,meteorimpactsand/orlongtermchangesintheglobalclimate.
10.A Adaptive radiation is the production ofecologically diverse species from a commonancestral stock or population, for example,Darwin’sfinches.
11.B Thetermmicroevolutionreferstoanincreaseordecreaseofallelefrequencieswithinagenepool.
12.C One form of reproductive isolation is spatialisolation. If members of two populations neverencounter each other, they willnever mate andnogeneflowwilloccur.
13.A Punctuated equilibrium does not invoke largeevolutionarystepsviamacromutationswithlargeeffectsonthephenotype.
14.D Behavioural isolation involves geneticmodifications to behaviour that lead to lackof mating between two groups of a species. Itscontinued presence may eventually result inspeciation.
15.A A pre-zygotic isolating mechanism preventsfertilisation (the fusionof the spermandovumnuclei).
16.D Transientpolymorphismisbeingexhibited:thereisclearlynobalancingselectionoccurringsincethere is random mating and both morphs areexhibitingequalreproductivesuccess.
17.D The maintenance of the frequency of bothdominantandrecessivegenesoveranumberofgenerations by selecting for the heterozygotes.Balancedpolymorphismexplainsthediscrepancyinnumberofindividualswithsickle-celldiseasecomparedtosickle-celltraitinAfricacomparedtoNorthAmericanblacks.
18.C Theincreaseinthefrequencyofthedarkmelanicform of Biston betularia is thought to be theresultofnaturalselection.Thedark formswerebetter camouflaged on the soot covered trunksand thus prone to less predation by birds. Themutation rate would have been approximatelyconstantduringthisperiodoftime.
19 (a) 2.6%=2.6/100=0.026(b) p2 + 2pq + q2=1LetqrepresentthefrequencyoftheHbSalleleq2=0.026;q=√0.026=0.161
(b) LetprepresentthefrequencyoftheHbAallelep+q=1;p=1–0.161=0.839Hence,frequencyofheterozygotes=2pq=2x0.839x0.161=0.270Numberofheterozygotesinpopulation
=600x0.270=162.(c) Balancedpolymorphismisbeingexhibited.(d) The force responsible for maintaining this
polymorphismistermedheterozygoteadvantage–theheterozygoteisfitterthaneitherhomozygote,sinceitisimmunetomalaria.
392
Answers
20 (a) LetprepresentthefrequencyofM;p=0.5andletqrepresentthefrequencyofm;q=0.5.
At equilibrium the genotype frequencies willbe 0.25 MM, 0.5 Mm and 0.25 mm, accordingto the Hardy-Weinberg equation. Therefore thephenotypicratiois0.75melanic:0.25pale.
(b) 500mmmothswillbeeatenleaving500MMand1000mm.
Numberofallelesremaining=(2x1500)=3000 NumberofMalleles=(2x500)+1000=2000 Numberofmalleles=1000 FrequencyofM=2000/3000=0.667 Frequencyofm=1000/3000=0.333(c) Outof2000moths,500mmwillsurvive.There
are no melanic forms, so the frequency of m =1.0.
(d) mandM.(e) Transientpolymorphismisbeingexhibited.
D3 HUMAN EVOLUTION
1. B A. erectus is not a species of Australopithecus.Homoerectusisinadifferentgenusandspecies.
2. C Anatomicalandbiochemicalevidence,especiallymaternalmitochondrialDNAsequenceanalysis,strongly suggest that modern man evolved inAfrica.Theso-called‘OutofAfrica’hypothesis.
3. A The species of Homo in order of evolutionaryappearance are: H. habils, H. erectus, H.neanderthalensisandH.sapiens.
4. A In modern humans the head is not kept inpositionbypowerfulneckmuscles.
5. A Culturalevolutionislearnedbehaviourwhichispassed from generation to generation via non-genetic processes, for example, transmissionof knowledge via books and the passing onof religious beliefs and cultural practices orlanguage.
6. B The first Australopithecine fossils date back toapproximately 5 million years ago. They haveonlybeenfoundinAfrica.
7. C Binocular vision is a characteristic of the orderprimates.
8. D The genus Homo lived with the genusAustralopithecus in East Africa for about 1millionyears.
9. C Homo sapiens made the most sophisticatedtools.
10.D All of the hominid skeletons show evidence ofbipedalismtosomedegree.
