... (1) ... (1) 2. Describe the consequence of a base substitution mutation with regards to sickle cell anaemia. (Total 7 marks)

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1 1. (a) Define sex linkage. (b) State one example of sex linkage. (c) Draw a simple pedigree chart that clearly shows sex linkage in humans. Use conventional symbols. Start with an affected woman and an unaffected man. (4) (Total 6 marks) 2. Describe the consequence of a base substitution mutation with regards to sickle cell anaemia. (Total 7 marks) 3. Describe, with the aid of a diagram, the behaviour of chromosomes in the different phases of meiosis. (Total 5 marks) 1

2 4. Explain how meiosis and fertilization can give rise to genetic variety. (Total 6 marks) 5. The diagram below shows the pedigree of a family with red green colour-blindness, a sex-linked condition. (a) Define the term sex-linkage. (b) Deduce, with a reason, whether the allele producing the condition is dominant or recessive. 2

3 (c) (i) Determine all the possible genotypes of the individual (2nd generation 1) using appropriate symbols.... (ii) Determine all the possible genotypes of the individual (3rd generation 4) using appropriate symbols.... (Total 5 marks) 6. Explain the use of two named enzymes in biotechnology. (Total 8 marks) 7. Which enzymes are needed to produce recombinant plasmids that are used in gene transfer? A. DNA polymerase and ligase B. DNA polymerase and restriction enzymes C. Restriction enzymes and ligase D. Helicase and restriction enzymes 3

4 8. Which response describes the behaviour of chromosomes in metaphase I and anaphase II of meiosis? Metaphase I Anaphase II A. Chromosomes line up at the equator Separation of homologous chromosomes B. Tetrads (bivalents) line up at the equator Separation of homologous chromosomes C. Chromosomes line up at the equator Separation of sister chromatids D. Tetrads (bivalents) line up at the equator Separation of sister chromatids 9. In garden peas, the pairs of alleles coding for seed shape and seed colour are unlinked. The allele for smooth seeds (S) is dominant over the allele for wrinkled seeds (s). The allele for yellow seeds (Y) is dominant over the allele for green seeds (y). If a plant of genotype Ssyy is crossed with a plant of genotype ssyy, which offspring are recombinants? A. SsYy and Ssyy B. SsYy and ssyy C. SsYy and ssyy D. Ssyy and ssyy 10. What constitutes a linkage group? A. Genes carried on the same chromosome B. Genes whose loci are on different autosomes C. Genes controlling a polygenic characteristic D. Alleles for the inheritance of ABO blood groups 4

5 11. The following diagram represents a two generation pedigree showing the blood groups of the individuals. The female has been married to two different individuals. (a) Define the term co-dominant alleles. (b) Deduce with a reason the probable father of 2nd generation 1. (c) If 2nd generation 3 marries a man with blood group AB, predict the possible genotypes of the children. (3) (Total 6marks) 5

6 12. Describe how sexual reproduction promotes genetic variation within a species. (Total 4 marks) 13. Polygalacturonase (PG) plays an important role in fruit softening by making the pectin of the cell wall more soluble. It is synthesized only when the fruit is ripe. In order to slow down the ripening of tomatoes (Lycopersicon esculentum), antisense RNA technology was used. Messenger RNA from untransformed and transformed fruit was hybridized to a radioactively labelled probe specific to the PG sense strand. The results of a gel electrophoresis of mrna are given below. (The size of the mrna strands is expressed in kilobases, kb.) The histogram shows these results expressed as the percentage of PG mrna in ripe untransformed fruit. Lane 1: Ripe untransformed fruit Lane 2: Unripe untransformed fruit Lane 3: Ripe transformed fruit Lane 4: Unripe transformed fruit [Source: Smith et al., Nature, (1988), 334, pages ] (a) State the percentage of PG mrna in ripe transformed fruit. 6

7 (b) Compare the results obtained for ripe and unripe fruit. (c) Using the information provided, explain how the antisense technology affects transformed fruit. (3) (Total 6 marks) 14. Outline the differences between the behaviour of the chromosomes in mitosis and meiosis. (Total 5 marks) 7

8 15. A cell with a diploid number of 12 chromosomes undergoes meiosis. What will be the product at the end of meiosis? A. 2 cells each with 12 chromosomes B. 4 cells each with 6 chromosomes C. 2 cells each with 6 chromosomes D. 4 cells each with 12 chromosomes 16. The following is a DNA gel. The results are from a single probe showing a DNA profile for a man, a woman and their four children. [Source: The Biology Project, University of Arizona] Which fragment of DNA is the smallest? A. I B. II C. III D. IV 8

9 17. The following is a DNA gel. The results are from a single probe showing a DNA profile for a man, a woman and their four children. Which child is least likely to be the biological offspring of the father? A. Child 1 B. Child 2 C. Child 3 D. Child A parent organism of unknown genotype is mated in a test cross. Half of the offspring have the same phenotype as the parent. What can be concluded from this result? A. The parent is heterozygous for the trait. B. The trait being inherited is polygenic. C. The parent is homozygous dominant for the trait. D. The parent is homozygous recessive for the trait. 9

10 19. Which process results in the greatest genetic variation in a population? A. Meiosis B. Mitosis C. Cytokinesis D. Natural selection 20. The allele for red flower colour (R) in a certain plant is co-dominant with the allele for white flowers (R ). Thus a plant with the genotype RR has pink flowers. Tall (D) is dominant to dwarf (d). What would be the expected phenotypic ratio from a cross of RR dd plants with R R Dd plants? A. 9:3:3:1 B. 50% pink 50% white, and all tall C. 1:1:1:1, in which 50% are tall, 50% dwarf, 50% pink and 50% white D. 3:1 21. Two genes A and B are linked together as shown below. If the genes are far enough apart such that crossing over between the alleles occurs occasionally, which statement is true of the gametes? A. All of the gametes will be Ab and ab. B. There will be 25% Ab, 25% ab, 25% ab and 25% AB. C. There will be approximately equal numbers of Ab and ab gametes. D. The number of Ab gametes will be greater than the number of ab gametes. 10

11 22. (a) Define the term co-dominance. (b) A man of blood type AB and a woman of blood type B are expecting a baby. The woman s mother had blood type O. Deduce the possible phenotypes of the offspring from the cross shown below. (4) (Total 5 marks) 23. Sickle cell anemia is a serious disease caused by a single base substitution mutation. Explain how a single base substitution mutation can have significant consequences for an individual. (Total 6 marks) 24. Outline a method for carrying out gene therapy, using a named example. (Total 8 marks) 11

