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Pearson Edexcel · International A-Level · Biology

  • 1

    Molekul, Diet, Transportasi dan Kesehatan

    1.1

    From molecules to a working circulation

    A red blood cell leaving your lung carries oxygen on haemoglobin, rides an artery, squeezes through a capillary narrower than itself, releases the oxygen to a respiring cell, and returns through a vein. Every structure in that journey is built from the molecules this unit starts with, and every exam question about it asks you to connect structure to function.

    WBI11 Molecules, Diet, Transport and Health is the first IAS unit: a written paper of 1 hour 30 minutes, 80 marks, all compulsory. It assesses Topic 1 (Molecules, Transport and Health) and Topic 2 (Membranes, Proteins, DNA and Gene Expression). Expect data questions on health studies, calculations with standard form, and extended answers that name structures and explain mechanisms.

    1.1

    Water, carbohydrates and food tests

    Silabus

    Pearson Edexcel IAL Biology Specification Issue 2 (February 2021), Topic 1 statements 1.1-1.4 with Core Practical 1 (spec pp.15-16). These are local teaching subdivisions of the official statements.

    Water as the transport solvent: dipole nature, hydrogen bonding between molecules, and why polar and ionic substances dissolve. Carbohydrates: mono/di/polysaccharides (glucose, fructose, galactose; maltose, sucrose, lactose; glycogen, amylose, amylopectin); condensation forming glycosidic bonds and hydrolysis splitting them; relating structure to energy supply and storage (alpha-glucose helix vs branched amylopectin/glycogen; cellulose excluded here). Core Practical 1: semi-quantitative Benedict's estimation of reducing-sugar concentration and iodine estimation of starch using colour standards.

    Sumber: Silabus Cambridge International

    Water is the transport medium because it is a polar solvent 极性溶剂. In a water molecule, oxygen pulls the shared electrons more strongly than hydrogen: each O–H bond has a slight negative end at O and a slight positive end at H. This is the dipole 偶极 nature of water. Opposite charges between neighbouring molecules form hydrogen bonds 氢键, and polar or charged particles (glucose, sodium ions, amino acids) dissolve because water clusters around them.

    Carbohydrates are joined and split by two reactions:

    • Kondensasi 缩合 removes a water molecule as a glycosidic bond 糖苷键 forms between two sugars.
    • Hidrolisis 水解 adds water back and breaks that bond.

    Two glucose molecules condense to maltose; glucose plus fructose give sucrose; glucose plus galactose give lactose. Long chains make polysaccharides: amylose 直链淀粉 (1,4-linked, coils into a helix, compact), amylopectin 支链淀粉 dan glycogen 糖原 (both 1,4-chains with 1,6-branches; glycogen is the animal store in liver and muscle). Branching means many chain ends, so glucose can be released quickly; the helix means a lot of energy fits in a small space.

    Core Practical 1: estimating sugar and starch concentration

    Reducing sugars reduce blue Benedict's reagent on heating, giving a brick-red precipitate. To estimate an unknown concentration, compare the colour against standards of known concentration, or read the solution in a colourimeter.

    Calibration of Benedict's reagent: absorbance of colour standards against glucose concentration, with an unknown read off the straight part of the line.

    Worked check. Standards of 0, 0.5, 1, 2 and 4 % glucose give absorbances 0.02, 0.09, 0.18, 0.35, 0.64. An unknown gives 0.58. On the straight section, each 1 % adds about 0.16 absorbance, so the unknown is near $0.58/0.16 \approx 3.6\%$. Reading beyond the standards (extrapolating past 4 %) is less reliable than measuring within them — a point the examiners like you to make. Iodine turns blue-black with starch; intensity of the blue-black gives the same kind of estimate for starch.

    1.2

    Lipids: triglycerides and ester bonds

    Silabus

    Pernyataan 1.5 (spes p.16): sintesis trigliserida melalui kondensasi gliserol dengan tiga asam lemak membentuk ikatan ester; asam lemak jenuh vs tak jenuh (tidak ada ikatan rangkap vs satu atau lebih C=C, rantai lurus yang rapat vs rantai bengkok, konsekuensi titik lebur). Kaitan ke risiko PJK (topik sub-1.5) melalui kimia lipid diet dan uji emulsi etanol.

    Sumber: Silabus Cambridge International

    A triglyceride 甘油三酯 is one glycerol plus three fatty acids, joined by three condensation reactions that form ester bonds 酯键. Saturated 饱和 fatty acids have no C=C bonds: straight chains pack closely, so animal fats are solid at room temperature. Unsaturated 不饱和 acids have one or more double bonds that kink the chain, packing is looser, and oils are liquid. Lipids store more energy per gram than carbohydrates because they are more reduced.

    1.3

    Why animals need a pump: the heart and vessels

    Silabus

    Pernyataan 1.6-1.8 (spes pp.16-17): mengapa hewan multiseluler memerlukan sirkulasi transportasi massa (batas difusi); struktur-fungsi kapiler, arteri, dan vena (lapisan dinding, lumen, katup, jaringan otot elastis dan otot polos); ruang jantung mamalia, katup, dan pembuluh utama; siklus jantung (sistol atrium, sistol ventrikel, diastol) dengan perubahan tekanan dan volume serta penutupan katup. Detail stimulasi miogenik TIDAK diperlukan pada IAS (catatan spes).

    Sumber: Silabus Cambridge International

    Diffusion supplies cells only a few tenths of a millimetre thick. A big organism needs mass transport 大规模运输: a pump and pipes that carry every cell's supply and waste. The three vessel types match three jobs:

    Vessel Wall Mengapa
    Arteri thick muscle and elastic fibres, small lumen carries blood at high, pulsing pressure; elastin smooths the pulse
    Kapiler one cell thick endothelium short diffusion path, huge total surface area; tissue fluid leaks out at the start
    Vena thin wall, wide lumen, valves low pressure return; valves and the surrounding muscles stop backflow

    The mammalian heart is two pumps in one organ. The right side sends blood to the lungs; the left side, with the thicker ventricle wall, sends blood to the body. Atria receive; ventricles deliver. The coronary arteries feed the heart muscle itself — blockage there is a heart attack.

    Siklus jantung

    One beat has three phases. Sistol serambi 心房收缩期 squeezes the last blood into the ventricles. Sistol ventrikel 心室收缩期 raises ventricular pressure above arterial pressure, the semilunar valves open, and blood is driven out. Diastole 舒张期 relaxes the muscle, pressure falls, the semilunar valves slam shut (the second heart sound), and the chambers refill from the veins.

    One cardiac cycle: ventricular and aortic pressure with ventricular volume. Valves change state whenever the two pressure curves cross.

    Read the figure the way an examiner wants: valve state is decided by pressure difference, never by an instruction. Atrio-ventricular valves open when atrial pressure exceeds ventricular pressure; semilunar valves open when ventricular pressure exceeds aortic pressure.

    1.4

    Haemoglobin and the oxygen journey

    Silabus

    Pernyataan 1.9 (spes p.17): struktur kuaterner hemoglobin (empat polipeptida, masing-masing memiliki gugus heme), pengikatan kooperatif yang ditunjukkan oleh kurva disosiasi sigmoid, transportasi oksigen dan karbon dioksida, dan efek Bohr - peningkatan tekanan parsial karbon dioksida menggeser kurva ke kanan seiring masuknya ion hidrogen ke hemoglobin. Bandingkan kurva hiperbola mioglobin di mana protein penyimpan rantai tunggal muncul.

    Sumber: Silabus Cambridge International

    Haemoglobin is a protein of four polypeptide chains, each folded around a haem group carrying one Fe²⁺ ion that binds one O₂ molecule — four in total. Binding the first oxygen changes the shape of the whole molecule and makes the next bindings easier: cooperative binding 协同结合 gives the oxygen dissociation curve its shallow-then-steep S shape.

    Oxygen dissociation curves: the sigmoid curve for normal blood and the rightward Bohr shift when carbon dioxide rises.

    In the lungs (high pO₂) the curve flattens near 100 % saturation. In resting tissue near 1–2 kPa the curve is steep, so a small fall in pO₂ unloads a large amount of oxygen — exactly where it is needed. When tissue respires hard it releases more carbon dioxide; hydrogen ions attach to haemoglobin and shift the whole curve to the right (the Bohr effect 波尔效应), forcing even more oxygen to unload in the busiest tissue.

    Worked check. At pO₂ 2 kPa the normal curve reads about 30 % saturation and the shifted curve about 15 % — haemoglobin holds only about a third as much oxygen at that partial pressure, so most of its four sites are released to the tissue sooner.

    1.5

    Cardiovascular disease: a chain of events

    Silabus

    Pernyataan 1.10-1.13 dengan Praktikum Inti 2 (spes pp.17-18): perjalanan aterosklerosis (kerusakan endotel, respons peradangan, deposisi LDL, pertumbuhan plak, risiko aneurisma dan trombose); kaskade pembekuan (pelepasan tromboplastin, protrombin menjadi trombin, fibrinogen menjadi fibrin); tekanan darah (sistolik/diastolik, hipertensi sebagai faktor risiko) dan Praktikum Inti 2 penyelidikan detak jantung (Daphnia atau setara). Faktor risiko saling mengalikan, bukan hanya menjumlahkan.

    Sumber: Silabus Cambridge International

    Atherosclerosis 动脉粥样硬化 begins with damage to the endothelium 内皮 lining an artery. The inflammatory response grows, low-density lipoproteins invade, smooth muscle and connective tissue build a plaque 斑块, the artery narrows and hardens, and a clot can finally block it or break loose.

    Clotting itself is a cascade. Damage releases thromboplastin 凝血酶原激酶, which (with calcium ions) converts the soluble plasma protein prothrombin 凝血酶原 into the enzyme thrombin 凝血酶; thrombin converts soluble fibrinogen 纤维蛋白原 into the insoluble mesh of fibrin 纤维蛋白 that traps platelets and cells.

    Blood pressure 血压 is reported as systolic over diastolic. Hypertension, smoking, high blood LDL-cholesterol, diet, inactivity, age, gender and genetics are risk factors 危险因素 — and they multiply, not add: two moderate risks can matter more than one severe risk.

