Life Sciences Grade 12 Paper 2 North West Mock 2025 Memo

education Department: Education North West Provincial Government REPUBLIC OF SOUTH AFRICA LIFE SCIENCES SEPTEMBER 2025 MARKING GUIDELINES MARKS: 150 These marking guidelines consists of 12 pages. Copyright reserved Please turn over

NW/September 2025 PRINCIPLES RELATED TO MARKING LIFE SCIENCES a. 10. 11. 12; 13. If more information than marks allocated is given Stop marking when maximum marks are reached and put a wavy line and ‘max’ in the right-hand margin. If, for example, three reasons are required and five are given Mark the first three irrespective of whether all or some are correct/incorrect. If whole process is given when only part of it is required Read all and credit relevant part. If comparisons are asked for but descriptions are given Accept if differences/similarities are clear. If tabulation is required but paragraphs are given Candidates will lose marks for not tabulating. If diagrams are given with annotations when descriptions are required Candidates will lose marks. If flow charts are given instead of descriptions Candidates will lose marks. If sequence is muddled and links do not make sense Where sequence and links are correct, credit. Where sequence and links are incorrect, do not credit. If sequence and links become correct again, resume credit. Non-recognised abbreviations Accept if first defined in answer. If not defined, do not credit the unrecognised abbreviation, but credit the rest of answer if correct. Wrong numbering lf answer fits into the correct sequence of questions but the wrong number is given, it is acceptable. If language used changes the intended meaning Do not accept. Spelling errors If recognisable, accept the answer, provided it does not mean something else in Life Sciences or if it is out of context. If common names given in terminology Accept, provided it was accepted at the provincial memo discussion meeting. Copyright reserved Please turn over

14. 15. 16. 17. 18. 19. NW/September 2025 If only fetter is asked for and only name is given (and vice versa) Do not credit. If units are not given in measurements Candidates will lose marks. The marking guideline will allocate marks for units separately. Be sensitive to the sense of an answer, which may be stated in a different way. Caption All illustrations (diagrams, graphs, tables, etc.) must have a caption. Code-switching of official languages (terms and concepts) A single word or two that appear(s) in any official language other than the learners’ assessment language used to the greatest extent in his/her answers should be credited, if it is correct. A marker that is proficient in the relevant official language should be consulted. This is applicable to all official languages. Changes to the marking guideline No changes must be made to the marking guideline without consulting the Provincial Internal Moderator who in turn will consult with the Provincial Internal Moderator (and the External moderators where necessary) Copyright reserved Please turn over

SECTION A QUESTION 1 1.1 Sfog cae oe ceree shy pauper ber per ber es ner Ge GE COONAN WNH = 12 NNNNNNNNNN SH OANDOAARWN—= 1.3 ied @ & & wro 14 a fe Dvv cv ByyY Dvv By AVY Cvv AVY Dvv NucleolusY¥ GenomevY CytokinesisY Centriole ¥/Centrosome Australopithecus ~ Stem cells¥ ContinuousY variation Species ¥ Homologous” structures PeptideY bond None ¥¥ A only V¥ B only VY (a) AlleleY (b) MonohybridY cross 1.4.2 OvaryY/Ovule 1.4.3 (a) Violet ¥ (b) Wrinkledv 1.4.4 Law of dominance vv 14.5 (a) 9:3:3:1v¥v (b) RrPpyv Copyright reserved NW/September 2025 (9x2) (18) (10x1) (10) (3x 2) (6) (1) (1) (1) (1) (1) (2) (2) (1) (10) Please turn over

1.5 1.5.1 Bone fragmentY 1.5.2 Yesv NW/September 2025 1.5.3. There are bands on the profile of B which correspond with those of the mother and those of son AVY 1.5.4 to: - identify suspects of crimesY - prove paternityY - determine probability of causes of genetic disorder v - to establish compatibility of tissues for organ transplantsY (Mark first TWO only) Copyright reserved Any TOTAL SECTION A: Please turn over (1) (1) (2) (2) (6) 50

NW/September 2025 SECTION B QUESTION 2 2.1 2.1.1 Ribosomev 2.1.2 AspartateY SerineY 2.1.3. - GAC and GAU code for the same amino acid” /aspartate - therefore there will no effect on the formation of proteinv/same protein is formed. 2.1.4 - Each tRNA carries a specific amino acidY - When the anticodon on the t(RNAY - matches the codon on the mRNAY - then tRNA brings the required amino acid to the ribosome. - Amino acids become attached by peptide bondsY - to form the required protein” Any 2.1.5 TY DNA Replication Transcription Two identical DNA molecules are One mRNA molecule is formedY formedv Two strands act as templates“ One strand act as a templatev Base pairs are formedv No base pairs formedY DNA nucleotides are used¥ RNA nucleotides are usedv (Mark first TWO only) Table 1 + Any (2 x 2) Copyright reserved Please turn over (1) (2) (2) (5) (5) (15)