12 25. A polygenic character is controlled by two genes each with two alleles. How many different possible genotypes are there for this character? A. 2 B. 4 C. 9 D A cross is performed between two organisms with the genotypes AaBb and aabb. What genotypes in the offspring are the result of recombination? A. Aabb, AaBb B. AaBb, aabb C. aabb, Aabb D. Aabb, aabb 27. (a) Define the term degenerate as it relates to the genetic code. 12

13 (b) Apart from international cooperation, outline two positive outcomes of the Human Genome Project. (c) State the catalytic activity of reverse transcriptase. (d) State one use of monoclonal antibodies in diagnosis and one use in treatment. Diagnosis:... Treatment:... (Total 6 marks) 28. Explain how meiosis results in great genetic variety among gametes. (Total 8 marks) 29. In Zea mays, the allele for coloured seed (C) is dominant over the allele for colourless seed (c). The allele for starchy endosperm (W) is dominant over the allele for waxy endosperm (w). Pure breeding plants with coloured seeds and starchy endosperm were crossed with pure breeding plants with colourless seeds and waxy endosperm. (a) State the genotype and the phenotype of the F 1 individuals produced as a result of this cross. genotype... phenotype... 13

14 (b) The F 1 plants were crossed with plants that had the genotype c c w w. Calculate the expected ratio of phenotypes in the F 2 generation, assuming that there is independent assortment. Use the space below to show your working. Expected ratio... (3) The observed percentages of phenotypes in the F 2 generation are shown below. coloured starchy 37% colourless starchy 14% coloured waxy 16% colourless waxy 33% The observed results differ significantly from the results expected on the basis of independent assortment. (c) State the name of a statistical test that could be used to show that the observed and the expected results are significantly different. 14

15 (d) Explain the reasons for the observed results of the cross differing significantly from the expected results. (Total 8 marks) 30. Discuss the ethical arguments for and against the cloning of humans. (Total 4 marks) 31. Outline the process of DNA profiling (genetic fingerprinting), including ways in which it can be used. (Total 6 marks) 15

16 32. Discuss the ethical issues for and against the use of transgenic plants (Total 6 marks) 33. (a) Define the term sex linkage. 16

17 (b) A male and female with normal colour vision each have a father who is colour blind. They are planning to have children. Predict, showing your working, the possible phenotypes and genotypes of male and female children. (3) (c) Explain the relationship between Mendel s law of segregation and meiosis. (3) 17

18 (d) Distinguish the differences between animal cells and plant cells undergoing mitosis and cytokinesis. (Total 9 marks) 34. The diagram below shows a cell during meiosis. How many chromosomes would each daughter cell have at the end of meiosis? A. 1 B. 2 C. 4 D. 8 18

19 35. The pedigree below shows which members of a family were Rhesus positive ( and ) and Rhesus negative ( and O). The allele for Rhesus positive blood (Rh + ) is dominant over the allele for Rhesus negative blood (R - ). Which are possible genotypes of the individuals numbered I, II and III? I II III A. Rh + Rh + Rh + Rh + Rh + Rh B. Rh + Rh + Rh + Rh Rh + Rh + C. Rh + Rh + Rh + Rh Rh + Rh D. Rh + Rh Rh + Rh Rh + Rh + 19

20 36. The diagram below shows the results of DNA profiling using gel electrophoresis. What conclusion can be drawn about the DNA in bands I and II? A. The DNA in the two bands has the same base sequence. B. The DNA in the two bands consists of fragments of the same length. C. The DNA in the two bands has the same ratio of bases. D. The DNA in the two bands came from the same source. 37. The diagram below shows chromosomes during prophase I of meiosis. How many chromosomes and chiasmata are visible? Number of chromosomes Number of chiasmata A. 2 2 B. 4 2 C. 2 4 D

21 38. In peas the allele for round seed (R) is dominant over the allele for wrinkled seed (r). The allele for yellow seed (Y) is dominant over the allele for green seed (y). If two pea plants with the genotypes YyRr and Yyrr are crossed together, what ratio of phenotypes is expected in the offspring? A. 9 round yellow : 3 round green : 3 wrinkled yellow : 1 wrinkled green B. 3 round yellow : 3 round green : 1 wrinkled yellow : 1 wrinkled green C. 3 round yellow : 1 round green : 3 wrinkled yellow : 1 wrinkled green D. 1 round yellow : 1 round green : 1 wrinkled yellow : 1 wrinkled green 39. What is a difference between autosomes and sex chromosomes? A. Autosomes are not found in gametes but sex chromosomes are. B. Sex chromosomes are found in animal cells and autosomes are found in plant cells. C. Autosomes are diploid and sex chromosomes are haploid. D. Sex chromosomes determine gender and autosomes do not. 40. (a) State the names of the parts of the chromosome labelled (i) and (ii) on the diagram below. [Source: adapted from Hartwell (editor) (2003), Genetics: from Genes to Genomes, 2nd edition, McGraw Hill, page 81] 21

22 (b) Explain how the inheritance of chromosome 21 can lead to Down s syndrome. (3) (c) Explain how meiosis promotes variation in a species. (Total 7 marks) 41. Outline a basic technique for gene transfer involving plasmids. (Total 5 marks) 42. Outline DNA profiling (genetic fingerprinting), including one way in which it has been used. (Total 5 marks) 43. Karyotyping involves arranging the chromosomes of an individual into pairs. Describe one application of this process, including the way in which the chromosomes are obtained. (Total 5 marks) 22

23 44. Outline two examples of the commercial application of enzymes in biotechnology. (Total 6 marks) 45. Discuss the potential benefits and possible harmful effects of genetic modification. (Total 7 marks) 46. The diagram below shows a cell in meiosis. What can be deduced from this diagram? [Source: J W Saunders, (1968), Animal Morphogenesis, MacMillan, page 7] Stage of meiosis shown Haploid number of chromosomes in this cell A. Metaphase I 6 B. Prophase I 3 C. Prophase I 6 D. Metaphase I 3 23

24 47. If red (RR) is crossed with white (rr) and produces a pink flower (Rr), and tall (D) is dominant to dwarf (d), what is the phenotypic ratio from a cross of Rr dd and rr Dd? A. 9:3:3:1 B. 50% pink, 50% white and all tall C. 1:1:1:1, in which 50% are tall, 50% dwarf, 50% pink and 50% white D. 3:1 48. Define the terms gene and allele and explain how they differ. (Total 4 marks) 49. Outline one example of inheritance involving multiple alleles. (Total 5 marks) 50. Using an example you have studied, explain a cross between two linked genes, including the way in which recombinants are produced. (Total 9 marks) 24