    Reading health-risk data

    Questions 8 and 9 of every paper live here. Four habits cover almost every mark:

    • Correlation is not causation. 相关不等于因果 Two variables moving together may share a hidden cause, or the link may be chance.
    • Judge the study: How large was the sample? How was it selected? What was controlled? Was it blind? Are there confounders 混杂变量 such as age or smoking?
    • Do the arithmetic carefully: percentage change, ratios in standard form, per-capita rates.
    • Ask who or what is missing: non-responders, unpublished studies, extrapolated lines.

    Worked check. A study finds gum disease correlates with endothelial damage. Before claiming one causes the other you would want a mechanism, a dose–response pattern, and confounders (diet, smoking, age) excluded — and the question "were the groups matched for blood pressure?" is exactly a confounder question.

    1.7

    Diet, cholesterol and treating CVD

    Silabus

    Pernyataan 1.19-1.20 (spesifikasi hlm.18): bagaimana pengetahuan ilmiah tentang diet (keseimbangan energi, obesitas diukur dengan IMT dan rasio pinggang-panggul, garam, lemak jenuh vs tak jenuh) informing pilihan; manfaat dan risiko pengobatan KJD - antihipertensi (beta-blokir, inhibitor ACE, diuretik), statin (peningkatan reseptor LDL, efek samping otot dan hati), antikoagulan (warfarin, aspirin) dan obat penghambat trombosit - bersamaan dengan intervensi gaya hidup.

    Sumber: Silabus Cambridge International

    Energy balance decides body mass. Overweight and obesity are screened by BMI (mass in kg divided by height in metres squared; 18.5–24.9 is the healthy band) and waist-to-hip ratio (WHR). Dietary saturated fat raises blood LDL-cholesterol; unsaturated fat and soluble fibre lower it.

    Worked check. A woman of mass 74 kg and height 1.66 m: BMI $= 74 / 1.66^2 = 74/2.76 = 26.8$ — just into the overweight band. Waist 84 cm and hips 98 cm give WHR $= 84/98 = 0.86$.

    Treatments all carry risks alongside benefits:

    • Antihypertensives 抗高血压药 (beta blockers, ACE inhibitors, diuretics) lower blood pressure.
    • Statins 他汀类 (for example simvastatin) block a liver enzyme in cholesterol synthesis, so cells take up LDL from blood.
    • Anticoagulants 抗凝剂 (warfarin) and antiplatelet drugs (aspirin) reduce clotting, so bleeding risk rises.

    A question asking you to "discuss" wants both sides plus a judgement anchored in the data given.

    1.8 1.9

    Membranes and transport

    Silabus

    Topik 2 pernyataan 2.1-2.3 dengan Praktikum Inti 3 (spesifikasi hlm.19-20): sifat permukaan pertukaran (luas permukaan besar, tipis, gradien konsentrasi terjaga); struktur membran dari model mozaik cair - bilayer fosfolipid, protein intrinsik dan ekstrinsik, kolesterol, glikolipid dan glikoprotein; sejarah model membran (Overton hingga Singer-Nicolson) dan Praktikum Inti 3 penyelidikan permeabilitas membran (bit pada suhu atau konsentrasi pelarut yang bervariasi).

    Pernyataan 2.4-2.5 (spesifikasi hlm.20): osmosis sebagai pergerakan molekul air bebas dari potensial air lebih tinggi ke lebih rendah melalui membran semi-permeabel; transpor pasif (difusi, difusi terbantu melalui protein pembawa dan saluran) vs transpor aktif (ATP, melawan gradien); endositosis dan eksositosis untuk material massal; Hukum Fick diterapkan pada luas permukaan, gradien konsentrasi, dan jarak difusi (laju sebanding dengan SA x gradien / ketebalan).

    Sumber: Silabus Cambridge International

    Every exchange surface is a membrane doing controlled business. The fluid mosaic model 流动镶嵌模型 is the accepted picture: a phospholipid bilayer (fluid because the tails can slide), studded with intrinsic proteins that span the bilayer (channels, carriers), extrinsic proteins, cholesterol that controls fluidity, and glycoproteins and glycolipids for recognition.

    Four transport routes:

    • Difusi 扩散 — net movement down a concentration gradient, no ATP.
    • Facilitated diffusion 易化扩散 — down the gradient through a channel or carrier protein.
    • Transportasi aktif 主动运输 — against the gradient, powered by ATP, through carrier proteins that change shape.
    • Osmosis 渗透 — water moving from higher to lower water potential 水势 across a partially permeable membrane. Endocytosis and exocytosis move bulk material in vesicles.

    For diffusion, Fick's law: rate is proportional to (surface area × concentration difference) ÷ diffusion distance. Gas exchange surfaces answer all three: huge area, maintained gradient, and a wall one or two cells thick. This is why alveoli, capillaries and fish gills all look the way they do.

    1.10

    Proteins and enzymes

    Silabus

    Pernyataan 2.6-2.8 dengan Praktikum Inti 4 (spes p.20): struktur asam amino (gugus amino, gugus karboksil, gugus R, pembentukan ikatan peptida melalui kondensasi); tingkat struktur protein (urutan primer hingga tersier dan kuaterner yang terikat hidrogen/ion/disulfida); aksi enzim - spesifisitas situs aktif, penyesuaian induksi, pengaruh suhu/pH/konsentrasi substrat terhadap laju, inhibisi kompetitif dan non-kompetitif; Praktikum Inti 4 menyelidiki aktivitas enzim (suhu atau pH atau [S] dengan contoh amilase-pati). Aplikasi enzim terimobilisasi dan berantai.

    Sumber: Silabus Cambridge International

    An amino acid carries an amino group, a carboxyl group, a hydrogen and an R group on one central carbon; two join by a peptide bond 肽键 in a condensation reaction. The primary structure 一级结构 is the sequence; hydrogen bonds coil it into the sekunder 二级结构; further hydrogen, ionic and disulfide bonds fold the tertiary 三级结构; separate chains assemble into the quaternary 四级结构 — haemoglobin is the classic example.

    An enzyme 酶 is a protein whose active site 活性部位 fits one substrate shape. Binding is not rigid: the induced-fit 诱导契合 model has both molecule and site flexing until complementary. This is why temperature and pH change rate — they act on the tertiary structure — and why the active site is specific.

    Enzyme rate curves: temperature rises then collapses as the protein denatures above its optimum; pH shows a sharp optimum; substrate concentration saturates, and a competitive inhibitor is out-competed at high concentration.

    Competitive inhibitors 竞争性抑制剂 bind the active site and are outnumbered as substrate concentration rises; non-competitive inhibitors 非竞争性抑制剂 bind elsewhere and distort the site, so rate falls whatever the concentration.

    Worked check. Q10 is quoted at 2.4: between 10 °C and 20 °C the rate is 2.4 times faster; between 45 °C and 55 °C the same factor means the denatured enzyme is collapsing at an increasing rate. Q10 only describes; it does not explain.

    1.11

    DNA, replication and the code

    Silabus

    Statements 2.9-2.12 (spec p.21): mononucleotide structure (pentose sugar, phosphate, nitrogenous base; deoxyribose vs ribose), DNA polymer formation by phosphodiester bonds, double helix with complementary base pairing and antiparallel strands; semi-conservative replication (helicase unwinding, DNA polymerase, free nucleotides, Meselson-Stahl evidence); a gene as a base sequence coding for an amino-acid sequence; the triplet code - non-overlapping, degenerate, universal, with start and stop codons.

    Sumber: Silabus Cambridge International

    A mononucleotide 单核苷酸 is a pentose (deoxyribose or ribose), a phosphate and a nitrogenous base. Condensation between phosphate and sugar of neighbours makes a strand held by phosphodiester bonds 磷酸二酯键. DNA is two antiparallel strands: A pairs with T (two hydrogen bonds), G with C (three), twisted into a double helix.

    Replication is semi-conservative 半保留: helicase unzips the two strands, each is a template, and DNA polymerase DNA聚合酶 joins free nucleotides in the 5' to 3' direction. Meselson and Stahl grew bacteria first on heavy ¹⁵N then light ¹⁴N nitrogen; after one round of replication all the DNA was one intermediate density, after two rounds half was light — exactly what semi-conservative predicts and the alternatives forbid.

    A gene is a sequence of bases coding for a sequence of amino acids. The code is a triplet 三联体 code: three bases per amino acid. It is non-overlapping 非重叠 (read in successive groups), degenerate 简并 (most amino acids have several codons, so some mutations are silent), and carries start and stop codons.

    Worked check. A gene of 300 base pairs coding a protein of 99 amino acids plus a stop codon: $99 \times 3 + 3 = 300$. ✓

    1.12

    From gene to protein — and when it goes wrong

    Silabus

    Pernyataan 2.13-2.18 (spes pp.21-22): transkripsi (RNA polimerase, untai templat, pemrosesan pre-mRNA) dan translasi (kodon mRNA, antikodon tRNA dengan asam amino, ribosom); mutasi akibat kesalahan replikasi DNA - substitusi, insersi, delesi dengan konsekuensi pergeseran kerangka baca; alel, genotipe, fenotipe, dominansi, dan kaitan seks sejauh tuntutan WBI11 (fibrosis kistik sebagai model efek mutasi gen terhadap transportasi klorida CFTR, akumulasi lendir, risiko infeksi); skrining genetik (identifikasi pembawa sifat, diagnosis genetik pra-implantasi, amniosentesis dan pengambilan vili korion) beserta isu etika dan sosial dari program skrining.

    Sumber: Silabus Cambridge International

    Transcription: RNA polymerase binds the gene, uses one strand as template and builds mRNA 信使RNA. Translation: in the cytoplasm, a ribosome 核糖体 reads the mRNA codons, and tRNA 转运RNA molecules whose anticodons 反密码子 pair with those codons deliver amino acids in order; peptide bonds join them.

    Mutasi 突变 arise from replication errors. A substitution 替换 swaps one base: one amino acid changes, or nothing changes (degeneracy). An insertion 插入 atau deletion 缺失 adds or removes a base and shifts every reading frame after it — a frameshift 移码 — usually destroying the protein.