11.B Tool use is probably restricted to the genusHomo–Homohabiliswas thefirstmemberofthisgenus.
12.D Humans and apes are presently classified inthe same category at all of the following levelsexceptgenus.Humansbelongintheirowngenus(Homo). Apes are divided among four genera.Humans and apes belong in the same Order:Primates, Class: Mammalia, Phylum: Chordata,Kingdom:Animalia.
13.D Ofthelivinggeneraofapes,theonemostcloselyrelated to humans are the chimpanzees (thatis, the common chimpanzee and the pygmychimpanzeeorbonobo).
14.C Opposablethumbsisafeaturethatsetsprimatesapart from all other mammals. The ability toproducemilkisacharacteristicofallmammalsthatdistinguishesthemfromtheothervertebrateClasses; placental embryonic development isa characteristic of the placental mammals thatdistinguishes them from the marsupials andmonotremes.
15.B Two important characteristics of the firsthominidswerebipedallocomotionandrelativelysmallbrains.
16.C TheoldesthominidfossilshaveallbeendiscoveredinAfrica.
17.B Earlyhominidsdatingtobetween6and3millionyears ago are considered primitive becausethey possessed many ancestral characteristicscommontoallhominids.
18.B The fossil specimen nicknamed “Lucy” is anexampleofAustralopithecusafarensis.
19.D ThefossilfootprintsatLaetolishowclearbipedalcharacteristicswhichinclude:non-divergentbigtoe,heelstrikeandawelldevelopedarchonthefoot.
20.A Thefossilrecordstronglysuggeststhatbipedalismevolved3.5millionyearsbeforetooluse.
Answers
393
21.C Neoteny refers to a delay in the maturationprocess,especiallythatofthecerebralcortex.Itisthereforepossible that intelligenceand learningareenhancedbeforetheorganismreachessexualmaturity.
22.D FossilsofHomoerectushaveonlybeenfoundintropicalAfricaandAsia,forexample,Indonesia.
23.B Culturalevolutionisarapidandpowerfulforce,forexample,thetransmissionofknowledgeandskillsvialearning.
24.C Whenfossilsareunearthed,itisgenerallyfoundthatthebonesorshellsofthedeadorganismhavebeen replaced by hard minerals, for example,calciumcarbonate.
25.B Waterwithdissolvedoxygenisapoorpreservativemedium for the remains of animals and plants.Theseconditionswouldencouragethegrowthofbacteriawhichwoulddigesttheremains.
26.C Thehalflifeofcarbon-14isrelativelyshortitcanonlybeusedtodateorganicremainsthatarelessthan70,000yearsold
27 (a) (i) The genus is Homo; the species isfloresiensis.
(ii) Todemonstrate thatLB1 isnotanaberrantindividual,butwasatypicalmemberofthepopulation.
(b) (i) Dwarfingmaybecommononislandsduetothe relatively limited amount of resources,forexample,foodsupplyandbreedingspaceandlackofpredators.Smalleranimalsmayalsobeabletoachievethermoregulationonsmalleramountsoffood.
(ii) Ageneralincrease.(c) It suggests that how the brain is ‘wired up’,
namely,itscomplexity(thenumberofneuronalsynapses), is perhaps more important than itsabsolutesizeorbraintobodysizeratio.
(d) Theuseoftools.
28 (a) There are fewer complementary base pairs andhence fewer hydrogen bonds to hold the twostrandstogether.
(b) TheDNAofdistantlygeneticallyrelatedspecieshas less complementary base pairs and hencetherewillagreaternumberofmismatchedbasepairs. Fewer hydrogen bonds are formed andhence less heat is required to separate or ‘melt’thestrands.Closelyrelatedspecieswillhavemorecomplementarybasedpairsandhenceagreater
numberofmatchedbasepairs.Alargernumberofhydrogenbondsare formedandhencemoreheatisrequiredtoseparateor‘melt’thestrands.
(c) 30/7.3=4.1millionyears(d) 4.1millionyearsx1.9=7.8millionyears(e) A=gibbons;B=orang-utan
D4 THE HARDY-WEINBERG PRINCIPLE
1. A For a Hardy-Weinberg equilibrium to bemaintained and for evolution not occur, therehas to be a large population, isolation fromother populations (that is, no immigration ormigration), no mutations, random mating andnoselectionpressure.
