biology stpm experiment 13

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Pra U STPM 2022 Penggal 1 - Biology

PELANGI BESTSELLER PRE-U STPM Text Biology Y.K. Richard TERM 1 David Tan SCAN ME For the 21st Century Learner 1 PREFACE Pre-U STPM Text Biology Term 1 is written based on the new syllabus made by the Malaysian Examinations Council (MEC). The book is well designed and organised with the following features to help students to understand the concepts taught. 1CHAPTER vPierwoCvoBoMidnfIOeOcthsLeLeapEOnctCGoMUoInCvaLcepAEerpSaLtlls STPM Scheme of Assessment learnt in the chapter Term of Paper Code Theme / Title Type of Test Mark Duration Study and Name (Weighting) First Administration Term 964/1 MoBlieocluolgeiscaalnd Written test Biology 60 Paper 1 Metabolism aqm1Snu5eusecwclsttotieiipomrolennepd-sAu.ctlhosoobirceye (26.67%) Concept Map ST1P5 M Scheme of Water Chemical properties Physiological Assessment Physical properties roles in Carbohydrates Section B organisms ALsasteessst mSTenPtMstSarcthinegm2e0o1f2bs2tercuaoncmtsuwprueelrdseodqr.uyesshtioornts to 15 Monosaccharide, disaccharide, polysaccharide 1—21 hours Central assessment Section C 30 2 out of 3 essay Triglycerides questions to be Phospholipids (lecithin) answered. Steroids (cholesterol) Polar, non-polar, acidic, basic Biological Molecules Functions oAnlltqoupeicssti1ontos are based CHAPTER 6. Lipids Second 964/2 Physiology Proteins Term Biology Written test 60 Paper 2 (26.67%) aqm1Snu5eusecwclsttotieiipomrolennepd-sAu.ctlhosoobirceye 15 Bilingual Keywords Nucleic acids Structures of nucleotides bs2SteercucaotncmitsouwpnrueeBlrdseodqr.uyesshtioornts to 15 A list of bilingual DNA & RNA (mRNA, tRNA and rRNA) terms is provided Analytical Paper 1—21 hours Central techniques chromatography Structure of DNA assessment Electrophoresis aq2Snueosecuwstttieiooronefnd3sC. teossbaey 30 Bilingual Keywords BNEDoluiefncfcrdlatercoo–ttpiiIodhkneaotr–a–ensPNiseumk–lbeEeoleltaikdutaraonforesis CCMoaonrnbsootishtauycedcnraht ta–eriJd–ueKzu–akrMboohniodsraatkarida All questions are based on topics 7 to 13. Triglyceride – Trigliserida vi Phospholipid – Fosfolipid Biology Term 1 STPM Chapter 1 Biological Molecules Hydrogen 1 bonds forming 2010 cross bridges Microfibril β -Glucose molecules CH2OH CH2OH CH2OH OO OO O OO OO CH2OH CH2OH Figure 1.12 The structure of cellulose 13. CIhannaeeldtlmhumelooelcsyceeshliplase.nwrIifatcolahlr,lmelyiltpswspahrisnoetntrteuhciectttsisusutrmhpaeplixorceoredltlelowfifnriotphtmlhalenibgtfsnootritnhhm.rpoohufygcsheilcltauwlraginloljruorpfirpeeslsasanuntrsde. Quick Check 1234....QWWWHuohhhiwyayctakddarooerCesactthleahllrleemuchdlcooiknasffeone2sdrfaeocgnrclmcyhecasmorbgiadeecentrwsomerfeieabdnkruieDlcsgi-naoggslouidscnuofgtoshaeoredsacnbredeulsltLewnr-gvaoelltulsac?oloflssdeoi?smacecphlaarnitdecse?lls? Provides short question for students to test their 1.3 Lipids Learning Outcomes Learning Outcomes S(atu) ddsaoeetnfrnsudttcrcsirdgtiusbilsyrheetcorseitubh,rluiedpdtrieobosnep, earbtileesto: A list of subtopics that understanding of the 12.3. .LatcsO(o(n(i(ooT(a(bpcddatmbl)h)ihhv)))d)eyeemsTiTToTdnrrTGhhaohxhrthpcreoeneyeehseheyniggyyuireyeeeproacmhsadnnrrhrhciraagece.yihevlcnaaaasleieyantlgsisUnmoc,liprihsctanmctepiorihihleclaogrclislesauoskppoohylypbearirmensovorlearaetepitprfrrnscioaetoiueeacadinrresulocermlwtsbtlisnswayeputiao,ardateiaryeryheodsvretes.tyfeessathdmlrahsfbdebesroasarfeieoaotrsffnihtlfoomnleaeedtilrlcgrhno,rleeoupbwaddbswtlneomeseohgnfl:strlweraaw:zuioerdseabemaentrolesmeeaeenrurft.gu.apiisdaontcptniohoyfifnftocealscofrocsaenowoirndn-obleptvsxorpeocyannaalagn,apttnedsehrend.gyraoo.dlhrrcarigaootevoahgsemn.eonilasc. students will learn in concepts learnt in the p(lheocsitphhino)liapniddssteroid (b) (sop(ctlfhehatocotresiiltgephtlshyhintecoe)elrifaorpuinldin)dd;ecsstsi,otenrsoids each chapter subtopics (c) (dsucianhfftseouarlreeatusntertteaidarteotaedl)n;bdfaetttwyeaecnids. 15 ii Biology Term 1 STPM Chapter 2 Structure of Cells and Organelles Biology Term 1 STPM Chapter 1 Biological Molecules Learning Outcomes 2.1 Prokaryotic and Eukaryotic Exam TipsCHAPTER tiPpsroEfvoxirdasemtsuTdhieepnlsptfsuiln(v) sbpgdFueeoiobfnllrfuyieenpnrskeeeitinxpenadattvimsdkotioeiplnnvsl,aeedfd,αishg,uaiah.αerenaem.deg2m1eβg.n2oαrr6eog-e.ulusoappnnbeidibtcnsetβifavohnergaldeeys.nf2oEefuaβofrcu-ohurronmfpisottihunsl.yebgpmTuethnwpecittoraisendteayofrophfereamssttwwtooaoof Cells 1 fpsaouRirnfmnoedcpmtpetrgiloeioenlmtonseabs;binn.uaedslnra:dcrtthohptenehrojeuffotiigrbeuairnrotsetuy,dspthees answering exam S((abt)u) dscoteofanmpttesrpotaskhrhaeeoryctuhoeldetllicbsthtearenuoadcbrtyule;retso: Past Year Questions2 (c) ecasuonekmdeanppryaulaornentditcteycrcpeeeilclllslesac;latsaronnimal EnabTleagstguidnegnts to(d) mdpeerlieicsncrccotrrisipobclneeospmtheoiecsfr;lboigashcstiocapny.d Cell AobTArychgccSaeeocnlolrhidsrlimeisynidtsgh.etneTo,hbaRiasubsiiodsctoaaulnpcncisoiVttr,idroacinfnhdgothwSteo,chastwtlhlrauecncecntleul,slrlaeatrzhaioesnoeodlrofyrfg,ouitmsnhtc.etpiroreens-ueolxtfisoatfillnstgluivdcienielglss Polypeptide questions Polypeptide 1. bAliypcoceeplllrliosdtmievianisdioemnue.pmobfrparnoet.oWplaitshminsuthrreoucenldl,etdhbeybaasseicleccthiveemlyispteryrmoefablilfee chain 4 (β2) chain 1 (β1) Prosthetic 2. group (haem) 3. Summary famieliaxraisme qthueesptaiotntesrn ofC12.e.LllAsoCeattfxrhecunialeeslclstlolgltinulraiisyvgrureiinsacaeagebnglftladrshoseinmfbicugynCuspcc.nrhteeiitol-lenofck division. exists. into two types, the prokaryotic cells and the Cells are divided 4. eukaryotic cells. Polypeptide chain 3 ( 2) Polypeptide Cell Theory chain 2 ( 1) VIDEO Summary SummaryFigure 1.26 The quaternary structure of haemoglobin cSounmtchemepastsreiscleetiaosrnnkteiynD1e. ncccrPhaooaratndnountiftarogeroteriainlmodtsntiaiho.ntceintaooTnsnhhaaianobsmpfedeoitsdhrRceeeaennbfnpieabrectarouaatotsurueuiesilrsynedasibntbticeshyoaoenbnlhurobebtohakleeftre,eansPelt.etarxeTtodtrehretieameedmsi.egnsenTlhoshobpieowuHnlnwa,arelicnashkshfetiamrgbpuuoiecrncteaducla1srsne.2tah7bnoa.edrt Comparison Between Structures of Prokaryotic and Denaturation and Language CheckEu12.k. aPnbArroeyolgtookehnnatagirevcyretaooCnltitisuechecledcllesekpulilrsnsogaakdnraeodrmypaorrtiePimcrfooictkueiCalvnelredllyiwscoiaenatlllaslbess..ahTcCothaeweprsnyiualaeianrneFdnigocuytartenruo2eb.1ac.cetllesr.iTa,hweyhidcoh 2013, 2015 321r.e..npbheHtamDHooynteeduneeaccnarrdrthoaghansatgsyatsiunouoenarlgpnnfgeanepepstdoailssroiinfoenbndtfptsuhreiretoneiikosonafcitnnH.kesptiieshic+orntoanictpedeinu. e – Language Check • Bacterium – singular AdditionalPlasma membrane language ibcoroneniacckebsnohtnryaddtsiro.ongeanlsoand + • Bacteria – plural Cytoplasm – ++ – Flagellum – singular Flagella – plural Ribosomes information. Eg. plural 4. Organic solvent Plasmid and singular ibnrteearakcstihoyndsr.ophobic Denaturation 5. rbpRbcetRrhoeafeoeoescnnknrsoamedabimtbnetueeulneerdbraraa.eoifttftuivunoiotnerdnhencrsesdtiesimcowpdanoeernanaotaolnlt.bkyeseinthat + Reacts 6. with Pili + other Globular structure Chain is – loosen Prokaryotic and Bacterial flagellum Precipitate Eukaryotic cell Nucleoid (circular DNA) VIDEO 2. (((cFba)a))ceHthpcaHoayaHnesrdneisadl.arnvytotoyhTc.gataaoHtmetutofnauescecaenabthhtsaucoiilgtstnnltioehihodFccndiernsgcie.eeuooaawngrntsneaaeeshctt1as.uiei.kvg2Entreh7haxtbretcaDitaohmoeketnmainniopndrangplitseeuneetsrcdariotlclotafuiigotkedunrHebnoeroree:mu+ef.rpapgoeksryror.cateounOFifnrdoHtyrhb–aerenebpxadrarkeomsatiwkelpvieslneear,kimoiebnoononiliczsensycfdocamursanml.neeedsds Figure 2.1 Structure of a generalised prokaryotic cell 30 52 Animal tissues Biology Term 1 STPM Chapter 2 Structure of Cells and Organelles Biology Term 1 STPM Chapter 6 Photosynthesis Homogenisation Exam Tips Homogenate CoaCttiAcpdesAmrhpaaamaeasehcMrlppnnlttbopptotishtoeceprptodtenophrtilosaaanrleytasndntaiconreyurttiteeatoisrlncdhosiaxevasttluennieroucesdcavtmicieeitiseisdhnimlatouogcotaanoifprclasnoriocnoanibcaucdoabcrgraratttelseealcl.nodaanntttrtsnaihoetrTibeldwgehamsishrytn.hoa,nieopvgotyCleeidcnemmrt3aa.PoarcyipmotCbEn.luoorfoP4iaerrSsxnhentecyccaogrseacatnotrthmrartetopbhrrbmavibooteorspeouoinasixrlditngapyopidhncllnrisaaadtootsnsmripioxdeoestlamisuoaxadpldrscniiaecokdaeett.apseeiwannstt.mrCiehgbaTdhtoe4lnohrereopsiee.lenfplpyaciftnreiuohtnaatcotoestrcoid.eronrnd.TeenowaiohdBmgsweueaoihynlncintltcehgd Exam Tips coRpRefreeliulnmmuscleaeeipmrsmlecbibnsoeemrtorhtfptehhcoeeiensneeboxtnlaraatisfstmuii.cogpnaletoisof n. 8. Rctlaceihogenofemhfmdnictpcpeicireemeneannntsrbettbserneaiaonnrsttnciiitotopehydnnh,aoioottpifestxpoovmivdliseanepyerntyrneytirtsnlhaowtewusirtiesh. Centrifugation at 600 g 9. for 10 minutes 10. 2 Nuclei and Supernatant Summary unbroken cells Centrifugation at 10,000 g for 20 minutes 1D.iffFerraecnttiioalncaetinotnrifbuygation homogenisation Mitochondria, ER Supernatant 2. cmfCmCoheirelitonnon2rtctot0rrohiipffmouuloanggibsndeettrast1i6oain00a,o0nnb0ugd0tca0flieongir 10 and Golgi bodies Centrifugation at 3. 100,000 g for 60 minutes Quick Check 3 carbon dioxide 4. Centrifuge light, temperature and for 60 min 100,000 g EcEoxxnppcllaeaininntrtfahatcieotsonir.gsnitfhicaatnacfefeocftcopmhopteonsysanttihoensipsoointth.er than Ribaonsdommeics,romfiilcarmoteunbtsules ribosomes to obtain 1. Nucleic acids and 5. opciUnfrooltdmvrtiaaefpfc-ienocursneue,ennmRnttrtNitSfoouAgfvsaaereilnbupwdoeaisstrDhaoNtmegAeel 2. proteins 6 3. wFuitrhthmerordeiftfhearneFni1gti0uar0le,02c0.e20n5ttSriimtefupegsbeygsritasevpuitclyet.lrl afr-accetinontratiifoungation using force 54. .((((TdpcDbHuamT))hr))nNoiheogiDGtTiAatTlhesmesvyheiehctconfelreeeueoaisticdinlrthsSshaesaormaapbnmnavddneadidawoqpdycdlsalueueeeereenfirdepaercdgdawsiduahicttstriulipottaooetaistrrhroatoseeteteicntoohcndoshtraheiepbmavufeptSrmesehattooatihvtewoivmrvDoixeaenaeetNutlbrmesuRduelneAtraoNeaomrerslnwtalAoec.hetcdinawoxc,eeSantunnesmidutnlrvutufebereraoocbetiedglalreflueeel.ouetcossileftgcecwsuapeemstianssserrac:etepdsariadahadacaitmrotrsnhiseoanuccde-mtlaalleiudiaonlvfnoelirnetrelbobrle..odsemaecf.-uiscnveFslneepaeontosdcahrtruoriermmaeufiftxtmdeuaeuanldigbmeftaofedfaetpr.ntiroloieteeomonsnsn.,.t STPM PRACTICE 6 WABCDhiFcFPPehellrararrssceettoododPcmoqoyxuxaibiinnnniionnnaetioPPPPnllllaaaasissssttttooootcqqQcryuuyuaiainnenniooinonnnfee P, Q and R? R STPM PracticeO1b.jecWpABCtDhihvoeatPPDotLrQhehsiaygaouoarnhctterkottotesihsatsoyeinynosstsdtaihnstesneets?mhmdeIpachanatnarhdsdbeeposIhnI ocfrtieaoxqralbuytosiiroinesndorfefadwixcuaatrtitieoinrongns Info Bio Plastoquinone taodnshs8rnuoueee0eebtypSn-aebthouorrqiaiantgbuvt8isotaeea0solsdn.SlotodsWoimntwrriot1hbeeeon0yaosre0m6s6scso0s0-a?usmmlsninnTaea+iebthnslsdleid4s.uw.0n4Siissti0ohtss, Plastocyanin Ferredoxin Ferredoxin 3. Wdiahgicrahmsotaftleimghetndt episenndeont t true in the Z reaction below? Ferredoxin A variety of examination-2. Tsthaegedoifagprhaomtossyhnothwessist.he light dependent Quinone Q P Info Bio typed questions to checkPrimary acceptor e– 4e– 2 NADP+ 2 NDPH2 e– P 4e– e– etrhlaeedctiboraoacncktgibvreeoaumDnNdaAnodfannddoiftfnelQorigernhmutt airbleceDfarkmaNcAtbCivebeheuisnseeeddcpeaxktroeastrewedsiotbhlvyet4huselmtroaabllcjeeercntotrbbiefjeulcgigtasht?itoenne? d Primary acceptor Provides extra1. R NADP+ + 2H+ Sunlight Sunlight How can students’ understandingH2O PS How can2. 2e– Cyctoomchprloemx e 2e–NA2DeP–N+AreDdPuHct+asHe+ INpIPehl6AeP8o0c6D8t0toraPopn+nhdsosiPspa7rh0e0oorrPxayaSrilideasI itesiodoePxdn70i0tdoisinehdighnaefotrner-ecnytehcrlegiicyr Why can Q I inrfeolramteastiotontthheat3. of the chapter learnt2H+ + —12O22e– or darkened? P680 II Light Light ATP P700 levels subtopics learnt 83 209 iii Biology Term 1 STPM Model Paper (964/1) STPM Model Paper (964/1) Answers Section A [15 marks] ANSWERS ComarpeleptreovaindsewdersOb11Sj6e.1t1.cr.6DutC.iAcv1t(e u7a(Q)(r.(b ced )u)(d )EabeT(c)Qn2sA1ibEh7E anTt.bz2 )tdiur.n1ubsdh ye o•o.