Life Sciences Grade 12 Paper 2 KZN Mock 2025 Question Paper

KWAZULU-MATAL PROVINCE EDUCATION REPUBLIC OF SOUTH AFRICA NATIONAL SENIOR CERTIFICATE GRADE 12 LIFE SCIENCES P2 PREPARATORY EXAMINATION SEPTEMBER 2025 MARKS: 150 TIME: 21/2 hour This question paper consists of 16 Pages. Copyright reserved Pl t 6 lease turn over PUL UE A

« Life Sciences/P2 Z Preparatory Examination September 2025 &STRUCTIONS AND INFORMATION Read the following instructions carefully before answering the questions. 10. 11. Answer ALL the questions. Write ALL the answers in the ANSWER BOOK. Start the answers to each question at the top of a NEW page. Number the answers Correctly according to the numbering system used in this question paper. Present your answers according to the instructions of each question. Do ALL drawings in pencil and label them in blue or black ink. Draw diagrams, tables or flow charts only when asked to do so. The diagrams in this question paper are NOT necessarily drawn to scale. Do NOT use graph paper. You may use a non-programmable calculator, protractor and a compass. Write neatly and legibly.

Life Sciences/P2 oe Preparatory Examination September 2025 SECTION A QUESTION 1 1.1 Various options are provided as possible answers to the following questions. Choose the answer and write only the letter (A to D) next to the question number (1.1.1 to 1.1.9) in the ANSWER BOOK, for example 1.1.10 D. 1.1.1 Sources of variation in a human population are ... mutation, mitosis, random mating and random fertilisation. replication, random mating, meiosis and mitosis. mutation, meiosis, random fertilisation and random mating. random mating, replication, random fertilisation and meiosis. G0OWDY 1.1.2 Meiosis is a significant type of cell division, as it is responsible for the ... production of genetically identical gametes. doubling the chromosome number from generation to generation. production of haploid gametes. production of diploid gametes. uawPY 1.1.3. A portion of DNA molecule has 140 guanine and 70 thymine nitrogenous bases. What is the total of deoxyribose sugars in this portion of DNA molecule? A 420 B 210 Cc 140 D 70 1.1.4 A homozygous purple flowering plant (P) is crossed with a pink flowering plant (p) to produce the F1-generation. One of the F1 generation offspring is crossed with a pink flowering plant to produce the F2-generation offspring. Which ONE of the following is the CORRECT phenotypic ratio of the F2- generation? 1 purple flowering plant: 1 pink flowering plant 1 purple flowering plant: 3 pink flowering plant 3 purple flowering plant: 1 pink flowering plant 1 purple flowering plant: 2 pink flowering plant 0QODWUDY Copyright reserved Please tum over

Life Sei ? ination September 2025 ENces/p2 preparatory Examina P' 4 Nse) ing of a 1:4; ible in the offspring 5 Which of the following genotypes is possible with blood group B? Omozygous male with blood group A and a fema A IAA BA Cc ii DB Fj an animal 1.4.6 The reproductive isolation mechanism Facalia aideni ats Mii displaying a specific action to attract a mating partner is A breeding at different times of the year. B species-specific courtship behavior. C infertile offspring. D prevention of fertilization. TA.7 Below is a list of statements in human evolution. (i) Australopithecus fossils were found in Africa only (ii) Homo habilis fossils were found in Africa only (iii) Homo sapiens fossils were found in Africa only (iv) The oldest Homo erectus fossil was found in Africa Which ONE of the following combinations is the CORRECT fossil evidence for the “Out of Africa hypothesis”? A (i), (ili) and (iv) only B (i), (ii) and (iii) only C (i), (ii) and (iv) only D (i), (ii), (iii) and (iv)

Life Sclences/P2 § NSC Preparatory Examination September 2025 1.4.8 The graph below shows a portion of DNA molecule containing 20 nitrogenous bases in each strand. C-NHOEh 0070953 Mlstrand 1 (Jstrana 2 Number of DNA nitrogenous bases (units) A T G Cc TA c G Strands The strand 2 was used as a template for the formation of mRNA during protein synthesis. The percentage difference between uracil and adenine on the mRNA strand is ... A 30 B 20 Cc 15 D oO Piouse lain ever Copyright reserved e

ratory Examination September 2025 Prepa' Life Sciences/P2 4 NSC brown in ite and others are 1.1.9 There is variation in tawny owls. Some are white a! colour. ars, from 1980 to 2010, Scientists studied these owls over a period of 30 ye d increasing global : Sse to determine the effect of climate change that cau temperatures with less snow each year. The scientists: ; * Conducted the investigation over the same four months of winter each year. * Observed the same population of tawny owls each year. ; * Determine the number of tawny owls of each colour that survived every year. The results are shown in the graph below: Number of tawny owls of each colour that survived every year from 1980 to 2010 80 70 —— White owls ~@- Brown owls Number of owls surviving Db oO 1980 1985 1990 1995 2000 2005 2010 Time (years) What conclusion can be made about the sui : tability of each colour owl to survive in more snow? A __ The brown owls are less suited to survive th B__ The white owls are less suited to survive th C Both brown and white owls are equ D__ The brown owls are more suited to an the white owls an the brown owls ally suited to Survive Survive than the White owls (9x2) (18) Copyright reserved Bloor tines

