# Life Sciences Grade 12 P1 NSC November 2013 QP (Eng) > Grade 12 Life Sciences Paper 1 question paper from the NSC November Exam 2013 (English), 15 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/life-sciences-p1-nov-2013-eng/06b11e92-7e89-48c3-b276-6b7ceaa5ac57 - Updated: 2026-09-27 - Subject: Life Sciences - Grade: Grade 12 - Document type: Question paper - Year: 2013 - Exam period: NSC November Exam - Paper: 1 - Curriculum: CAPS - Pages: 15 Breadcrumbs: [Home](https://www.hlayiso.com/) › [Library](https://www.hlayiso.com/library) › [Grade 12](https://www.hlayiso.com/library/grade-12) › [Life Sciences](https://www.hlayiso.com/library/grade-12/life-sciences) › [Life Sciences Grade 12 P1 NSC November 2013 QP (Eng)](https://www.hlayiso.com/library/life-sciences-p1-nov-2013-eng/06b11e92-7e89-48c3-b276-6b7ceaa5ac57) ## About this document Life Sciences · Grade 12 · NSC November Exam · 2013. Question paper, 15 pages. Read online or download the PDF. Read it free online on the page above; download the PDF from there. ## Document text Text extracted from the PDF. Layout, diagrams and tables may not survive extraction. ### Page 1 Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE GRADE 12 LIFE SCIENCES P1 NOVEMBER 2013 MARKS: 150 TIME: 2½ hours This question paper consists of 15 pages. Copyright reserved Please turn over ### Page 2 Downloaded from hlayiso.com Life Sciences/P1 2 DBE/November 2013 NSC INSTRUCTIONS AND INFORMATION Read the following instructions carefully before answering the questions. 1. Answer ALL the questions. 2. Write ALL the answers in the ANSWER BOOK. 3. Start the answer to EACH question at the top of a NEW page. 4. Number the answers correctly according to the numbering system used in this question paper. 5. Present your answers according to the instructions of each question. 6. Do ALL drawings in pencil and label them in blue or black ink. 7. Draw diagrams or flow charts only when asked to do so. 8. The diagrams in this question paper are NOT necessarily drawn to scale. 9. Do NOT use graph paper. 10. You must use a non-programmable calculator, protractor and a compass where necessary. 11. Write neatly and legibly. Copyright reserved Please turn over ### Page 3 Downloaded from hlayiso.com Life Sciences/P1 3 DBE/November 2013 NSC SECTION A QUESTION 1 1.1 Various options are provided as possible answers to the following questions. Choose the correct answer and write only the letter (A to D) next to the question number (1.1.1 to 1.1.10) in the ANSWER BOOK, for example 1.1.11 D. 1.1.1 The distribution of organisms is known as … A biochemistry. B biogeography. C archaeology. D palaeontology. 1.1.2 Which ONE of the following CORRECTLY describes the cells produced by meiosis? CELLS PRODUCED BY MEIOSIS Chromosome complement Genetic composition A Haploid Different B Diploid Identical C Diploid Different D Haploid Identical 1.1.3 Study the following statements: (i) Organisms in a population show a great deal of variation. (ii) Characteristics are passed on from parents to offspring. (iii) If an organism uses an organ frequently it becomes more developed. (iv) A large number of offspring are produced, but only few survive. Which ONE of the following combinations refers to observations upon which Darwin based his theory? A (i), (ii), (iii) and (iv) B (i), (ii) and (iii) only C (i), (ii) and (iv) only D (i), (iii) and (iv) only Copyright reserved Please turn over ### Page 4 Downloaded from hlayiso.com Life Sciences/P1 4 DBE/November 2013 NSC 1.1.4 Study the following list of molecules: (i) Sugar (ii) Phosphate (iii) Nitrogenous base (iv) Amino acid Which ONE of the following combinations represents components of a nucleotide? A (i), (ii) and (iv) only B (i), (ii) and (iii) only C (i), (ii), (iii) and (iv) D (ii), (iii) and (iv) only 1.1.5 Four different phenotypes are possible in the F 1 -generation if the parents' blood types are … A B and B. B A and B. C O and AB. D AB and AB. 1.1.6 In humans, brown eye colour is dominant over blue eye colour. A mother with blue eyes had two children, a boy with brown eyes and a girl with blue eyes. The eye colour of the father is ... A brown, because the allele for brown eye colour is sex-linked. B brown, because at least one of the parents must have brown eyes. C blue, because at least two other members of the family have blue eyes. D blue, because at