    Cystic fibrosis is the course's model: deletion of three bases removes one phenylalanine from the CFTR protein, so chloride transport fails, mucus stays thick, and airways clog. The allele is recessive: carriers are healthy. Genetic screening 基因筛查 can identify carriers, test embryos (pre-implantation genetic diagnosis) or foetuses (amniocentesis, chorionic villus sampling). Screening questions always carry ethics: who is tested, who is told, what happens to a positive result.

    1.12

    Cek diri sendiri

    1. Explain why amylopectin releases glucose faster than amylose, referring to bonds and ends of chains.
    2. Calculate the BMI of a 91 kg man 1.83 m tall and classify him.
    3. During ventricular systole, which valves are open and which are shut, and why?
    4. The oxygen saturation at 4 kPa falls from about 60 % to about 34 % when CO₂ rises. Explain the molecular cause.
    5. Describe the clotting cascade in order, from endothelial damage to fibrin.
    6. A study reports that people who eat more yoghurt have fewer heart attacks. List three checks you would make before believing a causal claim.
    7. Explain why a competitive inhibitor lowers the rate at low substrate concentration but not at very high concentration.
    8. Meselson and Stahl ruled out conservative replication. Which observation, after one round, did the job?
    9. A frameshift mutation is usually worse than a substitution. Explain why, using the words triplet and degenerate.
    10. Name two benefits and two risks of genetic screening for cystic fibrosis carriers.

    Answers: 1 many 1,6-branch points give many chain ends for enzymes to attack simultaneously; 2 $91/1.83^2 = 27.2$, overweight; 3 atrio-ventricular shut, semilunar open, because ventricular pressure exceeds aortic but not atrial-from-above pressure; 4 hydrogen ions from CO₂ bind haemoglobin, changing its shape and lowering oxygen affinity (Bohr effect); 5 damage exposes collagen, platelets and damaged cells release thromboplastin, thromboplastin + Ca²⁺ converts prothrombin to thrombin, thrombin converts fibrinogen to fibrin, fibrin mesh traps cells; 6 sample size and selection, confounders such as age/exercise/diet, blinded comparison, mechanism or dose–response; 7 the inhibitor competes for the active site, so at very high substrate concentration nearly every collision with the site is substrate; 8 after one round all DNA was a single intermediate band — conservative predicts one heavy and one light band; 9 a frameshift changes every triplet after the mutation, so the whole protein from that point is a different amino-acid sequence, while degeneracy lets many substitutions change nothing; 10 benefits — informed family planning, earlier treatment or lifestyle change; risks — anxiety, discrimination by insurers or employers, false positives or negatives.

  • 2

    Sel, Perkembangan, Keanekaragaman Hayati dan Konservasi

    2.1

    From one cell to a living world

    You began as one fertilised egg cell. This unit explains how that cell divided, how the copies were shuffled and halved to make gametes, how identical genes became different tissues, and how the variety that produces is counted, classified and — increasingly — defended. It is the unit of microscopes, calculations and definitions done precisely.

    WBI12 Cells, Development, Biodiversity and Conservation is the second IAS unit: 1 hour 30 minutes, 80 marks, all compulsory. It assesses Topic 3 (Cell Structure, Reproduction and Development) and Topic 4 (Plant Structure and Function, Biodiversity and Conservation). Expect labelling diagrams, microscope and magnification calculations, the index of diversity and Hardy–Weinberg questions, and long answers on conservation evaluation.

    2.1

    Cell ultrastructure and the microscope

    Silabus

    Topik 3 pernyataan 3.1-3.7 dengan Praktikum Inti 5 (spesifikasi hlm.22-23). Ultrastruktur eukariotik - nukleus dan nukleolus, ribosom, retikulum endoplasma kasar dan halus, aparatus Golgi dan vesikel, mitokondria, kloroplas, lisosom, sentriol, membran permukaan sel - setiap organel dikaitkan dengan fungsinya, dan ultrastruktur prokariotik (dinding sel, kapsul, plasmid, DNA melingkar tunggal, ribosom 70S, flagelum). Organisasi menjadi jaringan, organ, dan sistem (pernyataan 3.2). Perbesaran vs resolusi; mikroskopi cahaya dan elektron (TEM/SEM); perhitungan perbesaran M = gambar/sebenarnya; peran pewarnaan. Praktikum Inti 5 memeriksa sel dengan mikroskopi.

    Sumber: Silabus Cambridge International

    A eukaryotic 真核 cell is a set of specialised compartments:

    Structure Function
    Nucleus and nucleolus DNA stored behind a double membrane; nucleolus builds ribosomes
    Ribosomes (80S) site of protein synthesis
    Rough endoplasmic reticulum folds and transports proteins made on its ribosomes
    Golgi apparatus modifies, packages and ships proteins in vesicles
    Mitokondria site of aerobic respiration, ATP production
    Chloroplast (plants) site of photosynthesis
    Lisosom digestive enzymes for worn organelles
    Cell-surface membrane controls transport, recognition

    Cells organise into tissues 组织 (similar cells), organs 器官 (cooperating tissues) and organ systems 器官系统. A prokaryotic 原核 cell has none of the membrane-bound organelles: a cell wall, a slime capsule 荚膜, one circular DNA molecule plus plasmids 质粒, 70S ribosomes, and sometimes a flagellum 鞭毛.

    Pembesaran 放大倍数 is how much bigger the image is; resolution 分辨率 is the smallest separation still seen as two. Light microscopes resolve about 200 nm; electron beams, far shorter in wavelength, resolve to a few nanometres — but need thin stained sections (TEM) or gold-coated surfaces (SEM). $M = \text{image size} / \text{actual size}$, keeping units consistent.

    Worked check. A mitochondrion measures 14 mm on a photograph labelled ×20 000. Actual length $= 14\,000\ \mu\text{m} / 20\,000 = 0.7\ \mu$m. A nucleus that must be seen with its double membrane needs an electron microscope: the two membranes lie closer than a light microscope's resolution.

    2.2

    Meiosis: halving and shuffling

    Silabus

    Pernyataan 3.9-3.13 (spesifikasi hlm.23): lokus dan tautan; meiosis sebagai dua pembelahan menghasilkan empat sel haploid yang berbeda secara genetik; pengelompokan independen pasangan homolog dan crossing over kromatid sebagai dua sumber variasi; spesialisasi gamet mamalia (akrosom sperma, mitokondria bagian tengah, sitoplasma telur dan zona pelusida); fertilisasi pada mamalia (reaksi akrosom, reaksi kortikal memblokir polispermia) dan pada tumbuhan berbunga (tabung serbuk sari, fertilisasi ganda).

    Sumber: Silabus Cambridge International

    Meiosis 减数分裂 is two successive divisions after one DNA replication, producing four genetically different haploid 单倍体 cells from one diploid 二倍体 cell. Two processes create the variation:

    • Penataan bebas 独立分配 — each homologous pair lines up and separates at random in meiosis I; each gamete gets one of each pair, from either parent at random ($2^n$ combinations).
    • Percilangan silang 交叉互换 — in prophase I, homologous chromatids swap sections, mixing alleles within chromosome pairs.

    Mammalian gametes are specialised for delivery and provisioning: the sperm 精子 carries an acrosome 顶体 (digestive enzymes), a midpiece of mitochondria and a flagellum; the egg 卵子 contributes nearly all cytoplasm and a zona pellucida 透明带 that locks out extra sperm after the cortical reaction 皮质反应. In flowering plants the pollen tube delivers two male nuclei: one fertilises the egg, the other the polar nuclei — double fertilisation 双受精.

    DNA content per cell through the cell cycle, both meiotic divisions, gametes and fertilisation. Note the doubling in S phase and the two halvings.
    2.3

    Mitosis and the cell cycle

    Silabus

    Pernyataan 3.14-3.16 dengan Praktikum Inti 6 (spesifikasi hlm.23): siklus sel (fase interfase G1-S-G2, kemudian mitosis dan sitokinesis); profase, metafase, anafase, telofase dengan apa yang terjadi pada kromosom di setiap tahap; mitosis menghasilkan sel identik secara genetik untuk pertumbuhan, perbaikan, dan reproduksi aseksual; menghitung indeks mitosis (sel dalam mitosis dibagi total sel) dan penggunaannya secara klinis; Praktikum Inti 6 mempersiapkan dan mengamati irisan ujung akar.

    Sumber: Silabus Cambridge International

    Mitosis 有丝分裂 produces two genetically identical diploid cells: growth, repair, asexual reproduction. Interphase ($G_1$, S — DNA replication, $G_2$) fills most of the cycle; mitosis itself is prophase (chromosomes condense, spindle forms), metaphase (chromosomes line at the equator), anaphase (spindle fibres pull chromatids to poles), telophase (nuclear membranes reform), then cytokinesis 胞质分裂.

    The mitotic index 有丝分裂指数 $=$ cells in mitosis $\div$ total cells counted. High values in a tissue biopsy mean rapid division — a cancer signature.

    Worked check. A root-tip squash shows 36 of 450 cells in mitosis: index $= 36/450 = 0.08 = 8\%$. If the cycle takes 20 hours, mitosis occupies $0.08 \times 20 = 1.6$ hours.

    2.4

    Stem cells and switching genes on

    Silabus

    Pernyataan 3.17-3.21 (spesifikasi hlm.24): sel punca totipoten dan pluripoten, tahap morula dan blastosis; diferensiasi oleh ekspresi gen diferensial; perubahan pasca-transkripsi pada mRNA memberikan beberapa protein dari satu gen; interaksi genotipe dan lingkungan dalam fenotipe; alel ganda dan pewarisan poligenik menghasilkan variasi kontinu. Etika penggunaan sel punca贯穿整个考试问题.

    Sumber: Silabus Cambridge International

    A totipotent 全能性 stem cell can form a whole organism (the first divisions of the zygote); a pluripotent 多能性 cell (inner blastocyst 胚泡) can form any tissue but not the whole organism. Differentiation happens by differential gene expression 差异基因表达: every cell keeps the whole genome, but each type switches on only its own subset. One gene can yield several proteins by post-transcriptional modification 转录后修饰 — RNA splicing choices of the pre-mRNA. The final phenotype also reflects the environment (height needs both genes and nutrition) and, for many traits, multiple genes acting together (polygenic 多基因 inheritance giving continuous variation).