NW/September 2025 2.2 2.2.1 Biogeography ¥ 2.2.2 - Genetic evidence V - Modification by descentY/Homologous structures - Fossil evidenceY 2.2.3 Speciation is formation of new species from the existing ones” while Extinction is the complete disappearance of species from the Earth’ 224 - Copyright reserved A population of baobab tree species (in Madagascar) becomes separated by a geographical barrier, the seav¥ then the population splits into two sub-populations”, in Africa and in Australia There is now no gene flow between the two populations”. Since each population may be exposed to different environmental conditionsv /the selection pressure may be different natural selection occurs independently in each of the two populations” such that the individuals of the two populations become very different from each otherY genotypically and phenotypically”. Even if the two populations were to mix againv they will not be able to interbreedv. The two populations are now two different species, in Africa and in Australia¥ Any Please turn over (1) (3) (2) (6) (12)

2.3 2.3.1 2.3.2 2.3.3 2.3.4 2.3.5 2.3.6 24 2.4.1 2.4.2 24.3 2.4.4 NW/September 2025 (a) A Homologous chromosomevY (b) B Spindle fibrev Metaphase Iv - Double stranded chromosomes become visible ~ - the centrosome duplicates/divides into centrioles which will move to opposite polesY - spindle fibres are formedv - and attach to the centromerev/holds chromosomes - Crossing over occur Any - The exchange of genetic material” in crossing over - Results formation of genetically different gametesY - leading to increase in genetic variation among the offspringY - where some will have favourable characteristics, - and others to have unfavourable characteristicsY - When the environmental conditions change” /when there is - competition/selection pressures - those with favourable characteristics will surviveY - and those with unfavourable characteristics will die ¥ Any 23v chromosomes - Non disjunction occurredv at pair 21/chromosomes/chromatids at pair 21 failed to separatev - during anaphasev Any (a) Woman/female with X-linked Alport syndromev ¥ (b) XAYvv 3v HeterozygousY - She and Abe, a male with X*Y genotypev/ with X-linked Alport syndrome, - have sons with X4Y genotypev/without the condition - anda daughter with XX? genotypev /with the condition Any Copyright reserved Please turn over (1) (1) (1) (4) (5) (1) (2) (15) (2) (2) (1) (1) (2) (8) [50]

QUESTION 3 3.1 3.2 3.1.4 3.1.2 3.1.3 3.2.1 3.2.2 3.2.3 3.2.4 3.2.5 3.2.6 NW/September 2025 To determine the effect of Bt-gene on the yield of cotton plants Vv (a) (Varieties/type) of cotton plantsY (b) Yieldv of cotton plants - Equal number of the two varieties of cotton plants were planted in each plotY - The environmental conditions for the two plots were exactly the samev¥ - Insects of the same type were introduced in each plotv - Equal number of/500 insects were introduced in each plotv Any To increase the reliability of the results. ~ The cotton plants with the Bt-gene have more yield. vv (a) (Skull) AY (b) (Pelvis) lv The pelvis is shortv and widev - Large craniumyv - houses a large brainv - leading to more intelligencev Any the foramen magnum move from a backward positionY in quadrapedalism to a more forward positionv in bipedalism Species D have small/short caninesv, since they feed on soft/cooked foody Species E have large/long and pointed/sharp canines- to it to be able to tearY raw/uncooked foodv Any C-shaped ¥ Copyright reserved Please turn over (2) (1) (1) (1) (1) (2) (8) (1) (1) (2) (2) (2) (4) (1) (13)

NW/September 2025 3.3. 3.3.1 Phylogenetic treev 3.3.2 Australopethicus anamensis¥ 3.3.3 (a) 3v (b) 6Y 3.3.4 3¥ million years agoY/mya 3.3.5 They share the most recent common ancestor, Paranthropus aethiopicusv V 3.3.6 (a) Australopethicus afarensisY (b) Homo erectusY 3.3.7 - Fossils of Homo habilis are found in Africa onlyY - The oldest fossils of Homo erectus are found in AfricaY while younger fossils are found in other parts of the worldv - The oldest fossils of Homo sapien are found in AfricaY while younger fossils are found in other parts of the worldv Any Copyright reserved Please turn over (1) (1) (1) (1) (2) (2) (1) (1) (3) (13)

NW/September 2025 3.4 3.4.1 (12+9+6+12) : (0+3+6+4) v =39 13 /3:1v 3.4.2 - The parents in the P1 generation, one with black skin and another with brown skin “/homozygous with contrasting characteristics - butall the offspring they produced in the F1 generation have black skinY (2) 3.4.3 12 = o Number of lizards bh Oo ND 3rd breeding pair Breeeding pairs The number of black and brown lizards for the 3 and 4" breeding pair in the F, generation = Black lizards = Brown lizards 4th breeding pair Guideline for assessing the graph CRITERIA ELABORATION MARK Correct type of graph (T) | Bar graph drawn 1 Caption of graph (C) | Both variables, skin colour and 1 breeding pairs Axes labels (L) | X-and Y axis correctly labelled 4] with units Scale for X- and Y-axis (S) }- Equal width and spaces of bars 1 -__ Correct scale for Y-axis Plotting of co-ordinates (P) |- 1 to 3 bars plotted correctly 1 -__All4 bars plotted correctly 2 Copyright reserved Please turn over (2) (6) (10)