25 51. Rats were bred for several generations to prefer alcohol (ethanol) consumption. When tested, it was discovered that the brains of these rats possessed lower quantities of the chemical neuropeptide Y (NPY). To test the hypothesis that lower quantities of NPY leads to a preference for alcohol, rats were genetically engineered to be NPY deficient (genotype NPY / ), or to produce an excess of NPY (NPY-EX). In separate experiments, the two groups were compared to normal rats (in terms of their alcohol preference) possessing the genotype NPY +/+. The groups were offered solutions of increasing alcohol concentration. The quantity of each solution consumed per day was measured. Figure 1 Figure 2 [Source: adapted from Thiele et al, Nature, (1998), 396 pages ] (a) Calculate the difference in consumption of the 6% alcohol solution between the (i) NPY / and NPY +/+ rats (figure 1); (ii) NPY-EX and NPY+/+ rats (figure 2)

26 (b) Compare the alcohol consumption of the NPY / rats with the NPY-EX rats. (3) (c) Identify the relationship between NPY levels and alcohol consumption. An experiment was carried out to test the hypothesis that an increase in preference for alcohol might be related to a decrease in sensitivity to its effects. Rats were injected with a sample of alcohol and then assessed for the length of time it took for them to regain the righting reflex. (The righting reflex refers to the ability of the rat to return to its feet after being placed on its back.) Figure 3 [Source: adapted from Thiele et al, Nature, (1998), 396 pages ] 26

27 (d) Deduce the relationship between NPY levels and the time required to regain the righting reflex. (3) An additional experiment was carried out to determine whether differences in sensitivity to the effects of alcohol might be related to differences in the rats ability to remove alcohol from their blood. Rats were injected with alcohol and blood samples were taken one hour and three hours later to determine alcohol levels. The results are shown below. Figure 4 [Source: adapted from Thiele et al, Nature, (1998), 396 pages ] 27

28 (e) Evaluate the hypothesis that differences in sensitivity to the effects of alcohol might be related to differences in the ability of the rats to remove alcohol from their blood. (f) Using all the data, outline the relationship between preference for alcohol and sensitivity to the effects of alcohol. (g) (i) Define the term homozygous

29 (ii) State the phenotype of a rat with the genotype NPY +/ (iii) Using a Punnett grid, predict the fraction of offspring that would have the genotype NPY +/ if two rats were crossed, one homozygous for the NPY+ allele and one homozygous for the NPY allele.... (Total 17 marks) 52. Outline two examples of the commercial application of named enzymes in biotechnology. (Total 6 marks) 53. (a) State two examples of transgenic techniques in agriculture involving animals. 29

30 (b) Outline the use of antibiotics in livestock production. (Total 4 marks) 54. Outline the use of transgenic techniques in agriculture, using one named animal example (Total 3 marks) 55. What is a possible consequence of two base substitution mutations occurring in the same gene? A. Two amino acids coded for by the gene are changed. B. Amino acids in two polypeptides coded for by the gene are changed. C. All of the codons between the two mutations are changed. D. All of the codons from the first mutation onward are changed. 30

31 56. Hemophilia is caused by an X-linked recessive allele. In the pedigree shown below which two individuals in the pedigree must be carriers of hemophilia? A. I-1 and II-1 B. I-4 and II-2 C. II-1 and II-2 D. III-2 and III A single gene in humans causes blood to be either rhesus positive (dominant allele) or rhesus negative (recessive allele). A woman with rhesus negative blood has already had a child with rhesus positive blood. There could be complications during pregnancy if she has another child with rhesus positive blood. What is the probability of this, if the father is the same, and if his mother is known to have rhesus negative blood? A. 25% B. 50% C. 75% D. 100% 31

32 58. Potatoes with more starch have a lower percentage water content. This has an advantage in the transport, cooking and processing of potatoes. In a strain of Escherichia coli scientists found an enzyme which increases the production of starch. Using biotechnology, the gene for this enzyme was transferred to potatoes, increasing their starch content (transgenic potatoes). The gene was transferred to three potato varieties to create three transgenic lines. The table shows the mean amount of starch and sugar contained in three lines of transgenic potatoes and normal potatoes (control), after storage for four months at 4 C. Carbohydrate / % of fresh weight Potato Line Sugar Starch I Transgenic II III Mean I Control II III Mean [Source: Stark et al, (1999), Annals of the New York Academy of Sciences, 792, pages 26 36] (a) State which line of transgenic potato has the greatest amount of starch. (b) (i) Compare the levels of carbohydrate between the transgenic lines and the control potatoes

33 (ii) Suggest reasons for these differences Potato tubers were harvested from the field and stored in high humidity at 4 C for three months. After this period, the tubers were stored at 16 C, and samples were removed after 0, 3, 6 or 10 days, cut into strips, and fried. The colour of the fried potatoes was then measured and values reported using a 0 4 rating (light to dark), where a score of 2 or lower indicates acceptable colour. The results are shown in the table. [Source: Stark et al, (1999), Annals of the New York Academy of Sciences, 792, pages 26 36] 33

34 (c) Evaluate the effect of transferring the E. coli gene on the suitability of the potatoes for frying. An important part of storage management is to delay sprouting of potatoes. A second sample of potatoes was harvested from the field and stored at high humidity for three months at 4 C. Storage temperature was then raised to 16 C and a sample of potatoes were examined daily and scored for the number of sprouts longer than 0.5 cm. The number of days it took for 50% to sprout is indicated in the graph for control varieties (C1, C2 and C3) and three transgenic varieties (T1, T2 and T3) of potatoes. [Source: Stark et al, (1999), Annals of the New York Academy of Sciences, 792, pages 26 36] 34

35 (d) Deduce how the E. coli gene affects the storage of potatoes. (e) Discuss three possible harmful effects of genetically modified potatoes. (3) (Total 12 marks) 35

