Life Sciences Grade 12 P2 North West Mock 2023 Paper & Memo

——————— ——— Downloaded from Stanmorephysics.com education Department: Education North West Provincial Government REPUBLIC OF SOUTH AFRICA MARKS: 150 TIME: 2% hours IMD 12832E X10 CL This question paper consists of 16 pages. Please turn over Copyright reserved

DoWhISEGET from Stanmorephysics.com INSTRUCTIONS AND INFORMATION Read the following instructions carefully before answering the questions. 1. Answer ALL the questions. 2. LL the answers in the ANSWER BOOK. ° he answers 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 for each question. 6. 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. 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 compass where necessary. 11. Write neatly and legibly. Copyright reserved Please turn over

Dowritéedee?2From Stanmorephysics.com Nw/September 2023 SECTION A QUESTION 1 S options are provided as possible answers to the following questions. se the answer and write only the letter (A-D) next to the question bers (1.1.1 to 1.1.9) in the ANSWER BOOK, for e.g. 1.1.10 D. Q : ‘ fica During which phase of meiosis do spindle fibres start to form? A Prophase | B- Metaphase | C Anaphase | D_ Telophase | 1.1.2. The following list includes some features that are studied in evolution. (i) Bipedalism (ii) Dentition (iii) Brow ridges (iv) Stereoscopic vision Which ONE the following key features separate the human line of evolution from that of the African Apes? A (i) only B (i) and (iv) only C (ii) and (iv) only D (i), (ii) and (iii) only 1.1.3. Asmall section of mRNA has the following sequence of bases that codes for different amino acids: ccc CGU UAA Which ONE of the following is the representation of the anticodons and number of amino acids coded for by this section? ANTICODONS NUMBER OF AMINO ACIDS A GGG | GCA AUU 9 B GGG | GCA AUU 3 Cc GGG | GCA ATT 9 D GGG | GCA ATT 3: Taar fanah Copyright reserved Please turn over

DowhaNes from Stanmorephysics.com 1.1.4. Ina family with four children, each child has a different blogg Brou: The genotypes of the parents must be ... A Wiand [Bi B 1B and ii C Biand Ie D iandii Which ONE of the following monohybrid crosses wh ; ere complete dominance applies, will result in a phenotypic ratio of 3 21? A Both parents are heterozygous B One parent is heterozygous, and the other is homozygous recessive C Both parents are homozygous for the dominant characteristic D One parent is heterozygous, and the other is h jomozygous dominant QUESTIONS 1.1.6 AND 1.1.7 ARE BASED ON SHOWING THE POSSIBLE EVOLUTION OF S$: THE TIMELINE BELow OME HOMINIDS: Australopithe cus afarensis Ss Homo erectu Len 3 i 1 0 it (| Homo habilis Ardipithecus ramidus Q Homo Sapiens years ago did this species appear? A Australopithecus afarensis; 4 mya B Homo erectus; 2 mya C Homo habilis; 2.2 mya D Ardipithecus ramidus; 5mya o0QnaPY Na0og Copyright reserved Which species inhabited the Earth for the lon gest time, and how many How many genera are represented on the timeline above?

Dowhideaxed’?from Stanmorephysics.com NW/September 2023 NSC 1.1.8 The diagram below represents different phases of meiosis. Identify the correct phase in diagram 2 A Anaphase II B- Metaphase II C_ Prophase | D- Metaphase | 1.1.9 Ifa DNA molecule contains 20 000 nitrogenous bases, of which 20% are cytosine, how many adenine molecules will be present? A 1000 B 2000 Cc 6000 D 8000 (9x2) (18) Please turn over Copyright reserved

DoWiseaed from Stanmorephysics.com —_** NW/Sentembét202: 132 1.3 Give the correct biological term fon each tof tthe following descriptions. Write only the term next to the question numbers (1.2.1 to 1.2.9) in’the ANSWER BOOK. 1.2.1___ The mode of locomotion where an animal uses four limbs 1.2, The bond formed between amino acids 1.2180 1 The part of the skull that houses the brain — 1.2.4 Achange in the sequence of nitrogenous bases 1.2.5 A disorder due to the presence of an extra chromosome on the 21°" pair of chromosomes in somatic cells of humans 1.2.6 Genetic evidence that is used to support the ‘Out of Africa’ hypothesis 1.2.7 The process by which genetically identical organisms are formed using biotechnology 1.2.8 Type of dominance in which both alleles are expressed in the phenotype 1.2.9 Similar structures that are inherited from a common ancestor and modified for different functions (9x1) (9) Indicate whether each of the statements 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 numbers (1.3.1 to 1.3.3) in the ANSWER BOOK. COLUMN | COLUMN II 1.3.1. The genotypeofaperson |A: XY with haemophilia B; Xhxh 1.3.2 Variation in human height | A: Continuous variation B: Discontinuous variation 1.3.3 Involved in the discovery of |A: Eldredge DNA molecule B: Gould (3x2) (6) Copyright reserved

