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JABATAN PELAJARAN NEGERI SEMBILAN TRIAL STPM 2011 MARKING SCHEME BIOLOGY 964 Paper 1 (964/1) Paper 2 (964/2) PEPERIKSAAN PERCUBAAN BERSAMA STPM PAPER 1(96411) Marking Scheme- BIOLOGY +- ANSWER B NO 1 2 -"~ -4 5 6 7 -- ~- A B D 8 c 9 A 1 D A B D ~____1__8 D ,--- B D 22 r---23 j24 c- ____2s -- -- _f____ I -T _ - c - B B A I 32 i\ - . --12--1 19 20 21 - c cA ---- c B A 13 14 ____ 15 ___._I' 16 17 · - - - - - - · - -- --t -t c ___[ - 49 so c A 29 30 31 "" ~~ 34 35 36 37 38 39 40 41 42 43 44 45 46 47 ~---- 48 10 I1 ANSWER -- 28 c A B NO 26 27 D c D B D -- c ~- D B c c D A D B -· c -· . ·-- 1-- -- 1-------- -- A B A 1-------t--- I 1 .MARKING SCHEME PAPER 2 SECTION A ANSWER NO - Amino acid 1(a) 'J(b) . ~~) (i) Hydrogen bond r-- (ii) • Secondary structure/ 0-pleated sheet • Many hydrogen bonds are formed between C=O and NH- group from the peptide bond regions. SUBTOTAL TOTAL 1 1 1 1 1 I 1 These maintain the stable structure of 0-pleated structure • 1(d) Polypeptide chain is folded to form ~-pleated sheet Hydrogen bonds break; the secondary structure II ~-pleated sheet will unfold II l Fibrous protein :Keratin I collagen I myosin I silk I 1 Globular protein : any named enzyme e.g amylase I any 1 • • Is linked by peptide bonds Contains sequence ( any 1) 2 Polysaccharide Polypeptide • 1 (Any 1) named protein hormone I antibodies 1(f) 2 l become denatured I ! (e) 1 I Has glycosidic j 2 bonds • of difference Has identical/same monomers 1 Some have 1 monomers/amino I L__ acids • Linear/straight chain • branches (any 2) total 10 ·-- 2 NO ANSWER SUBTOTAL TOTAL 2 (a) • Both X and Y axes with correct labels and unit Both curves with labels correct • • %of Hb saturation for PC0 2 (40mmHg)- 60%, 75% % ofHb saturation for PC0 2 (80mmHg)- 95%, 98% · (c) • (d) • At higher partial pressure of C0 2 , the affinity of Hb for 0 2 is reduced Hb is comprised of 4 protein subunits I 4 polypeptide chains Has 2 a chains and 2 p chains Each globular subunit contains a haem molecule At the centre of each haem molecule is a ferrous ion • (b)rul (i i) 1 - - · -+-- • • • '-------·· 1/0 110 2 2 Max2 3 • (e) • • • RBC has no nucleus -more space for more haemoglobin thus more 02 can be transported Round, disc-like shape, flexible/elastic- allow cells to be squeezed through narrow capillaries Thin plasma membrane- give a short diffusion distance into and out of cells Biconcave- increases the surface area to volume ratio/ providing larger surface area for diffusion of gases I gaseous exchange l I I l Max 3 Total 0 0 (a) (b) • • Allometric growth Different parts of the body grows at different rates compared to the overall growth rate of the organisms • P-Thymus/ Lymphoid tissues Q-Brain/head R-Absolute growth rate of human S-Reproductive organs • • • l 2 l I l l l • (c) • • • • P-Lymphoid tissues grow rapidly in early childhood compared to adolescent. This is because the risk of infection is high in early life when immunity has not yet been acquired S-The reproductive organ grow slowly In early life but rapidly at puberty To ensure that the reproductive system starts to develop after other parts of the body have completed their development 10 4 l l l l I Max 4 1---- L Total I - 10 4 ·~~-----~~--------------~--~~~~~--,-----, 4 (a) GameteS : 3 Gamete T: I 2 6 (b) (i) Meio~is I Meiosis II = GameteS Gamete S Gamete T Gamete T Non-disjunction (ii) 7 Aneuploidy ( c) Total 5 ANSWER SCHEME PAPER 2 BIOLOGY TRIAL STPM 2011 Section B NO - ANSWER SUBTOTAL I TOTAL 5(a) ~ }G I Drawing and structures correct 5 labels(all): 1. Intrinsic(A)(D)/lntegral(A)(D)//Transmembrane(A)(D)/ Pore protein( A)/Extrinsic protein (G) 2. Phospholipid bilayer (B) 3. Cholesterol (E) 4. Glycoprotein (C) 5. Glycolipid (F) Structure and function: • the biological cell membrane acts as barrier and are selectively permeable • the membrane consists of a fluid bilayer o f phospholipids and various protein molecules act as ion c hannels, CatTier protein or pumps embedded in it • the phospholipids bilayer has a hydrophobi c fatty acids tail and a hydrophilic head • the phospholipids bilayer is permeable to very small uncharged molecules like oxygen and carbon dioxide, steroid based hormone, fatty acids and alcohol (simple diffusion) -~----+-=------'cc_:_.:_:ccL.__.