    2.5 2.6

    Plant transport tissues

    Silabus

    Topik 4 pernyataan 4.1-4.6 dengan Praktikum Inti 7-8 (spesifikasi hlm.26-27): ultrastruktur sel tumbuhan (dinding sel selulosa, plasmodesmata, kloroplas, amiloplast, vakuola); kontras struktur pati dan selulosa dengan peran penyimpanan vs penunjang; mikrofibril selulosa dan penebalan sekunder memberikan kekuatan tarik; pembuluh xilem (dilignifikasi, mati, transportasi air dan penunjang) dan tabung floem dengan sel pendamping (translokasi); distribusi di akar, batang, dan daun; Praktikum Inti 7 menyelidiki struktur jaringan tumbuhan dan Praktikum Inti 8 transportasi air (pekerjaan potometer).

    Pernyataan 4.7-4.13 (spesifikasi hlm.27-28): keberlanjutan serat tumbuhan dan pati; pentingnya air dan ion anorganik (nitrat untuk asam amino, kalsium untuk struktur dinding dan sinyal, magnesium untuk klorofil); zat antimikroba dan terapeutik dari tumbuhan (Praktikum Inti 9 pengujian antimikroba); kondisi pertumbuhan bakteri; pengembangan pengujian obat dari protokol historis hingga kontemporer termasuk tahapan uji coba.

    Sumber: Silabus Cambridge International

    Plant cells add a cellulose cell wall 细胞壁 joined to neighbours by plasmodesmata 胞间连丝. Cellulose chains bundle into microfibrils 微纤维 whose criss-cross laying gives tensile strength; secondary thickening with lignin strengthens and waterproofs.

    Tissue Built from Function
    Xilem 木质部 dead hollow cells, lignified walls water and mineral transport; support
    Floem 韧皮部 living sieve-tube elements + companion cells translocation of sugars
    Schematic cross-sections: in the root the xylem forms a central star with phloem between its arms; in the stem, vascular bundles arrange phloem outside xylem in a ring.

    Water and ions matter beyond drinking: nitrate 硝酸盐 builds amino acids, kalsium 钙离子 cross-links pectin in walls and signals inside cells, magnesium 镁离子 sits at the heart of chlorophyll — a shortage yellows the leaves.

    2.7

    Measuring life: classification and biodiversity

    Silabus

    Pernyataan 4.14-4.18 (spes p.28): klasifikasi mengorganisir kehidupan berdasarkan karakteristik bersama, dengan sistem tiga domain (Arkaea, Bakteri, Eukaria) dan bukti dari sekuens molekuler; kekayaan spesies dan indeks keanekaragaman (D) dalam suatu habitat; indeks heterozigositas dalam suatu spesies; endemisme; ancaman terhadap keanekaragaman hayati. Perhitungan dengan rumus D diujikan setiap tahun.

    Sumber: Silabus Cambridge International

    Classification arranges life by shared characteristics; the three-domain system 三域系统 (Bacteria, Archaea, Eukarya) rests on molecular evidence — ribosomal RNA sequences — that revealed Archaea are closer to us than to bacteria. Within a habitat, diversity has two measures:

    • Species richness 物种丰富度 — how many species.
    • Index of diversity 多样性指数 — $D = \dfrac{N(N-1)}{\sum n(n-1)}$, where $N$ is all individuals and $n$ each species' count. Higher $D$, more diverse.

    Worked check. Habitat A: 3 species with counts 40, 35, 25 ($N = 100$). $D = 100 \times 99 / (40 \times 39 + 35 \times 34 + 25 \times 24) = 9900 / (1560 + 1190 + 600) = 9900/3350 = 2.96$. Habitat B: 3 species, counts 90, 6, 4: $D = 9900/(8010 + 30 + 12) = 1.23$. Same richness, very different balance — $D$ sees what richness cannot.

    Within one species, the heterozygosity index 杂合度指数 $H =$ (number of heterozygotes) $\div$ (number of individuals) measures genetic variety. A species found nowhere else is endemic 特有种 — and its island home makes it both precious and fragile.

    2.8

    Niches, Hardy–Weinberg and conservation

    Silabus

    Pernyataan 4.19-4.21 (spes pp.28-29): konsep niche dan contoh adaptasi (anatomi, fisiologi, perilaku); persamaan Hardy-Weinberg p^2 + 2pq + q^2 = 1 untuk frekuensi alel dalam populasi dan asumsi yang dimilikinya; mengevaluasi metode konservasi kebun binatang dan bank benih, pembiakan penangkaran, dan koridor habitat.

    Sumber: Silabus Cambridge International

    A niche 生态位 is a species' way of life: its role, its requirements, its tolerances. Adaptations come in three registers — anatomical (thick fur), physiological (enzyme variants), behavioural (migration timing).

    The Hardy–Weinberg equation 哈迪–温伯格方程 tracks allele frequencies in a large, randomly breeding population with no selection, migration or mutation:

    $$p^2 + 2pq + q^2 = 1$$

    $q^2$ = frequency of the homozygous recessive genotype. If 4 % of a population shows the recessive phenotype, $q^2 = 0.04$, so $q = 0.2$, $p = 0.8$, carriers $2pq = 0.32$.

    Worked check. Of 500 foxes, 45 are homozygous recessive: $q^2 = 0.09$, $q = 0.3$, $p = 0.7$. Expected carriers $= 2pq \times 500 = 0.42 \times 500 = 210$.

    Conservation questions want balance. Zoos and seed banks 种子库 keep species against extinction and run captive breeding, but cost space, lose natural behaviour and serve few individuals; habitat protection and corridors 走廊 preserve ecology at scale but need land, money and political will. Credit the specific argument, not the sentiment.

    2.6

    Culturing and growing microbes

    Silabus

    Pernyataan 4.7-4.13 (spesifikasi hlm.27-28): keberlanjutan serat tumbuhan dan pati; pentingnya air dan ion anorganik (nitrat untuk asam amino, kalsium untuk struktur dinding dan sinyal, magnesium untuk klorofil); zat antimikroba dan terapeutik dari tumbuhan (Praktikum Inti 9 pengujian antimikroba); kondisi pertumbuhan bakteri; pengembangan pengujian obat dari protokol historis hingga kontemporer termasuk tahapan uji coba.

    Sumber: Silabus Cambridge International

    For the microbiology here (and Unit 4's): grow bacteria on nutrient agar or in broth, using aseptic technique 无菌技术 — sterilise loops by flaming, flame bottle necks, work near an updraught, seal plates, incubate below body temperature so human pathogens cannot grow. Count colonies on spread or pour plates (each colony from one cell or a clump: colony-forming units), or count cells with a haemocytometer, or follow turbidity 浊度 in broth.

    A microbial growth curve on a log scale: lag, exponential (log), stationary and death phases.

    Worked check. In the exponential phase, the growth-rate constant $k = (\log_{10} N_t - \log_{10} N_0)/(0.301 \times t)$. From $1 \times 10^4$ ke $4 \times 10^7$ cells in 4 hours: $\log$ difference $= 3.602$, so $k = 3.602/(0.301 \times 4) \approx 3.0$ divisions per hour — about a 20-minute generation time.

    2.6

    Cek diri sendiri

    1. Calculate the actual width of a nucleus that appears 24 mm wide at ×4 000. Give your answer in µm.
    2. Name the two processes in meiosis that create genetic variation, and state when each happens.
    3. A cell has 2 a.u. of DNA in G1. Sketch how DNA content changes through S phase, meiosis I and meiosis II.
    4. Explain why all cells of a blastocyst can become any tissue, but a mature muscle cell cannot.
    5. Explain two ways xylem is adapted to its function, one structural and one chemical.
    6. Two habitats each contain 5 plant species. Explain why the index of diversity can still differ between them.
    7. In a population of 1 000, 16 people show a recessive condition. Calculate q, p and the expected number of carriers.
    8. A species of snail is endemic to one island. Explain why endemic species face higher extinction risk.
    9. State three conditions a population must meet for the Hardy–Weinberg equation to hold.
    10. Give one benefit and one limitation of seed banks compared with protecting habitat.

    Answers: 1 $24\,000/4\,000 = 6\ \mu$m; 2 crossing over in prophase I, independent assortment in metaphase/anaphase I (and II); 3 rises to 4 in S, halves to 2 after meiosis I, halves to 1 after meiosis II, restored to 2 at fertilisation; 4 blastocyst cells are pluripotent — all genes still switchable; a muscle cell has switched off other pathways permanently in ordinary conditions; 5 hollow, lignified dead cells form continuous water columns with no cytoplasm to resist flow; lignin waterproofs and strengthens walls against collapse; 6 D weighs the evenness of abundance, not just the species count; 7 $q^2 = 0.016$, $q \approx 0.126$, $p \approx 0.874$, carriers $2pq \approx 0.22$, about 220 people; 8 a single local event (storm, disease, introduced predator, habitat loss) removes the whole species at once; 9 large population, random mating, no selection/mutation/migration; 10 seed banks store genetic diversity cheaply and safely against habitat loss (benefit) but species exist out of their ecology — no interactions, no evolution in place, and recolonisation is uncertain (limitation).

  • 3

    Keterampilan Praktis Biologi I

    • 3.1 Perencanaan penyelidikan (Spesifikasi Unit 3, bagian perencanaan)

      Program pembelajaran segera hadir

    • 3.2 Implementasi dan pengukuran (Spesifikasi Unit 3, bagian implementasi)

      Program pembelajaran segera hadir

    • 3.3 Pengolahan hasil, grafik, dan ketidakpastian (spesifikasi Unit 3, bagian pengolahan)

      Program pembelajaran segera hadir

  • 4

    Energi, Lingkungan, Mikrobiologi dan Imunitas

    4.1

    Energy in, energy through, energy lost

    A wheat field intercepts sunlight, fixes carbon, feeds a food chain — and loses energy at every step. This unit follows the energy from photon to ecosystem, then turns to the smallest participants: the microbes that recycle, the pathogens that take, and the immune system that defends. WBI14 is also the most calculation-heavy paper after Unit 1: expect NPP, growth constants, diversity indices and time-of-death arithmetic.