Dowrilereed? from Stanmorephysics.com NWiSeptember 2023 1.4 Three ladies each gave birth to a baby boy in a hospital. The babies were accidentally swopped. The diagram below shows a technique used in paternity testing and the results of each parent and their baby boys. SMITH STEVENS JONES BABY MR MRS MR MRS MR] MRS 41 2 3 aay 1.4.1 Identify the technique shown above. (1) 1.4.2. Which baby (1, 2 or 3) belongs to the Smiths? (2) 1.4.3 State TWO other uses for the technique named in QUESTION 1.4.1. (2) (5) Copyright reserved Please turn over

Déowreaded from Stanmorephysics.com 41.5 on bacterial resistance. NWiSentermber, 202: An investigation was carried out to determine the-éffect. oF ar’antibion< The diagram below shows the effect of an antibiotic on a bacterial population. ulation before tment with an antibiotic Key Key Resistant Bacteria Bacteria population (2) | Bacteria Non-Resistant Cc) G Bacteria population after treated with an antibiotic New bacterial population 1.5.1. Give the: (a) Dependent variable (1) (b) Independent variable (1) 1.5.2 Name the type of evolution that occurs in this bacterial population. (1) 1.5.3 What is the change in the environment that brings abut the change in the bacteria population? donar (1) 1.5.4. Which variation can be found in greater proportion na population? = (1) Copyright reserved —= Please turn over

7Py , OoWnisae, ded from Stanmorepfiysics.com heii Septarriner Beas 455° Whatpercentage of the bacieriain the original papulatio7 was ‘resistant to antibiotics” (1) (6) 1.6 Agenetic disorder is caused by a recessive allele (a). An ‘individual with the disorder is described as affected, and an individual fit is described as unaffected. The affected individual is indicated by circle or square. ‘pedigree diagram below shows the inheritance of this disorder in a HO 1.6.1 How many of each of the following are present in the diagram? (a) Males (1) (b) Generations (1) (c) Children of individuals A and B (1) 1.6.2 Give the: (a) Possible genotype of individual D (2) (b) Phenotype of individua! C (6) TOTAL SECTION A: 50 Copyright reserved Please turn over

Downleaded from Stanmorephysics.com NWiSeptember 2023 NSC SECTION B QUESTION 2 2.1 “e below represents chromosomes in a cell that is undergoing m . fon) =e ——> B Cc (—__—___ A 2.1.1 Identify: (a) Chromosome pair A (1) (b) Structure B (1) (c) Partc (1) 2.1.2 Which process is illustrated in the diagram? (1) 2.1.3 During which phase of meiosis does the process named in QUESTION 2.1.2 take place? (1) 2.1.4 What is the importance of the process named in QUESTION 2.1.2? (1) 2.1.5 Describe ONE other process in meiosis which will ensure the same outcome as the one stated in QUESTION 2.1.4. (3) (9) 2:2 Tabulate THREE differences between meiosis | and meiosis II. (7) (7) Copyright reserved Please turn over

Downloaded from Stanmorephysics.com | NW/September 2023 2.2> “Lamarckisminefers tg imvarck Answer the following questions on Lamarck's theory 2.3.1_State the Two 28volution, Principles (laws) that Lamarck used to explain 6 Onn... a8 pePlain ONE reason why most scientists reject this theory. (2) dt (4) 2.4 Human blood types are determined b i : : Y multiple alleles. Each pe ha: only two alleles in their DNA that code for thelr blood type. es Explain how an analysis of blood groups can be used to determine paternity. (5) (5) 2.5 Scientists wanted to determi ne which type of inheritance accounted for most of the selected genetic disorders in dogs. To do this they sequenced the genomes of many dogs of the same breed that suffered from the genetic disorders. The table below represents the type of inheritance for genetic disorders in dogs. TYPE OF PERCENTAGE OF GENETIC INHERITANCE DISORDERS (%) Autosomal recessive 71 Autosomal dominant 11 Sex-linked 10 Other 8 2.5.1 Define the term genome. (2) 2.5.2 State ONE factor that was kept constant in this investigation. (1) 2.5.3 Draw a pie graph to represent the information provided in the 6) table. ; Please turn over Copyright reserved

