MARKSCHEME blogs 4j lane edu

Markscheme Blogs 4j Lane Edu-PDF Download

  • Date:18 Feb 2020
  • Views:58
  • Downloads:0
  • Pages:11
  • Size:267.87 KB

Share Pdf : Markscheme Blogs 4j Lane Edu

Download and Preview : Markscheme Blogs 4j Lane Edu

Report CopyRight/DMCA Form For : Markscheme Blogs 4j Lane Edu


4 N10 4 BIOLO HP2 ENG TZ0 XX M, 1 Follow the markscheme provided award only whole marks and mark only in RED. 2 Where a mark is awarded a tick check must be placed in the text at the precise point where it. becomes clear that the candidate deserves the mark One tick to be shown for each mark awarded. 3 Sometimes careful consideration is required to decide whether or not to award a mark In these. cases write a brief annotation to explain your decision You are encouraged to write comments. where it helps clarity especially for moderation and re marking It should be remembered that the. script may be returned to the candidate, 4 Unexplained symbols or personal codes notations are unacceptable. 5 Record marks in the right hand margin For Section A this should be against each mark allocation. shown in square brackets e g 2 The total mark for a question must equal the number of ticks for. the question, 6 Do not circle sub totals Circle the total mark for the question in the right hand margin at the end. of the question, 7 Where an answer to a part question is worth no marks put a zero in the right hand margin next to. the square bracket, 8 Where work for Section A is submitted on additional sheets the marks awarded should be shown as.
ticks and a note made to show that these marks have been transferred to the appropriate. square bracket in the body of the script, 9 Section A Add together the total for each question and write it in the Examiner column on the. cover sheet, Section B Insert the total for each question in the Examiner column on the cover sheet. Total Add up the marks awarded and enter this in the box marked TOTAL in the. Examiner column on the cover sheet, 10 After entering the marks on the cover sheet check your addition to ensure that you have not made an. error Check also that you have transferred the marks correctly to the cover sheet All scripts are. checked and a note of all clerical errors will be given in feedback to examiners. 11 If an answer extends over more than one page and no marks have been awarded on a section draw a. diagonal line through that section to indicate that it has been marked. 12 If a candidate has attempted more than the required number of questions within a paper or section. of a paper mark all the answers and use the marks of those answers that have the highest mark. unless the candidate has indicated the question s to be marked on the cover sheet. 13 A mark should not be awarded where there is contradiction within an answer Make a comment to. this effect in the left hand margin,5 N10 4 BIOLO HP2 ENG TZ0 XX M. Subject Details Biology HL Paper 2 Markscheme,Mark Allocation.
Candidates are required to answer ALL questions in Section A 32 marks and TWO questions in. Section B 2 20 marks Maximum total 72 marks, 1 A markscheme often has more marking points than the total allows This is intentional Do not award. more than the maximum marks allowed for part of a question. 2 Each marking point has a separate line and the end is signified by means of a semicolon. 3 An alternative answer or wording is indicated in the markscheme by a slash Either wording can. be accepted, 4 Words in brackets in the markscheme are not necessary to gain the mark. 5 Words that are underlined are essential for the mark. 6 The order of marking points does not have to be as in the markscheme unless stated otherwise. 7 If the candidate s answer has the same meaning or can be clearly interpreted as being of. equivalent significance detail and validity as that in the markscheme then award the mark. Where this point is considered to be particularly relevant in a question it is emphasized by. writing OWTTE or words to that effect, 8 Remember that many candidates are writing in a second language Effective communication is. more important than grammatical accuracy, 9 Occasionally a part of a question may require an answer that is required for subsequent. marking points If an error is made in the first marking point then it should be penalized. However if the incorrect answer is used correctly in subsequent marking points then. follow through marks should be awarded Indicate this with ECF error carried forward. 10 Only consider units at the end of a calculation Unless directed otherwise in the markscheme. unit errors should only be penalized once in the paper Indicate this by writing 1 U at the. first point it occurs and U on the cover sheet,6 N10 4 BIOLO HP2 ENG TZ0 XX M.