(r•ss8Ates•emim ec•yunC roan,•s .C) a• Iany rst•Ipge otIetp ifp•t I (d thIeuofn Qtp mia t•tIe h r•Osehnti tzr)ehh ic•Ia uoa yet meyatn•asi tioantiIosn eilnensN esmsu ttzsee s•Hn nm s dotdat ayrpaClat Aiho•tAaeduzsdehsuc+ms sih ot•b yhoTys 3DTioct1eypersrhcts i8anbaodsyEmmlelhed.etninosns3rryPietes Hr.birglTmreedoeaee.bg.tin.iuloseohlioeoDoni mcahaeh c ot.aeyaCttsmbpti(illeostlshmpAeehhlstneiogae ynPeelchi iedhTnt.su e dtcoelepnshsgcadr.iooiaime)e Hahitdumrsmiltersoneu ms.rrv.sserinroeeearlfrnsfic+puooeceoop. pdmyn racaeootbfsulshglnsryTecueiso edeflfieibrmlaoiedtsncthrrlsaie datuvfotsrosyclsb4geu t1trrrseeedtii9eeu.nrihupe reayy.onb4rly nc sic.sreedgenle uofmer.gubtteretmor osDdpraesnofhcpo 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O CH3 H H2N C C CNC OH A aAncaceurmobuilcatrieosnpiraotfionlactic acid during CH3 H Pengumpulan asid anerobik H C laktik semasa respirasi 217 iv CONTENTS Chapter 4.4 Inhibitors 146 1 BIOLOGICAL MOLECULES 1 4.5 Classification of Enzymes 150 ••••••••••••••••••••••••••••••••••••••••••••• 4.6 Enzyme Technology 152 1.1 Water 2 STPM Practice 4 156 1.2 Carbohydrates 6 Answers 160 1.3 Lipids 15 1.4 Proteins 22 Chapter 1.5 Nucleic Acids 33 5 CELLULAR RESPIRATION 163 1.6 Analytical Techniques 39 ••••••••••••••••••••••••••••••••••••••••••••• STPM Practice 1 42 5.1 The Need for Energy in Living 164 Organisms Answers 48 5.2 Aerobic Respiration 168 Chapter STRUCTURE OF CELLS AND 5.3 Anerobic Respiration 177 2 ORGANELLES 51 STPM Practice 5 180 ••••••••••••••••••••••••••••••••••••••••••••• Answers 185 2.1 Prokaryotic and Eukaryotic Cells 52 2.2 Cellular Components 59 Chapter 2.3 Specialised Cells 84 6 PHOTOSYNTHESIS 187 STPM Practice 2 102 ••••••••••••••••••••••••••••••••••••••••••••• Answers 107 6.1 Autotrophs 188 6.2 Light-dependent Reactions 193 Chapter MEMBRANE STRUCTURE AND 6.3 Light-independent Reactions 197 3 TRANSPORT 111 6.4 Limiting Factors 206 ••••••••••••••••••••••••••••••••••••••••••••• STPM Practice 6 209 3.1 Fluid Mosaic Model 112 Answers 214 3.2 Movement of Substances across 116 Membrane STPM Practice 3 128 STPM Model Paper (964/1) 217 Answers 131 Answers 223 Chapter Glossary 225 4 ENZYMES 133 ••••••••••••••••••••••••••••••••••••••••••••• 4.1 Catalysis and Activation Energy 134 4.2 Mechanism of Action and Kinetics 136 4.3 Cofactors 144 viii 1CHAPTER BIOLOGICAL MOLECULES Concept Map Chemical properties Physiological Physical properties roles in Water Carbohydrates organisms Lipids Monosaccharide, disaccharide, polysaccharide Proteins Biological Molecules Triglycerides Functions Phospholipids (lecithin) Steroids (cholesterol) Polar, non-polar, acidic, basic Nucleic acids Structures of nucleotides DNA & RNA (mRNA, tRNA and rRNA) Analytical Paper techniques chromatography Structure of DNA Electrophoresis Bilingual Keywords Bond – Ikatan Nucleotide – Nukleotida Constituent – Juzuk Electrophoresis – Elektroforesis Carbohydrate – Karbohidrat Diffraction – Pembelauan Monosaccharide – Monosakarida Triglyceride – Trigliserida Phospholipid – Fosfolipid Biology Term 1 STPM Chapter 1 Biological Molecules CHAPTER Learning Outcomes 1.1 Water Students should be able to: 1 (a) describe the chemical properties (solvent, Water o(Hrg2aOn)isims sa. clear, colourless, odourless and tasteless liquid essential bond angles and for all There are other properties of water that enable it to play hydrogen bond) of its roles in organisms. water and relate its physiological roles in the organisms; (b) describe the physical properties (polarity, Chemical Properties of Water cohesiveness, density, 1. Each water molecule is composed of two hydrogen atoms and an surface tension, oxygen atom as shown in Figure 1.1. specific heat capacity, and latent heat of vaporisation) of water and relate its O– Electron physiological roles in organisms. – – O Covalent H+ H+ bond 104.5° H H + + Why is Water a O — Oxygen Polar Molecule? H — Hydrogen δ+ — Small positive charge INFO δ– — Small negative charge Figure 1.1 Structure of water molecule, a polar molecule 2. Two hydrogen atoms are attached to an oxygen atom at an angle of 104.5°. Each hydrogen atom shares a pair of electrons with the oxygen atom, so there are two covalent bonds. 3. Water is polar, the oxygen atom is slightly negative and the two hydrogen atoms are slightly positive. This is because the oxygen atom draws the electrons in the bond towards itself, giving a partial negative charge to the oxygen atom. The water molecule, as a whole is neutral. 4. Water is an excellent solvent to dissolve polar substances like electrolytes and non-electrolytes such as hydrophilic organic gcoromuppos.uWndast,erwmhioclhecuhlaevseca–nOsHu,rr–oCunOdOpHo,la–r NgrHou2,p–s CafOte–r waneadke–nPiOng4 and separating inter-molecular or inter-ionic bonds within a substance such as in sodium chloride in Figure 1.2. Cl– Na + Water molecules –+ Figure 1.2 Shell of orientated water molecules surround ions 2 Biology Term 1 STPM Chapter 1 Biological Molecules 5. Similarly, organic substances such as sucrose is soluble in water H+ H+ CHAPTER as water molecules interact and surround the sugar molecules, breaking down their inter-molecular bonds. Water molecules Hydrogen O– 1 are linked through hydrogen bonds. The hydrogen bond is an bonds H+ electrostatic attraction between the positively charged hydrogen atom of one molecule and the negatively charged oxygen atom O– H+ O– H+ H+ of another as shown in Figure 1.3. H + O– 6. Water is a liquid at room temperature. Each molecule of water H+ O– H+ can form a maximum of four hydrogen bonds with different molecules of water. Under room temperature, about 20% H+ hydrogen bonds exist in water. The lower the temperature, the more hydrogen bonds are formed. At 0 °C, it freezes where all Figure 1.3 Polarisation of water the molecules are involved in forming the three-dimensional molecules causes them to be structure of ice. This explains why ice expands and has a lower linked together by hydrogen bonds density. i.e. ice floats on water. These bonds are shown in Figure 1.4 below. Exam Tips Remember that water acts as reactant in hydrolysis and is chemically inert. Weak Oxygen Hydrogen Stable hydrogen Hydrogen Oxygen hydrogen bonds bonds Water Water molecules molecules Water Ice Figure 1.4 Hydrogen bonds in water and ice 7. Because of its chemical properties, the physiological roles of water Summary are as follows: (a) Water provides a medium for reactions to take place. This can Chemical properties of be seen in seeds that absorb water and start germinating almost water immediately. Enzyme and substrate molecules in the cytoplasm 1. Water contains two of cells within the seeds are activated and will start reacting with each other to bring about faster respiration and to start cell hydrogen atoms and divisions. one oxygen atom in a (b) Water acts as a solvent to transport substances in the blood bent molecule of 104.5° of animals, or in xylem and phloem of plants. Inorganic and bond angle. organic nutrients dissolve in water and are transported in soluble 2. Water molecules exhibit forms in the blood of animals. In plants, inorganic ions such as polarity and form nitrate are transported in the xylem. Organic substances such as hydrogen bonds with sucrose, amino acids and some plant hormones are transported one another. in the sieve tubes of the phloem. 3. Water is inert and a good solvent for reactions and transport. 3 Biology Term 1 STPM Chapter 1 Biological Molecules CHAPTER (c) Water ionises and acts as substrates for photolysis during photosynthesis and hydrolysis reactions during digestion of 1 food. The products of its ionisation i.e. hydrogen and hydroxyl ions are then used to form products during the light-dependent stage of photosynthesis. During hydrolysis, water is split into hydrogen ions and hydroxyl ions too to break bonds between complex biopolymers into monomers such as starch into glucose. (d) Water interacts with macromolecules such as proteins, nucleic acids and molecules in the lipoprotein membrane structure. The water molecules surround these macromolecules, making their structures more stable and thus maintaining their three dimensional structures to perform their functions. For example, a protein can act as an enzyme (catalyst) to speed up reactions when a substrate molecule binds to the enzyme molecule. Physical Properties of Water 1. Water has a high specific heat capacity of 4.2 kJ/K/kg, which means 4.2 kJ of heat is required to raise the temperature by 1 K for 1 kg of water. (a) This means that a lot of energy is required to raise the temperature of water. This is because the breaking of the hydrogen bonds within water molecules requires a large amount of energy before the temperature can be raised. (b) Temperature is raised when the individual molecules or atoms vibrate vigorously. (c) The physiological roles of a high specific heat capacity are as follows: • Water in our body can absorb a lot of heat from the muscles or the environment before our body temperature rises. This prevents our body temperature from rising too fast. • Water provides a more constant environment in the sea for aquatic organisms to live in. 2. Water has a high latent heat of vaporisation of 2260 kJ/kg, that is 2260 kJ of heat is required to vaporise 1 kg of water. (a) This means that a lot of heat is required to change water to vapour as the water molecules require a lot of heat energy to break all the hydrogen bonds between them to escape as gas. (b) Vaporisation always results in the cooling of the surroundings. There are two sorts of vaporisation which are evaporation, the slow vaporisation and boiling, the fast vaporisation. Evaporation occurs at the water surface which provides maximum cooling effect with minimum loss of water. Boiling occurs under the surface of water at the boiling point of water. (c) Water’s physiological roles of high heat of vaporisation are as follows: 4 Biology Term 1 STPM Chapter 1 Biological Molecules • This enables many land invertebrates such as earthworms Exam Tips CHAPTER and snails to survive. These invertebrates are nocturnal. They come out at night to forage for food in order to cut down Water in the new syllabus 1 water loss through evaporation during the day. has chemical properties of 104.5°C bond angle, • This also helps to lower our body temperature when we sweat. with polarity and solvent When sweat evaporates, it absorbs heat from our skin and properties. It has physical cools down our body. properties of high specific capacity; its highest • Panting helps to get rid of the excessive heats in dogs and density at 4°C; it has high birds. Water evaporating from the lungs is more effective to latent heat of vaporisation; cool body temperature than the skin. Heat is absorbed within high cohesive forces and the body rather than the body surface. high surface tension. (2006 essay question) • Similarly, water evaporating during transpiration in plants helps to lower the temperature of leaves. This cuts down overheating Summary when they are exposed to bright sunlight during photosynthesis. Physical properties of 3. Density or specific gravity of water is highest at 4 °C. water (a) This means that when water is cooled to 4 °C, it sinks. Further 1. Water has high specific cooling will cause it to float. At 4 °C, the water molecules are most compact with the shortest intermolecular bonds. heat capacity to (b) At 0 °C, water freezes. Ice expands during freezing as hydrogen prevent quick change bonds occupy more space in solid medium. Thus, this results in of temperature in a lower density than water causing it to float. organisms or in the (c) Freezing starts from the top of the water surface downwards as earth. the surface is subjected to low temperature first. 2. Water has high latent (d) After a certain thickness is formed, ice insulates the water below. heat of vaporisation Heat cannot escape from the water preventing further freezing to lower temperature as shown in Figure 1.5. during transpiration or sweating. Air = –20 °C 3. Water’s highest density Ice = 0 °C at 4°C allows it to freeze from top to bottom Water = 4 °C Lake permitting organisms to live below and circulate Figure 1.5 Water in a lake would not be totally frozen nutrients during winter. 4. Water has high cohesive (e) The physiological roles of water by having the highest density at force allowing transport 4 °C are as follows: of water in xylem • Water at the bottom of lakes or rivers will not freeze. This vessels. enables aquatic organisms to survive during winter or in 5. Water has high surface the tundra ocean. This happens especially when the depth of tension allowing insects water is more than a meter. to carry out activities at • Nutrients can circulate in the lake, helping in the colonisation the surface. of organisms into a greater depth. When the atmospheric temperature decreases, the surface water sinks at 4 °C. When the temperature of water at the bottom increases after winter, water moves up bringing soluble salts along. 5 Biology Term 1 STPM Chapter 1 Biological Molecules CHAPTER 4. Water has a high cohesive force. (a) This means that there is a high attractive force that exists among water molecules due to the hydrogen bonds. 1 (b) Its physiological roles of high cohesive force are as follows: • Together with adhesive force, water creates a transpirational pull in the xylem vessels when evaporation occurs in the leaves. This helps to transport water and mineral ions up the leaves for photosynthesis. • Cohesion and adhesion of water molecules enable water to stay in the upper layer of soil called topsoil. Good quality soil of fine grain sizes together with humus retain the right amount of water so that plants and other soil organisms can live well in it. 5. Water has a high surface tension. (a) High surface tension means that there is a strong inward pull of water forming a skin-like layer at the surface of water. This is caused by the high cohesive forces of water molecules as a result of hydrogen bonding. (b) Surface tension creates a habitat on the water surface. It allows insects to stay on the surface where they can gather food or catch their preys like water-skaters do. (c) Surface tension allows female mosquitoes to stand on water to lay eggs. The tiny eggs can float on the surface of water and the larvae can attach to the surface so that their syphons can breathe in air. Quick Check 1 1. What are the other properties of water that are important to organisms? Learning Outcomes 1.2 Carbohydrates Students should be able to: 1. Carbohydrates (hydrates of carbon) are a large group of biochemicals (a) classify carbohydrates which consist of carbon, hydrogen and oxygen in the proportion, 1 : 2 : 1. into monosaccharides, disaccharides and 2. Carbohydrates are sugar-containing compounds. These are the most polysaccharides with common organic compounds found on earth and in the largest respect to their physical amount as cellulose in plants. and chemical properties; (b) classify monosaccharides 3. Their general formula is Cx(H2O)y, such as glucose (C6H12O6) and according to the number sucrose (C12H22O11). of carbon atoms and the functional groups: (i) triose e.g. glyceraldehydes, (ii) pentose e.g. ribose and deoxyribose, (iii) hexose e.g. glucose and fructose; 6 Biology Term 1 STPM Chapter 1 Biological Molecules Classification of Carbohydrates (c) illustrate the molecular CHAPTER structure of a Carbohydrates are divided into three groups with respect to their physical monosaccharide and 1 and chemical properties: differentiate between the reducing and non- (a) Monosaccharides (one sugar unit) reducing ends; 1. Monosaccharides are simple sugars, containing only one basic unit (d) describe the formation which can be used to form disaccharides and polysaccharides. They of glycosidic bond in are the smallest carbohydrate molecules. disaccharides (maltose and sucrose) and 2. Monosaccharides have the following physical properties: polysaccharides (starch, (a) They are sweet in taste. glycogen and cellulose); (b) They dissolve in water. (c) They can be crystallised. (e) relate the structure of disaccharides and polysaccharides to their functions in living organisms. 