NW/September 2025 3.5 Pi Phenotype Man with blood x Woman with blood group A group BY Genotype \Aj x [Bi v Meiosis Gametes ie i x 18 iv Fertilization Pe — Fi Genotype iA Ai 15) iiv Phenotype Blood Group Blood Group Blood group Blood Group AB A B Ov All blood groupsv* were possible in F1 generation Pi and Fiv Meiosis and fertilisationv “4 compulsory mark + Any 5 OR Pi Phenotype Man with blood x Woman with blood group A group BY Genotype Vi x [Biv Meiosis Fertilization Gametes |A i 18 le Bj i Ai ii 1 mark for correct gametes 1 mark for correct genotypes Fi Phenotype Blood Group Blood Group Blood group Blood Group AB A B Ov All blood groupY* were possible in F1 generation P1 and Fi¥ Meiosis and fertilisationv “4 compulsory mark + Any 5 (6) [50] TOTAL SECTION B: 100 GRAND TOTAL: 150 Copyright reserved

education Department P rivate Bag X2044 Eaucaton Mmabatho 2735 North West Provincial Governmen * et. (O18) 383-5805 REPUBLIC OF SOUTH AFRICA o-mait: Kiamologonwra gov 2a hief Directorate: Examinations and Assessment Directorate: Assessment - the nucleolusY and nuclear membrane disappearsY ae a ee eee eee ee | ee rT oe SUBJECT LIFE SCIENCES | PAPER 2 DATE 22 September 2025 TIME OF ERRATA QUESTION MARKS QUESTION MARKS QUESTION MARKS QUESTION MARKS TOTAL MARKS AFFECTED TOTAL MARKS FOR THE PAPER AFTER ADJUSTMENT 0 “| 6 2) g g 5 CORRECTIONS/ADDITIONS/ADJUSTMENT 4 2|°3 . 1_| 1.1.8 | Correct Answer is C not A (2) 2 | 1.3.2 | Correct Answer is B only not A only (2) 1.5.3 | Rephrase: There are DNA bands/pattern on the profile of B which correspond (2) with those of the mother and those of son Av (split ticks) 2:15 DNA Replication Transcription The whole molecule unwind and A small portion of molecule unwinds unzipY and unzipv Thymine pairs with adeninev Adenine pairs with uracil¥ (5) 2.2.2|- Geneticv evidence - Modification by descent’ /Homologous structures -__Fossilv evidence/records (3) 2.2.3 | Speciation is formation of new species from the existing/original ( move tick) (1) 2.2.4 - A population of baobab tree species (in Madagascar) becomes separated by a geographical barrier, the seaw/ ocean/continental drift - then the population splits into two” sub-populations/population, in Africa and in Australia - There is now no gene flow between the two populations”. - Since each population may be exposed to different environmental conditions” /the selection pressure may be different - natural selection occurred independently in each of the two populationsY - such that the individuals of the two populations become very differentY from each other - genotypically and phenotypicallyv. - Even if the two populations were to mix againY/cross pollinate they will not be able to interbreedv. (don’t accept reproduce) - The two populations are now two different species”, in Africa and in Australia. Any (6) 2.3.3 |- Double stranded chromosomes become visible v

spindle fibres are formedv and attach to the centromerew/holds chromosomes Crossing over occur” Any (4) | 2.3.6 Non disjunction occurredv at pair 21 at position 21 failed to separatev during anaphasev/ Any (2) 24.4 She and Abe, a male with XY genotypev/ with X-linked Alport syndrome, have sons with X4Y genotypev/without X-linked Alport syndrome and a daughter with X®X* genotypev/ with X-linked Alport syndrome Any (2) (a) Presence and absence of Bt-geneY NOT Varieties of cotton plants (1) Equal number of the two varieties of cotton plants“ were planted in each plot Equal number of /500 insectsv were introduced in each plot Equal planting fieldsY Same type of insectsY¥ Same environmental conditions for the two plotsY Any (1) 3.4.3 The number of black and brown Lizards for 3" and 4* breeding pairs in F, generation my & nn is} w ° Cy @ Black Lizards a ™ Brown Lizards Number of Lizards S N 3rd Breeding Pairs 4th Breeding Pairs Breeding Pairs (6) Note: If a histogram or line graph is drawn instead of a bar graph, marks will be lost for: e Type of graph (T) e Scale (S) If axes are transposed: e learner can get all marks if labels are also swapped and bars are of equal width and spaces are equal e Learner will loose marks for label and scale : if labels are swapped or not corresponding.

e But bars can be credited if 1 to 3 or ALL are correct Bars may not attched together but must be close to each other per breeding pair.