Extended response questions quality of construction. Extended response questions for HL P2 carry a mark total of 20 Of these marks 18 are. awarded for content and 2 for the quality of construction of the answer. Two aspects are considered,expression of relevant ideas with clarity. structure of the answers, 1 quality mark is to be awarded when the candidate satisfies EACH of the following criteria. Thus 2 quality marks are awarded when a candidate satisfies BOTH criteria. Clarity of expression, The candidate has made a serious and full attempt to answer all parts of the question and the. answers are expressed clearly enough to be understood with little or no re reading. Structure of answer, The candidate has linked relevant ideas to form a logical sequence within at least two parts. of the same question e g within part a and within part b or within part a and within part c. etc but not between part a and part b or between part a and part c etc. It is important to judge this on the overall answer taking into account the answers to all parts of. the question Although the part with the largest number of marks is likely to provide the most. Candidates that score very highly on the content marks need not necessarily automatically gain. the 2 marks for the quality of construction and vice versa. The important point is to be consistent in the awarding of the quality points For sample scripts. for moderation the reason why quality marks have been awarded should be stated. Indicate the award of quality marks by writing Q2 Q1 or Q0 in red at the end of the answer. 7 N10 4 BIOLO HP2 ENG TZ0 XX M,1 a i gid1 1 1, ii between 10 8 and 10 7 mol dm 3 units required 1.
iii breakdown of starch to maltose 1, b 25 1 in 4 1 3 seeds produced would be homozygous recessive. no response to inhibits gibberellin in homozygous recessives results in less. germination, less growth dwarf plants produced must be in context. would produce plants with infertile flowers that cannot produce rice grains. would lower rice production less yield because infertile plants cannot produce. seeds that humans can eat 3 max,c i Sub1C 1, ii Sub1A is expressed strongly the most Sub1A produces the most RNA. Sub1B always has the lowest expression produces least mRNA. Sub1A expressed produces mRNA for the longest time days 1 to 10. Sub1C expressed produces mRNA for the shortest time days 3 to 7 2 max. is only expressed in indica Sub1B and SubC are expressed in both rice varieties. indica is the variety showing submersion tolerance vice versa for japonica 2 max. e i it increases the length of time before flowering 1. ii long day light exposure increases time before flowering only if. is not overexpressed in WT and, long day light exposure decreases time before flowering for and or. length of day does not make much difference makes least difference for. overexpression for reduces time before flowering, acts as a control has nearly the same length of time before flowering as WT 2 max.
Accept numerical answers if they are making a clear comparison. iii is a short day plant because WT has shortest time shorter time before. flowering in shorter days than longer days as it takes less time to flower. under short day conditions 1,8 N10 4 BIOLO HP2 ENG TZ0 XX M. f codominant alleles show intermediate phenotype when both present. could be codominant because homozygous shows longer time before. flowering than heterozygous, or homozygous not overexpressed has a slightly longer time before flowering. than WT so factors other than codominance could be influencing flowering. dominance shown with short day light exposure while codominance in long day. light exposure, because presence causes overwhelming difference compared with absence in. short day light exposure, OsGI could be dominant because its presence always causes longer time. before flowering 2 max, g the mutant gid1 1 would not be useful because it produces sterile plants.
genetically modified rice rice with Sub1A is more tolerant to submersion can. withstand seasonal flooding torrential rain, OsGI varieties adapted to different latitudes day length could be produced. to overcome food shortages 2 max,9 N10 4 BIOLO HP2 ENG TZ0 XX M. 2 a I sepal,II ovary receptacle,III petal 3, b i Angiospermophyta Angiospermophytes Angiosperms 1. Do not accept flowering plants,ii confirms the hypothesis must be qualified. stigma anther inside the flower ring of petals so as visiting animal enters. it brushes past them, colourful petals provide contrast so that flowers can be seen by animals.