Dowsoaded from Stanmorephysics.com NWiSeptember 2023 2.6 The diagram below shows parts of two different processes that take place in the nucleus and a process taking place at the ribosome. Bg Ribosome 2.6.1 Identify: (a) Molecule A (1) (b) Organelle B (1) (c) Molecule C (1) (d) The number of codons shown in the diagram on the mRNA (1) 2.6.2 Give the anticodon for the second codon from left to right. (2) 2.6.3 Name AND describe the process that is taking place at X. (6) 264 Describe the role of mRNA in protein synthesis. (4) 20 Copyright reserved Please turn over

2 Life Sciences/P* Ne .. N Downloaded from Stanmorephysics.com alana QUESTION & 3.1 The baobab tree belongs to the genus Adansonia. Different species are found in Africa and Australia. Originally these two continents formed part of Gondwanaland, as shown in the diagram below. Gondwanaland \ Australian baobab Explain how the different species of baobab could have evolved on the different continents. (6) | Copyright reserved Please turn over

DoWnidaded from Stanmorephiysics.com en calli 3.2. Read the extract below. Warblers/songbirds belong to a large group of birds grouped together in the same bird family. In Siberia, Russia, two distinct species of greenish blers coexist, one in the west and one in the east. A large region of t, where they cannot survive, separates the two groups of warblers. istributions only narrowly overlap in central Siberia, where they do terbreed. The two species differ, in the songs that males sing to t females. Also, females of each species do not recognise the song other species but respond strongly to the song of the males from their own species. i 3.2.1. Name AND describe the reproductive isolation mechanism described in the extract. (3) 3.2.2 Mention TWO other reproductive isolation mechanisms that keep species separated. (2) (5) 3.3. The diagrams below show the pelvic structure of two species not drawn to scale. A B WIC 3.3.1 Which diagram represents the pelvis of a hominid? (1) 3.3.2 Explain ONE reason why Diagram B is better suited for a bipedal mode of locomotion. (2) 3.3.3 Explain TWO advantages of bipedalism. (4) (7) aan] Copyright reserved Please turn over

Downl 34 iloac estPy . loaded from Stanmorephysics.com naiSeptember 2023 The diagrams below r , epresert! the side and ventral views of the skulls of three prmat sapiens a re A gorilla. Australopthecus africanus and Homo -_ he lagrams showing the site views are not necessarily in e same order as the diagrams showing the ventral views Copyright reserved 3.4.1 Name the family that all these primates belong to. (1) 3.4.2 Which diagram (I, Il or Ill) represents the ventral view of skull B? (1) 3.4.3 Give TWO visible reasons for your answer to QUESTION 3.4.2. (2) 3.4.4. Describe THREE visible differences between the jaws in diagrams II and Ill, which show trends in human evolution. (6) 3.4.5 Based on the differences in dentition, what conclusion.can be made about the difference in the diet of the ogres in diagrams II and III? (2) He Please turn over

Doverfkradéa from Stanmorephiysics .com NwiSeptember 2023 3.5 In guinea pigs, black fur (B), is due to a dominant gene and white fur (b) is due to a recessive gene. Short fur (H) is due to a dominant gene and long fur (h) is due to a recessive gene. A bygeder made the following cross: BbHh x bbhh 3.940 State the phenotypes of the parents that the breeder crosse (2) yi Give the genotypes of ALL the possible gametes of the parents. (2) Give: 3.5:3 a) All the possible phenotypes of the offspring (2) b) The dominant allele for fur colour (1) 3.5.4 Explain Mendel's law of independent assortment. (3) (10) 3.6 Study the extract below. Female fruit flies are approximately 2,5 mm long. Males are smaller and possess a distinct black patch on their bodies. Females lay up to 400 eggs, which develop into adults in 7 to 14 days. Fruit flies can survive and breed in small flasks containing a simple nutrient medium consisting mainly of sugar. The gonosomes of fruit flies are similar to that of humans (XX and XY). 3.6.1. Explain TWO reasons why the fruit fly is a useful organism for studying genetic crosses. (4) 3.6.2 In fruit flies, the gene for eye colour is sex-linked and is carried on the X chromosome. The allele for red eyes R is dominant to the allele for white eyes r. A male with white eyes is crossed with a homozygous female with red eyes. Use a genetic cross to show the phenotypic ratio of the Fi offspring. (8) (10) [50] TOTAL SECTION B: 100 GRAND a 150 Uf i Copyright reserved