    WBI14 Energy, Environment, Microbiology and Immunity is the first IA2 unit: 1 hour 30 minutes, 90 marks, all compulsory. It assesses Topic 5 (Energy Flow, Ecosystems and the Environment) and Topic 6 (Microbiology, Immunity and Forensics).

    4.1

    Photosynthesis: two stages, two places

    Silabus

    Topic 5 statements 5.1-5.8 with Core Practical 10 (spec pp.29-30). The overall photosynthesis reaction; light-dependent reactions (photolysis of water, photophosphorylation of ADP, reduction of NADP) in the thylakoid membranes; the light-independent reactions (Calvin cycle: carbon dioxide fixation by RuBP catalysed by rubisco, GALP production, ribulose bisphosphate regeneration) in the stroma; chloroplast structure matched to both stages; absorption and action spectra; chromatography of pigments with Rf values (Core Practical 10); limiting factors (light intensity, carbon dioxide concentration, temperature) and agricultural manipulation of them.

    Sumber: Silabus Cambridge International

    The overall reaction: $6\mathrm{CO_2} + 6\mathrm{H_2O} \rightarrow \mathrm{C_6H_{12}O_6} + 6\mathrm{O_2}$, driven by light energy. In the thylakoid 类囊体 membranes, the light-dependent reactions 光反应 photolyse water (releasing O₂ and H⁺), photophosphorylate 光合磷酸化 ADP and reduce NADP. In the stroma 基质, the light-independent reactions 暗反应 (Calvin cycle) fix carbon dioxide onto ribulose bisphosphate with rubisco, reduce the products using the ATP and reduced NADP from stage one, and make GALP — some to glucose, most to regenerate RuBP.

    Structure matches function: thylakoids stacked for absorbing surface; stroma packed with enzymes for the cycle.

    Pigments and spectra

    Chlorophyll a and b absorb red and blue light; carotenoids widen the range and protect. The absorption spectrum 吸收光谱 shows what each pigment takes up; the action spectrum 作用光谱 shows what rate of photosynthesis each wavelength drives — their agreement is evidence that these pigments do the work. Chromatography separates pigments by solubility; $\mathrm{Rf} = \text{distance moved by pigment} / \text{distance moved by solvent}$.

    Absorption spectra of the chloroplast pigments, with the shaded action spectrum tracking their combined absorption.

    Faktor pembatas 限制因素: light intensity, CO₂ concentration, temperature. A greenhouse raises the weakest of the three — and exam answers say which and why.

    4.2

    Productivity and the short food chain

    Silabus

    Pernyataan 5.9-5.10 (spes p.30): produktivitas primer kotor (GPP), produktivitas primer bersih (NPP = GPP dikurangi kehilangan respirasi) dan satuan yang digunakan (kJ m-2 yr-1); efisiensi transfer biomassa antar tingkat trofik dan perhitungannya; mengapa transfer kehilangan energi (respirasi, egestasi, ekskresi, bagian yang tidak dimakan) dan mengapa rantai makanan pendek.

    Sumber: Silabus Cambridge International

    Gross primary productivity 总初级生产力 (GPP) is the energy fixed by photosynthesis per area per time (kJ m⁻² yr⁻¹). Plants spend much of it on their own respiration; what remains is net primary productivity 净初级生产力:

    $$\mathrm{NPP} = \mathrm{GPP} - R$$

    Only about 10 % of the energy at each trophic level passes upward — lost to respiration as heat, egestion, excretion and parts never eaten. That is why chains rarely exceed four or five levels and why top predators are rare.

    Energy leaving a food chain: GPP split by plant respiration, then roughly a tenth passing at each step upward.

    Worked check. A grassland fixes 45 000 kJ m⁻² yr⁻¹ and respirers 19 000. NPP $= 26\,000$ kJ m⁻² yr⁻¹. Cattle eating that grass keep about 10 %: 2 600 kJ m⁻² yr⁻¹. Humans eating the cattle keep 260 — the energy arithmetic behind "eating lower on the chain feeds more people".

    4.3

    Ecosystems, populations and succession

    Silabus

    Statements 5.11-5.15 with Core Practical 11 (spec pp.30-31): the terms habitat, population, community and ecosystem; biotic and abiotic factors controlling distribution and abundance; the niche concept applied; predator-prey cycles; methods of measuring abundance (quadrats, transects, capture-mark-recapture); stages of succession from colonisation by pioneer species to climax community (Core Practical 11 surveys a habitat).

    Sumber: Silabus Cambridge International

    Four definitions, precisely: habitat 栖息地 (where), population 种群 (one species' individuals), community 群落 (all populations), ecosystem 生态系统 (community + environment). Abiotic factors (light, water, temperature, pH) and biotic factors (competition, predation, disease) set who lives where and how abundantly — together, the niche 生态位.

    Predator and prey cycle with a delay: hares rise, lynx follow, hares fall, lynx starve — a lag of about a quarter cycle.

    A classic predator-prey cycle: the lynx population tracks the hare population with a delay.

    Succession 演替 begins when pioneer species colonise bare ground; each stage changes the soil and microclimate, enabling the next, until the climax community 顶极群落 stabilises. Measuring abundance uses quadrats and transects (random sampling for estimates, systematic along a gradient) and capture-mark-recapture for animals ($N \approx \text{first catch} \times \text{second catch} / \text{marked recaptures}$).

    4.4

    Climate change: evidence, causes, effects

    Silabus

    Pernyataan 5.16-5.22 dengan Praktikum Inti 12 (spes pp.31-32): bukti perubahan iklim (rekaman suhu, inti es, dendrokronologi, serbuk sari dalam gambut); gas rumah kaca dan efek rumah kaca diperkuat (CO2, metana, oksida nitrous - sumber, masa tinggal di atmosfer, potensi pemanasan global); siklus karbon dan metode mengurangi karbon atmosfer; ekstrapolasi data dan keandalannya; dampak perubahan iklim pada pola curah hujan, distribusi spesies, ekosistem yang bergantung enzim; Praktikum Inti 12 menyelidiki pengaruh suhu terhadap aktivitas enzim/aktivitas relevan habitat.

    Sumber: Silabus Cambridge International

    Evidence comes from direct temperature records, ice cores (trapped air giving ancient CO₂), tree rings (dendrochronology 树轮年代学) and pollen layers in peat. The enhanced greenhouse effect 温室效应: outgoing long-wave radiation is absorbed by CO₂, methane and nitrous oxide and re-emitted downward. The gases differ — methane is potent but short-lived; CO₂ lasts centuries; nitrous oxide combines both — so policy weighs global warming potential 全球增温潜势 against lifetime.

    Extrapolating a trend beyond its data is a judgement, not a fact: state the assumption (the trend continues) and its weakness. Effects already examined: shifting rainfall and species ranges, and — close to this course — temperature's effect on enzyme rates in cold-blooded organisms and whole ecosystems.

    Worked check. Methane's warming potential is about 28× CO₂'s per kg, but it lasts ~12 years against centuries. A one-tonne methane leak matters 28× a one-tonne CO₂ leak this decade, but fades while the CO₂ persists — the arithmetic of "warming potential vs cumulative load".

    4.5

    Evolusi dan spesiasi

    Silabus

    Pernyataan 5.23-5.26 (spes p.32): evolusi sebagai perubahan frekuensi alel yang disebabkan mutasi, seleksi (arah dan stabilisasi), aliran gen dan drift; isolasi mengurangi aliran gen (geografis = allopatric; reproduktif = sympatric) yang mengarah pada spesiasi; menafsirkan kesimpulan ilmiah yang kontroversial (iklim, evolusi) sesuai standar bukti; reboisasi, sumber daya berkelanjutan, dan biofuel.

    Sumber: Silabus Cambridge International

    Evolution is a change in allele frequency 等位基因频率. It needs variation (mutation, meiosis) and a filter: directional selection 定向选择 shifts the mean when the environment moves; stabilising selection 稳定选择 trims the extremes in a stable one. Drift changes frequencies by chance in small populations; gene flow (migration) mixes them back.

    Spesiasi 物种形成 needs isolation: allopatric 异域的 (a physical barrier — allopatric = different homeland) or sympatric 同域的 (reproductive isolation in the same place — behavioural, temporal or genetic). Isolated populations diverge until they can no longer interbreed.

    4.6

    Culturing microorganisms

    Silabus

    Pernyataan 6.1-6.4 dengan Praktikum Inti 13 (spes p.33): media kultur (agar nutrisi, kaldu); teknik aseptik secara rinci; metode mengukur pertumbuhan mikroba (hitung sel, kekeruhan, hitung viabel dengan pelat pengenceran); empat fase kurva pertumbuhan bakteri (lag, log/eksponensial, stasioner, kematian) dan konstanta laju pertumbuhan eksponensial k; Praktikum Inti 13 menyelidiki laju pertumbuhan dalam kultur cair dengan kurva kalibrasi.

    Sumber: Silabus Cambridge International

    Aseptic technique protects you and the culture: flame the loop to red heat, flame bottle necks, lift lids briefly, seal plates with tape, incubate at 25 °C (never body temperature, so human pathogens cannot multiply). Measure growth by direct cell count (haemocytometer), turbidity 浊度 in broth, or viable counts by dilution plating 稀释涂布 (colonies counted × dilution factor).

    The growth curve: lag 延迟期 (enzymes synthesising), log 对数期 (exponential doubling), stationary 稳定期 (deaths = divisions), death 衰亡期. The exponential constant:

    $$k = \frac{\log_{10} N_t - \log_{10} N_0}{0.301 \times t}$$

    Worked check. From $2\times10^4$ ke $1.6\times10^7$ cells in 5 hours: log difference $= 2.903$; $k = 2.903/(0.301\times5) = 1.93$ generations per hour — a doubling time of about 31 minutes.