3. Monosaccharides have the following chemical properties: (a) They are either aldehydes which contain –CHO groups or ketones which contain C=O groups. Both ketones and aldehydes have carbonyl groups, C=O. Summary (b) They are all reducing sugars. They can carry out reduction since carbonyl groups donate electrons. Classification of (c) They react with Benedict’s or Fehling solution to give a brick-red carbohydrates precipitate. They reduce alkaline copper(II) sulphate (CuSO4) into insoluble copper(I) oxide (Cu2O). Carbohydrates Cu2+ + e– → Cu+ Monosaccharides Blue solution Brick-red precipitate (Single sugar unit) Disaccharides (d) During reduction, they themselves become oxidised. (Two sugar units) Polysaccharides H (Many sugar units) |  Oxi+dOation→C ar–bCoOxyOliHc acid Carbohydrates: –C=O or –CHO Definition, Aldehyde Classification and Functions (e) They can be reduced in the cell, forming alcohols. H INFO A– lCd| =ehOyde Re+du2Hct+ion →P–rCimHa2rOyHalcohol – KCe| =toOne Re+du2Hct+ion →Se–coC| nHdOarHy alcohol 7 Biology Term 1 STPM Chapter 1 Biological Molecules SummaryCHAPTER 4. Monosaccharides have a general formula of (CH2O)n. They can be classified into trioses, pentoses and hexoses according to the number 1 Classification of of carbon atoms and the type of functional groups. Functional groups can be either aldehyde group (–CHO) or keto group (–CO). monosaccharides –Trigolsyecser(aCld3 eshuygdaers) 5. Trioses (3C) – Cm3oHn6oOs3a.ccharides that contain three atoms of carbon (aldose) (a) Trioses are e.g. glyceraldehyde and dihydroxyacetone. –Perinbtoossees(a(lCd5osseu)gars) (b) Their molecular structures are shown below: and deoxyribose (aldose) H H CO H C OH H– egxluocsoesse(C(a6ldsousgea)rs) H C OH and fructose CO (ketose) H C OH H C OH Exam Tips H H The molecular structure of Glyceraldehyde Dihydroxyacetone glyceraldehydes, ribose, (Aldose) (Ketose) deoxyribose and glucose should be remembered to (c) Both trioses are the intermediate products of metabolism during show that you know the the breakdown of glucose (glycolysis) or synthesis (photosynthesis) presence of carbonyl group of it. Glyceraldehyde exists as phosphoglyceraldehyde and for reducing. dihydroxyacetone exists as phosphodihydroxyacetone during glycolysis and carbon fixation during photosynthesis. They seldom exist long in the free state in cells. 6. Pentoses (5C) – C5H10O5. (a) They are monosaccharides that contain five atoms of carbon. Examples include ribose, deoxyribose and ribulose. H H C1 O H C1 OH C2 O H C2 OH H C3 OH H C3 OH H C4 OH H C4 OH H C5 OH H C5 OH H H Ribulose Ribose (Ketose) (Aldose) 8 Biology Term 1 STPM Chapter 1 Biological Molecules CHAPTER (b) Pentose can either exist in an open chain form or a ring structure e.g. ribose and deoxyribose are shown in Figure 1.6 below. HO HO 1 1C 1C CH2OH CH2OH H 2C OH O H 2C H O OH OH H 3C OH H 3C OH H 4C OH HO OH H 4C OH HO Pentose ring form Pentose ring form 5CH2OH 5CH2OH Open chain form Open chain form Ribose Deoxyribose Figure 1.6 The straight-chain and ring form of ribose and deoxyribose (c) The structure of deoxyribose is the same as ribose except that the oxygen in carbon number 2 is missing. (d) The physiological roles of pentoses are as follows: • Pentoses are used for the synthesis of nucleotides in which the ribose is for the formation of ribonucleotides in RNA and the deoxyribose is for the formation of deoxyribonucleotides in DNA. • They are used for the synthesis of coenzymes such as NAD, NADP, FAD and coenzyme A. • They can be used for the synthesis of polysaccharides. • Ribulose biphosphate is used as a receptor for the fixation opfhoCtoOs2ynatthethsies. beginning of the light independent stage of 7. Hexoses (6C) –suCc6hH12aOs6. glucose, and are (a) Hexoses galactose fructose 2013 monosaccharides that contain six atoms of carbon. H H C1 O H C1 OH H C2 OH HO C3 H C2 O H C4 OH HO C3 H H C5 OH H C4 OH H C6 OH H C5 OH H H C6 OH Glucose H (Aldose) Fructose (Ketose) 9 Biology Term 1 STPM Chapter 1 Biological Molecules CHAPTER Exam Tips (b) Fructose and other ketoses have ketone (–CO) group as Remember examples of functional groups at one end. Both aldehyde and ketone groups 1 monosaccharides (STPM have carbonyl (–CO) group as donor of electrons for reducing, 2013) at the reducing end. (c) All of them are isomers, which have the same empirical formula cohf aCin6Hfo12rOm6.oTrhaerignlgucsotrsuecmtuoreleacsulsehocwann either exist in an open in Figure 1.7. 1CHO reducing end 6CH2OH OH 6CH2OH O OH H 2C OH H 5C H 5C HO 3C H H 1C H 1C H 4C OH H 2C OH H 2C H H 5C OH 4C OH OH 4C OH OH OH 3C OH 3C 6CH2OH non-reducing end H H D-glucose α -glucose β -glucose Figure 1.7 Straight chain structure and ring structure of glucose Summary As shown in the open chain structure, glucose like all aldoses have aldehyde (–CHO) as a functional group at one end. Molecular structure of monosaccharide (d) As shown in the straight chain structure, the glucose is Reducing end – has polyhydroxyaldehyde or aldohexose. D-glucose and L-glucose carbonyl group. are optical isomers or stereoisomers. Non-reducing end – has no carbonyl group. (e) Hexose or pyranose ring structure of glucose can exist in two forms i.e. α-glucose and β-glucose depending on whether the OH group at the carbon 1 atom is below or above the plane respectively. (f) The physiological roles of hexoses are as follows: • Hexoses act as a source of ATP in respiration especially glucose when completely broken down into water and carbon dioxide. • They can balance water potential between the inside and outside of cells, so as to maintain the shape of cells e.g. red blood cells. • They are used to synthesise disaccharides such as sucrose and lactose. • They are used to synthesise polysaccharides such as starch, glycogen and cellulose. • They are used to synthesise other substances such as fats and amino acids in the body when required. Disaccharides (two sugar units) 1. Disaccharides are double-sugars formed from two units of monosaccharides through condensation reaction with the removal of a water molecule. 10 Biology Term 1 STPM Chapter 1 Biological Molecules CHAPTER 2. The bond between the monosaccharides is called glycosidic bond. 1 The glycosidic bond can be α or β depending on the disaccharide. 3. Their physical properties are as follows: (a) They are all sweet. (b) They are all soluble in water. (c) They are all crystalline substances. 4. Their chemical properties are as follows: (a) They are reducing sugars except sucrose. (b) They can be hydrolysed to form monosaccharides by heating in dilute acid or by specific enzymes. Sucrase can hydrolyse sucrose, maltase can hydrolyse maltose and lactase can hydrolyse lactose. 5. Examples of disaccharides are maltose, sucrose and lactose. 6. Maltose is formed from two molecules of glucose, which are bonded by α-1,4-glycosidic bond as shown in Figure 1.8. α –Glucose α –Glucose Maltose 6CH2OH 6CH2OH Condensation 6CH2OH 6CH2OH 5O 5O Hydrolysis 5O 5O 4 1 4 1 4 14 1 HO OH 2 OH OH 2 HO OH O OH OH OH HO 3 OH 3 2 3 2 3 OH OH H2O 1,4-glycosidic bond Figure 1.8 The formation and hydrolysis of maltose 7. The reaction is condensation when a molecule of water is removed between two hydroxyl groups from cCat1aloyfseodnbey gslyunctohseetaasned(liCga4 soe)f another glucose unit. The reaction is enzyme. 8. This occurs in the stroma of chloroplast, in amyloplast of parenchyma cells and cytoplastin of liver and muscle cells. 9. Sucrose is formed from the bonding of glucose and fructose by an α-1,2-glycosidic bond as shown below. Glucose Fructose Sucrose O 6CH2OH CH2OH CH2OH O 5O O HOH2C OH Condensation HOH2C CH2OH HO Hydrolysis OH 4 1 OH O OH HO OH 2 OH OH 3 OH OH 1,2-glycosidic bond H2O Figure 1.9 The formation and hydrolysis of sucrose 11 Biology Term 1 STPM Chapter 1 Biological Molecules CHAPTER 10. This is also a condensation reaction that occurs in the stroma chloroplast and cytoplasm of plant cells catalysed by synthase. This occurs in the mesophyll cells and the sucrose is transported away by 1 phloem. This is a common reaction in sugar cane, sugar beet, flower and fruit cells, where sucrose is stored. Polysaccharides (many sugar units) 1. Polysaccharides are complex carbohydrates, a type of biopolymer i.e. long chains of molecules which are made of repeating units of monosaccharides. 2. The monosaccharides in polysaccharides may be all hexoses hence they are called hexosan, wanhdichthethirencoamremcoanllefdorpmenutlaosiasn(C(C6H5H108OO54))nn. They can be all pentoses or a mixture of both. 3. Their physical properties are as follows: (a) Polysaccharides are not sweet. (b) They are insoluble in water. If dissolved in hot water, they form colloids because they are large molecules. (c) They cannot be crystallised but form amorphous mass if desiccated. (d) They are compact and not osmotically active in cells. (e) They can be extracted and purified to form white powder. 4. Their chemical properties are as follows: (a) All of them have no reducing power. (b) Polysaccharides like starch and glycogen are easily hydrolysed to maltose by amylase. (c) Polysaccharides like cellulose can be hydrolysed by cellulase, which is only produced by microbes and snails. 5. Examples of polysaccharides are starch, glycogen and cellulose. 6. Starch is formed by a-1,4-glycosidic bonds in amylose, an unbranched component. So, one amylose synthetase is involved. The other branched component, amylopectin is formed by both a-1,4-glycosidic and a-1,6-glycosidic bonds. So, two amylopectin synthetases are involved. 7. The reaction is polymerisation by adding one glucose monomer to maltose at a time to elongate in chloroplast and amyloplast. 8. Glycogen is a branched biopolymer with both α-1,4-glycosidic and α-1,6-glycosidic bonds. It is formed in the smooth endoplasmic reticulum of both liver and muscle cells, also by polymerisation catalysed by two synthetases. After forming a straight chain of more than 12 units, a short chain is cut by a hydrolase. Then, the chain is transferred to another chain to form a branch forming a-1,6-glycosidic bond catalysed by another synthetase. 12 Biology Term 1 STPM Chapter 1 Biological Molecules 9. Cellulose is formed by β-1,4-glycosidic bonds into a straight Exam Tips CHAPTER biopolymer. The β-glucose units are catalysed by synthetase to form long chains at the plasma membrane. Remember that reducing 1 sugars (monosaccharides 10. The enzymes involved are synthesised in the rough endoplasmic and disaccharides) have reticulum packaged into vesicles. The enzymes are released at free carbonyl (aldehyde or the plasma membrane to form cellulose molecules, and then ketone) groups. microfilaments to form new layer of cell wall. Remember the bonds in maltose and sucrose with 11. After plant cells divide, vesicles gather in the central plate and 2 monosaccharide units. fuse. The enzymes in the vesicles form cellulose molecules and They can be used only microfilaments that form the new cell wall. The membrane of vesicles when the glycosidic bond is form new plasma membrane at both sides of the new cross wall. hydrolysed. So, there is a control in its usage. Functions of Disaccharides and Polysaccharides in Living 2015 Organisms Summary 1. Disaccharides like maltose consists of two glucose units and is an intermediate in the synthesis of polysaccharides. It also acts as the Formation of glycosidic intermediate when starch or glycogen is broken down. Thus, it acts as bonds in disaccharides a control in both directions i.e. to form polysaccharides or to release and polysaccharides glucose. Disaccharides (One synthetase) 2. Disaccharides like sucrose acts as a medium to transport carbohydrate 1. Maltose – α-1,4- in phloem. It can lower water potential and create hydrostatic flow in the sieve tube. It can be stored in plant cells especially sugar cane and glycosidic bond sugar beet as food reserve. Sucrose is found in fruits and nectar to 2. Sucrose – α-1,2- attract animals and it is not easily destroyed like glucose or fructose as it needs sucrase to digest into glucose and fructose. glycosidic bond Polysaccharides (More 3. Another disaccharide, lactose that is found in milk acts as food. Like than one synthetase) all disaccharides, lactose needs to be hydrolysed before it can be 1. Starch – amylose – used by babies. Thus, the control of glucose and galactose release is determined by the rate of hydrolysis of the disaccharide. α-1,4-glycosidic bond 2. Starch – amylopectin 4. Starch is a polysaccharide stored as food reserve that is formed from α-glucose in the form of amylose and amylopectin. – α-1,4 and α-1,6- glycosidic bonds 5. Amylose is unbranched and coiled to form an a-helix but is easily 3. Glycogen – α-1,4 & hydrolysed by amylase to form maltose then by maltase to release α-1,6-glycosidic bonds glucose for entry into cells for respiration. 