36 59. Studies were carried out on the leaves of the wild type of the flowering mustard plant (Arabidopsis thaliana) to determine the pigment composition of the thylakoid membrane. The results were compared with the leaves of a mutant strain of the plant. These studies were repeated using thylakoid membranes from leaves grown in cultures deficient in either iron or magnesium. Strain of Plant Mineral Deficiency Chlorophyll / µg g 1 leaf Carotenoids / µg g 1 leaf None Wild Type Iron Magnesium None Mutant Iron Magnesium [Source: Lu et al, (1995), Botanical Bulletin of Academia Sinica, 36, pages ] (a) State the concentration of chlorophyll found in the leaves of the mutant strain of plant when deficient in iron. (b) (i) Calculate the percentage increase in carotenoids found in the wild type when magnesium is deficient.... (ii) Suggest why magnesium deficiency causes the changes shown in the pigment content of the leaves

37 (c) Using the data in the table, outline the effects of iron deficiency on the pigment content of the leaves. (Total 6 marks) 60. If a person inherited an allele with the same base substitution mutation from both parents, what sequences could be altered from normal in the person s cells? A. One mrna base sequence only B. Two mrna base sequences only C. One mrna base sequence and one polypeptide amino acid sequence only D. Two mrna base sequences and two polypeptide amino acid sequences only 61. What is the locus of a gene? A. The proportion of the population that have the gene B. The part of the phenotype that is affected by the gene C. The position of a gene on a chromosome D. The predicted effect of natural selection on the frequency of the gene 37

38 62. Which processes always occur in meiosis but not normally in mitosis? I. Chiasmata formation II. III. Recombination of genes Separation of homologous chromosomes A. I and II only B. II and III only C. I and III only D. I, II and III 63. The diagram below shows the life cycle of a moss. The haploid chromosome number is shown as n and the diploid number as 2n. At which stage in the life cycle does meiosis take place? A. I B. II C. III D. IV 38

39 64. Mendel crossed pure breeding (homozygous) tall pea plants that had coloured flowers with pure breeding dwarf pea plants that had white flowers. All of the resulting F 1 plants were tall and had coloured flowers. If Mendel had crossed these F 1 plants with a pure breeding strain of dwarf pea plants with coloured flowers, what proportion of tall coloured plants would be expected in the offspring? A. B. C. D. 65. Up to two additional marks are available for the construction of your answers. (a) (b) Define the term gene linkage and outline an example of a cross between two linked genes. Describe the inheritance of ABO blood groups including an example of the possible outcomes of a homozygous blood group A mother having a child with a blood group O father. (8) (5) (c) Outline sex linkage. (5) (Total 20 marks) 39

40 66. What are the chromosomes of fungi made of? A. DNA only B. DNA and protein only C. DNA and RNA only D. DNA, RNA and protein 67. What is a sex-linked gene? A. A gene whose locus is on the X chromosome only. B. A gene whose locus is on the X or Y chromosomes. C. A gene whose locus is on the both X and Y chromosomes. D. A gene whose locus is on the Y chromosome only. 40

41 68. The pedigree chart below shows the inheritance of a genetic disease in a family. What is the nature of the allele that causes this disease? A. Dominant and sex linked B. Dominant and non-sex linked C. Recessive and sex linked D. Recessive and non-sex linked 41

42 69. Which characteristics are used to identify chromosomes when constructing a karyotype? I. The length of the chromosome. II. III. IV. The position of the centromere on the chromosome. The pattern of bands on the chromosome. The position of the chromosome on the spindle. A. I only B. I and II only C. I, II and III only D. I, II, III and IV 70. What is always a difference between the alleles of a gene? A. Their position on the chromosome B. Their amino acid sequence C. The number of codons that each contains D. Their base sequence 42

43 71. Hypophosphataemia is a disorder involving poor re-absorption of phosphate from glomerular filtrate in humans. It shows a sex-linked dominant pattern of inheritance as illustrated in the following pedigree. Which row in the table correctly identifies the genotypes of individuals 1 and 2? Individual 1 Individual 2 A. X H X h X H Y B. X h Y X H X H C. X h Y X H X h D. unaffected affected 72. Why is it possible for a gene from one organism to be introduced and function in a different organism? A. All organisms are made of cells. B. All organisms have nuclei. C. The genetic code is universal. D. All organisms have ribosomes. 43

44 73. Weeds growing with crop plants can reduce yields because they compete for nutrients, water and sunlight. Synthetic chemical herbicides are often used to control these weeds. Herbicides are classified by the kinds of plants they kill and their mechanism of action. Broad-spectrum herbicides kill many different kinds of plants, but often kill the crop plant as well. Genetic engineering can create resistance to specific broad-spectrum herbicides which may solve the problem in crop plants. Different genes from bacterial sources known to protect against the effects of individual herbicides were engineered into corn plants (Zea mais). The resistance of normal and genetically engineered corn plants was measured and compared by calculating the percentage of plants that survived for 200 days with regular herbicide treatments. Graph 1 Exposure of Normal and Resistant Plants to Different Herbicides Herbicide Resistant Genes GP Glyphosate GP-R BR Bromozymil BR-R GU Glufosinate GU-R SU Sulfonylurea SU-R (a) (i) Calculate the difference between the survival of engineered plants and normal plants treated with Glyphosate (GP)

45 (ii) Identify the engineered plant which shows the greatest difference in resistance to herbicide treatment (iii) Suggest a reason for the difference in survival of the normal plants treated with Glyphosate (GP) and Bromozymil (BR) (b) (i) Define the term genetically modified crop (ii) State an example of a genetically modified plant other than corn

46 The graph below represents data from experiments in which plants were genetically engineered with more than one resistance gene. Graph 2 Exposure of resistant plants to combinations of herbicides (c) (i) Using both graphs, compare the data for BR-R with the data for SU-R, and for BR- R + SU-R in the same plant (ii) Suggest a possible reason for these results

47 (d) Evaluate the effects on survival when combining two herbicide resistance genes in the same plant. (3) (Total 11 marks) 74. Up to two additional marks are available for the construction of your answers. (a) (b) Outline how the process of meiosis can lead to Down s Syndrome. Discuss the advantages and disadvantages of genetic screening for chromosomal and genetic disorders. (4) (8) (c) Describe the technique for the transfer of the insulin gene using E. coli. (6) (Total 20 marks) 75. A gene has three alleles. How many different genotypes can be found for this gene? A. 3 B. 6 C. 9 D

48 76. A cross is carried out between two heterozygous individuals (AaBb) where the genes A and B are not linked genes. What would be the proportions of genotypic recombinants amongst the offspring of this cross? A. 0% B. 2% C. 7% D. 100% 77. What is the genetic cross called between an individual of unknown genotype and an individual who is homozygous recessive for a particular trait? A. Test-cross B. Hybrid cross C. Dihybrid cross D. F1 cross 78. What is the usual cause of Down s syndrome? A. 21 pairs of chromosomes B. Trisomy 21 C. Non-disjunction of sex chromosomes D. Fertilization of the egg by two sperm 48