2. A p2+2pq+q2=1representstheHardy-Weinbergequation.
3. D In the Hardy-Weinberg equation, p representsthefrequencyofadominantallele.
4. B In the Hardy-Weinberg equation, q representsthefrequencyofarecessiveallele.
5. C TheHardy-Weinbergprincipleassumesthatthepopulationisnotexperiencingmutation.
6. B p2+2pq+q2=1;p=0.7;q=0.3;2pq=(2x0.7x0.3)=0.42
7. D Significant deviations from the frequenciespredictedbytheHardy-Weinbergprinciplemayindicate that natural selection is occurring andhence altering allele and genotype frequenciesovertime.
8. D p2+2pq+q2=1;q2=0.09;q=0.09=0.3
9. B p2+2pq+q2=1;q=0.3,hencep=0.7; 2pq=(2x07x0.3)=0.42
10.C TheHardy-Weinbergprinciplecanbeappliedtothreeormorealleles.Forexample,theexpressionwhenappliedtothreeallelestakestheformof
(p+q+r)2,whichwhenexpandedgives p2+q2+r2+2pq+2rq+2rp=1.
11.A Random mating will maintain the proportionof dominant to recessive characteristics in apopulation from generation to generation. ThisisbecausethepopulationisinHardy-Weinbergequilibrium and allele frequencies remainconstantovertime.
394
Answers
12 FrequencyofgenotypeBB=p2=0.82=0.64 FrequencyofgenotypeBb=2pq=2x0.8x0.2=0.32
Frequencyofgenotypebb=q2=0.22=0.04
13 p2=frequencyofAA2pq=frequencyofAa q2=frequencyofaa,whosefrequencyis1in20000q2
=1/20000=0.00005 Therefore,q=√0.00005=0.007p=1–q=0.933 Hence,thefrequencyofheterozygotes =2x0.993x0.007=0.014
14 (a) Letq2representthefrequencyofthehomozygousrecessivegenotype.
q2==0.23 Inmalesq=0.48. q2==0.16. Infemalesq=0.40.(b) Thesamplesofmalesandfemalesaresmalland
thedifferencesincalculatedallelefrequenciesisduetosamplingerror.
15 (a) q2=1/1000 = 0.0001, therefore, q = √0.0001 =0.01
p+q=1;p=1–0.99 Let 2pq represent the frequency of the
heterozygotes. 2pq=2x0.01x0.99=0.0198 2000x0.0198=39.6 Henceabout40peoplein2000wouldbeexpected
tobeheterozygous.(b) Ifthecouplehavealreadyhadonephenylketonuric
child, they must both be heterozygous. Theprobability of two heterozygotes having aheterozygouschildisor50%.
16 (a) The53%resistantratshavethegenotypesRRorRr.The47%non-resistantratshavethegenotyperr.
q2=0.53,thereforeq=0.73;p+q=1, thereforep=1–q;p=0.27 ThereforefrequencyofRallele=0.27.(b) p=0.27andq=0.73;p2=0.073(RR);2pq=0.39
(Rr);q2=0.53 Outof100rats,7areRR,40areRrand53arerr. Non-resistant rats have the genotype rr. If all
theseratswerekilled, therewouldbe7RRand40Rrratsremaining.
(c) 7 + 40 = 47 animals remaining. 2 x 47 = 94alleles.
Numberofralleles=47;frequencyofrallele==0.5.
Substituting q = 0.5 into the Hardy-Weinbergequation,thefrequencyofnon-resistantrats(rr)=q2=0.25.
D5 PHYLOGENY AND SYSTEMATICS
1. C The universality of the genetic code is strongevidenceforasingleorunitaryoriginoflifeonEarth. It is very conserved and arose early inevolution.
2. C Homologyreferstotheappearanceofatraitthatwaspresentinacommonancestoroftwoormorerelatedgroupsoforganisms.
3. D This phenomenon is known as an evolutionaryor molecular clock. It is a consequence of theneutral theoryof evolutionwhichholds that inanygivenDNAsequence,mutationsaccumulateatanapproximatelyconstantrateas longastheDNAsequenceretainsitsoriginalfunction.
4. D Homologous features imply structural anddevelopmentalsimilarities,wherethestructuresconcernedmaynotperformthesamefunction.The human hand and whale fin both share thesamebasicpentadactyllimbstructure.