4. Cellulose – β-1,4- glycosidic bond CH2OH O CH2OH CH2OH O CH2OH O O O O O OH OH O OH OH O OH OH OH OH Amylose Figure 1.10 The structure of amylose 13 Biology Term 1 STPM Chapter 1 Biological Molecules 6. Amylopectin is branched, consisting of a-1,4 and a-1,6-glycosidicCHAPTER bonds. Thus, it needs isomaltase to hydrolyse a-1,6-glycosidic bonds 1 causing the release dependent on the control of another enzyme. CH2OH CH2OH O O O OH O OH OH HO O CH2OH CH2OH CH2 CH2OH O O O O O O OH O OH O OH O OH OH OH OH OH Amylopectin Exam Tips Figure 1.11 The structure of amylopectin Remember the differences 7. Starch is insoluble, compact and not osmotically active making it a in structures related to the good food reserve in the cell. This is especially so in seed and storage functions of starch and organs where glucose can be released for growth when conditions are cellulose. (STPM 2009) suitable. Summary 8. Inter-conversion of starch and maltose maintains water potential in the cells especially in the guard cells for the opening and closure of Functions of stomata. disaccharides and polysaccharides 9. Glycogen is an ‘animal starch’ which is found in the liver and muscles as food reserve. Glycogen is formed from α-glucose Disaccharides and its molecular structure is the same as amylopectin but more branched. (a) Maltose 10. In the liver and the muscles, glycogen can be broken down to glucose – two unit stage to form phosphate by phosphorylase. Glucose phosphate can be used directly for respiration. starch and glycogen 11. Glycogen in the liver can be converted to free glucose and released – two unit stage to into the blood in between meals when blood glucose falls. After meal, glucose is absorbed and stored as glycogen. Thus, glycogen acts release glucose for to control blood sugar level. respiration or change to 12. Cellulose is a polysaccharide, which is used to make the cell wall of plants. The molecules are cross-linked by hydrogen bonds to form other compounds microfibril as shown in Figure 1.12. (b) Sucrose – to transport carbohydrate in phloem so can generate hydrostatic pressure – to store food that can release glucose and fructose under the control of sucrase Polysaccharides (a) Starch – to store food in compact forms in plants cells (b) Glycogen – to store glucose in compact forms and easily release glucose for respiration in animals (c) Cellulose – to form cell walls for protection and support in plants 14 Biology Term 1 STPM Chapter 1 Biological Molecules Hydrogen CHAPTER bonds forming 1 cross bridges 2010 Microfibril β -Glucose molecules CH2OH CH2OH CH2OH OO OO O OO OO CH2OH CH2OH Figure 1.12 The structure of cellulose 13. Cellulose performs a structural role in the form of cell wall of plants. In the cell wall, it protects the cell from both physical injuries and haemolysis. It helps in the support of plants through turgor pressure and mechanically when it is mixed with lignin. Quick Check 2 1. Why are all monosaccharides reducing sugars but not all disaccharides? 2. What are the differences between D-glucose and L-glucose? 3. Why do starch and glycogen make good food reserves? 4. How do cellulose form macrofibrils as in the cell wall of some plant cells? 1.3 Lipids Learning Outcomes 1. Lipids are organic compounds that are made up of carbon, hydrogen Students should be able to: and oxygen. Unlike carbohydrate, the ratio of oxygen atoms (a) describe the to hydrogen atoms in a lipid molecule is much lower and have a common physical property of being soluble in non-polar organic structures, properties solvent such as chloroform, ether or benzene. and distribution of triglycerides, 2. Other physical properties are as follows: phospholipids (a) They are insoluble in water and polar organic solvents. (lecithin) and steroid (b) Their densities are always less than water. (cholesterol); (c ) They have high viscosity. (b) state the functions (d) They are greasy and leave behind grease spots on paper. of triglycerides, phospholipids 3. Their chemical properties are as follows: (lecithin) and steroids (a) Their chemical structures differ between different categories. (cholesterol); (b) Generally, they are esters formed from fatty acids and alcohol. (c) differentiate between saturated and unsaturated fatty acids. 15 Biology Term 1 STPM Chapter 1 Biological Molecules CHAPTER Summary 4. Examples of lipids are triglycerides (fats), phospholipids, steroids, waxes and terpenes. 1 Structures, properties and Structure, Properties and Distribution of Triglycerides, distribution of triglycerides (TG), phospholipids (PL) Phospholipids and Steroids and steroid (cholesterol) 1. Structures: (a) Triglycerides (a) TG: Ester of 1 glycerol and 3 fatty 1. Triglycerides are fats or oils formed from one molecule of glycerol acids and three molecules of fatty acids. The process of their formations (b) PL: Ester of 1 is called esterification, which is a condensation between an alcohol group and a carboxylic group. glycerol, 2 fatty acids, 1 phosphoric acid & 1 choline (c) Cholesterol: 4 2. Esterification is also called lipogenesis in which one molecule of fused rings with 1 glycerol reacts with three molecules of fatty acids as shown in Figure hydroxyl group and 1 hydrocarbon tail 1.13. 2. Properties: (a) All soluble in non- O HO polar organic solvent. H (b) Lower density than H C OH +H O C R H C O C R +H O H water. O O (c) PL more soluble than cholesterol H C OH + H O C R H C O C R +H O H with TG as the least O O soluble in water. (d) TG – hydrophobic, H C OH + H O C R H C O C R +H O H PL – amphipathic H & cholesterol little H amphipathic 3. Distribution Glycerol Fatty acids Triglyceride Water (a) TG – in adipose tissue under skin Figure 1.13 Esterification or around internal organs (a) The process involves condensation of the three hydrogen atoms – in seeds and from the hydroxyl (OH) groups of the glycerol, each forming an ester bond with a hydroxyl group from three fatty acids. It is cuticle catalysed by a ligase. (b) This process takes place in the adipose tissue and can happen (b) PL – in membrane and blood (c) cholesterol – in membrane & fatty tissues also in the liver. In plants, the process can occur in the chloroplast and seeds. (c) The three fatty acids in a triglyceride are usually the same type forming products such as tristearin, tripalmitin and triolein. (d) Tristearin, therefore, is made up of one glycerol and three stearic acids. 3. The physical properties of triglycerides are as follows: (a) Triglycerides are insoluble in water. (b) Their specific gravity is less than water, therefore they float on water. (c) They are soluble in non-polar organic solvents like acetone. (d) They form emulsion if shaken with alcohol. (e) They leave behind grease spots on paper. 16 Biology Term 1 STPM Chapter 1 Biological Molecules CHAPTER 4. Their chemical properties are as follows: 1 (a) Triglycerides react with atmospheric oxygen and become rancid if kept for too long a period. (b) They react with Sudan III reagent to form a dark red complex. (c) They can be hydrolysed by lipase or boiling with dilute alkali to form glycerol and fatty acid. 5. In animal, triglycerides are stored as oil droplets in the cytoplasm of adipose cells. The adipose cells are found underneath the skin, in the mesentery or surrounding the intestines, kidney and even the heart. It can be found in between muscle fibres. 6. In plants, they are found in seeds and fruit walls such as in oil palm fruit, seeds of groundnuts, rapeseeds, sunflower seeds and kernels of all cereals. (b) Phospholipids Exam Tips 1. Phospholipids are lipids that contain phosphorus atoms usually Remember the structure of in the phosphate form that can be removed by hydrolysing with lecithin and its importance phosphatase. They are the membrane lipids forming the basic in cell membrane structure. bimolecular layers within all cell membranes. 2. There are two major types, the derivatives of phosphatidic acid and sphingolipid, for which the former is more common. 3. An example of the derivatives of phosphotidic acid is lecithin, which is phosphatidylcholine. 4. The molecular structure of lecithin is shown in Figure 1.14. H O | || Gl ycerol → H –C–O–C–C17H35 O → || →Fatty acid chains H –C – O – C – C17H35 C| H3 H| H| O– | Hydrophobic  tail Hydrophilic head Hydrophobic tail H ydrophilic head Hydrophobic tail CH 3 – N+ – C–C–O – P —— O – CH | | | || | CH3 H H O H ↑ ↑ Choline Phosphate Hydrophilic head Figure 1.14 The structure of lecithin 17 Biology Term 1 STPM Chapter 1 Biological Molecules CHAPTER 5. As shown in Figure 1.14, phospholipids are derived from one molecule of glycerol bonded with two molecules of fatty acids, one molecule of phosphoric acid forming phosphatidic acid through ester 1 bonds. Choline is an alcohol complex bonded to the phosphatidic acid forming lecithin. 6. The physical properties of phospholipids are as follows: (a) They are amphipathic, containing hydrophobic and hydrophilic groups. (b) They form a layer of micelle at the water surface if poured slowly on water as shown in Figure 1.15. Hydrophobic Lipid tail droplet Hydrophilic Surface Spherical "Detergent" head Water micelle effect of micelle Exam Tips Figure 1.15 Micelle of phospholipid Remember the structure of phospholipid (STPM (c) They are soluble in water forming emulsion of spherical micelles 2007 structure question) if shaken in water. and its functions compared to those of triglyserides. (d) They act as ‘detergent’, helping to clear droplets of lipid in Remember the structure of aqueous medium especially in the blood. cholesterol. (e) They can combine with protein, forming lipoprotein units to transport cholesterol. Low density lipoproteins transport cholesterol from the liver to tissues including arterial walls, forming blockage in arteries. High density lipoproteins transport cholesterol from tissues to the liver to be excreted. Cholesterol Less cholesterol Phospholipid Protein More protein Phospholipid Low density lipoprotein (LDL) High density lipoprotein (HDL) Figure 1.16 Phospholipids combined with proteins forming lipoprotein unit 7. Phospholipid is found in the cell membrane as bilayer. The membrane surrounds the cell and organelles and also within some organelles in both plant and animal cells. 18 Biology Term 1 STPM Chapter 1 Biological Molecules CHAPTER (c) Steroids 1 1. Steroids are lipids that are made of four hydrocarbon rings as shown in Figure 1.17. Hydrophobic part of molecule Hydrophilic 12 17 part of 11 16 molecule 13 1 15 2 9 14 10 8 HO 357 46 Cholesterol Figure 1.17 The structure of steroid and cholesterol 2. There are 17 atoms of carbon within the fused rings. They are labelled according to standard practice. Different steroids differ from one another from the different organic groups attached to the different carbon atoms. 3. Examples of steroids include cholesterol, testosterone, estrogen, progesterone, aldosterone, vitamin D, bile salts i.e. sodium glycocholate dan taurocholate and those found in plants known as phytosterol. 4. Properties of steroids: (a) They are insoluble in water as they are hydrophobic in nature. (b) Most steroids are soluble in organic solvents such as petroleum, ether and acetone. (c) They can be solified at low temperature. (d) They can form micelles with phospholipids and can be soluble in water or alcohol as emulsion. Functions of Triglycerides, Phospholipids and Steroids Function of Triglycerides 1. They can act as energy sources, in which they yield 38 kJ g –1 of energy compared to that of 17 kJ g –1 for carbohydrates. 2. They act as energy reserves in the adipose tissue of animals and seeds of many plants. They offer advantages of more quantity of energy, less space involved, insolubility in water and easily respire. 3. They can be used for the formation of other chemicals including glucose to maintain blood glucose level and amino acids to make proteins. They can be used to form all other chemicals especially in plants and herbivores after dormancy during winter. 19 Biology Term 1 STPM Chapter 1 Biological Molecules CHAPTER 4. They insulate the animal body from cold in the form of subcutaneous fat layer, which is a bad conductor of heat. This is to ensure survival in cold temperature especially aquatic mammals like 1 seals as they are not covered with fur. 5. They protect organs such as the kidney from physical damage. 6. They waterproof both animals and plants. Our skin is covered with oil from sebaceous glands. Fur and feather of animals are covered with oil to make better insulators. Plants and other animals have cuticles of fat to reduce excessive evaporation. Functions of phospholipid 1. Phospholipid is the main component and forms the basic structure of the membrane with hydrophilic heads facing outside, hydrophobic tails opposite each other as in Figure 1.18. Glycocalyx Outside of cell Protein Phosphate head Lipid tail Inside of cell Cholesterol Summary Figure 1.18 The membrane structure with phospholipid as basic component Functions of Triglycerides 2. It forms the bimolecular layer in the membrane in which other (TG), Phospholipids (PL) components are bonded to it. and cholesterol 1. TG – food reserve to 3. The phospholipid molecules in the membrane are dynamic which means each of the molecules can move freely or they are fluid-like. release glycerol + fatty acids for respiration 4. The molecules ensure stability in the membrane. The outer surfaces or conversion to other interact with aqueous medium on both sides whereas the inner layers compounds are locked in non-polar bonding. The bilayer can easily reform even – insulation, protection if the structure is temporarily broken. and water proofing 2. FL – Basic structure 5. Its lipid nature permits hydrophobic substances especially those of membrane and of lower molecular size like methanol to diffuse easily across the transport lipids membrane. It acts as a barrier for polar molecules and ions to cross 3. Cholesterol – regulatory the membrane. function in membrane and precursor for 6. The bimolecular nature of the phospholipid in the membrane allows steroids (e.g. steroid reorientation of their molecules to form vesicles or the vesicles fuse hormones) with membrane become part of it. Thus, endocytosis and exocytosis can easily take place. 