    4.7

    Pathogens and the body's barriers

    Silabus

    Pernyataan 6.5-6.7 (spes p.33): struktur bakteri dan virus (asam nukleat, kapsid, amplop) dibandingkan; jalur infeksi (pencernaan, pernapasan, kontak seksual, luka, vektor); penghalang infeksi (kulit, lendir, asam lambung, pembekuan darah); bagaimana Mycobacterium tuberculosis dan HIV menginfeksi dan merusak tubuh, serta bagaimana HIV memperparah TB.

    Sumber: Silabus Cambridge International

    Bacteria: cell wall, plasmids, ribosomes, circular DNA, some with capsule and flagellum. Viruses: nucleic acid (DNA or RNA) in a capsid 衣壳, some with an envelope — no organelles, no metabolism of their own; they hijack host cells. Mycobacterium tuberculosis 结核分枝杆菌 infects lungs, surviving inside macrophages; HIV 人类免疫缺陷病毒 destroys T helper cells, so the immune system collapses and TB reactivates — the classic AIDS-defining illness.

    Barriers come first: skin, mucus and cilia, stomach acid, lysozyme in tears, and the clotting cascade sealing wounds.

    4.8

    Respons imun

    Silabus

    Statements 6.8-6.12 (spec p.34): non-specific responses (inflammation, lysozyme, interferon); antigens; the humoral response - B effector cells differentiating into plasma cells that secrete antibodies, B memory cells giving secondary response; the cellular response - T helper, T killer and T memory cells; the antibody structure (four polypeptide chains, variable region binding site) and antigen-antibody complex; natural and artificial, active and passive immunity; vaccination and herd immunity; the evolutionary race between pathogens and hosts.

    Sumber: Silabus Cambridge International

    Non-specific: inflammation (vasodilation and fluid leak bringing phagocytes), phagocytosis, interferon. Specific — antigens 抗原 trigger it:

    • Humoral 体液: B effector cells → plasma cells → antibodies 抗体; B memory cells B记忆细胞 remain for the faster, stronger secondary response.
    • Cellular 细胞: T helper cells coordinate, T killer cells T杀伤细胞 destroy infected cells, T memory cells persist.

    An antibody is four polypeptide chains (two heavy, two light) with a variable region 可变区 forming one specific binding site — the shape that fits one antigen. Immunity splits two ways: aktif 主动 (your own antibodies — natural infection or vaccination) vs passive 被动 (given antibodies — mother to baby, antiserum); natural vs artificial in each. Vaccination works at the scale of populations too (herd immunity 群体免疫) — but pathogens evolve back: the evolutionary race.

    Worked check. A vaccine's second dose produces antibody levels ten times the first and within days rather than weeks: clonal selection finds the memory B cells, which divide rapidly into plasma cells — the secondary response the primary response built.

    4.9

    Antibiotics and hospital-acquired infections

    Silabus

    Statements 6.13-6.15 with Core Practical 14 (spec p.34): bacteriostatic vs bactericidal antibiotics; how antibiotic resistance arises and spreads (selection); Core Practical 14 investigates antibiotic effects on bacteria; how an understanding of the causes of hospital-acquired infections (hygiene, invasive procedures, resistant strains) reduces their incidence.

    Sumber: Silabus Cambridge International

    Bacteriostatic 抑菌的 antibiotics stop bacterial growth; bactericidal 杀菌的 ones kill. Resistance arises by mutation and spreads by selection whenever exposure kills the susceptible and spares the resistant — finishing a course and avoiding unnecessary use both slow it. Hospital-acquired infections thrive where vulnerable patients, invasive devices and resistant strains meet; hand hygiene, sterile procedure and isolating carriers break the chain.

    4.10

    Decomposition, PCR, profiling and time of death

    Silabus

    Pernyataan 6.16-6.20 (spes pp.34-35): peran mikroorganisme dalam dekomposisi dan daur nutrisi; rantai polimerase (denaturasi, annealing primer, ekstensi) memperbanyak DNA; elektroforesis gel memisahkan fragmen DNA berdasarkan panjang; profil DNA untuk identifikasi dan hubungan genetik; menentukan waktu kematian dari suhu tubuh, kaku mayat, tingkat kontraksi otot, dan suksesi serangga di atas tubuh.

    Sumber: Silabus Cambridge International

    Decomposers recycle carbon and nitrogen — the ecosystem's waste system. The polymerase chain reaction 聚合酶链式反应 cycles denaturation (95 °C), primer annealing (50–60 °C) and extension (72 °C), doubling the target DNA each cycle: $n$ cycles give $2^n$ copies. Gel electrophoresis 凝胶电泳 drags negatively charged DNA fragments through gel — short fragments run furthest. The profile 电泳图谱 of band positions identifies individuals (enough loci differ between unrelated people) and measures relatedness (shared bands).

    Time of death narrows by four clocks: body temperature (falls ~1 °C per hour, modified by size, clothing, air), rigor mortis 尸僵 (sets then passes), decomposition degree, and insect succession on the body — each stage's arrivals time-stamp the interval.

    Worked check. A body found at 22 °C core temperature in a 15 °C room: $(37-22)/(1\ ^\circ\text{C per hour}) \approx 15$ hours — but state the assumptions (still air, average build) before trusting it.

    4.10

    Cek diri sendiri

    1. Name the products of the light-dependent reactions used by the Calvin cycle.
    2. A crop fixes 60 000 kJ m⁻² yr⁻¹ and respirers 20 000. Calculate NPP and the energy reaching a third trophic level at 10 % transfer each step.
    3. Explain why the action spectrum is evidence that chlorophyll carries out photosynthesis.
    4. Distinguish directional and stabilising selection with one example of each.
    5. Give two reasons a food chain rarely exceeds five trophic levels.
    6. A culture grows from $5\times10^3$ ke $4\times10^6$ in 6 hours. Calculate k and the doubling time.
    7. Explain why a second vaccine dose raises antibody titre far faster than the first.
    8. State the difference between bacteriostatic and bactericidal, and why the distinction matters for a patient with a weak immune system.
    9. Three PCR cycles from one double-stranded template: how many copies?
    10. A body's core is 30 °C. Estimate the time since death under standard assumptions and name one factor that would extend the true interval.

    Answers: 1 ATP and reduced NADP; 2 NPP $= 40\,000$; third level $= 400\,\text{kJ m}^{-2}\text{yr}^{-1}$ (10 % of 4 000); 3 wavelengths chlorophyll absorbs best are the wavelengths that drive photosynthesis best; 4 directional — antibiotic resistance shifting the mean; stabilising — human birth weights; 5 energy lost at each step to respiration/heat/egestion, so little remains; producers also lose to respiration before the first transfer; 6 log difference 2.903, $k = 2.903/(0.301\times6) = 1.61$ h⁻¹, doubling ≈ 37 min; 7 memory B cells persist; the second encounter selects clones that divide rapidly into plasma cells — the secondary response; 8 bacteriostatic stops growth (immune system must clear), bactericidal kills outright — a weak immune system needs bactericidal support; 9 $2^3 = 8$ copies; 10 $(37-30)/1 = 7$ hours; heavy clothing or a warm room slows cooling, so true time is longer.

  • 5

    Respirasi, Lingkungan Internal, Koordinasi dan Teknologi Gen

    5.1

    Di dalam tubuh: energi, keseimbangan, dan kontrol

    Lari sprint mengejar bus dan tubuh Anda memainkan simfoni: otot membakar glukosa lebih cepat, jantung menggandakan outputnya, ventilasi menjadi lebih dalam, keringat mulai mengalir, dan ginjal menyesuaikan kembali pemulihan airnya — semua dikendalikan tanpa kesadaran sadar. Unit ini mengikuti mesin tersebut: respirasi pada skala molekuler, otot dan jantung pada skala jaringan, saraf dan hormon sebagai pengontrol, serta teknologi genetik sebagai jendela modern untuk memahaminya.

    WBI15 Respirasi, Lingkungan Internal, Koordinasi, dan Teknologi Genetik adalah unit IA2 kedua: 1 jam 45 menit, 90 skor. Unit ini mengevaluasi Topik 7 (Respirasi, Otot, dan Lingkungan Internal) dan Topik 8 (Koordinasi, Respons, dan Teknologi Genetik), dan satu pertanyaan 20-skor dibangun di atas artikel ilmiah yang dirilis sebelumnya — Anda telah mempelajarinya terlebih dahulu.

    5.1

    Respirasi: empat tahap, satu tujuan

    Silabus

    Topic 7 statements 7.1-7.8 with Core Practicals 15-16 (spec pp.35-36). The overall aerobic reaction and respiration as a stepped, enzyme-controlled process; glycolysis in the cytoplasm (phosphorylation of hexoses, substrate-level phosphorylation of ATP, reduced NAD, pyruvate; lactate in anaerobic conditions); the link reaction and Krebs cycle in the mitochondrial matrix (decarboxylation, ATP, reduced NAD, reduced FAD); oxidative phosphorylation on the cristae - the electron transport chain pumping hydrogen ions into the intermembrane space and chemiosmosis through ATP synthase, oxygen as the final electron acceptor; lactate metabolism after anaerobic exercise; the respiratory quotient RQ = CO2 produced / O2 consumed for carbohydrate (1.0), lipid (0.7) and protein (0.9); Core Practicals 15-16 (respirometers and artificial hydrocarbonate-indicator respiration).

    Sumber: Silabus Cambridge International

    Respirasi melepaskan energi secara bertahap, masing-masing dikatalisis oleh enzim spesifik. Ringkasan respirasi aerob:

    $$\mathrm{C_6H_{12}O_6} + 6\mathrm{O_2} \rightarrow 6\mathrm{CO_2} + 6\mathrm{H_2O}\ (+\ \text{ATP})$$
    Tahap Lokasi Apa yang terjadi Hasil bersih
    Glikolisis sitoplasma heksosa difosforilasi lalu dipecah; piruvat terbentuk 2 ATP, 2 NAD tereduksi
    Reaksi penghubung matriks mitokondria piruvat teroksidasi dan dekarboksilasi menjadi asetil 1 CO₂, 1 NAD tereduksi per piruvat
    Siklus Krebs matriks asetil teroksidasi sepenuhnya; substrat regenerasi CO₂, ATP, NAD tereduksi, FAD tereduksi
    Fosforilasi oksidatif krista rantai transpor elektron memompa H⁺; kemiosmosis melalui sintase ATP; O₂ sebagai akseptor akhir sebagian besar ATP

    Kondisi anaerob menghentikan rantai — NAD tereduksi tidak dapat didaur ulang — sehingga piruvat menerima hidrogennya dan berubah menjadi laktat di otot. Setelah berolahraga, laktat dibawa ke hati dan direkonstruksi menjadi glukosa (membutuhkan oksigen: utang oksigen).