--;:---;------;------;----c:-----Transportation of substances into the cell: • simple diffusion of water molecules across the semi permeable cell membrane is called osmosls. • some integral mt;mbrane protein fom1 hydrophilic ion cham1els enable diffusion of various charged ions e.g.K+, Na+, Ca+, and HC0 3- down their concentration gradient • some of this ion protein channels can open or close and are called gated channels e.g. voltage-gated channels and ligand-gated channels • large sized hydrophilic molecules such as glucose are twnsportcd across the cell membrane th;ough facilitated dr>IusJOn usmg a canier pwtem ,_____ "_ir1__i_:lC!lttatscl_ c]~ffusron, the_QJ_11dmg_ __o__L_s_t.llStdt1C_es_t_o__tl1e Diagram :1 I Label: 210 3 Max 5(b) 1 1 1 l T l I 4 - I 1 1 i I I -~_}__-~'----~ 6 ANSWER NO • • • • f-- 6(a) I specific protein carrier causes the change in its shape and the substance is released into the cell Ill active transport, the shape of protein carrier changes using energy (ATP) to transport substances across the cell membrane in endocytosis the substances are transported into the cell through the invagination of the cell membrane pinocytosis occurs when the cell membrane invaginates to actively transport a small amount of fluid into the cell all these structures and its related process enable the cell membrane to function as semi-permeable membrane as well as enable the cell membrane to regulate the movement of substances in and out of the cell • Phloem tissue consists mainly of sieve tubes and companion cells • Sieve elements are arranged end to end to form a long cylindrical sieve tubes • The end walls of sieve tubes are perforated fanning a sieve plates with sieve pores • Mature sreve tubes are Living cells with no nucleus, ribosomes or Golgi apparatus • Companion cells connected to the sieve elements through plasmodesmata contains a large nucleus, dense cytoplasm and numerous of mitochondria (b) • Translocation IS the movement of orgamc solutes I sucrose/an1ino acids/soluble Qroducts of Qhotosynthesis • From the leaves/source through the sieve tubes to be carried to other garts of the Qlant/sink/root. (c) The Mass flow hypothesis state that : • The hydrostatic pressure gradient formed between the leaf cells (source) and root cells (sink) drives • the passrve mass flow of water and dissolve solutes downwards • In leaves, orgamc substances /sucrose synthesized Ill the mesophyll cells are actively transQorted into sieve tube • Low water QOtential is created in sieve tubes, • Thus water enters the sieve tubes tLu·ough osmosis from the ;:cylem • The entry of water generates a high hydrostatic Qressure in the sieve tubes • In the roots/sink, su~ars/sucrose are activelv transQorted into the tissue for cellular respiration or converted to starch for storage • As sugars are ren1oved, the water [JOtential of sieve tubes in roo!,;_b;_ increased, causing the water molecules to diffuse out through osmosis into xyler~1 roots_______.__ SUBTOTAL TOTAL 1 1 1 1 (Any 8) Max 8 TOTAL 15 l 1 1 1 1 Max4 1 1 2 1 1 I 1 I 1 I 1 L - __ _ _j 7 NO ANSWER • Low hvdrostatic pressure is created in the sieve tubes of the roots • Antibody is globular protein that reacts with specific antigen " it is made up of 2 heavy chains and 2 light chains • the structure are held together by disulphide bonds " each chain has constant and variable parts • the variable part is the binding site of the antibody onto a sgecific antigen (b) "Humoral I antibody-mediated immune response involves B lymphocyte cell • pathogens bearing foreign antigens invade the body • macrophages 111 the body engulf I phagoc ytocizes these pathogens I digest the antigen into fragments by hydrolytic enzymes •MHC class II molecules bind to the macrophages to the antigen fragment to fonn MHC complex • this foreign antigen , MHC complex is then displayed on the cell surface • this macrophage is called antigen-presenting cell (APC) • helper T cell binds to APC and secretes IL-2 • The activated B cell then divides by mitosis and a clone of B cell is produced • B cells differentiate, becoming plasma cells and memory B cells • The plasma cells secrete antibodies SUBTOTAL I TOTAL TOTAL 15 I I I I I 5 7(a) I 1 1 I l I I I I I 10 I 1---- r-- 8(a)(i) 1ardy-Weinberg's Law • The frequency of (dominant and recessive) alleles ll1 a population will rem am constant from one generation to generation/ allele and genotype frequencies do not change from generation to generation 111 a population at genetic equilibrium • provided these four conditions are met: 1. the population is large ]I. mating is random Ill. no mutation occurs IV. no migration occurs (all four conditions must be correct) (ii) Sickle cell anaemia • is a result of point/gene mutation/base substitution • Glutamine is replaced by Valine. • This causes the abnormal haemoglobin to be pulled into a I sickk shape ~___j__._c::ausingJE~S~_()XjigCn to be carried to tbe tisstJe 9 TOTAL IS 1 1 2/0 2 I I I I ' 1-- Max 3 I I ' _ _ _ _ _ __j 8 ANSWER NO - SUBTOTAL TOTAL (b) r' + p+q=l 2pq + q' = 1 1 requency of allele h = q, frequency of allele H = p, q' = 0.085% q = q = p = p = p = v = 0.00085 0.00085 0.029* I - q I - 0.029 0.971 * %genotype HH 0.971 2 X 100 = 94.28% q· = 4% = 0.04 v q = q = 0.2 p p p = = = .. (c)(u) ~ 1 Max4 I - q I - 0.2 0.8 % genotype Hh = 2 X 0.8 X 0.2 X 100 = 32% Normal blood cell sickle cell anemia blood cell (hh) (HH, Hh) 4% 64% + 32% 4% 96% 24 I I I I I I Max 3 I I 2 -- TOTAL 9(a) I I 1 0.04 % genotype HH = 0.8 2 X 100 = 64% -, 1 = % genotype Hh = 2 X 0.971 X 0.029 X 100 = 5.63% (c)(i) 1 ~- Amniocentesis • indirect screening for genetic defects in a foetus. • a long, sterilized needle is pierced through the abdominal and uterus walls into the amniotic cavity. • a small sample of amniotic fluid together with cells sloughed off from the foetus' body are extracted. • the foetal cells are cultured in laboratory. • biochemical analysis of DNA and products of defective genes I (such as) u-foctoprotcin (AFP) for the spina bifida defect 15-- I 1 1 1 1 I I I' I I I 9 NO ANSWER and karyotypic study are carried out. • this test is especially important for mothers above the age of 35 • if the foetus is handicapped, measures such as gene therapy, genetic counseling or abortion can be carried out (b) DNA fingerprinting. • DNA is isolated from samples of blood, sperm or skin • and is amplified using the polymerase chain reaction (PCR) teclmique. • restriction enzyme is used to cut the DNA into fragments that are different in length and base sequences (the fragments are placed in agarose gel). • the fragments are then separated according to s1ze and charged by using the technique of gel electrophoresis. • DNA fragments are heated to split double-stranded DNA into single strands (and then denatured tlu·ough chemical treatment.) • then transferred to a nylon membrane (through the process of Southem blotting). • radioactive probes are added to bind to complementary base sequences in the DNA and are placed on an X-ray film. • radioactive probes in DNA fragments produce dark bands on the film/autoradiography SUBTOTAL TOTAL 1 1 Max 6 1 1 1 1 1 I 1 I _____ --c,--c;------;.:~-------;---:-.---------------+-'"(a=n:y_y_7:1)--+__::M.:.:a::-x:...:7~ Use of fingerprinting: • DNA fingerprint can be used in forensic science for testing 1 specimens and identification of individuals in criminal cases I accidents, immigration cases. • determination of kinship I patemity. I • identification of defective gene. I (any 2) Max 2 --lO(a) Biogeochemical cycle means: • Natural cycle of essential chemical elements in (various forms) by the geological and biological processes • chemical elements flow from abiotic reserve to the biotic components and back to the abiotic pool II Involves interactions between living things and non-living things • The cycle prevent depletion of the resource. 1 1 3 10 ANSWER NO (b) SUBTOTAL TOTAL erosion\!' Inorganic mining, industrial phosphate I>' production of fertilisers Dissolved in rocks decomposition by phosphate detmposecs v ions in the soil inorganiC phosphates in r ~ inorganic f-+ fr shwater, teans active uptake by aquatic organ1sms J land t uplih V death, decomposition death active ion upt ke bones, teeth, ,). shells. excretorywas tes Sedimentation V c> •cue" i Organic phosphates eaten by animals V Organic phosphates Inorganic phosphates on ocean floors in animals in plants - ~- D iagrarn co nect 2 210 Max4 All 6 ticked (,J) labels must be conect c------- L Sedimentation of phosphates on ocean floor from the rocks. 2. After millions of years, geological land uplift raised the rocks containing phosphorns above sea leveL Throngh erosion, phosphates returned to soil, rivers, lakes ~and oceans from the rocks. 4. Mining of rocks for the manufacture of fertilizers supply inorganic phosphates to the soiL 5. Inorganic phosphate ions are absorbed by plants, 6. then assimilated to synthesise important compounds/amino acids. 7. Transfen-ed to herbivores when eaten and then to other trophic levels in the food chain through assimilation. 8. When organisms die, remains of bones , teeth , shells and excretory wastes area acted on by bacteria to release inorganic phosphates into soil through decomposition. 0 I I I I c-_··--- ~- - I 1 I I I I I 1 1 TOTAL 8 15 ~ ~