Downloaded from Stanmorephysics.com education Department: Education North West Provincial Government REPUBLIC OF SOUTH AFRICA NATIONAL SENIOR CERTIFICATE a” © i © i if ie ie jennie uu ty " LIFE SCIENCES P2 " " SEPTEMBER 2023 1" " MARKING GUIDELINES : “Sc cccccesdccccccccsesceccc we eccccceccceeeeese’ MARKS: 150 These marking guidelines consists of 12 pages. Copyright reserved Please turn over

Rowsiended from Stanmorephysics.com Nwiseptember 2023 NSC — Marking Guidelines PRINCIPLES RELATED TO MARKING LIFE SCIENCES 1. 10. 11. 12. 13. 14. 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 theYfirst three irrespective of whether all or some are correct/incorrect. If the whole process is given when only a part of it is required Read all and credit the relevant part. If comparisons are asked for, but descriptions are given Accept if the 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 the 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 the 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 the answer if correct. Wrong numbering If the answer fits into the correct sequence of questions, but the wrong number is given, it is acceptable. If the 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 are given in terminology Accept, provided it was accepted at the national memo discussion meeting. If only the letter is asked for, but only the name is given (and vice versa) Do not credit. Copyright reserved Please turn over

Rowsiended from Stanmorephysics.com Nwiseptember 2023 15. 16. 17. 18. 19. NSC — Marking Guidelines If units are not given in measurements Candidates will lose marks. Marking guidelines will allocate marks for units separately. Be sensitive to the sense of an answer, which may be stated differently. 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 learner's 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 memorandum No changes must be made to the memorandum. The provincial internal moderator must be consulted. Copyright reserved Please turn over

Bownleaded from Stanmorephysics.com SECTION A QUESTION 1 1.10 1.1.1 1.1.2 1.1.3 1.1.4 1.1.5 1.1.6 1.1.7 1.1.8 1.1.9 1.2 1.2.1 1.2.2 1.2.3 1.2.4 1.2.5 1.2.6 1.2.7 1.2.8 1.2.9 1.3 1.3.1 1.3.2 1.3.3 14 1.4.1 1.4.2 1.4.3 1.5. 1.5.1 1.5.2 1.5.3 1.5.4 1.5.5 NSC — Marking Guidelines AVY Biv v BY v AvY Avy BY Byv Dvv Cvv QuadrupedalismY Peptidev Craniumy Gene mutation” Trisomy 21v/Down Syndrome Mitochondrial DNAY /mt-DNA CloningY Co-Dominancev Homologous structures” Both A and B¥v A onlyv¥v Nonevv DNA profiling’ Baby 3vv — Identifying suspects in a crimev — Tracing missing personsY — Identifying genetic disordersv — Matching tissues for organ transplantsv — Identifying dead personsY (Mark first TWO only) (a) Bacterial resistance” (b) Effect of antibioticsv Natural Selection’ /Punctuated Equilibrium AntibioticsY Resistant Bacteriav 25% Copyright reserved NW/September 2023 (9x2) (18) (9x1) (9) (3x2) (6) (1) (2) Any (2) (5) (1) (1) (1) (1) (1) (1) (6) Please turn over

Rowsiended from Stanmorephysics.com Nwiseptember 2023 NSC — Marking Guidelines 16 161 (a) 5Y (1) (b) 3v (1) (c) 3v (1) 1.6.2 Jtfa) Aav/AAY (2) (b) Affected femalev (1) (6) TOTAL SECTION A: 50 SECTION B QUESTION 2 2.1. 2.1.1. (a) HomologousY chromosomes/bivalent (1) (b) Centromere” (1) (c) Chromatidy (1) 2.1.2 Crossing overY (1) 2.1.3. Prophase Iv (1) 2.1.4 Ensures genetic variation’ (1) 2.1.5 — Chromosomes/chromatids arrange themselves randomly” — oneither side of the equatory” — during metaphase lv/II (3) (9) Copyright reserved Please turn over