    Kuotien respirasi: $\mathrm{RQ} = \mathrm{CO_2}\ \text{produced} / \mathrm{O_2}\ \text{consumed}$. Karbohidrat 1.0, protein 0.9, lipid ≈ 0.7 — RQ organisme mengungkapkan apa yang sedang dibakar.

    Kiri: laju respirasi mencapai puncak pada optimum enzim mendekati suhu tubuh dan runtuh saat protein terdenaturasi. Kanan: kuotien respirasi berdasarkan substrat.

    Cek contoh terkerjakan. Respirometer menunjukkan 60 cm³ CO₂ yang dihasilkan dan 60 cm³ O₂ yang dikonsumsi dalam interval yang sama: RQ = 1.0 — karbohidrat. Benih yang berkecambah dengan RQ 0.7 sedang membakar cadangan lipid.

    5.2

    Otot: filamen meluncur

    Silabus

    Pernyataan 7.9-7.11 (spesifikasi hlm.36): bagaimana otot, tendon, kerangka dan ligamen berinteraksi dalam pergerakan (otot menarik tulang melintasi sendi; tendon menghubungkan otot ke tulang; ligamen menahan tulang bersama); struktur serat otot rangka (sel multinuklear yang menyatu, miofilamen aktin dan miosin); teori filamen geser kontraksi (ion kalsium mengekspos situs pengikat, kepala miosin dengan ATP menempel dan menarik, aktin meluncur); serat cepat-twitch dan lambat-twitch serta adaptasinya terhadap lari sprint dan kerja daya tahan.

    Sumber: Silabus Cambridge International

    Tendon menghubungkan otot ke tulang; ligamen menghubungkan tulang ke tulang melintasi sendi; otot bekerja dalam pasangan antagonis. Serabut otot rangka adalah sel yang menyatu, multinukleat, penuh dengan miofilamen aktin dan miosin.

    Kontraksi adalah siklus filamen geser: potensial aksi melepaskan Ca²⁺; kalsium mengekspos situs pengikatan aktin; kepala miosin yang teraktivasi (ATP-nya telah terurai) berikatan, menarik aktin, lalu terlepas ketika ATP baru berikatan — ribuan kepala per detik, setiap langkah beberapa nanometer.

    Serat serabut cepat: tebal, sedikit mitokondria, cepat dan kuat, lelah dengan cepat (lari sprint). Serat serabut lambat: banyak mitokondria, suplai darah kaya, mioglobin — daya tahan.

    5.3

    Jam jantung sendiri dan intervensi medula

    Silabus

    Pernyataan 7.12-7.13, 7.15 dengan Praktikum Inti 17 (spesifikasi hlm.36): sifat miogenik otot jantung; bagaimana aktivitas listrik jantung dimulai dari nodus sino-atrial, melewati nodus atrioventrikular dan sepanjang serabut Purkyne; kontrol laju jantung dan laju ventilasi oleh pusat kontrol kardiovaskular dan pusat ventilasi di medula oblongata; output jantung = volume stroke x laju jantung dan perhitungannya; variasi ventilasi dan output jantung selama olahraga; Praktikum Inti 17 menggunakan jejak spirometer untuk menyelidiki efek olahraga terhadap volume tidal, laju napas dan ventilasi menit.

    Sumber: Silabus Cambridge International

    Otot jantung bersifat miogenik: berdetak tanpa input saraf. Nodus sinoatrial menetapkan tempo; gelombang menyebar di seluruh atrium menuju nodus atrioventrikular, kemudian melalui serat Purkinje ke ventrikel — atrium berkontraksi lebih dulu, ventrikel beberapa saat kemudian.

    Pusat kontrol kardiovaskular dan pusat ventilasi di medula oblongata menyesuaikan kedua sistem sesuai kebutuhan: reseptor kimiawi mendeteksi CO₂ dan pH, baroreseptor mendeteksi tekanan.

    $$\text{cardiac output} = \text{stroke volume} \times \text{heart rate}$$

    Cek terarah. Saat istirahat: 70 cm³ × 72 min⁻¹ ≈ 5.0 dm³ min⁻¹. Selama latihan: 120 cm³ × 150 min⁻¹ = 18 dm³ min⁻¹ — tabel yang disukai ujian. Ventilasi meningkat seiring, dibaca dari jejak spirometer (volume tidal × frekuensi napas = ventilasi menit).

    5.4

    Homeostasis dan umpan balik

    Silabus

    Pernyataan 7.14, 7.16-7.17 (spesifikasi hlm.36): homeostasis sebagai mempertahankan lingkungan internal dalam kesetimbangan dinamis; umpan balik negatif memegang variabel dalam batas sempit (dan umpan balik positif memperkuat perubahan); peran sistem saraf otonom, adrenalin dalam respons fight-or-flight, dan hipotalamus dalam termoregulasi (vasodilatasi, berkeringat, menggigil, vasokonstriksi, ereksi rambut).

    Sumber: Silabus Cambridge International

    Homeostasis mempertahankan lingkungan internal dalam kesetimbangan dinamis — bukan ketenangan, melainkan variabel yang berfluktuasi dalam batas sempit. Umpan balik negatif mengoreksi penyimpangan ke arah mana pun (pola termostat); umpan balik positif memperbesarnya (spiral, bukan loop — seperti pada potensial aksi atau demam yang tidak terkendali).

    Termoregulasi berjalan melalui hipotalamus: hilang panas melalui vasodilatasi dan berkeringat; konservasi panas melalui vasokonstriksi, menggigil, dan ereksi rambut. Adrenalin mempersiapkan respons fight-or-flight — detak jantung, glukosa darah, dan saluran udara semuanya merespons bersamaan.

    5.5

    Ginjal: filter, reabsorpsi, penyesuaian halus

    Silabus

    Pernyataan 7.18-7.21 (spesifikasi hlm.36-37): struktur makroskopis dan mikroskopis ginjal mamalia (korteks, medula, pelvis; nefron dengan glomerulus, kapsul Bowman, tubulus konvolusi proksimal, loop Henle, saluran pengumpul); ultrafiltrasi (tekanan tinggi dari arteriol aferen, podosit, membran basal) dan reabsorpsi selektif di PCT (glukosa, asam amino, beberapa garam dan air); loop Henle dan gradien potensial air di medula; ADH dari hipofisis meningkatkan permeabilitas air saluran pengumpul; produksi urea di hati dari kelebihan asam amino (deaminasi, siklus ornitin tidak diperlukan secara rinci).

    Sumber: Silabus Cambridge International

    Darah masuk glomerulus nefron bertekanan tinggi — arteriol aferen lebih lebar daripada eferen — dan molekul kecil dipaksa melewati podosit dan membran basal ke dalam kapsul Bowman (ultrafiltrasi: semua kecuali sel dan protein besar). Tubulus konvolutus proksimal mereabsorpsi semua glukosa dan asam amino serta sebagian besar garam dan air (transport aktif lalu osmosis). Lingkungan Henle membangun gradien potensi air di sumsum melalui perkalian contre-courant. Duktus kolektiva mengambil keputusan akhir: ADH dari hipofisis menyisipkan aquaporin agar lebih banyak air direabsorpsi — urine pekat saat dehidrasi, encer saat tidak.

    Potensi air filtrat turun tajam melalui lingkungan Henle; permeabilitas duktus kolektiva — diatur oleh ADH — menentukan konsentrasi urine akhir.

    Asam amino berlebih dideaminasi di hati; amonia beracun berubah menjadi urea untuk diekskresikan.

    Cek terarah. ADH ada → dinding duktus kolektiva permeabel → air keluar顺着 gradien ke darah sumsum → volume kecil, urine pekat. Alkohol menekan ADH → volume besar, urine encer.

    5.6

    Mengaktifkan gen: faktor transkripsi dan epigenetik

    Silabus

    Pernyataan 7.22 (spesifikasi hlm.37): bagaimana gen diaktifkan dan dinonaktifkan oleh faktor transkripsi yang mengikat DNA (termasuk hormon steroid yang masuk ke dalam sel untuk bertindak sebagai faktor transkripsi); epigenetik - metilasi DNA dan modifikasi histon yang menonaktifkan gen serta konsekuensinya (misalnya, penekanan tumor supresor, imprinting Prader-Willi); operon lac sebagai model regulasi gen jika relevan.

    Sumber: Silabus Cambridge International

    Faktor transkripsi berikatan dengan promotor gen dan mengaktifkan atau menonaktifkan transkripsi. Hormon steroid menembus membran dan bertindak sebagai faktor transkripsi itu sendiri — satu gen dapat melayani beberapa jaringan. Epigenetik membungkam gen tanpa mengubah urutan basa: metilasi DNA dan modifikasi histon (asetilasi) mengemas DNA menjauh dari mesin transkripsi. Kegagalan penting: gen penekan tumor yang termetilasi tidak dapat menahan pembelahan sel; kesalahan imprinting menjelaskan sindrom Prader–Willi.

    5.7 5.8

    Impuls saraf dan sinaps

    Silabus

    Pernyataan 8.1-8.7 (spesifikasi hlm.37-38): struktur neuron sensorik, relai, dan motor; potensial istirahat yang dijaga oleh pompa natrium-kalium; bagaimana potensial aksi muncul (depolarisasi, ambang batas semua-atau-tidak sama sekali, repolarisasi, periode refrakter); konduksi sepanjang akson tanpa mielin dan konduksi saltatoris pada akson bermielin; struktur dan fungsi sinaps (pelepasan neurotransmiter, pengikatan reseptor, unidireksionalitas); bagaimana obat memengaruhi impuls saraf di sinaps (agonis, inhibitor, blokir reuptake SSRI).