slightly cone shaped flowers so animals come in 2 max. c first name Campanula for genus second name persicifolia for species. all members of Campanula persicifolia share special unique features. two names make a unique combination to designate species worldwide. recognized nomenclature 2 max, 3 a the acquisition of antibodies from another organism 1. b an example e g detection of antibodies to HIV reject AIDS isoenzyme in. heart attack HCG in pregnancy test kits blood and tissue typing 1. detection of malarial parasites,Accept any other valid examples. c an organism virus that causes a disease 1,Accept specific examples of. d antibiotics block inhibit specific metabolic,inhibition such as cell protein. pathways cell functions found in bacteria,synthesis cell wall formation.
viruses must use host eukaryotic cell metabolism viruses do not have their. own metabolic pathways, host eukaryotic cell metabolism pathways not blocked inhibited by antibiotics 2 max. 10 N10 4 BIOLO HP2 ENG TZ0 XX M, Remember up to TWO quality of construction marks per essay. 4 a structure collagen,transport transthyretin hemoglobin. enzyme catalyst lysozyme,movement actin tubulin,hormones insulin. antibodies immunoglobulin,storage albumin 4 max, Accept any other valid function of proteins with a named example.
For example sodium potassium pump but do not accept simply in membranes. without a clear function, To award 4 max responses need a function of protein and a named example. Only accept the first four answers,b made of protein. made of rRNA,large subunit and small subunit,three tRNA binding sites. Aminacyl A Peptidyl P and Exit E,mRNA binding site on small subunit. 70S in prokaryotes 80S in eukaryotes,can be free bound to RER in eukaryotes 6 max.
c RNA polymerase polymerase number is not required. binds to a promoter on the DNA,unwinding the DNA strands. binding nucleoside triphosphates,to the antisense strand of DNA. as it moves along in a 5 3 direction,using complementary pairing A U and C G. losing two phosphates to gain the required energy, until a terminator signal is reached in prokaryotes. RNA detaches from the template and DNA rewinds,RNA polymerase detaches from the DNA.
many RNA polymerases can follow each other, introns have to be removed in eukaryotes to form mature mRNA 8 max. Plus up to 2 for quality,11 N10 4 BIOLO HP2 ENG TZ0 XX M. 5 a Award 1 for each of the following clearly drawn and correctly labelled. head and midpiece mid section body, tail flagellum at least four times length of the head and containing fibres. acrosome shown as distinct structure near front of head. nucleus occupying more than half the width or length of head. mitochondria as repetitive structures inside membrane of mid piece. centriole between head and midpiece, plasma membrane shown as single line covering whole cell. microtubules in 9 plus 2 array 4 max,b FSH promotes development of a new follicle.
also leads to the production of estrogen, estrogen brings about repair and growth of uterine lining. estrogen causes negative feedback of FSH,estrogen brings about LH production. LH stimulates follicle growth,LH triggers ovulation. estrogen contributes to the proliferative phase of the uterine cycle triggers. progesterone contributes to the secretory phase of the uterine cycle maintains. uterus lining, lowered level of progesterone due to degeneration of corpus luteum leads to. menstruation 6 max,c cause 4 max,AIDS caused by HIV.
penetrates T lymphocytes, envelope glyco protein and cell receptors involved. reverse transcriptase enables DNA to be produced reject DNA transformed. from viral RNA into RNA,number of lymphocytes reduced over years. results in lower immunity, other illnesses develop as result of lower immunity. AIDS is the observed syndrome when final stages of infection develop OWTTE. transmission 3max, HIV transmitted through blood sexual contact body fluids placenta childbirth breastfeeding. distribution transmission uneven around the world, transmission risk increased depending on society s traditions beliefs behaviour.
rare minority of individuals do not have cell receptors and do not develop AIDS. condoms latex barriers only protection against transmission through sexual contact. Thus 2 quality marks are awarded when a candidate satisfies BOTH criteria Clarity of expression The candidate has made a serious and full attempt to answer all parts of the question and the

Related Books