49 79. Which enzyme is used to produce complementary DNA (cdna) from mrna? A. Restriction endonuclease B. Reverse transcriptase C. DNA ligase D. RNA primase 80. Which human trait shows a pattern of polygenic inheritance? A. ABO blood type B. Sickle cell anemia C. Skin colour D. Co-dominant alleles 81. If the haploid number of an organism is 8, how many different varieties of gametes are possible, not considering the effects of crossing over? A. 16 B. 64 C. 128 D (a) A farmer has rabbits with two particular traits, each controlled by a separate gene. Coat colour brown is completely dominant to white. Tailed is completely dominant to tail-less. A brown, tailed male rabbit that is heterozygous at both loci is crossed with a white, tailless female rabbit. A large number of offspring is produced with only two phenotypes: brown and tailed, white and tail-less, and the two types are in equal numbers. 49

50 (i) Deduce the pattern of inheritance of these traits (ii) State both parents genotypes and the gametes that are produced by each during the process of meiosis. Male genotype:... Female genotype:... Male gametes:... Female gametes:... (iii) Predict the genotypic and phenotypic ratios of the F2 generation. Show your working

51 (b) Outline the biotechnology used to transfer genes from one organism to another. (3) (Total 9 marks) 83. Up to two additional marks are available for the construction of your answers. (a) (b) (c) Outline the structure of DNA. Describe the effects of polygenic inheritance using two specific examples. Explain the process of transcription in eukaryotes. (5) (5) (8) (Total 20 marks) 84. What are the components of a eukaryotic chromosome? A. One DNA molecule and one large protein B. Many DNA molecules and many proteins C. One DNA molecule and many proteins D. Many DNA molecules and one large protein 51

52 85. What is a karyotype? A. Maternal and paternal autosomes arranged in pairs. B. Chromosomes arranged in pairs according to the number of their genes. C. Chromosomes arranged in pairs according to their size and shape. D. Chromosomes arranged in pairs according to their size. 86. What is the cause of sickle cell anemia? A. Errors in the translation of mrna B. A base substitution mutation in DNA C. A transcription error that replaces A with U D. A mutation that leads to glutamic acid instead of valine 87. What is the relationship between Mendel s law of segregation and meiosis? A. Only one of a pair of alleles appears in a gamete. B. The separation of paternal and maternal chromosomes shows no pattern. C. Gametes contain all dominant or all recessive alleles. D. Variation only results from two divisions. 52

53 88. What are the possible applications of DNA profiling? I. Solving paternity suits II. III. Aiding certain criminal investigations Identifying people who died last century A. I only B. I and II only C. II and III only D. I, II and III 89. Which fluid is sampled to try to detect chromosomal abnormalities in a fetus? A. Placental B. Umbilical C. Amniotic D. Spinal 53

54 90. (a) Draw and label a simplified structure of a nucleotide. A genetic cross was made between pure-breeding snapdragon plants with red flowers and purebreeding snapdragon plants with white flowers. The cross produced F 1 offspring that had only pink flowers. When the F 1 plants were self-pollinated, the resulting F 2 generation had some red, some white and some pink flowers. (b) (i) Identify the relationship between the red and white alleles for flower colour.... (ii) Deduce the genotype of the F 1 plants. (iii) (iv)... Construct a Punnett grid to show the cross between two F 1 plants. Deduce the proportion of the different phenotypes of the F 2 offspring (c) Discuss two advantages of genetic screening. (Total 9 marks) 54

55 91. Up to two additional marks are available for the construction of your answers. (a) (b) Draw and label a generalized prokaryotic cell as seen under the electron microscope. Outline the process of meiosis. (4) (6) (c) Explain the role of the following hormones in the menstrual cycle: estrogen, progesterone, follicle stimulating hormone (FSH) and luteinizing hormone (LH). (8) (Total 20 marks) 92. How does the X chromosome differ from the Y chromosome in humans? A. The Y chromosome is longer. B. Some genes on the X chromosome are absent from the Y chromosome. C. The genes are the same but some on the Y chromosome are not expressed. D. The X chromosome determines sex. 93. What are the functions of the polymerase chain reaction? I. Copy fragments of DNA II. III. Amplify fragments of DNA Translate fragments of DNA A. I and II only B. I and III only C. II and III only D. I, II and III 55

56 94. How does recombination normally occur for unlinked genes? A. Crossing-over in Prophase I B. Random chromosome assortment C. Failure of spindles to form D. Random gene mutations 95. Why is it sometimes difficult to identify how certain characteristics are inherited in humans. A. Most genes are linked. B. Rates of mutation are high. C. The inheritance may be polygenic. D. The environment varies so little. 56

57 96. Phenylketonuria (PKU) is a disease caused by a gene mutation that makes too much phenylalanine which may cause brain damage. The enzyme phenylalanine ammonia lyase (PAL), converts phenylalanine into harmless products. Mice with PKU were injected with PAL. The levels of phenylalanine in blood plasma were measured immediately after the injection (0 hour) and every hour for the next three hours. Different groups of mice with PKU were injected with three different doses of PAL. The results are shown below as a percentage of the levels of phenylalanine before the PAL injection. [Source: C. Sarkissian et al., A different approach to treatment of phenylketonuria: Phenylalanine degradation with recombinant phenylalanine ammonia lyase, Proceedings of the National Academy of Sciences (1999), vol. 96, pp Copyright (1999) National Academy of Sciences, U.S.A.] (a) Calculate the approximate percentage reduction in phenylalanine at 0 hour when the mice were injected with a dose of two units of PAL.... % (b) Outline the effect of a dose of twenty units of PAL on phenylalanine levels. 57

58 (c) Discuss the effectiveness of the different doses of PAL to treat PKU mice. (3) (d) Outline how the type of mutation that causes PKU differs from Klinefelter s syndrome. (Total 7 marks) 97. Which of the following represents a test cross to determine if phenotype T is homozygous or heterozygous? (Note: allele T is dominant to allele t.) A. Phenotype T crossed with another phenotype T B. Phenotype T crossed with a phenotype T which is homozygous C. Phenotype T crossed with a phenotype T which is heterozygous D. Phenotype T crossed with phenotype t 58