5. A This suggests that reptiles, birds, and mammalshaveacommonevolutionaryancestor.
6. B The toad, the amphibian, is the most distantlyrelated,inevolutionaryterms,oftheorganisms.The common ancestor of all toads and all theorganismsevolvedbeforethecommonancestoroftheotherorganisms.
7. B The term phylogeny describes the branchingsequenceoftheevolutionbynaturalselectionoforganisms.
Chapter 16
1 (a) retina(b) rodsandcones(c) rods Several rods are linked to one bipolar neuron.
This means that their impulses are ‘added up’and this explains their higher light sensitivity.Howeveritalsoreducesaccuracy.
(d) Rodsandconescontainphotopigmentsthatarebrokendownwhenexposedtolight.Thiscausesthecellstosendanactionpotentialtothebrain.
2 (a) behaviourwhichnormallyoccursinallmembersofaspecies,includinginyoungandinexperiencedanimals.
Answers
395
(b) innatebehaviouriscontrolledbygenes(c) taxesandkineses Euglenagoingtowardsthelightisaphototaxis. Woodlicemovingfasterandturningmoreoften
isakinesis.
3 • learningmaybenefitindividualand/orgroupanimalscanlearnhowtocatchfoodefficientlyoravoidfoode.g.bearcatchingsalmonandbirdavoidingtoeatwaspswhichmakesitmoreefficientwarningcallsincreasesafetyofallindividualsinthegroupe.g.monkeysothercooperativebehavioure.g.huntingofdolphins
4 • psychoactive drugs influence the brain byaffectingitattheleveloftheneurotransmitters.They can increase or decrease postsynaptictransmission.psychoactivedrugscancausedependenceexcitatory drugs work by releasing theneurotransmittor,mimicingitseffectorreducingordelayingthere-uptakeoftheneurotransmittor,prolongingitseffect
nicotine stimulates receptors that respond toacetyl choline causes increase in adrenalinemakingpeoplefeelmorealert
cocaineblocksthere-uptakeofdopaminecausingpeopletofeelhappy
amphetamines release dopamine causingimprovedconcentrationandperformanceinhibitory drugs block neurotransmittorreceptors
benzodiazepines increases the inhibitory effectofGABAslowdownthebrainandcalmpeopledown
alcohol increases the inhibitoryeffectofGABAandreducesinhibitions
tetrahydrocannabinol THC blocks cannaboidreceptorsandcauserelaxationandpainrelief
5 (a) The sympathetic nervous system is part of theautonomous nervous system and not undervoluntary control. The sympathetic nervoussystemisinvolvedinactione.g.fright,fight,flightresponse
(b) Itincreasesheartrate,contractstheradialmusclesoftheirisandreducethebloodflowtothegut.
(c) Increasingheartratewillpumpmorebloodroundthebody,alsotothemuscles.Bloodcarriesfoodandoxygen,neededtoreleaseenergybycellularrespiration. Contracting radial muscles in iris
•
•
••
•••
•
••
•
willincreasesizeofpupilwhichwillallowmorelightintotheeyeandbettervision.Bloodwillberedirectedfromtheguttoe.g.muscles,increasingspeedorstrength.
6 (a) Altruism is behaviour which benefits anotherindividualbutatcosttotheperformer
(b) (i) Aworkerbeebeingkilledwhiledefendingthehive
(ii)A male blue throated side blotched lizardsformpartnerships;occasionallyoneofthemspendssomuchtimedefendingtheterritorythat he does not mate - the other one canmatebecausetheterritoryisdefended
(iii)Vampirebatsregurgitatingbloodforanotherbatthatdidnotfeedforsomedays
(c) Geneticallyrelatedindividualsbenefitifanotherindividualcarryingsomeoftheirgenesincreasesitschancesofsurvival.
Reciprocal altruism in bats: individuals benefitfrom blood regurgitated by others only if theythemselvesalsofeedotherswhenneeded
Chapter 17
F1 DIVERSITY OF MICROBES
1.D,2C,3B,4D,5A,6A,7D,8B,9D,10D,11D,12A,13C,14B,15C,16C,17C,18B,19B,20A,21D,22A,23D,24B,25D,26B,
27 (a) RefertoFigures215and1707intextbook(b) Gram negative bacteria have a much thinner
cellwallthanGrampositivebacteria.Hence,thecellwalldoesnotretainthestain(crystaliodinecomplex).