least one of the parents must be heterozygous for eye colour. Copyright reserved Please turn over ### Page 5 Downloaded from hlayiso.com Life Sciences/P1 5 DBE/November 2013 NSC QUESTIONS 1.1.7 TO 1.1.9 ARE BASED ON THE INFORMATION AND GRAPH BELOW. One way to measure the resistance of an insect population to an insecticide is to determine the time taken for the insecticide to kill 90% of the population. This is called the resistance factor (RF). The graph below shows the results from a resistance investigation in a single population. The same amount of each of three insecticides was used. 100 A 80 B Mortality % KEY 60 C Insecticide A 40 Insecticide B 20 Insecticide C 0 0 10 20 30 40 50 Time (min.) 1.1.7 What is the approximate RF value for insecticide B? A 0,5 B 1 C 2 D 24 1.1.8 Which ONE of the insecticides has the fastest killing action given that the same quantity of each insecticide was used? A Insecticide A B Insecticide B C Insecticide C D All the insecticides were equally fast 1.1.9 A general conclusion from the data shown in the graph above, at 30 minutes, is that … A both insecticide B and insecticide C have been 100% effective. B both insecticide A and insecticide B have been 100% effective. C both insecticide A and insecticide C have been 100% effective. D all the insecticides have been 100% effective. Copyright reserved Please turn over ### Page 6 Downloaded from hlayiso.com Life Sciences/P1 6 DBE/November 2013 NSC 1.1.10 Numbers 1, 2, 3 and 4 below refer to four populations of frogs. These populations are represented diagrammatically by circles. Overlapping circles show populations that are capable of interbreeding to produce fertile offspring. 1 2 1 3 4 It would be reasonable to conclude that ... A if populations 2 and 4 were to die out, two different species would remain. B populations 1, 2, 3 and 4 represent four different species. C if population 2 were to die out, two different species would remain. D if population 3 were to die out, only one species would remain. (10 x 2) (20) 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 1.2.8) in the ANSWER BOOK. 1.2.1 The triplet of bases found in a tRNA molecule 1.2.2 The condition where an organism has more than two complete sets of chromosomes in a cell 1.2.3 A genetic disorder characterised by the absence of a blood clotting factor 1.2.4 Organisms that have different alleles at a given locus 1.2.5 The physical and/or functional expression of a gene 1.2.6 An arrangement of black bars representing DNA fragments that can be used to determine whether people are related 1.2.7 All the chromosomes found in a cell except the sex chromosomes 1.2.8 The biotechnological production of genetically identical offspring (8) Copyright reserved Please turn over ### Page 7 Downloaded from hlayiso.com Life Sciences/P1 7 DBE/November 2013 NSC 1.3 Indicate whether each of the statements in COLUMN I 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 1.3.8) in the ANSWER BOOK. COLUMN I COLUMN II 1.3.1 A trait such as height in A: Polygenic inheritance humans that has a range of B: Codominance intermediate phenotypes 1.3.2 Alternative forms of the same A: Alleles gene in a population B: Chromosomes 1.3.3 A type of gene mutation in A: Frame-shift mutation which adenine is lost/deleted B: Point mutation from a DNA base triplet 1.3.4 The structure that moves A: Centromere chromosomes/chromatids to B: Spindle the poles during cell division 1.3.5 Inheritance of acquired A: Lamarck characteristics B: Darwin 1.3.6 Influence(s) the inheritance of A: Multiple alleles blood groups B: Codominance 1.3.7 A characteristic that humans A: Opposable thumb share with other primates B: Binocular vision 1.3.8 A characteristic that is only A: Recessive expressed when in the B: Dominant homozygous state (8 x 2) (16) Copyright reserved Please turn over ### Page 8 Downloaded from hlayiso.com Life Sciences/P1 8 DBE/November 2013 NSC 1.4 A scientist gathered information about the affected and unaffected individuals for a certain genetic disorder. The genetic disorder is caused by a dominant allele. The findings of the scientist are represented in the pedigree diagram below. A B C D X Key Affected male Affected female Unaffected male Unaffected female 1.4.1 Using the letters R and r to represent the dominant and recessive alleles respectively, state the genotype of individual: (a) A (1) (b) D (1) 1.4.2 What is the percentage chance of individual X having a genetic disorder? (2) 1.4.3 If individual C marries an unaffected male, state the possible genotype(s) of their offspring. (2) (6) TOTAL SECTION A: 50 Copyright reserved Please turn over ### Page 9 Downloaded from hlayiso.com Life Sciences/P1 9 DBE/November 2013 NSC SECTION B QUESTION 2 2.1 Study the diagram below, which shows a section of a DNA molecule during a process taking place in a cell. A A – Adenine T G – Guanine T – Thymine G C – Cytosine C 2 1 T T G 2.1.1 Name the process represented in the diagram. (1) 2.1.2 When exactly will this process take place in a cell? (1) 2.1.3 Give labels for 1 and 2. (2) 2.1.4 State TWO ways in which the structure of RNA differs from the DNA structure shown in the diagram. (2) (6) Copyright reserved Please turn over ### Page 10 Downloaded from hlayiso.com Life Sciences/P1 10 DBE/November 2013 NSC 2.2 The diagram below shows a part of the process of protein synthesis. 1 2 3 C AU G G G AU G Molecule X Molecule Y Ribosome 2.2.1 Name the stage of protein synthesis represented in the diagram above. (1) 2.2.2 Identify: (a) Molecule X (1) (b) Molecule Y (1) (c) Structure 1 (1) 2.2.3 The table below shows the DNA base triplets that code for different amino acids found in human proteins. AMINO ACID BASE TRIPLET IN DNA Leucine GAA Proline GGG Lycine TTT Histidine GTA Serine TCA Methionine TAC Glycine CCC Glutamine GTC Using the information in the table and the diagram above, write down the sequence of the amino acids that correspond with structures 1, 2 and 3. (3) (7) Copyright reserved Please turn over ### Page 11 Downloaded from hlayiso.com Life Sciences/P1 11 DBE/November 2013 NSC 2.3 The diagrams below show chromosome pair 21 in the nucleus of a cell of the ovary of a woman. The chromosomes are involved in a process that takes place in a phase of meiosis. Diagram X Diagram Y Diagram Z B A 2.3.1 Give labels for A and B. (2) 2.3.2 Rearrange the letters X, Y and Z to show the correct sequence in which the events take place in this phase. (1) 2.3.3 Explain why the chromosomes in Diagram X and Diagram Y are different in appearance. (3) 2.3.4 The diagram below shows the nuclei of the four cells that resulted from meiosis involving the chromosomes in Diagram X above. Nuclei of the cells formed at the end of meiosis II M N O P Daughter chromosomes (a) Explain why nuclei O and P do NOT have chromosomes. (2) (b) Name and explain the disorder that will result if diagram M represents an egg cell that fuses with a normal sperm cell. (3) (11) 2.4 When flies with grey bodies were crossed with flies with black bodies, all the offspring in the F 1 had grey bodies. Use the letters G and g to represent a genetic cross to show the F 2 genotypes and phenotypes if the F 1 were interbred. (6) [30] Copyright reserved Please turn over ### Page 12 Downloaded from hlayiso.com Life Sciences/P1 12 DBE/November 2013 NSC QUESTION 3 3.1 Read the extract below. Genetic engineering involves a process whereby a gene is isolated from one organism and transferred into another organism. This gene can become part of the new host's genome. Usually the gene transfer takes place between organisms from different kingdoms. For example, a gene from a certain bacterium codes for an enzyme that deactivates a herbicide (a weedkiller). This gene is isolated from the bacterium and inserted into the chromosome of a crop plant. The resulting plant will now be herbicide-resistant. Before the products of genetic engineering can be sold, many tests must be done. Some seed companies have exclusive rights to sell the seeds that they have genetically engineered. Farmers cannot use seeds harvested from the crops that they have grown. Farmers must buy the seeds from the seed companies every time they want to plant the crop. [Adapted from Microbiology and Biotechnology, 1994] 3.1.1 What is meant by the term genome referred to in the extract? (1) 3.1.2 State ONE way in which the genetic engineering described in the extract differs from selective breeding. (2) 3.1.3 Give ONE reason why the products of genetic engineering must undergo many tests before they can be sold. (1) 3.1.4 Explain the value of growing herbicide-resistant crops. (3) 3.1.5 State THREE advantages of genetic engineering in crop production other than those mentioned in the extract above. (3) 