59 98. Which of the following blood group phenotypes always has a homozygous genotype? A. A B. B C. AB D. O 99. In the pedigree shown below, the female, labelled I-2, is a carrier for colour blindness, however neither male (I-1 or II-1) is colour blind What is the probability that offspring III-1 will be colour blind? A. 50% B. 25% C. 12.5% D. 0% 59

60 100. What happens to the unfertilized egg used in the cloning process of a differentiated cell? A. It becomes fertilized. B. Its nucleus is replaced by the nucleus of the differentiated cell. C. Its nucleus is fused with the nucleus of the differentiated cell. D. Its nucleus is exchanged with the nucleus of the sperm A tiny amount of DNA was obtained from a crime scene and amplified. Following digestion with restriction enzymes, which laboratory technique would be used to separate the fragments of DNA? A. Karyotyping B. Genetic screening C. Gel electrophoresis D. Polymerase chain reaction 102. What was the original goal of the Human Genome Project? A. To determine the function of genes B. To determine the nucleotide sequence of all human chromosomes C. To determine how genes control biological processes D. To understand the evolution of species 60

61 103. Why is amniotic fluid collected during prenatal testing for abnormal chromosomes? A. To obtain uterine cells B. To obtain fetal cells C. To obtain dissolved chemical by-products of fetal development D. To replace it with fluid containing special growth hormones 104. Up to two additional marks are available for the construction of your answers. (a) (b) Draw and label a simple diagram to show how DNA is constructed from sugars, phosphates and bases. Define the terms gene and gene mutation. (6) (4) (c) Genetic modification involves the transfer of DNA from one species to another. Discuss the potential benefits and possible harmful effects of one example of genetic modification in a named organism. (8) (Total 20 marks) 105. What procedure is used to determine whether a chromosome is in excess or missing in an organism? A. X-ray B. Karyotyping C. Centrifugation D. DNA fingerprinting 61

62 106. What feature demonstrates codominance in the inheritance of ABO blood groups? A. When A antigens and B antigens are present on red blood cells. B. When A antibodies and B antibodies are present in blood serum. C. When I A and i alleles are expressed in homozygotes. D. When I A and i alleles are expressed in heterozygotes Which event occurs first in meiosis? A. Centromere appearance B. Chiasmata formation C. Crossing over D. Synapsis 108. Alleles S and T are both dominant. In the theoretical cross ttss Ttss, which of the following offspring would show recombination? A. TS, ts B. TS, Ts C. ts, Ts D. TS, ts 109. A gene in humans called APC is located on chromosome 5. This gene controls cell division and is known as a tumour suppressor gene. Mutations of APC cause a genetic disease called FAP (Familial Adenomatous Polyposis). 62

63 (a) State, with a reason, whether FAP is a sex-linked genetic disease or not. 50% of the gametes produced by a person with FAP have an APC gene with the mutation. (b) Identify, with a reason, whether FAP follows a dominant or recessive pattern of inheritance. In a person with FAP, each cell contains a copy of the APC gene with the mutation. If a mutation occurs on the cell s other copy of the APC gene, the cell becomes a tumour cell. Almost everyone with FAP develops cancer before the age of 50. (c) Explain why almost everyone with FAP eventually develops cancer. In 2004, doctors in Britain were given permission to test embryos to see whether an APC gene with the mutation is present. This test can be used where one of the parents is known to have FAP. The procedure involves the parents using in-vitro fertilisation (IVF) to produce embryos, testing the embryos for the gene and implanting only embryos that do not have the mutation. 63

64 (d) (i) State the name of this type of test.... (ii) State one advantage and one disadvantage of testing embryos in this way. Advantage: Disadvantage: (Total 8 marks) 64

65 110. In humans, Duchenne Muscular Dystrophy (DMD) is a lethal X-linked recessive disorder caused by mutations in the dystrophin gene. Affected individuals show a decline in muscle mass over time along with a decline in muscle strength. One promising area of research in the treatment of DMD involves inhibiting the activity of myostatin, a naturally occurring protein that regulates muscle growth by limiting the development of new muscle cells. Researchers investigating the disorder in mice predicted that inhibition of myostatin would increase muscle mass. Over a period of three months one group of DMD mice (treated) were given injections of anti-myostatin antibody that inhibited myostatin. A second group of DMD mice were untreated (control). Figure 1 below shows the differences in body mass during the test period for both groups. [Source: Reprinted with permission from Macmillan Publishers Ltd: Sasha Bogdanovich et al., Functional improvement of dystrophic muscle by myostatin blockade, Nature (28 November 2002), vol. 420, issue 6194, pp , 2002] (a) Outline the relationship between body mass and time in the treated group of mice. 65

66 (b) Compare the changes in body mass in the two groups of mice over the test period. (3) (c) Predict the results that the researchers would have expected if the experiment was continued beyond 13 weeks in (i) the treated group.... (ii) the control group.... Figure 2 below shows the change in mass over the course of the experiment on a particular muscle called the EDL muscle in samples of the treated and control groups of mice. [Source: Reprinted with permission from Macmillan Publishers Ltd: Sasha Bogdanovich et al., Functional improvement of dystrophic muscle by myostatin blockade, Nature (28 November 2002), vol. 420, issue 6194, pp , 2002] 66

67 (d) Calculate the percentage increase in the average mass of the EDL muscle between the treated group and the control group. Further tests were conducted to see whether myostatin inhibition influenced the muscle function of the mice in the study. Figure 3 below shows the effect of treatment on muscle strength. [Source: Reprinted with permission from Macmillan Publishers Ltd: Sasha Bogdanovich et al., Functional improvement of dystrophic muscle by myostatin blockade, Nature (28 November 2002), vol. 420, issue 6194, pp , 2002] (e) Determine the difference in peak contraction force between the treated group and the control group. 67

68 (f) Evaluate the effectiveness of myostatin inhibition as a treatment for DMD in humans. (3) (g) Another proposed treatment for DMD in humans is gene therapy using the dystrophin gene. Outline, in general, the process of gene therapy. (3) 68

69 (h) Monoclonal antibodies are laboratory-produced identical antibodies that can target specific antigens, such as the myostatin protein. Describe how monoclonal antibodies are produced. (3) (Total 19 marks) 111. Up to two additional marks are available for the construction of your answers. (a) (b) (c) Describe the structure and function of the placenta. Draw and label a diagram of the adult male human reproductive system. Discuss the implications of genetic screening. (6) (5) (7) (Total 20 marks) 69