(c) EscherichiaandSalmonella.(d) Both gram negative and gram positive bacteria
both exhibit quorum sensing. Briefly outlinewhathappensduringthisprocess.
(e) The membrane lipids are composed of straightcarbon chains attached to glycerol by an esterlinkage (-CO-O-); histones and introns areabsent.
F2 MICROBES AND THE ENVIRONMENT1A,2C,3C,4A,5C,6C,7B
8 RefertoFigure1726inthetextbook
396
Answers
F3 MICROBES AND BIOTECHNOLOGY1A,2B,3D,4A,5B,6D,7D,8A,9A,10B
F4 MICROBES AND FOOD RODUCTION1B,2D,3D,4B,5A
F5 METABOLISM OF MICROBES1B,2A,3C,4D,5A
F6 MICROBES AND DISEASE1A,2C,3C,4A,5B,6B,7C,8C,9C,10C,11C,12D,13A,14A,15C,16A,17C.
Chapter 18
1 SimilaritiesBoth
arelongtermrelationshipsbetweenorganismsofdifferentspecieswhereatleastoneofthembenefits
Differences:
parasitism mutualism
one organims benefits, the other is harmed
both organisms benefit
e.g. athlete’s foot: fungus growing on damp skin; fungus obtains nutrients from host which loses nutrients
e.g. algae and fungus in lichenalgae photosynthesis, fungus takes up water and minerals
e.g. fleas on a dogfleas drink blood from dog which loses blood
sea anemone and clown fishclown fish chases away fish that eat anemones; tentacles of anemone protect clownfish
2 (a) grossproductionistheamountoforganicmatterproducedbyphotosynthesisinplants.
(b) net production is gross production minusrespiration.
(c) diversityincreasesassuccessionproceedsuntilamaximum, then it reduces some until a climaxcommunity is reached. Production increasesuntilamaximumisreachedandstabilizes.
•••
3 (a)
SITE 1
n n-1 n(n-1)
water beetle 19 18 342
water snail 66 65 4290
mosquito fish 19 18 342
leech 11 10 110
diptera larvae 11 10 110
bivalve 786 785 617010
Ostracada 5 4 20
tadpoles 8 7 56
true worms 74 73 5402
sum 999 627682
SITE 2
n n-1 n(n-1)
Ostracada 16 15 240
tadpoles 32 31 992
true worms 39 38 1482
water hoglouse 157 156 24492
swimming mayfly nymph 102 101 10302
dusky mayfly nymph 16 15 240
non biting midge larva 614 613 376382
‘Hawker’ dragon fly 16 15 240
sum 992 414370
site 1 site 2total number of individuals at each site N
999 992
N-1 998 991
N(N-1) 997002 983072
sum of n(n-1) for all species at one site
627682 414370
simpson’s index 1.588 2.372
(b) Simpson’s indexforsite2 isbigger thanforsite1.Thismeansthatthediversityatsite2isgreaterthanatsite1despitethefactthatsite2hasfewerspecies. The smaller divesity index at site 1 iscausedbythefactthatonlyonespeciesispresentinlargenumberswhileinsite2morespecieshavealargernumberofindividuals.
(c) Site 1 could be more polluted than site 2.Only some species are capable of competingsuccessfully in a polluted environment so thediversityofapollutedareaisgenerallylessthanthatofanunpollutedarea.Ifbothsitesarefrom
Answers
397
thesamestream,itislikelythatsite2washigherupstreamthansite1.
Another possiblity is that succession hasproceededfurtheratsite2thanatsite1,unlesssite1wastheclimaxcommunity.