    Pernyataan 8.8-8.10 (spesifikasi hlm.38): organisasi sistem saraf mamalia (SSN = otak dan sumsum tulang belakang; SSPN, dibagi menjadi somatik dan otonom, simpatis dan parasimpatis); bagaimana reseptor mendeteksi rangsangan (sel batang dan kerucut di retina, fotoreseptor, termoreseptor, kemoreseptor, baroreseptor, propioreseptor, nosiseptor); busur refleks spinal (zat abu-abu dan zat putih) serta jalur tiga-neuron; habituatasi sebagai penurunan respons terhadap rangsangan berulang.

    Sumber: Silabus Cambridge International

    Neuron diam memompa Na⁺ keluar dan K⁺ masuk (pompa natrium-potassium), menjaga interior sekitar −70 mV. Stimulus melebihi ambang batas (sekitar −55 mV) membuka saluran natrium: potensial aksi all-or-nothing melonjak ke +30 mV, kemudian saluran kalium melakukan repolarisasi; masa refraktori mencegah perjalanan mundur dan menetapkan frekuensi maksimum.

    Potensial aksi: depolarisasi saat natrium masuk, repolarisasi saat kalium keluar, lalu masa refraktori.

    Pada neuron bermyelin, impuls melompat dari node ke node (konduksi saltatorik) — jauh lebih cepat. Di sinaps, masuknya kalsium memicu vesikel neurotransmiter melebur dengan membran presinaptik; transmitter berdifusi dan berikatan dengan reseptor postsinaptik; enzim kemudian membersihkan celah. Obat bekerja di sini: SSRIs memblokir reuptake serotonin, sehingga sinyal bertahan; lidokain memblokir saluran natrium; ecstasy (MDMA) membanjiri celah dengan serotonin lalu mengurasnya.

    Sistem saraf terbagi: SPP (otak, sumsum tulang belakang) dan SPB — somatik (sukarela) dan autonom (tidak sukarela; simpatetik mempercepat, parasimpatetik menenangkan). Reseptor adalah transduser: batang (cahaya redup, banyak per sel bipolar — sensitivitas), kerucut (warna, satu-ke-satu — ketajaman), serta termoreseptor, kemoreseptor, baroreseptor, nosiseptor (nyeri). Busur refleks spinal melewati neurone sensorik → relai → motorik di materi abu-abu sumsum tulang belakang — cepat, tidak sukarela, protektif. Habituasi adalah penurunan respons terhadap stimulus berbahaya berulang.

    5.9

    Tumbuhan: sakelar cahaya dan pertumbuhan

    Silabus

    Pernyataan 8.11-8.12 dengan Praktik Inti 18 (spesifikasi hlm.38): bagaimana fitokrom (Pr menyerap cahaya merah dan Pfr menyerap jauh-merah) mengontrol berbunga dan perkecambahan; auksin (IAA) dan gibberelin dalam pertumbuhan dan perkecambahan (elongasi sel, induksi enzim pada lapisan aleuron); penyelidikan praktik respons tumbuhan dengan Praktik Inti 18 dan praktik tambahan yang disarankan mengenai habituatasi.

    Sumber: Silabus Cambridge International

    Fitokrom berubah antara Pr (menyerap merah) dan Pfr (menyerap jauh-merah; bias merah matahari membuat Pfr dominan, mengontrol perkecambahan dan pembungaan berdasarkan panjang hari). Aksin memanjangkan sel dengan asam dinding sel; gibberelin menginduksi enzim (amilase di lapisan aleuron) yang memobilisasi pati dalam perkecambahan.

    5.10

    Otak dan kimianya

    Silabus

    Pernyataan 8.13-8.16 (spesifikasi hlm.38-39): koordinasi melalui sistem saraf dan hormonal; lokasi dan fungsi hemisfer serebral, hipotalamus, hipofisis, serebellum dan medula oblongata; bagaimana citra MRI, fMRI, CT dan pemindaian PET terbentuk dan penggunaannya; ketidakseimbangan neurotransmiter (dopamin pada Parkinson, serotonin pada depresi) dan bagaimana pengetahuan tentang kimia otak mendasari pengobatan obat (L-Dopa menembus sawar darah-otak).

    Sumber: Silabus Cambridge International

    Landmark dan fungsi: belahan serebral (sukarela, pikiran), serebelum (koordinasi, keseimbangan), hipotalamus dan hipofisis di bawahnya (homeostasis, hormon), medula oblongata (jantung, pernapasan). MRI menunjukkan struktur; fMRI menunjukkan aktivitas melalui aliran darah beroksigen; CT pemindaian struktur cepat; PET metabolisme dengan pelacak radioaktif.

    Kimia terlebih dahulu: Parkinson adalah defisit dopamin — diobati dengan L-Dopa, yang menembus sawar darah-otak dan dikonversi menjadi dopamin di otak. Serotonin rendah terkait depresi — target dari SSRI.

    5.11 5.12

    Teknologi gen

    Silabus

    Pernyataan 8.17-8.19 (spesifikasi hlm.39): bagaimana obat dapat diproduksi menggunakan organisme yang dimodifikasi secara genetik (insulin dari bakteri, protein manusia dari hewan GM); memproduksi DNA rekombinan - enzim restriksi memotong pada situs pengenalan dengan ujung lengket, ligase DNA menyambungkan, vektor plasmid dengan gen penanda; mentransformasi sel inang (guncangan panas, elektroporasi); seleksi gen penanda.

    Pernyataan 8.20-8.21 (spesifikasi hlm.39): mikroarray untuk mengidentifikasi gen aktif (hibridisasi mRNA); istilah bioinformatika dan penggunaannya dalam membandingkan urutan DNA antar spesies dan individu; risiko dan manfaat organisme GM dalam pertanian dan kedokteran; aplikasi seperti hormon pertumbuhan manusia rekombinan dan zebrafish fluoresen.

    Sumber: Silabus Cambridge International

    DNA rekombinan: enzim restriksi memotong DNA pada sekuens pengenalannya, meninggalkan ujung lengket; gen manusia (dibuat tanpa intron dari mRNA) disisipkan ke dalam plasmid dengan ligase DNA; plasmid mentransformasi bakteri inang; gen penanda (resistensi antibiotik atau fluoresensi) menyeleksi sel yang menyerapnya. Skala besar dalam fermenter dan bakteri mensekresikan insulin manusia — lebih bersih dan aman daripada mengekstraksi dari babi. Hewan GM menghasilkan protein manusia dalam susu; risiko dan manfaat tanaman GM (resistensi, hasil panen versus aliran gen ke kerabat liar) adalah evaluasi standar.

    Mikroarray mengidentifikasi gen aktif: sampel mRNA berhibridisasi dengan probe komplementer pada chip; titik yang berfluoresensi menandai gen yang ditranskripsi. Bioinformatika membandingkan sekuens lintas spesies dan individu — bagaimana kekerabatan kini diukur.

    5.11 5.12

    Periksa diri Anda

    1. Sebutkan tahap respirasi yang terjadi di matriks mitokondria dan salah satu inputnya.
    2. Otot pelari menghasilkan laktat. Jelaskan mengapa, dan apa yang terjadi pada laktat setelah itu.
    3. Kepala miosin tidak dapat berlepas tanpa ATP. Jelaskan apa artinya ini untuk otot setelah kematian.
    4. Hitung output jantung untuk volume stroke 85 cm³ dan denyut jantung 140 min⁻¹.
    5. Jelaskan mengapa arteriol aferen lebih lebar daripada arteriol eferen.
    6. Nyatakan dua perbedaan antara umpan balik negatif dan positif, dengan satu contoh masing-masing.
    7. Mengapa mielinasi meningkatkan kecepatan konduksi?
    8. Ambang akson adalah −55 mV. Stimulus memindahkannya ke −60 mV. Apa yang terjadi, dan mengapa?
    9. Cahaya merah mengubah fitokrom menjadi bentuk mana?
    10. Jelaskan mengapa L-Dopa diberikan alih-alih dopamin itu sendiri.
    11. Titik mikroarray berfluoresensi kuat. Apa yang ditunjukkan hal itu?
    12. Mengapa gen insulin manusia harus dibuat dari mRNA daripada DNA genomik saat disisipkan ke dalam bakteri?

    Jawaban: 1 reaksi penghubung (atau siklus Krebs) — input piruvat (asetil); 2 oksigen terbatas sehingga NAD tereduksi didaur ulang dengan mereduksi piruvat menjadi laktat; setelah itu laktat menuju hati dan dikonversi kembali menjadi glukosa, menggunakan oksigen; 3 tanpa ATP, miosin tetap terikat pada aktin — kaku jenazah; 4 $85 \times 140 = 11\,900\ \text{cm}^3\,\text{min}^{-1} = 11.9\ \text{dm}^3\,\text{min}^{-1}$; 5 perbedaan tekanan mendorong ultrafiltrasi plasma ke kapsul Bowman; 6 umpan balik negatif mengembalikan variabel ke titik setelnya (termoregulasi); umpan balik positif mendorongnya semakin menjauh (demam naik); 7 impuls melompat antar nodus Ranvier alih-alih mendepolarisasi seluruh membran — langkah yang lebih sedikit namun lebih besar; 8 tidak ada apa-apa — ambang batas tidak tercapai, sehingga tidak ada potensial aksi (semua atau nothing); 9 Pfr; 10 dopamin tidak dapat menembus sawar darah-otak tetapi L-Dopa dapat, dan dikonversi menjadi dopamin di dalam otak; 11 gen tersebut sedang ditranskripsi secara aktif — mRNA-nya hadir dan berhibridisasi; 12 bakteri tidak dapat melakukan splicing intron, sehingga gen harus berasal dari mRNA matang yang sudah tidak memilikinya.

  • 6

    Keterampilan Praktis Biologi II

    • 6.1 Perencanaan investigasi A2 (Spesifikasi Unit 6, bagian perencanaan)

      Program pembelajaran segera hadir

    • 6.2 Implementasi dan kritik pengukuran (Spesifikasi Unit 6, bagian implementasi)

      Program pembelajaran segera hadir

    • 6.3 Analisis, statistik, dan kesimpulan yang beralasan (Spesifikasi Unit 6, bagian analisis)

      Program pembelajaran segera hadir

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