70 112. Plants have developed defence mechanisms against pathogens such as bacteria, fungi, and viruses. Chemicals released by these pathogens can trigger a defence response in infected plant cells. For example, the production of hydrogen peroxide (H 2 O 2 ) which reacts with pathogen membranes and cellular chemicals, eventually kills both the cell and the pathogen. The OSRac1 gene was isolated and introduced into a number of rice plant (Oryza spp.) lines to study its role in disease resistance of plants to Blast fungus. Experiments were carried out to see if the OSRac1 gene was part of the signalling pathway for hydrogen peroxide production. A control and four other genetically modified rice plant lines were exposed to chemicals known to initiate a defence response and the production of hydrogen peroxide. The results are shown in the graph below. Key: C: control A1 and A5: rice plants with the OSRac1 gene always turned on D41 and D42: rice plants with the OSRac1 gene suppressed [Source: E. Ono et al., Essential role of the small GTPase Rac in disease resistance of rice, PNAS (16 January 2001), vol. 98, issue 2, pp , 2003 National Academy of Sciences, U.S.A] (a) Identify the concentration of H 2 O 2 at time 0 for the control plants. 70

71 (b) Compare the change in H 2 O 2 production in the control and genetically modified plants two hours after the chemical was applied. (3) (c) Evaluate whether the data supports the hypothesis that OSRac1 gene is involved in disease resistance. (d) Suggest one possible concern about the use of transgenic plants with the disease resistant gene. (Total 6 marks) 113. What is a test cross? A. Crossing a possible heterozygote with a homozygous recessive B. Any genetic cross to determine genotype C. Crossing a possible homozygote with a homozygous dominant D. Crossing a possible heterozygote with another heterozygote 71

72 114. What is the aim of the Human Genome Project? A. Identify human infectious diseases B. Make improvements to the human genome C. Allow transfer of genes from other species to humans D. Sequence genetic information in humans 115. What does a karyotype show? A. Gel electrophoresis bands from DNA B. The number and appearance of chromosomes C. A pair of alleles controlling a specific character D. All the genes possessed by a living organism 116. A diploid cell in a gorilla has 48 chromosomes. How many chromosomes will be present in a haploid gorilla cell? A. 96 B. 48 C. 24 D

73 117. A woman has a heterozygous genotype for blood group B. She is expecting a baby with a man who is homozygous Group A. What are the possible blood groups for their baby? I. Group O II. III. Group A Group AB A. II and III only B. I and II only C. I and III only D. I, II and III 118. What enzymes are used in gene transfer techniques? A. Endonucleases and lipases B. Ligases and amylases C. Ligases and lipases D. Restriction enzymes and ligases 73

74 119. The diagram below shows a DNA profiling of a family with five children. Segments of the DNA inherited by some members of the family are shown as two dark bands in each column. The DNA fragments are labelled A to F. (a) State two properties of the fragmented pieces of DNA which allow them to be separated in gel electrophoresis. (b) Determine which DNA fragment Son 2 inherited from his mother and which from his father. From his mother:... From his father:... (c) Identify the child that genetically most resembles one of the grandparents. (d) Apart from determining family relationships, outline one other application for DNA profiling. (Total 4 marks) 74

75 120. Colour blindness in humans is caused by an X chromosome linked recessive allele. In the pedigree chart below which two individuals must, for certain, be carriers of colour blindness? A. II and IV B. I and III C. II and III D. I and II 121. The diagram below shows a cell undergoing meiosis. What is this stage of meiosis? A. Anaphase I B. Prophase I C. Anaphase II D. Telophase II 75

76 122. A pure breeding tall plant with smooth seeds was crossed with a pure breeding short plant with wrinkled seeds. All the F 1 plants were tall with smooth seeds. Two of these F 1 plants were crossed and four different phenotypes were obtained in the 320 plants produced. How many tall plants with wrinkled seeds would you expect to find? A. 20 B. 180 C. 60 D (a) State one advantage and one disadvantage of genetic modification technology for crop plants. Advantage: Disadvantage: (b) Explain the effect of base substitution mutation in sickle cell anemia. (3) (Total 5 marks) 76

77 124. What can be concluded on the basis of the following karyotype? A. Female with a normal set of chromosomes B. Male with Down syndrome C. Female with Down syndrome D. Male with a normal set of chromosomes 125. What are homologous chromosomes? A. Two chromosomes with differing sets of genes, in the same sequence, with the same alleles B. Two chromosomes with the same set of genes, in a different sequence, with the same alleles C. Two chromosomes with a different set of genes, in the same sequence, with different alleles D. Two chromosomes with the same set of genes, in the same sequence, sometimes with different alleles 77

78 126. Which features of DNA fragments are used to separate them in the process of gel electrophoresis? A. Their charge and their size B. Their charge and base composition C. The sequence of their bases and their charge D. Their base composition and their size 127. Which feature of a genetic pedigree chart demonstrates that a characteristic is sex linked? A. Numbers of offspring carrying the characteristic decreased over several generations. B. One gender is more commonly affected than the other. C. Equal numbers of males and females inherit the characteristic. D. Boys and girls only inherit the characteristic from their mothers What does the genotype X H X h indicate? A. A co-dominant female B. A heterozygous male C. A heterozygous female D. A co-dominant male 78

79 129. If a purple flowered (Pp) and a white flowered pea plant (pp) are crossed, what will the offspring be? A. 1 : 1 ratio of purple and white flowers B. 3 : 1 ratio of purple to white flowers C. 1 : 3 ratio of purple to white flowers D. All purple flowers 130. In what way are eukaryotic chromosomes different from prokaryotic chromosomes? Eukaryotic chromosomes Prokaryotic chromosomes A. Protein is present Protein is absent B. DNA is present DNA is absent C. RNA is present RNA is absent D. RNA is absent RNA is present 131. Which processes result in the greatest amount of genetic variation in a population? A. Natural selection and meiosis B. Meiosis and mutation C. Mutation and mitosis D. Mitosis and natural selection 79

80 132. Hemophilia is sex-linked and is caused by a recessive allele. A woman s father has hemophilia, but her husband does not. What is the probability of the women and her husband having a child with hemophilia? Probability of a son having hemophilia Probability of a daughter having hemophilia A. 50% 0% B. 0% 0% C. 100% 0% D. 0% 50% 133. Brachydactyly, abnormal shortness of the fingers, was the first human genetic disorder found to be caused by a dominant allele. The pedigree below shows a family with affected males, unaffected males, affected females and unaffected females. What are the genotypes of the father and mother in the first generation, using the symbol B for the dominant alleles and symbol b for recessive allele? A. bb and BB B. bb and Bb C. Bb and BB D. BB or Bb and bb 80