Rowsiended from Stanmorephysics.com Nwiseptember 2023 NSC — Marking Guidelines 2.2 Meiosis | Meiosis Il Crossing over takes placev | No crossing over takes placev In metaphase, the (In metaphase) chromosomes align on the chromosomes align singly~ equator in homologous on the equator pairsv During anaphase, During anaphase, chromosomesvY move chromatids” move towards towards the poles the poles Homologous chromosomes | Chromosomes occur singly in prophase | ” in prophase Il v7 The chromosome number is_ | The chromosome number halved” /changes from remains the samev diploid to haploid Results in two daughter Results in four daughter cellsv cellsv (Mark first THREE only) Any 3 x 2 +1 for table (7) 2.3 2.3.1 'Law' of use and disusev ‘Law' of the inheritance of acquired characteristics” (2) 2.3.2 — Acquired characteristics” — are not inheritedv/do not cause any change to the DNA of an organism's gametes (2) (4) 2.4 — The blood the of mother, possible father and the child must be comparedv —_ If this shows that it is not possible that these parents can produce a child with his/her blood group” — then this man is not the father” — If this shows that it is possible that these parents can produce a child with his/her blood groupY — then he may/ may not be the fatherY — because other males have the same blood group” (5) Any (5) 2.5 2.5.1 The mapping of the exact position of all the genes in allthe chromosomes of an organism/humanv ~ (2) 2.5.2 The breed of the dogsv (Mark first ONE only) (1) Copyright reserved Please turn over

oawnended from Stanmorephysics.com — Nwiseptember 2023 NSC — Marking Guidelines 2.5.3 Calculations Total: 71 + 11+ 10+ 8 = 100 Autosomal recessive: 71/100 x 360 = 255,6° Jautésomal dominant: 11/100 x 360 = 39,6° dOn, an 10/100 x 360 = 36° er: 8/100 x 360 = 28,8° PERCENTAGE GENETIC DISORDERS CAUSED BY DIFFERENT TYPES OF IN Other 8% Sex -linked 10% Autosomal dominant 11% Criteria for marking graph HERITANCE IN DOGS CRITERIA Pie chart drawn (T) Caption showing the relationship between the two variables(H) Correct calculation (C) to determine the proportions for the labelled sectors Plotting of Sectors (P) Correct proportions for labelled sectors 1-3 sectors drawn and labelled correctly All 4 sectors of the graph are drawn and labelled correctly Copyright reserved Autosomal recessive 71% MARK 1 1 2 1 loon — 2 (6) (9) Please turn over

Rowsiended from Stanmorephysics.com Nwiseptember 2023 NSC — Marking Guidelines 2.6 2.6.1 (a) DNAY (1) (b) NucleolusY¥ (1) (© tRNAY (1) (a) 3v (1) 2.6.2—UCCvv (2) 2.6.3 DNA Replicationv* — The DNA (double helix) unwinds” and — _unzipsv/hydrogen bonds break — to form two separate strandsY — Both DNA strands serve as the templatesY — to build a complementary DNAv’/(A to T and C to G) — using free (DNA) nucleotides” from the nucleoplasm — This results in two identical (DNA) moleculesY - Each molecule consists of one original strand and one new strandv *4 Compulsory + Any 5 (6) 2.6.4 -— mRNA forms during transcription” — by copying the coded message from DNAY — and moves out of the nucleusv — and attaches to the ribosomev — the anticodons on the tRNA match the codons on the mRNAY Any (4) (16) [50] Copyright reserved Please turn over