3.1.6 Give a reason why seed companies insist that they must have the exclusive rights to the selling of seeds. (2) (12) Copyright reserved Please turn over ### Page 13 Downloaded from hlayiso.com Life Sciences/P1 13 DBE/November 2013 NSC 3.2 Gregor Mendel conducted breeding experiments with pea plants to study the inheritance patterns of four different traits (plant height, seed shape, seed colour and seed coat colour). For each trait, for example plant height, he crossed homozygous tall plants with homozygous dwarf plants. The offspring obtained in the F 1 -generation were then interbred to form the F 2 -generation. He did the same for each of the other traits. The results obtained for the F 2 -generation are shown in the table below. TRAIT RESULTS OF RATIO F 2 -CROSSING Plant height Tall: 787 2,84 : 1 (Tall or dwarf) Dwarf: 277 Seed shape Round: 5 474 X (Round or wrinkled) Wrinkled: 1 850 Seed colour Yellow: 6 022 Y (Yellow or green) Green: 2 001 Seed coat colour Grey: 705 3,15 : 1 (Grey or white) White: 224 rd [Adapted from Basic Concepts in Biology, 3 edition., C Starr, 1997] 3.2.1 What is the expected phenotypic ratio for a trait involving two heterozygous parents? (1) 3.2.2 From the results, calculate X and Y. Also state which trait provided a ratio closest to the expected phenotypic ratio mentioned in QUESTION 3.2.1. Show ALL working. (3) 3.2.3 Give a possible reason why the ratio selected in QUESTION 3.2.2 was closest to the theoretical ratio. (2) 3.2.4 Using the results, state whether the allele for round seeds or for wrinkled seeds is dominant. (1) 3.2.5 State TWO factors that Mendel controlled during these breeding experiments. (2) 3.2.6 Write down Mendel's law of segregation. (2) (11) 3.3 Describe how allopatric speciation occurs. (7) [30] TOTAL SECTION B: 60 Copyright reserved Please turn over ### Page 14 Downloaded from hlayiso.com Life Sciences/P1 14 DBE/November 2013 NSC SECTION C QUESTION 4 4.1 Although copper is an essential nutrient for the growth of plants, it is toxic and may kill the plants if present in very high amounts. High amounts of copper in the soil may be a result of ore deposits from old copper mines. Some grass species are tolerant to high levels of copper. An investigation was carried out to determine the effect of copper on the growth of a species of grass, Eragrostis sp., found near an old copper mine. • Seeds were collected from plants growing in five sample areas that were 10 metres apart, starting from an old copper mine. • Forty seeds from each sample area were planted under laboratory conditions in soil containing nutrients for normal growth. • The seeds were allowed to germinate into seedlings. • The 40 seedlings from each sample area were divided into two groups (Group 1 and Group 2). • A dilute copper solution was added to the soil of the Group 1 seedlings. • Distilled water was added to the soil of the Group 2 seedlings. After two weeks, the number of seedlings surviving was counted and the average height of the surviving seedlings was measured. The results are shown in the table below. Sample Distance Number of Average height of area from the old seedlings surviving surviving seedlings copper mine (cm) where the Group 1 Group 2 Group 1 Group 2 seeds were collected (m) 1 10 20 20 3,9 4,0 2 20 20 20 3,8 4,1 3 30 12 20 4,0 4,9 4 40 4 20 3,7 4,8 5 50 2 20 3,8 4,9 [Adapted from Tropical Grasslands, Journal Volume 24, 1990] 4.1.1 Draw a line graph to represent the results obtained for the number of seedlings surviving in Group 1. (7) 4.1.2 Identify the dependent variable(s) in the investigation. (2) 4.1.3 Identify TWO independent variables in this investigation. (2) Copyright reserved Please turn over ### Page 15 Downloaded from hlayiso.com Life Sciences/P1 15 DBE/November 2013 NSC 4.1.4 In terms of natural selection, explain why seedlings from seeds collected closer to the mine for Group 1 had a high survival rate. (6) (17) 4.2 Describe how a study of the Y-chromosome provides evidence for the 'Out of Africa' hypothesis. (3) 4.3 Describe the structural changes to the skull that characterise the evolution of modern humans from their ape-like ancestors, and explain the significance of these changes. Content: (17) Synthesis: (3) (20) NOTE: NO marks will be awarded for answers in the form of flow charts or diagrams. 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