20 Biology Term 1 STPM Chapter 1 Biological Molecules 7. The phospholipid layer allows globular proteins called intrinsic CHAPTER proteins, with corresponding hydrophobic and hydrophilic surfaces, to span across the membrane. 1 8. Thus, the bilayer with such intrinsic proteins such as enzymes and transport proteins can respectively perform enzymatic functions and allow the passage of hydrophilic substances to cross the membrane. 9. Similarly, some extrinsic proteins that are attached to the outer or inner surface of the phospholipid layer can perform structural and enzymatic functions. 10. Short polysaccharides can bind with phospholipids or proteins. They serve as receptors or cementing substances. Such carbohydrates can also stabilise the membrane structure by forming hydrogen bonds with water molecules both outside and within the cell. Functions of steroids Exam Tips 1. Steroids such as sex hormones like testosterone, oestrogen and Remember an example of progesterone are required for maintaining sexual health. saturated fatty acid is stearic 2. Other steroid hormones from adrenal cortex, for example, the acid and unsaturated fatty acid is oleic acid. Ester bond corticoids are required for glucose and mineral metabolism. Steroid is formed between alcohol hormones being lipids can pass through the membrane into the and acid. Esterification is the nucleus to activate genes. process of forming the ester 3. Cholesterol is important in the synthesis of membrane structures as bond. such it is for the general well being of the skin cells and body. 4. Steroids in the form of vitamin D in the skin is needed for the metabolism of calcium and phosphate. 5. Steroids such as bile salts, emulsify lipids in our food in the small intestine. This is due to the solubility of the hydrocarbon rings in lipid and the ionic heads in water. Emulsified lipids can be easily digested and absorbed. Differences between Saturated and Unsaturated Fatty Acids 1. Saturated fatty acids are those that contain only single bonds with no double bonds within their carbon linkages. Examples of saturated fatty acids are as follows: (a) Palmitic acid, CH3(CH2)14COOH (b) Stearic acid, CH3(CH2)16COOH, as shown below: 21 Biology Term 1 STPM Chapter 1 Biological Molecules SummaryCHAPTER 2. Unsaturated fatty acids contain at least one double bond within their hydrocarbon chains. An example of unsaturated fatty acid is Oleic 1 Differences between acid as shown below: saturated and unsaturated fatty acids Saturated Unsaturated 3. Saturated fatty acids have higher ratio of H:C compared to Single unsaturated fatty acids. –C–C– Double bonds –C=C– 4. Saturated fatty acids solidified more readily when cooled. bonds 5. Saturated fatty acids are for energy storage and they easily associate More H:C ratio Less H:C with cholesterol. ratio 6. Saturated fatty acids are found in animal fats except fish oil. Easier to 7. Unsaturated fatty acids are found in plant oils such as corn and solidify Less tendency to rapeseed. Less solidify reactive Easier to Found metabolise more in animals More in plants Quick Check 3 1. Differentiate fat and oil. 2. What are the functions of phospholipids other than for the formation of membrane? 3. What are the advantages and disadvantages of having low levels of cholesterol in the diet? 1.4 Proteins Learning Outcomes 1. Proteins are polypeptides, biopolymers that are formed from amino acids. Students should be able to: (a) classify amino acids 2. Amino acids are carboxylic acids, which have amino groups. There are 20 amino acids used as basic units for protein synthesis. However into four main classes there are over a hundred others. based on their side chains: polar, non-polar, Structure and Classification of Amino Acids acidic and basic; 1. The basic structure of amino acid is as shown below: (b) describe the structure of an amino acid and H H O the formation of peptide | | || bonds in polypeptides; H–N–C–C–O–H (c) explain the properties | of protein (amphoteric, R isoelectric point, buffer and colloid); 2. Each amino acid is different from the other by having different (d) differentiate the various R-side chains. levels of organisation of protein structure (primary, secondary, tertiary and quaternary) and relate the functions of each structure to the organisation of proteins; 22 Biology Term 1 STPM Chapter 1 Biological Molecules 3. The 20 amino acids are divided into four main classes depending on (e) explain the denaturation CHAPTER their R-side chains. and renaturation of 2010 protein; (a) Non-polar / aliphatic group: glycine, alanine, valine, leucine, (f) classify proteins according 1 isoleucine, methionine and phenylalanine. to their structures, (b) Polar group: serine, threonine, cysteine, asparagine, tyrosine and compositions (simple and glutamine. conjugated) and functions. (c) Acidic group: aspartic acid and glutamic acid. (d) Basic group: lysine, arginine, histidine, tryptophan and proline. Summary 4. The four classes of amino acids are as shown below. (a) non-polar class Structure of amino acid HO HO H H O H3N+ C C H3N+ C C | H OH H3N+ C C CH2 O– H3C CH O– NH2 – C – COOH C H3N+ C O CH CH2 | O– CH H3N+ C C R H O– CH3 O– CH3 CH3 CH3 CH3 CH3 Carboxylic acid with amine Glycine (Gly) Valine (Val) Leucine (Leu) Isoleucine (Ile) Alanine (Ala) H O H3N+ H O Summary H3N+ C C O– CC O– H O CH2 Classification of amino CC O– CH2 acids H3N+ CH2 H O 1. Polar – e.g. serine with NH H2N+ C C O– CH2 H2C CH2 hydroxyl side chain 2. Non-polar – e.g. glycine S with hydrophobic side CH3 CH2 chain 3. Acidic – e.g. glutamic Phenylalanine (Phe) Methionine (Met) Tryptophan (Trp) Proline (Pro) acid with carboxylic acid side chain (b) Polar class 4. Basic – e.g. Lysine with basic amine side chain H O H3N+ CC O– CH2 H O H O H3N+ CC O– H3N+ CC O– CH2 CH OH OH CH3 OH 2013 Threonine (Thr) Tyrosine (Tyr) Serine (Ser) H O H3N+ H O H CC O– CC O– CC H3N+ CH2 CH2 H3N– CH2 O O– CH2 C C SH NH2 O NH2 O Glutamine (Glu) Cysteine (Cys) Asparagine (Asa) 23 CHAPTER Biology Term 1 STPM Chapter 1 Biological Molecules (c) Acidic class 1 HO HO H3N+ C C H3N+ C C CH2 O– CH2 O– CH2 CC –O O –O O Aspartic Glutamic acid (Glu) acid (Asp) (d) Basic class HO H O H3N+ CC CC O– CH2 H3N+ CH2 O– CH2 CH2 HO CH2 CH2 H3N+ CC CH2 CH2 NH O– NH C = NH2+ NH3+ NH2 NH+ Lysine (Lys) Arginine (Arg) Histidine (His) Formation of Peptide Bond in Polypeptides 1. Two amino acids can be linked by a peptide bond by condensation catalysed by enzyme to form dipeptide as shown in Figure 1.19. HHO HHO Peptide bond HHOHHO H N C COH+H N C CO HNCCNCCOH + HOH R1 R2 R1 R2 Water Amino acid 1 Amino acid 2 Dipeptide Figure 1.19 Formation of a dipeptide Since there are 20 different types of amino acids, the number of different combinations of dipeptides is 20 × 20 = 400. Tripeptides and polypeptides can be formed from dipeptides. Exam Tips 2. Formation of polypeptide: (a) This is the protein synthesis in which amino acids are linked Remember how amino together to form a long chain of polypeptide or protein. acids form peptide bond (b) The process happens in cells. A gene controls the synthesis of and polymerisation. each type of protein. (c) Human cells contain about 30, 000 such genes, which correspondingly control the synthesis of such large number of different proteins. 24 Biology Term 1 STPM Chapter 1 Biological Molecules (d) The sequence of amino acids in each type of protein is determined CHAPTER by the base sequence within a gene, a part of DNA. 1 (e) The part of DNA has to be transcribed or copied into another form called mRNA before it is translated into protein by the Summary ribosomes. Formation of peptide bond (f) The transcription occurs in nucleus catalysed by RNA polymerase in polypeptide protein and translation occurs in the cytoplasm involving tRNA catalysed synthesis → transcription by a ligase enzyme. → translation → amino acids are condendesed in Properties of Proteins fix sequence on ribosome and linked by peptide 1. Globular protein are soluble in water. However, as macromolecules, bond. these proteins exhibit a hydrophillic surface preventing them to dissolve entirely. Instead, they disperse evenly throughout water Brownian Motion forming colloid. Colloid exhibits the following effects: (a) Colloid if added to water form sols. When the amount of water INFO is reduced by heating, it forms gel. (b) Colloid cannot be crystallised but form amorphous mass when Summary dried and the protein is denatured by heating. (c) Colloid exhibits Tyndall effect i.e. if shone with light, cloudiness Properties of proteins is observed. This is due to the reflection of light by the 1. Colloid – solution of macromolecules. (d) Colloid exhibits Brownian movement i.e. if observed under the large molecules microscope, vibrations can be seen as the protein molecules 2. Amphoteric – with both obtain kinetic energy. positive and negative 2. Proteins are amphoteric in water. This means protein molecules can charges due to side behave like acid and alkali. Proteins contain carboxylic side chains chains of component that can behave like an acid. amino acids 3. Isoelectric point – least –COOH → –COO– + H+ soluble due to inter- They also contain amino side chains that can behave like an alkali. molecular bonding of positive and negative –NH2+ + H+ → NH3+ charges Therefore, the protein molecules are zwitterions in water, exhibiting 4. Buffer – can resist pH changes due to side both positive and negative charges. chains that bind with H+ 3. At isoelectric point, which is when the total amount of positive or OH– ions. charges is equal to the total amount of negative charges at a certain pH, proteins are least soluble in water or become insoluble precipitate. This is caused by the intermolecular bonding of the two charges. 4. Protein molecules can act as buffers i.e. they can resist pH changes when a small amount of acid or alkali is added. The presence of amino side chains neutralise acid. –NH2 + H+ → –NH3+ 25 CHAPTER Biology Term 1 STPM Chapter 1 Biological Molecules The carboxylic side chains, on the other hand, neutralise alkali. –COOH + OH– → –COO– + H2O 1 Levels of Protein Structure 1. This refers to the four levels of organisation of the protein structures. The protein molecules are large, complex and can be divided into primary, secondary, tertiary and quaternary levels. (a) Primary structure 2011 (i) This refers to the sequence of amino acids in the protein linked together by peptide bonds. This also includes the number of amino acids, the number of chains and the position of the disulfide bonds that link the chains as in insulin. HO H – N+ – gly–ile–val–glu–gln–cys–s–s–cys–ser–leu–tyr–gln–leu–glu–asn–tyr–cys–asn–C Chain A: 21 amino acids Chain B: 30 amino acids H cys val O– s ala–ser s Hs s O H–N–phe–val–asn–gln–his–leu–cys–gly–ser–his–leu–val–glu–ala–leu–tyr–leu–val–cys–gly–glu–arg–gly–phe–phe–tyr–thr–pro–lys–ala–C H O– Figure 1.20 The primary structure of bovine insulin, a protein (ii) The primary structure of each type of protein is different and unique. This is controlled by the base sequence of the gene. (iii) The primary structure determines the function of a protein as it controls the other higher levels of the structure i.e. it determines the shape of the protein molecule. So, the primary structure also determines the properties of a protein. (iv) Any changes in the primary structure can cause changes in the other levels. (v) For example, the primary structural change of the 6th amino acid in the β gene of haemoglobin from glutamic acid to valine, which is caused by a mutation leads to a genetic disease called sickle-celled anaemia i.e. the secondary, tertiary and quarternary levels are changed. Thus, the properties of the protein is also affected. (vi) The haemoglobin formed cannot fit in the valine, which is non-polar in nature. This results in the linking of haemoglobin molecules by non-polar interactions and elongation of red blood cells to become sickle-shaped. 26 Biology Term 1 STPM Chapter 1 Biological Molecules Valine CHAPTER Sickle-celled haemoglobin 1 Normal haemoglobin Chain of haemoglobin Normal red blood cell Sickle-shape red blood cell Figure 1.21 Formation of sickle-celled haemoglobin as a result of the replacement of one amino acid (b) Secondary structure (i) This refers to the α-helix coiling and the β-pleated sheet of the primary structure. ( ii ) The a-helix structure is formed by hydrogen bonding between the carbonyl group ( C=O) in one amino acid and the amide group ( N–H) of another amino acid four positions away. HO OO C N CC N CC C C CN OC O HO H H C O C H C O C N C N C C C N C N OH OH β - pleated sheet α-helix Exam Tips Figure 1.22 The secondary structure of proteins Remember the primary, secondary, tertiary and (iii) Such regularity may exist in the entire length of the primary quaternary structures of structure as in fibrous proteins or just certain portions as proteins with examples and in globular proteins like enzymes. Therefore, the proportion bonds involved. Relate the of a-helix structure in different protein varies. protein structures to their functions. (STPM 2005 (iv) Fibrous protein like keratin has very high proportion of essay question) a-helix structures. The a-helix structures in keratin found in tendon, horn and hoof are cross-linked by hydrogen bonds or disulfide bonds to form hard and tough structures. (v) Likewise, the β-pleated sheet structure is formed by regular hydrogen bonding of adjacent primary structure that lies close to each other in an anti-parallel way. It also happens in intra-chain bonding after the primary structure is folded in a ‘z’ shape. 27 Biology Term 1 STPM Chapter 1 Biological Molecules CHAPTER (vi) β-pleated sheet may exist in small portions in globular proteins like enzymes. It may exist in its entire primary 1 structure as in silk protein called fibrion. 