Lite Sciences'P2 Ae Proparatory Examination September 2025 1.2. Give the correct biological term for each of the following descriptions. Write only the term next to the question number (1.2.1 to 1.2.9) in the ANSWER BOOK. 1.2.3 1.2.4 1.2.5 1.2.6 V2 1.2.8 1.2.9 The number, shape and arrangement of all the chromosomes in the nucleus of a somatic cell The opening in the base of the skull through which the spinal cord passes The study of the distribution of living organisms found in different regions The permeant disappearance of species from earth An inherited disorder where blood fails to clot properly Act or ability of using two feet for walking or locomotion An organelle that is responsible for protein synthesis The type of evolution characterised by long periods of little or no change followed by short periods of rapid change A phase in the cell cycle where DNA replication occurs (9x1) (9) 4.3. Indicate whether each of the descriptions in COLUMN | applies to A ONLY, B ONLY, BOTH A AND B or NONE of the items in COLUMN II. Write A only, B only, both A and B, or none next to the question number (1.3.1 to 1.3.3) in the ANSWER BOOK. COLUMN | COLUMN Il 1.3.1 | A point where two chromatids A: Chiasma overlap B: Locus 1.3.2 | There is a range of different A: Continuous variation B: Discontinuous variation phenotypes for a particular characteristic 1.3.3 Little Foot Taung Child Fossils found in South Africa QP Copyright reserved (3x2) (8) Please tum over

Hie Sven engyy Ki NSC i eiosis in no order. 14 The diag below representa coll in two different phases of m Preparatory Examination September 2025 Diagram 1 Diagram 2 1.4.1 Identify the phase of meiosis represented in diagram: (a) 1 (b) 2 1.4.2 Give the LETTER and NAME of the part that: (a) Disappears at the end of the phase in diagram 1 (b) Pulls chromatids towards opposite poles 1.4.3 Give the chromatid number for each cell in diagram 2, 1.4.4 Name the: (a) Process that caused the appearance of chromosomes in diagram 2. (b) Phase that follows the one shown in diagram 1, (c) Organ in the female body where the Process of meiosis occurs. Copydulit reserved (1) (1) (2) (2) (1) (1) (1) (1) (10)

1.5 The diagram below illustrates the evolutionary relationships and shared ancestry NSC Homo sapiens Homo erectus Australopithecus afarensis Homo habilis Australopithecus africanus Ardipithecus vamidus Australopithecus aethiopicus of 45.40 35 30 25 20 15 1.0 06 Million years ago 1.5.1 1.5.2 1.5.3 1.5.4 Identify: (a) The diagram shown above (b) TWO genera that existed between 2.5 and 1.5 million years ago When did Homo habilis become extinct? How many million years did Australopithecus africanus live? Name the: (a) Most recent ancestor of the Homo habilis (b) Family to which all the species in the diagram above belong TOTAL SECTION A: Please turn over Preparatory Examination September 2025 (1) (1) (7) 50

Life Sciencesp2 40 Preparatory Examination September 2025 NSC SECTION B QUESTION 2 2.1 The diagram below shows part of the process involved in protein synthesis. A B AAG A A U UUCUCUUCA fz eee a ie i 2p py X Y 2.1.1 Name the stage of protein synthesis that is shown in the diagram. (1) 2.1.2 Identify: (a) A F (1) (b) B (1) 2.1.3 Give the: (a) Give the collective name of nitrogenous bases X (b) DNA base triplet that code for Y 2.1.4 The table below shows amino acids and DNA triplet base. AMINO ACIDS DNA TRIPLET BASE Leucine AGT Glycine AAG Glutamic Acid AAT Proline AGA With reference to the diagram and the table above: (a) | Write down the codon that codes for Proline (1) (b) Name the last TWO amino acids coded f Or by molecule Z j diagram (from left to right) y le Z in the (2) (8) Copyright res Copyright reserved Please turn avor

Life Sciencesiré 11 : NSC Proparatory Examination September 2026 2.2 Describe the process of transcription. (6) 2.3. Read the passage below, Komodo dragon and Timor monitor are two different types of monitor lizards. They evolved from @ population of the common ancestor and adapted in differant islands over time. These diffe i Komodo dragon have a larger bod feeds on larger prey such small prey such as inserts. y size than the Timor monitor. Komodo dragon as deer and pigs, whereas Timor monitor feeds on 2.3.1 2.3.2 2.3.3 Define the term population. (2) From the passage, name the species that feeds on insects. (1) Describe how the Komodo dragon and Timor monitor could have evolved from a common ancestor. (6) (9) 2.4 The diagram below shows the DNA profile of blood sample found in a crime scene, victim and suspect P, Q, R and S. 2.4.1 2.4.2 2.4.3 2.4.4 Blood Victim P Sample | State the conclusion that can be drawn from the analysis in the DNA profiles @) of Qand S. Give TWO reasons why DNA profiling does not always provide a conclusive evidence in criminal investigations. (2) State TWO benefits of using DNA profiling other than that of solving crimes. (2) Explain why DNA profiling is a reliable method for identifying suspects in criminal investigation. ' (8) Please turn over Copyright reserved