Rowsiended from Stanmorephysics.com Nwiseptember 2023 NSC — Marking Guidelines QUESTION 3 3.1 — The common ancestor of the original baobab population/the baobab population of Gondwanaland was separated into two populationsY 2) by the ocean/seav* “| There was no gene flow between the African baobab and Australian baobabv /two baobab populations. — Each baobab population were exposed to different environmental conditions ~/other selection pressures. — Natural selections occur independentlyv in each of the two baobab populations. — Such that the individuals of the two baobab populations become very different from each other. — genotypically and phenotypically” — even if the two baobab populations were to mix again” — they will not be able to interbreed. v The two baobab populations are now different baobab species (in the question) (No marks for this statement) *41 Compulsory + Any 5 (6) 3.2 3.2.1 -— They use species-specific courtship behaviour”* - Males sing songs to attract female matesY — and females do not respond/recognise songs from other males¥ *1 Compulsory +2 (3) 3.2.2 — Breeding at different times of the yearvY — Plant adaptations to different pollinators~ — Producing infertile offspringy’ — Prevention of fertilisation’ (Mark first TWO only) Any (2) (5) 3.3 3.3.1 BY (1) 3.3.2 — The pelvis is short and wideY — to support the body weight of the upper body” (Mark first ONE only) (2) Copyright reserved Please turn over

Rowsiended from Stanmorsphysics.com — Nwiseptember 2023 NSC — Marking Guidelines 3.3.3 — Frees the armsv so that they could carry offspringv /tools/food/manipulate things — Allows the ability to see further” to spot dangerv /food 4! Reduces the surface area exposed to the sun,” so, less heat is absorbedv /less heat lost/improved thermoregulation — Expose the genitalsv to attract the opposite sexv — Efficient locomotion’ allows travelling longer distancesY (Mark first TWO only) Any2x2 (4) (7) 3.4 3.4.1. Hominidaey (1) 3.4.2 Diagram Illv (1) 3.4.3. — The foramen magnum is in a more forward position’ — The palate is more roundedv’ (Mark first TWO only) (2) 3.4.4 — The jaw/palate is rectangular in Diagram Il” and rounded in Diagram IIlv — Large spaces between the teeth in Diagram IIV and small/ no spaces between the teeth in Diagram III — Large canines/teeth in Diagram II Y and small canines/ teeth in Diagram III v (Mark first THREE only) Any3x2 (6) 3.4.5 — This organism in Diagram II had a diet of hard, raw foodv - and the organism in Diagram III has a diet of softerv/ cooked food (2) (12) 3.5 3.5.1 Phenotype Black short fur” x White long fur” (2) 3.5.2 Gametes BH, Bh, bH, bhvY (2) 3.5.3 (a) Black Short fur; Black Long fur; White Short fur; White Long fury ¥ (2) (b) Black fur (1) Copyright reserved Please turn over

Rowsiended from Stanmorephysics.com — Nwiseptember 2023 NSC — Marking Guidelines 3.5.4 — Because of the random arrangement of chromosomes at the equator during meiosis“~/any one of two alleles of a characteristic can sort with any two of another characteristic — the alleles of different genes move independently of each other into the gametes” “! They can therefore appear in the gametes in different combinations” (3) (10) 3.6 3.6.1 = Females lay many eggs” which will ensure that the ratios obtained will be reliablev” — The eggs develop very quicklyv/7 to 14 days therefore, results will be obtained fast” — Can clearly differentiate between male and femalev making it easy to breedv — Fruit flies survive and breed in small flasks~ making it easy to observe and work withv — The food medium required is simplev/cheap making it easy/cheap to conduct the investigationv (Mark first TWO only) (Any 2 x 2) (4) 3.6.2 Copyright reserved Please turn over

Rowsiewded from Stanmorephysics.com — Nwiseptember 2023 NSC — Marking Guidelines Pi Phenotype: White-eyed male x red-eyed femalev Genotype: x'Y x XRXR Vv Meiosis G/gametes x, Y XR XR v Fertilisation [ = S ] F1 Genotype XRXr XRXT XRY XRY v Phenotype: 2 red-eyed males : 2 red-eyed females 1 red eyed male : 1 red-eyed femalev* P; and Fiv Meiosis and fertilisation v (*1 Compulsory + Any 5) OR P1 Phenotype: White-eyed male x Red-eyed femaleY Genotype: x'Y x XRXR v Meiosis Gametes XR XR x XR xr XR xr Fertilisation sain Y XRY XRY 1 mark for correct gametes” 1 mark for correct genotypesY Fi Phenotype: 2 red-eyed males: 2 red-eyed females 1 red-eyed male: 1 red-eyed femalev* Px and Fi ¥ Meiosis and fertilisation Vv (6) (*1 Compulsory + Any 5) (10) [50] TOTAL SECTION B: 100 GRAND-TOTAL: 150 Copyright reserved