2014 (vii) The higher the proportion of secondary structure, the more fibrous the proteins become. Fibrous proteins are insoluble in water except fibrinogen. They are tough and hard. (c) Tertiary structure (i) This refers to the folding and coiling of the already coiled primary structure of the polypeptide to form a globule, which has a three dimensional shape. Hence, tertiary structure is found in all globular proteins. With – pleated sheet With –helix coil Figure 1.23 The tertiary structure of protein (ii) Each type of protein has a unique three dimensional shape called conformation, which has a specific function. For examples, the functions of enzyme as a catalyst, insulin as a hormone and myoglobin as storage of oxygen are determined by the tertiary structure. (iii) Tertiary structure determines the globular shape of proteins. This also results in the solubility of the proteins. The outer surface has hydrophilic groups to interact with water and the inner part usually has hydrophobic groups. (iv) The bonds that are involved in the folding include the followings: • Disulphide bond, which involves two side chains of amino acid cysteine. The two SH groups form a disulfide (–S–S–) covalent bond with the removal of the two hydrogen atoms as shown in Figure 1.24. • Ionic or polar bond, which involves the side chains that are of opposite charges. For example, ionised –NH3+ bonds with –COO–. • Hydrogen bond, which involves the sharing of an electron of hydrogen atom between two side chains of different polar groups. For example, NH (amide) group with C=O (carbonyl) group. 28 Biology Term 1 STPM Chapter 1 Biological Molecules • Hydrophobic interactions which involve two side chains CHAPTER that are hydrophobic (non-polar). 1 • Van der Waal’s forces which involve two side chains that are of very weak opposite charges or of very weak hydrophobic nature. Disulphide bond Ionic or polar bond HC H HC H C Ionic bond S S OO H+ H H Side chain SH HC H N of cysteine SH HC H Disulphide link Figure 1.24 Disulphide bond and ionic or polar bond involved in tertiary and quaternary structures Hydrogen bond Hydrophobic interaction and Van der Waal’s forces N Hδ+ Oδ– Oδ – CH3 Oδ – H Hδ+ H Hδ + Hδ + CH3 Water excluded C Oδ – H C CH3 in hydrophobic N Van der Waal’s force interaction of attraction Hydrogen bonds Figure 1.25 Hydrogen bond, Van der Waal’s forces and hydrophobic interactions involved in tertiary and quaternary structures (v) The tertiary structure is determined by the primary Summary structure which determines the position of amino acid. Therefore, any replacement of amino acid in the primary Levels of protein structure structure can affect the tertiary structure and its function. 1. Primary – sequence of (d) Quarternary structure amino acids, determines (i) This refers to protein with more than one polypeptide, each overall function. with its own tertiary structure, to form a functional macro- 2. Secondary – α-helix unit. The overall structure is usually globular in shape. or β-pleated sheet, (ii) This structure is found in protein with more than one determines regular subunit. They can be of the same or different primary coiling or folding structures. especially in fibrous (iii) Such primary structures may involve one or more genes. proteins. (iv) The bonds involved are the same as those involved in 3. Tertiary – 3-D shape, tertiary structure, which include disulfide bond, ionic determines function of bond, hydrogen bond, hydrophobic interactions and van globular proteins likes der Waals forces. enzymes. 4. Quarternary – 2 or more polypeptides, determine the 3-D shape of some globular proteins. 29 Biology Term 1 STPM Chapter 1 Biological Molecules CHAPTER Exam Tips (v) For example, haemoglobin has four subunits of two different kinds i.e. 2 α-units and 2 β-units. There are two Remember the four types genes involved, α gene and β gene forming two types of 1 of proteins: the fibrous polypeptides, α and β respectively. Each polypeptide has to and globular proteins, the simple and conjugated be linked to a haem group before each of them can form a proteins; and their subunit as in Figure 1.26. functions. 2018 Polypeptide Polypeptide chain 4 (β2) chain 1 (β1) Prosthetic group (haem) Polypeptide Polypeptide chain 3 ( 2) chain 2 ( 1) Summary Figure 1.26 The quaternary structure of haemoglobin Denaturation and Denaturation and Renaturation of Protein renaturation of protein 1. Denaturation of protein 1. Proteins can be denatured by heat, extreme pH, chemicals and radiation. This is because the three dimensional structure or means less function due conformation of the proteins can be altered. The weak bonds to change in its shape. that control the shape can easily be broken. The globular shape 2. Heat supplies kinetic can unwind and each bond with one another and becomes more energy and breaks insoluble or precipitate as shown in Figure 1.27. hydrogen and ionic bonds of protein. – 3. pH changes the H+ concentration also + breaks hydrogen and – ++ – ionic bonds. 4. Organic solvent Denaturation breaks hydrophobic interactions. + Reacts 5. Renaturation means that with the denatured protein + other can be reversed to – become functional. Globular structure Precipitate 6. Renaturation is only Chain is loosen possible if the weak broken bonds can be is formed reformed. Figure 1.27 Denaturation of protein 30 2. Factors that cause denaturation include: (a) Heat. Heat increases the kinetic energy of the protein molecules and causes the weak bonds to break. For example, enzymes cannot function at high temperature. (b) pH. Too high concentration of H+ or OH– breaks ionic and hydrogen bonds. (c) Heavy metallic ions. Examples like mercury and silver ions can easily attach to negative charged groups and break weak bonds. (d) Organic solvents. Examples like benzene and petroleum either form bonds with non-polar groups in the protein or disrupt hydrophobic interactions. Biology Term 1 STPM Chapter 1 Biological Molecules (e) Radiation. X-ray and radioactive radiations can break weak Exam Tips CHAPTER and covalent bonds in the proteins. 3. Some proteins including globular proteins can withstand temperatures Remember the amphoteric, up to 80°C like those found in hot spring bacteria. Their tertiary buffering and colloidal 1 properties of proteins. structure contains more disulfide bonds and would not change shape Factors causing and denature so easily. denaturation of proteins 4. Renaturation of protein is possible when the denaturation does include heat, extreme pH, not occur beyond the critical stage. This is especially so when the chemicals and radiations. temperature is not too high or when the pH changes are minimal by lowering the temperature or reversing the pH changes respectively. Summary 5. Protein molecules and amino acids have their own isoelectric point and could be charged. Hence, protein molecules and amino acids can Classification of proteins. be separated and identified through electrophoresis. Based on: 2018 1. Structure Classification of Proteins • Globular – globule shape 1. Proteins can be classified according to their structures. – soluble in water (a) Globular proteins. They are spherical in shape and soluble e.g. enzymes in water. Examples include enzymes, insulin, antibodies and haemoglobin. Their hydrophobic side chains are found within • Fibrous the ball-like structure. Their hydrophilic side chains are found – fibre-like on the surface interacting with water. They have tertiary or – insoluble in water quaternary structure. e.g. nail (b) Fibrous proteins. They are hair-like in shape and some are in sheet or block forms as found in nail, hoof and horn. Another 2. Composition example includes collagen found in skin, tendons, cartilage, • Simple bones, teeth and the walls of blood vessels. Other examples – Only consists of include fibrion in silk, keratin found in hair, actin and myosin amino acid e.g. found in muscles. They are insoluble in water and most of them insulin cannot be digested by proteinase. • Conjugated 2017 – + non-protein prosthetic group (c) Fibrous proteins that are soluble. An example is fibrinogen, a like haem e.g. plasma protein used for blood clotting. haemoglobin 2. Proteins can also be classified according to their composition. 3. Function (a) Simple protein.They are formed only from amino acids and • Enzymic no other non-protein component. Examples include insulin, – Proteins catalysts enzymes like pepsin, fibrous proteins like myosin and collagen. e.g. amylase (b) Conjugated proteins. They are formed not just from amino • Non-enzymic acids but a non-protein component called prosthetic group. – Non-catalysts Conjugated proteins can be divided into several categories include hormones depending on the types of prosthetic groups. Most of them form and antibodies globular proteins and are soluble. The prosthetic groups include: (i) phosphate, forming phosphoproteins such as casein in Exam Tips milk. (ii) carbohydrate, forming glycoproteins such as mucus. Collagen is an example of (iii) lipid, forming lipoproteins in the membrane. a fibrous protein, so it is a (iv) haem, forming haemoproteins such as haemoglobin and non-globular protein. myoglobin. 31 Biology Term 1 STPM Chapter 1 Biological Molecules CHAPTER (v) flavin, forming flavoproteins such as dehydrogenase with FAD. (vi) nucleic acid, forming nucleoproteins in the chromosomes 1 and ribosomes. 3. Proteins can be classified according to their functions. (a) Enzymic proteins. All enzymes are protein catalysts with function to speed up biochemical reactions. Each enzyme has an active site to bind to a substrate which is then converted into a product. (b) Hormonal proteins. Some hormones like insulin, glucagon and growth hormone are proteins. They are required in small amounts to activate enzymes and control metabolic processes. (c) Transport proteins. They are found in the membrane like channel proteins and carrier proteins for transport of substances across the membrane. (d) Complement proteins. These are proteins involved in immune response. These include antibodies, interleukins, histamine and perforin. They are produced by white blood cells. (e) Structural proteins. They are part of the structure of cells or tissues. Examples of such proteins include soluble globular proteins in the cytoplasm. More noticeable ones are fibrous proteins such as collagen found in skin; and keratin found in hair. (f) Contractile proteins. These fibrous proteins are found in muscles. Examples are actin and myosin that enable muscles to contract. (g) Storage proteins. An example is casein found in milk, albumin found in eggs and gluten in aleurone found in cereal grains. Quick Check 4 1. What are the major differences among amino acids? 2. Why do globular proteins have hydrophilic amino acids on the outside and hydrophobic amino acids on the inside? 3. What makes fibrous proteins so hard? 4. Why does albumen turn white on boiling? 32 Biology Term 1 STPM Chapter 1 Biological Molecules 1.5 Nucleic Acids Learning Outcomes CHAPTER Nucleic acids are polynucleotides, biopolymers of nucleotides that include Students should be able to: 1 DNA and RNA. Nucleic acid is the important genetic material of all living (a) describe the structures organisms. of nucleotides and Nucleotides and Formation of Phosphodiester Bonds the formation of phosphodiester bonds 1. Nucleotides are biochemical compounds that are made of a base, a in a polynucleotide; pentose sugar and phosphate (Figure 1.28). (b) distinguish between DNA and RNA and the P Base three types of RNAs (mRNA, tRNA and rRNA); (c) describe the structure of DNA based on Watson and Crick model. 5ʹ N-glycosidic 4ʹ Sugar 1ʹ bond 3ʹ 2ʹ Figure 1.28 Bacis structure of a nucleotide Summary 2. The bases are nitrogenous biochemicals that are grouped as purines Nucleotides and prymidines as shown in Figure 1.29. Structure (a) Purine bases – Bases (A, C, G, T/U) – Pentose (ribose and O NH2 deoxyribose) – Phosphate (1, 2 or 3) N N N N N N N Types H2N – Ribonucleotides (A, C, G and U) N – Units in RNA Deoxyribonucleotides H H – units in DNA (A, C, G & Adenine Guanine T) (b) Pyrimidine bases Bases O O NH2 – Purines (A & G) – Pyrimidines (C & T/U) HN CH3 HN N Exam Tips O O O N N N Remember the structure of nucleotide in general H H H (consisting a base, Thymine Uracil Cytosine a pentose and three (DNA only) (RNA only) phosphates), specific examples (ATP, CTP, GTP, Figure 1.29 The nitrogenous bases of DNA and RNA TTP, UTP) and other forms (ADP and AMP). 3. Purines, i.e. adenine (A) and guanine (G), are bases with two fused hydrocarbon rings. 33 Biology Term 1 STPM Chapter 1 Biological Molecules CHAPTER Exam Tips 4. Pyrimidines, i.e. cytosine (C), thymine (T) and uracil (U), are bases with one hydrocarbon ring. Remember that the 5. The pentose sugars include ribose and deoxyribose. 1 phosphodiester bond links 6. There are two types of nucleotides i.e. ribonucleotides and nucleotides within DNA deoxyribonucleotides. Ribonucleotides comprise of bases of A, C, and RNA. Phosphodiester G and U with ribose and phosphate while deoxyribonucleotides comprise of A, C, G and T with deoxyribose and phosphate. bond is between a pentose 7. There may be 1, 2 or 3 phosphate groups bonded to it. Therefore, nucleotides can exist as nucleotide monophosphates e.g. AMP; and phosphate or a nucleotide diphosphates e.g. ADP or nucleotide triphosphates e.g. ATP. They are all soluble in water. nucleoside and phosphate 8. The bond between the base and the sugar is called the N-glycosidic bond, while the bonds between pentose and phosphates are called within a mono-phosphate the phosphoester bonds. 9. Nucleotides can be hydrolysed to form phosphates and nucleoside nucleotide. by phosphatase or nucleotidase. 