Life Sciences/P2 2.5 Fabry disease is a sex-linked disorder that affects kidney, Preparatory Examination September 2025 12 NSC heart, brain and skin. A recessive allele X° causes Fabry disease in humans. ‘ isease in a family. The pedigree diagram below shows the inheritance of Fabry disea y KEY (> Unaffected female | Affected male Unaffected male © O 2.5.1 2.5.2 2.5.3 2.5.4 How many generations are represented in the pedigree above? (1) Give the genotype of individual 2. (1) Explain why males are most likely to be affected by Fabry disease. (3) Individual 3 is heterozygous and is planning to : . husband (individual 4). 9 to have a child with her Use a genetic cross to show ALL the possible Phenotypes of the Offspring. (6) (41,

[fe Sclencesd2 13 NSG Proparatory E xatnination September 202% a6 The diagrams below show the parts of Homo sapiens and Chimnpanzco 2.6.1 2.6.2 2.6.3 Write down the LETTER of the diagram that represents the pelvic structure of a quadrupedal organism. Tabulate TWO visible differences between jaw C and D which shows trend in human evolution. Explain the significance of the change in pelvis from B to A. (2) (8) [50

preparatory Examination September 2025 re| Life Sciences/P2 14 NSC QUESTION 3 3.1 Read the passage below. od and immune Stem cell transplants are used to treat cancers that affect the blo damaged by system such as lymphoma and leukaemia. It replaces the oe ot . chemotherapy. The treatment uses stem cells from embryos an adult tissues to restore healthy blood cell production. 3.1.1 From the passage, give: (a) TWO sources where stem cells can be collected (2) (b) TWO types of cancer that can be treated using stem cells (2) 3.1.2 Stem cells from bone marrow of an adult can be used to replace cells that are damaged by chemotherapy. Explain why stem cells are used in replacing these damaged cells. (2) (6) 3.2 Read the passage below. During the Industrial Revolution in England, pollution from factories blackened trees with soot, leading to a shift in the peppered moth population from primarily light -coloured to dark-coloured because the dark moths were better camouflaged against the darkened bark and were thus less likely to be preyed upon by birds. 3.2.1 Explain why the population of dark- coloured moth increased after the Industrial Revolution in England using: (a) Lamarckism 6) (b) Darwinism 7) (13)

Le gciences/P2 3.3 A farmer collected 20 fish characteristics. 15 Nsc Preparatory Examination September 2025 Speci , Pecies (yellow fish and blue fish) with similar : oO 5 species of fish. ut to determine the dominant colour in these The The 3.3.1 3.3.2 3.3.3 3.3.4 3.3.5 3.3.6 procedure was as follows: Healthy yellow fish and blue fish were selected. 10 fish Species belonged to yellow fish and another 10 to blue fish. Each fish species consisted of 2 male fish. All fish species were of the same reproductive age. These fish species were bred for five years. 100 offspring were collected each year to observe the dominant colours. table below shows the average results of the offspring. PHENOTYPE AVERAGE NUMBER OF FISH OVER FIVE YEARS Yellow 19 Green 65 Blue 16 Give TWO planning steps that would have been taken for this investigation. (2) State TWO: (a) Ways in which the reliability of this investigation was ensured (2) (b) Ways in which the results of the investigation were made valid (2) Name the type of dominance shown in this investigation. (1) Give a reason for your answer in QUESTION 3.3.3. (2) wn in the table, explain why the green phenotype Based on the results sho vn and blue phenotype in the offspring. (2) is more common than yel show the average number of fish over five years for (6) in the table above. © Draw a bar graph to each phenotype given

(ite Seloncennts ta Preparatory Examination September 2025 Nuw 840 Deaerbe how non dlafimetion lence to Down syndrome. (5) 3. gtape planta, tod {tll colour (R) ts doniinant over green fruit colour (r), and broad leaf (B) ta dominant over nariow leaf (b), A grape plant that la hotorozygous for both red frull and broad leaf is crossed with aplant that is hotorozygoun for red frult and narrow leaf. Tho following table shows tho results of the genetic cross between the two plants. PHENOTYPE NUMBER OF OFFSPRING Rod frult and broad loaf a 288 Groon frult and Narrow leaf = 108 Groen frult and broad loaf 96 Red frult and narrow leaf” 296 3.5.1 Identify the type of crossing referred to above. (1) 3.5.2 Give a reason for your answer in QUESTION 3.4.1. (1) 3.5.3. Give the: (a) Genotypes of the parent plants that were crossed (2) (b) Possible genotypes in the gametes of the heterozygous red fruit and narrow leaf (2) (c) Phenotypic ratio of red fruit and broad leaf to green fruit and broad leaf (2) 3.5.4 Two grape plants that are homozygous for both characteristics were crossed. How many grape plants in the next generation are likely to have green fruit and narrow leaf? (1) (9) [50] TOTAL SECTION B: 100 GRAND TOTAL: 150 Copyright rosorved