4 (a) in situ: nature reserves e.g. Great Barrier ReefMarine Park in Australia and YellowstoneNationalParkintheUSA.
ex situ: zoos,e.gLondonZoo botanicgardense.g.NewBotanicGardenofthe
UniversityofHelsinkiatKumpulaManor seedbankse.g.MilleniumSeedBankProjectin
Kew(UK)(b) benefit of in-situ
some species are very hard to breed incaptivitythepopulationremainsadaptedtoitsoriginalhabitat,e.g.continueitnaturaldietindividualsmaintaintheirnaturalbehaviourspeciesinteractswithothersandfulfillsitsroleintheecosystemthehabitatremainsavailablefortheendangeredspecies(andothers)itrequiresalargergenepoolbutthenconservesthisvariationbetweenindividualsspeciescontinuestoevolveindividuals can have more space, e.g. for aterritoryindividualsmaynotneedtobetransported
uses of ex situprotectindividualsofanearlyextinctspeciescaptive breeding programmes can increasenumber of individuals and maintain somebiodiversityincreasingawarenessaboutexoticorendangedspecies
5
r-strategy K-strategybody size small large
life span short long
maturity early late
offspring many few
care for offspring little extensive
population size fluctuates stable
environment unstable stable
description opportunistic stable
example pathogens and pests e.g. mice
whales and elephants
•
•
••
•
•
••
•
••
•
6 PopulationsizeN=n1xn2/n3=809.25 Sotheestimateofthetotalpopulationsizewouldbe
justover800woodlice.Beawarethatifthenumberswerethesamebut9ofthewoodlicehadbeenmarked,the estimated population size would have been justover700.
Chapter 19
1 (a) Anti-DiureticHormone(ADH)(b) ADHisproducedbycellbodiesofneurosecretory
cellsinthehypothalamus(c) ADHisstoredintheposteriorlobeofthepituitary
gland and released from the nerve endings ofneurosecretorycellsthere
(d) cellsofthewallofthecollectingduct(e) The mechanism is an example of negative
feedback: Osmoreceptors in the hypothalamusdetect an increase in the concentration ofdissolvedparticlesinthebloodplasmaandcausethereleaseofADH.ADHtravelstothekidneysvia theblood,makes thewallsof thecollectingductmorepermeablesomorewasisreabsorbedand less urine is produced which decreases theconcentrationofdissolvedparticlesinthebloodsostopsthereleaseofADH.
2 (a) Pepsinandtrypsinareproteases,ifproducedinactive form would digest the cell that producesthem so they are made as pepsinogen andtrypsinogen
(b) Pepsinogen isactivatedbyHCl in the stomach.Trypsinogen is activated by enterokinase in thesmallintestine.
HCl and enterokinase are made by differentcells than those who make pepsinogen andtrypsinogen.
3 (a) microvilli, mitochondria, pinocytotic vesicles,mitochondria.
(b) microvilliincreasethesurfacearea,allowingmorefoodtobeabsorbedatthesametime
mitochondria provide energy since some of theprocesses of absorption are active transportwhichrequiresenergy
pinocytotic vesiclesasaresultofendocytosis tight junctionstopreventparticlesfromentering
inbetweencells
398
Answers
4 (a) glucose, amino acids, hormones, vitamins,cholesterol,
(b) glucose is taken up, converted to glycogen andstored
amino acids are used to make (blood) proteins,surplusaminoacidsarebrokendownforenergyandureaproducedasawasteproduct
vitamins AandDarestoredintheliver cholesterollevelsarecontrolled
5 • geneticfactors-ifoneormorefamilymembershaveCHD,theriskisincreasedage-CHDismorecommonwithincreasingagegender-CHDismorecommonamongmalessmoking - smoking increases the chance ofCHDobesity-obesityincreasesthechanceofCHDdiet-adiethighincholesterolandsaturatedfatsincreasesthechancesofCHD;thereisevidencethattransfattyacidsareparticularyunhealthylackofexercise-exerciseseemstodecreaselevelsofLDLcholesterolandreduceschancesofCHD.
6 (a) As the partial pressure of oxygen increases, thepercentagesaturationofhemoglobinwithoxygenincreases as shown by the sigmoid curve. Firstoxygenismoredifficulttoattachbutitfacilitatesbinding of subsequent oxygen molecules. Inlungs, pO2 is high so hemoglobin is (almost)saturated with oxygen. In tissue, pO2 is low sohemoglobinreleasesoxygen(torespiringcells).
(b) Bohr shift = Bohr effect takes place when CO2levelsarehigh
e.g.inactivelyrespiringtissueslikemuscle HighCO2levelscausesdropinpH,dropinpH
makesoxygendissociationcurvegototheright.So at a specific pO2, hemoglobin will have lessoxygen with the Bohr effect, that means moreoxygenisreleasedtothetissuewhichisavailableforcellularrespiration
•••
••
•