10. Nucleosides can be hydrolysed by nucleosidase to form bases and pentoses. Formation of phosphodiester bond in a polynucleotide 1. This is a process of polymerisation in which nucleotides are linked together to form polynucleotides by condensation with the formation of phophodiester bonds. 2. Figure 1.30 shows a condensation reaction between the hydroxyl (–OH) group of Cnu3 colefotoidnee nucleotide and the phosphoric acid group of another forming a dinucleotide. The repeated condensation process later forms a polynucleotide. OH OH P Sugar O P OH O P OH Phosphate O O T Base CH2 O Base CH2 O Base Sugar and phosphate P C HH HH join here by P H H α-phosphodiester bond OH H H2O O H Polynucleotide OH Phospho OP OH chain diester O P OH bond O O A G CH2 O Base CH2 O Base Sugar-phosphate P backbone H HH H H H OH H Dinucleotide OH H 2 Nucleotides Figure 1.30 Formation of polynucleotide 34 Biology Term 1 STPM Chapter 1 Biological Molecules 3. There are two types of polynucleotides or nucleic acids formed i.e. CHAPTER SummaryDeoxyribonucleic acid (DNA) and Ribonucleic acid (RNA). 4. The formation of DNA is known as replication catalysed by DNA Formation of 1 polymerase while that of RNA is known as transcription catalysed phosphodiester bond. by RNA polymerase. (They will be discussed in Chapter 3.) 1. Between –OH group 5. This occurs during the synthesis of DNA before cell division or 1ofnCu3cloefoptiednetoasned of synthesis of RNA especially messenger RNA before protein or phosphoric acid group polypeptide can be formed. of another nucleotide. 2. It is a condensation DNA Structure and RNA Structure reaction catalysed DNA structure by polymerase, DNA template is required to synthesis DNA (replication) and RNA (transcription) 1. The DNA structure is based on the Watson and Crick model of Exam Tips double helix (Figure 1.31). Remember that the DNA 2. It consists of two strands of polynucleotides, which coil around each structure is based on the other on the same axis and interlock. Watson and Crick’s model of double helix. You need to 3. The nucleotides are of deoxyribonucleotide type, which compose of know how to draw it. bases A, C, G and T. They are bonded by phosphodiester bond with deoxyribose and phosphates act as the backbone. 4. The two strands of polynucleotides are complementary to each other 2014 i.e. A in one strand is bonded by hydrogen bonds with T of another 2011 strand; and C bonded with G. Thus, A is paired with T with two hydrogen bonds and C is paired with G with three hydrogen bonds. 5. The two strands are anti-parallel i.e. in one strand, the carbon in the Summary pentose is arranged in a 5’ to 3’ pattern while the other in a 3’ to 5’ pattern. DNA Structure 1. Two strands coil to form 6. Each strand is very long, and can contain up to millions of nucleotides. α-helix. 2. Bases are A, C, G and 7. One coil of the strand is 34 nm long with ten pairs of bases. The diameter of the coil is 2 nm. T. 3. Complementary pairing 8. The structure of the molecule is very stable as the whole of both strands are linked by hydrogen bonds. of A = T and C G by hydrogen bonds. 9. There are hydrophobic interactions between the bases in the 4. Backbone of centre aided by water molecules surrounding it. These interactions deoxyribose and contribute to the stability of the molecule. phosphate is anti- parallel. 5. Each strand contains more than a million nucleotides 35 Biology Term 1 STPM Chapter 1 Biological Molecules CHAPTER O 5’ end Hydrogen bond 5’ 3’ 3’ end 1O H HO P O Base CG 4 NH O AT pairs AT HO P O CC Sugar- CG ON O HCH P phosphate TA HCH O C 3 C C CH backbones HN1 C N T N C2 N S CG AT S N C2 CG CG 1 H OH Deoxyribose T HN A O P sugar CG HO H Cytosine AT H Guanine O CG TA O H HO P O TA 3’ 5’ HO P O CH 3 4C O HN N O P Phosphate HCH O C5 C6 C CH O HCH S 3 HC NH N1 CN S HO C N C2 N GC H OH 1 Sugar - phosphate backbone of one O DNA strand H P CG 3’ end 5’ end Thymine Adenine Organic base (guanine) Hydrogen bonding of Arrangement of α-helix structure nucleotide pairs components in DNA of DNA Figure 1.31 The structure of DNA RNA structure 1. RNA consists of one strand of polynucleotide and is made up of A, C, G and U bases. The pentose is ribose. 2. There are three types of RNA i.e. ribosomal RNA (rRNA), transfer RNA (tRNA) and messenger RNA (mRNA). (a) rRNA (i) rRNA is found within the ribosome and consists about 80% of the total RNA inside the cell. They have long lifespan of several days before they are hydrolysed. (ii) Molecules of rRNA are of various sizes, some with more than 1000 nucleotides. (iii) They consist of intra-chain single-coiled sections and also double-coiled sections that are bound to protein molecules, forming the subunits of ribosomes. (iv) There are two types of ribosomes, the bigger eukaryotic (80S) ribosome and the smaller prokaryotic (70S) ribosome. The smaller ones are found in bacteria and inside mitochondria and chloroplasts. (v) The ribosomes in turn consist of two subunits. Eukaryotes: 80S → 60S + 40S Prokaryotes: 70S → 50S + 30S 36 Biology Term 1 STPM Chapter 1 Biological Molecules (vi) The bigger ribosomal subunits of eukaryotes contain about CHAPTER 3 types of rRNA and the smaller subunits contain only 1 type. In that of prokaryotes, the 50S also contain 3 types of rRNA and only one in the 30S subunits. 1 (vii) The rRNAs are transcribed from specific genes. These genes become the organisers for the nucleolus. (viii) The ribosomes in the eukaryotic cells are formed in the nucleolus. (ix) The role of rRNA is to form the ribosome. It enables the ribosome to combine with mRNA, to ‘read’ the codons in the mRNA and ‘translate’ the codons into sequences of amino acids in the primary structure of protein with the help of tRNA. So, rRNA forms ribosomes that provide two sites for tRNA to bind. Each tRNA can carry one specific amino to the site. (xi) Recent findings indicate that the rRNA in the ribosome functions as enzyme (ribozyme) where it catalyses the formation of peptide bond between the two amino acids brought to the ribosome. Summary (b) tRNA 2014 RNA Structure 3 types i.e. rRNA, tRNA (i) tRNAs are transcribed from genes in the DNA in the & mRNA – all single nucleus and are moved to the cytoplasm. They make up strand, consists of A, C, 10-15% of the total RNA in the cell. They have a shorter life G & U with ribose – All are span of several hours. “copied” (transcribed) using (ii) They have a specific shape like a clover leaf. This is formed DNA as template. by the intra-chain coiling to form double strands, aided by rRNA – Mixed with hydrogen bonds where A pairs with U and C pairs with G. (iii) Their molecules are the smallest with about 80 nucleotides. protein to form (iv) There are 61 types of tRNA, 3 less than the number of ribosome genetic codes. – No fixed shape – May contain OH Amino acid 1,000 nucleotides attaches – No codons or 3’ CA here anticodon 3’ C tRNA – Not mixed with G protein Amino acids – Fixed clover leaf Hydrogen bonds attaches here shape – About 80 5’ 3’ nucleotides – Anticodon is Unpaired found in central bases loop mRNA – Not mixed with CAU Anticodon protein Anticodon – No fixed shape – Few hundreds to Two dimensional structure Three dimensional structure thousand bases – Contains Figure 1.32 The structure of tRNA codons used for determining primary structure of protein 37 Biology Term 1 STPM Model Paper (964/1) ANSWERS Objective Questions 3. D 4. D 5. C • Such collagen fibres can cross-linked together 1. D 2. A 8. C 9. A 10. D to form even larger fibres as found in the 6. C 7. C 13. A 14. B 15. B tendon and ligaments to produce high tensile 11. A 12. C strength. • Collagen fibres also cross-linked with other fibrous proteins for attachment firmly between two bones at the joints. Structured Questions (b) • The procedure to separate organelles is by differential centrifugation after the tissue such 16. (a) Enzyme immobilisation is the technique of as liver is homogenised. binding enzyme molecules to solid medium. Thus, the enzyme can be used continuously to • Homogenisation is by sophisticated ultra- carry out industrial process. sound homogeniser to break up the cells without breaking organelles. (b) Adsorption (c) Entrapment • Chilled buffer is added before homogenisation (d) The enzyme molecules vibrate less as they are to maintain the structures and functions of organelles and enzymes. bound to solid. They are protected when the temperature is higher. They do not move about to • The homogenate obtained is centrifuged for 10 be affected by other factors minutes at 600 g force so nucleic will be spun down. 17. (a) • It is a nucleotide. • It has adenine. • The supernatant is then centrifuged at • It has three phosphate groups. 10,000 g force for 20 minutes to spin down • It has a ribose. mitochondria, ER and Golgi bodies. (b) • It is synthesised from ADP and inorganic • Then, the supernatant is further spun with • pIthiossaphsoaltueb(lPei)m. olecule and diffuses rapidly. 100,000 g force for 60 minutes to obtain • On hydrolysis of the third phosphate energy ribosomes, microtubules and microfilaments. released. • It acts as an intermediary between energy • The nuclei obtained may be mixed with yielding and energy requiring reactions. unbroken cells that need further centrifugation [maximum 3] to obtain higher concentration. (c) • It is synthesised by oxidative phosphorylation. • Similarly, mitochondria need to be separated • NADH moves to crista and transfers electrons from ER and Golgi bodies. to the electron transport chain. • H+ are pumped into the inter-membrane space • Ribosomes are most difficult to obtain as they exist as separated subunits or combined • To separate the two subunits of ribosomes, magnesium free buffer has to be used by ultra- centrifugation using gel. • to create H+ gradient. ADP Pi forming 19. (a) • At body temperature, the barrier of activation Chemiosmosis resulting and energy prevents reaction e.g. the breakdown of ATP. glycogen to occur. Essay Questions • Enzyme e.g. phosphorylase lowers the barrier 18. (a) • The structure of collagen consists of fibrous and permits the reaction to occur at that temperature. protein forming fibre to hold different tissues together as under the skin. • Enzymes allow substrate to bind to their active • Each molecule of collagen is made up of only sites to form complex. secondary structure twisted together with two similar molecules to have greater strength . • Then the enzymes act directly or indirectly on • The tripolypeptide unit is further lengthened the substrate molecules. by disulfide bonds with similar units to form longer fibre for stronger binding of tissues. • They break and reform certain bonds of the substrate to form the product. • An enzyme may bring two reactive substrate molecules together in its active site and a larger product is formed. • The enzyme molecule orientates two reacting substrate molecules so that the product can be easily formed. 223 Biology Term 1 STPM Model Paper (964/1) 20. (a) • Photosystems have pigments arranged in light harvesting clusters. • The enzyme provides a micro-environment at the active site to change substrate to product. • The primary pigment chlorophyll a is at the reaction centre. (b) • Both have active sites to bind to the substrate to form complexes. • P700 is found in PI, absorbs at 700 (nm). • P680 is found in PII, absorbs at 680 (nm). • Both involve at least two substrate molecules in • Accessory pigments, chlorophyll b and which sub-atomic particles, atoms or molecules are transferred. carotenoids, surround, primary pigment. • All accessory pigments can absorb light and • Both will break bonds of substrate before atoms or molecules are transferred. pass energy to primary pigment. • PI is involved in cyclic photophosphorylation. • Oxidoreductase transfers electron, H+ ion or • Light is absorbed results in electrons getting oxygen atom whereas transferase transfer any molecule. excited. • Electrons are emitted from P700. • An example of oxidoreductase is cytochrome • The electrons are passed to the chains of oxidase that transfers and combines electron, H+ ion and oxygen to form water whereas electron carriers. an example of transferase is hexokinase that • ATP synthesis occurs. transfers phosphate from ATP to glucose. • The electron returns to P700. (b) • Photolysis of water occurs. • Another example is succinate dehydrogenase in • The process releases H+. which H+ ions are removed whereas another • This occurs at PII. example of transferase is transaminase that • Electrons are released by PI. transfers amino group from amino acid to a • Both combine with NADP to form NADPH. keto acid. • NADPH is used to reduce PGA to become • Dehydrogenase requires a hydrogen acceptor PGAL. such as a coenzyme NAD+ or FAD whereas ordinary transferase does not require a coenzyme. 224 PRE-UTERM PRE-U STPM Text Biology CC039142a PRE-U STPM Text Term 1 specially designed for STPM Text students who are sitting for the STPM Biology examination. The comprehensive Biology notes and practices are based on the latest syllabus and exam format 1 set by Majlis Peperiksaan Malaysia. FEATURES This book will provide you with the necessary skills and strategies to ■ Comprehensive Notes and Practices excel in the subject. ■ Useful Features like Concept Maps, Learning Outcomes, Exam Tips, Info Our Pre-U & STPM Titles: Bio and STPM Taggings › Success with MUET ■ Summary › MUET My Way ■ STPM Practices › Pengajian Am Penggal 1, 2, & 3 ■ STPM Model Paper Term 1 › Bahasa Melayu Penggal 1, 2, & 3 ■ Complete Answers › Biology Term 1, 2, & 3 › Physics Term 1, 2, & 3 › Chemistry Term 1, 2, & 3 TERM › Mathematics (T) Term 1, 2, & 3 › Sejarah Penggal 1, 2, & 3 1 › Geograﬁ Penggal 1, 2, & 3 › Ekonomi Penggal 1, 2, & 3 › Pengajian Perniagaan Penggal 1, 2, & 3 eBook! Available W.M: RM30.95 / E.M: RM31.95 CC039142a ISBN: 978-967-2878-77-3 PELANGI