81 134. There are many different views on the ethics of reproductive cloning in humans. Which is a valid argument against cloning in humans? A. It involves the use of donor sperm which is unethical. B. It happens naturally when identical twins are conceived. C. Only females can be cloned. D. The life expectancy of children produced by cloning might be lower than normal Up to two additional marks are available for the construction of your answers. (a) (b) Explain why enzymes are substrate specific and why their activity is affected by substrate concentration. Outline the use of restriction enzymes (endonucleases) and DNA ligase in gene technology. (8) (6) (c) Outline the role of two enzymes found in the digestive system of humans. (4) (Total 20 marks) 81

82 136. The mold Penicillium expansum causes serious crop loss of apples and pears during storage and packing. The graphs below show the density of P. expansum spores on normal skin, damaged skin and lenticels (small air pores allowing gaseous exchange) of apples. The experiments show the results at two humidity levels over a four day period. [Source: This article was published in Physiological and Molecular Plant Pathology, Volume 67, Issue 1, authors Achour Amiri, Danielle Cholodowski and Gilbert Bompeix, Adhesion and germination of waterborne and airborne conidia of Penicillium expansum to apple and inert surfaces, pp , copyright Elsevier (2005).] (a) State the time taken for spores on the skin to reach a density of 100 spores/cm 2 at high humidity.. (b) Compare the density of spores on normal skin with spores on lenticels for apples stored at high humidity

83 (c) Discuss whether apples should be stored at high humidity or low humidity (3) (Total 6 marks) 137. In Drosophila the allele for normal wings (W) is dominant over the allele for vestigal wings (w) and the allele for normal body (G) is dominant over the allele for ebony body (g). If two Drosophila with the genotypes Wwgg and wwgg are crossed together, what ratio of phenotypes is expected in the offspring? A. 9 normal wings, normal body : 3 normal wings, ebony body : 3 vestigal wings, normal body : 1 vestigal wings, ebony body B. 3 normal wings, normal body : 3 normal wings, ebony body : 3 vestigal wings, normal body : 1 vestigal wings, ebony body C. 3 normal wings, normal body : 1 normal wings, ebony body : 3 vestigal wings, normal body : 1 vestigal wings, ebony body D. 1 normal wings, normal body : 1 normal wings, ebony body : 1 vestigal wings, normal body : 1 vestigal wings, ebony body 138. What constitutes a linkage group? A. Genes whose loci are on different chromosomes B. Genes carried on the same chromosome C. Genes controlling a polygenic characteristic D. Genes for the inheritance of ABO blood groups 139. What are the possible outcomes of recombination? 83

84 I. A different combination of unlinked genes not seen in the parents II. A different combination of linked genes not seen in the parents III. The same combination of genes seen in the parents A. I and II only B. I and III only C. II and III only D. I, II and III 140. A gene in cattle controls whether horns develop or not. When cattle without horns are mated together, none of the offspring ever has horns. A male with horns is mated with females without horns. If half of the offspring have horns and half do not, what is the conclusion? A. The male is homozygous dominant. B. The male is homozygous recessive. C. The male is heterozygous. D. Only males have horns Humans are in blood group M, N or MN. The alleles for blood group M (M) and blood group N (N) are co-dominant. Humans are also in blood group A, B, AB or O. The alleles controlling these blood groups are I A, I B and i. If two parents have the genotypes ii MM and I A i MN what is the ratio of possible phenotypes of their offspring? A. 9 group A, group M 3 group A, group N 3 group O, group M 1 group O, group N B. 9 group O, group M 3 group O, group N 3 group A, group M 1 group A, group N C. 3 group O, group M 3 group O, group MN 1 group A, group M 1 group A, group MN D. 1 group A, group M 1 group A, group MN 1 group O, group M 1 group O, group MN 84

85 142. A cell replicates its DNA and then starts to divide by meiosis. What is the expected arrangement of chromosomes if crossing over has taken place between the two genes shown? 143. (a) (i) Define allele (ii) Outline the consequences of a base substitution mutation

86 (b) (i) Mendel crossed tall, round-seeded plants with short, wrinkled-seeded plants. All F 1 produced were tall, round-seeded plants. When F 1 plants were crossed with other F 1 plants, the F 2 generation produced many more than 1/16 short, wrinkledseeded plants. Deduce, with reasons, the inheritance of these genes (ii) The same cross was later repeated but gave fewer F 2 short, wrinkled-seeded plants although still more than 1/16. Outline a named statistical test that could indicate if your deduction about the inheritance of these two genes is likely to be correct (Total 7 marks) 144. Up to two additional marks are available for the construction of your answers. (a) (b) Draw and label a diagram of a dicotyledonous animal-pollinated flower as seen with the naked eye and a hand lens. Describe how meiosis results in an enormous genetic variety in the production of pollen. (5) (5) (c) Using the theory of natural selection, explain how new species of dicotyledonous plants develop. (8) (Total 20 marks) 86

87 145. How are plasmids used in biotechnology? A. For respiration in prokaryotes B. For photosynthesis in eukaryotes C. For protein synthesis in prokaryotes and eukaryotes D. For gene transfer 146. The diagram below shows the cell of an organism going through the first division of meiosis. How many different combinations are possible for these chromosomes in the haploid cells formed by meiosis? A. 2 B. 6 C. 8 D. 9 87

88 147. If the amount of DNA in a haploid gamete is represented by, what is the net quantity of DNA in a cell from the same organism at the start of meiosis? A. 0.5 B. C. 2 D A parent organism of unknown genotype is mated in a test cross. Half of the offspring have the same phenotype as the parent. What can be concluded from this result? A. The parent is homozygous dominant for the trait. B. The trait being inherited is polygenic. C. The parent is heterozygous for the trait. D. The parent is homozygous recessive for the trait If a man has blood group O and a woman has blood group AB, what is the probability that their child will be blood group O? A. 0% B. 25% C. 50% D. 100% 150. (a) State two procedures used for the preparation of a DNA profile. 88

89 The following part of a DNA profile was used as evidence in a criminal investigation. DNA profiles of two suspects labelled S1 and S2 were compared to the DNA profile taken from the scene of the crime labeled E. [Source: Solomon and Berg, (1995), The World of Biology, Saunders Harcourt Brace College, Publishers Orlando, page 238] (b) Analyse the profiles to determine which suspect was present at the crime scene. (Total 3 marks) 151. (a) List two roles of testosterone in males