Student Study Notes

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SPM Biology

STPM Biology

As a Biology lover and Biology tuition teacher in Malaysia, I understand that the heavy load of homework and school projects cause the students to lose their interest in Biology. All they know about Biology is a "memorizing work" and "hard to score in exam". Therefore, I decide to figure out on how to improve the quality of the Biology notes to suit the students nowadays to gain back their curiosity on our natural world. Currently, there are significant improvements of my students on understanding the Biology. I am feel pleasure and will keep on going to make better study materials and available for everyone.

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FORM 6 / STPM BIOLOGY

First term: biological molecules and metabolism.

1. Biological Molecules

Part 1 Syllabus
Part 2 Water
Part 3 Carbohydrate - Monosaccharides
Part 4 Carbohydrate - Disaccharides
Part 5 Carbohydrate - Polysaccharides
Part 6 Carbohydrate - Polysaccharides - Starch
Part 7 Carbohydrate - Polysaccharides - Glycogen
Part 8 Carbohydrate - Polysaccharides - Cellulose
Part 9 Lipid
Part 10 Lipid - Triglycerides
Part 11 Lipid - Phospholipids
Part 12 Lipid - Steroids
Part 13 Amino Acids
Part 14 Amino Acids - Structure and Functions of Protein
Part 15 Amino Acids - Levels and Composition of Protein Structure
Part 16 Amino Acids - Properties of Protein
Part 17 Nucleic Acids
Part 18 Nucleic Acids - DNA and RNA
Part 19 Mineral Ions and Vitamins
Part 20 Osmotic, Turgor, Wall Pressure and Water Potential
Part 21 Movements of Substances through Membrane 1
Part 22 Movements of Substances through Membrane 2
Part 23 Movement of Substances through Membrane 3
Part 24 Analytical Techniques - Chromatography
Part 25 Analytical Techniques - Electrophoresis
Part 26 Analytical Techniques - X-ray diffraction

2. Structure of Cells and Organelles

Part 2 Cell
Part 3 Structure of Eukaryotic Cells as Seen under Electron Microscope
Part 4 Nucleus
Part 5 Endoplasmic reticulum
Part 6 Ribosome
Part 7 Golgi Bodies
Part 8 Lysosomes
Part 9 Peroxisomes
Part 10 Mitochondria
Part 11 Plastid
Part 12 Chloroplast of Higher Plants
Part 13 Microtubules
Part 14 Microfilaments
Part 15 Centriole
Part 16 Cilia and Flagella
Part 17 Vacuole
Part 18 Cell Wall
Part 19 Cell Membrane
Part 20 Prokaryotic Cells and Eukaryotic Cells
Part 21 Plant Tissues
Part 22 Meristem
Part 23 Parenchyma
Part 24 Collenchyma
Part 25 Sclerenchyma
Part 26 Simple Pits and Bordered Pits
Part 27 Xylem
Part 28 Phloem
Part 29 Epithelium
Part 30 The Development of Glands
Part 31 Nervous Tissues
Part 32 Connective Tissue
Part 33 Blood
Part 34 Hyaline Cartilage
Part 35 Bones
Part 36 Smooth Muscles
Part 37 Striated Muscles
Part 38 Cardiac Muscles
Part 39 Analytical Techniques - Light Microscope
Part 40 Analytical Techniques - Phase Contrast Microscope
Part 41 Analytical Techniques - Electron Microscope
Part 42 Analytical Techniques - Cell Fractionation

3. Membrane Structure and Transport

5. Cellular Respiration

6. Photosynthesis

Second Term: Physiology

1. Gas Exchange

2. Transport in Animals and Plants

3. Control and Regulation

4. Reproduction, Development and Growth

5. Homeostasis

6. Immunity

7. Infectious Diseases

Third Term: Ecology and Genetics

1. Taxonomy and Biodiversity

3. Selection and Speciation

4. Inheritance and Genetic Control

5. Gene Technology

6. Biotechnology

9 comments:

Thank you for updating ,aunty happy! love your notes so much. cant wait to read the sem2 notes!

You are welcome. Hope these will help you in your exams.

This comment has been removed by the author.

Do you have SEM 2 notes?

Hi. Im currently busy with family matters. Once settled down will continue update the learning materials. Thank you and sorry for any inconvenience.

when will you update? i really like your blog

hi aunty happy, could you provide me your email address? i would like to pay you tuition fees to attend bio course from you :)

Hello teacher, I have been studying your notes and it's really helpful for me. So please continue your notes for all the chapters as well , I'm really depending on it. Hope your family issues have resolved. Thank you very much .

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Tuesday, August 29, 2006

Stpm practical biology – assessment and summary of experiments.

Data, presentation, analysis, and others
Creativity/Innovation
Overall quality of report

5 comments:

Well, I have this great fear for frogs. I really really am very afraid of amphibians. But we do have to dissect frogs right for the biology practical? So, is this carried out individually or in a group? Oh my god, So, i am going to fail huh?

hi..em last practcl,we done individually on hamster..dnt skip that practical,u cn do that..it is only for about an hour...all the best!chayo2 fighting!!hehe

where can i find readily preserved insects?

well u can find ppreserve insect in Batu Caves. There are few shops which presere insects. Hope this info help you :D

Just a matter of asking. May I know whether should STPM Biology consists of hypothesis, variables and problem statement? If we put this into the report, what will happen?

Malaysia Students Blog is a team blog on Malaysian major examinations, secondary, pre-university & tertiary education, scholarship Malaysia , student resources, students' thoughts and everything relating to students & undergraduates at schools, colleges & universities in Malaysia - Student Education Malaysia Comment Policy: Comments posted at Malaysia Students blog should be on-topic, constructive and add value to the discussion . Comments that are off-topic, one-sentence, abusive or offensive will be removed. Please use proper English with correct spelling and grammar in your comment. For general questions, please post them at SPM Student Malaysia . For enquiries, please email the administrator of this blog: Student at Malaysia-Students dot com.

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Vgrow Art Learning Center

Stpm biology sem 1 b5: cellular respiration, course curriculum, 5: cellular respiration.

5.1 Glycolysis

5.2 Krebs Cycle

5.3 Electron Transport System (Part I)

5.4 Electron Transport System (Part II) & ATP Calculation

5.5 Anaerobic Respiration & Ending

About this course

7 hours of video content

Biology Specialist

STPM Term 1 Biology Notes (5 books - single sided 177 pages)

https://shopee.com.my/PR6-STPM-TERM-1-BIOLOGY-NOTES

* Click this link to purchase the notebooks

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Monday 30 July 2012

Stpm biology experiment 2 (onion).

http://education.cambridge.org/media/113248/biology_practical.pdf
http://waynesword.palomar.edu/lmexer1.htm#onion
http://www.uq.edu.au/_School_Science_Lessons/UNBiol1.html#9.5 6

2 comments:

I coul not open the link 1 and 2

Sorry, those links become invalid now. I don't have backup links.

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STPM Biology Projects! , Plants, Insects, Ecology...

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ms. sunflower	wow... y so cheap de??? mine 20 ald rm70 lo. Tmn Ehsan industrial park? sure no lo... i everyday pass by oso nvr c... wen u bought de?
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D3V1L	Hey, any1 can introduce few websites as reference to let me do my ecology project??? This post has been edited by : Feb 20 2009, 09:05 PM
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New Member	NEUROPTERA - Ant lions PHASMATODEA - Stick Insects MANTODEA - Praying Mantis Students can reach me at malayainsects(at)yahoo(dot)com <<at=@ dot=.>> or my facebook page at : Inquires are most welcome, perhaps you just need to identify your insects feel free to drop me a mail and I will try my best to help at my free time. Students from other states who is keen on getting insects can also contact me as we can post insect specimens to you.
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I'm just a human who wants to share my knowledge with the hope that I can help at least one person with what I know. I was - still do, search on internet before wanting to do anything and it would help a lot to have some sort of insight before starting it so here you go people

STPM Semester 1 Biology Experiment 2 : Preparation and observation of slides of animal and plant cells

Thankyou it is very helpful

Biology stpm sem 1 experiment 3 answers plss

My teacher said answer in part B 1st questions is protoxylem, epidermis and parenchyma.

Thank you so much

Animal and plant cells are fundamental units of life with key differences. Animal cells lack a cell wall and chloroplasts, but possess centrioles. Plant cells have cell walls, chloroplasts for photosynthesis, and large central vacuoles. Both contain organelles like the nucleus, mitochondria, endoplasmic reticulum, Golgi apparatus, and lysosomes. These organelles perform specific functions vital for cell survival. Animal cells are typically round and irregularly shaped, while plant cells have a more defined rectangular shape due to the cell wall. Both types of cells are essential building blocks for complex organisms.

STPM Semester 1 Chemistry Experiment 1 : Volumetric Analysis - Stoiciometry

MusicChocolateFizz

Monday 24 september 2012, comparison between c3 and c4 leaves biology practical(experiment 7) stpm 2012.

Thank you so much 😀

STPM BIOLOGY

Friday, april 13, 2012, stpm biology - experiment 8: dissection of the mammalian digestive system.

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Biology Specialist. Mr. Khor is a SPM Biology Trainer under The Star Media Group, whereby he provides workshops and tuition to SPM students in the Klang Valley. Attached to Yayasan Perak, he conducts SPM Biology Seminars in all the 10 Perak districts, educating about 3000 Biology students annually.
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November (13) October (21) September (9) August (11) July (3) Stpm biology experiment 2 (onion) STPM mathematics T chapter 1 note (Function) Modular System For STPM 2012; Blogger profile ... Stpm biology experiment 2 (onion) To determine the size of onion scale leaf epidermal cells.
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Hi guys or any new students searching insects for their biology projects for their 3rd Semester. I have been an established insect seller, and I am able to provide some of the following orders of insects: LEPIDOPTERA - Butterflies and Moth. COLEOPTERA - Beetles. ORTHOPTERA - Grasshoppers, Katydids and Locusts.
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