Pra-U STPM Biology Term 3 2018 CB039148c

192 Biology Term 3 STPM Chapter 17 Inheritance and Genetic Control 17 STPM PRACTICE 17 Objective Questions 6. The progeny of F1 from a cross between a homo pure breeding dominant smooth pea plant with a wrinkled pea plant were self-pollinated. What is the result in the F2 generation. A All peas are smooth B All peas are wrinkled C 1 4 of the peas are smooth D 1 4 of the peas are wrinkled 7. When red-flowered and white-flowered snapdragons are crossed, all progeny are pink-flowered. What is this condition called? A Epistasis B Codominance C Complete dominance D Incomplete dominance 8. The coat colour of mice is controlled by several alleles, C = normal, Cch = Chinchilla, Ch = Himalayan and c = albino with a dominance series: C > Cch > Ch > c. When two Chinchilla mice are crossed, the offspring produced is three Chichilla mice to one Himalayan. Which of the following is the phenotype of their parents? A CchCch × CchCh or CchCch × Cchc B CchCch × Cchc or CchCh × CchCh C CchCh × Cchc or CchCch × Cchc D CchCh × Cchc or CchCh × CchCh 9. In a dihybird cross, the phenotypic ratio of F2 generation is 9 : 3 : 4. What does this result indicate? A Epistasis B Lethal gene C Codominance D Incomplete dominance 1. Pure breed can be produced by A selfing B test cross C back cross D reciprocal cross 2. Two different pure strains when crossed will produce A trait B hybrids C mutants D phenotype 3. Which statement best explains the Mendelian’s Principle of Segregation? A It provides a suitable explanation for the result of dihybrid crosses. B During gamete formation, pairs of unit factors segregate independently of each other. C The alleles for each character segregate into gametes independently from alleles of other character. D If an individual is heterozygous for a gene, each gamete has 50% probability of receiving either one of the alleles. 4. A plant of CRCR genotype produces red flowers, CRCr genotype produces pink flowers and Cr Cr genotype produces white flowers. What percentage of the progeny has pink flowers if a cross is made between CRCr and CRCr ? A 0% C 50% B 25% D 75% 5. Which is the reason for a monohybrid cross producing F2 in a ratio of 2 : 1? A Epistasis B Lethal gene C Codominance D Incomplete dominance

193 Biology Term 3 STPM Chapter 17 Inheritance and Genetic Control 17 10. Which is true of epistasis? A A condition in which the expression of phenotype for both alleles is shown in heterozygote. B A condition in which a phenotype is controlled by more than one alleles of the same gene. C Two or more genes on the same chromosome which control different characteristics. D A gene suppresses the effect of a different gene at another locus. 11. Which progeny have different phenotypes from those of the original parents if their gametes and offsprings are shown in the table? Gametes AB Ab aB ab AB a b c d Ab e f g h A d and g B f and g C f and g D f and h 12. Alleles for purple stems and red fruits are dominant in tomatoes over those of green stems and yellow fruits. If two double heterozygous plants are crossed, what proportion of the progeny has purple stems and yellow fruits? A 1 16 C 4 16 B 3 16 D 9 16 13. In Drosophila, short leg (s) is recessive to normal leg (S) while hairy body (h) is recessive to normal body (H). What is the proportion of obtaining normal fly with respect to the above two characters if a cross SSHh × Sshh is made? A 1 16 C 1 4 B 1 8 D 1 2 14. What is the phenotypic ratio of a random crossing of F1 produced from the parental cross of AABB and aabb? A 3:1 C 1:1:1:1 B 1:2:1 D 9:3:3:1 15. If two pure breeding plants of two different traits are crossed, what proportion of the F2 is similar to that of the grandparents? A 1 16 C 9 16 B 3 16 D 10 16 16. In garden peas, allele Y for yellow is dominant to allele y for green and allele S for smooth is dominant to allele s for wrinkled. A cross between yellow smooth and yellow wrinkled produces the following result. Yellow smooth 302 Yellow wrinkled 298 Green smooth 98 Green wrinkled 102 What is the genotype of yellow smooth? A YYSS C YySS B YYSs D YySs 17. In humans, free ear lobe is dominant to attached ear lobe while curly hair is dominant to straight hair. If the husband and wife are double heterozygous, what is the chance of their children having free ear lobes and straight hair? A 1 16 C 3 16 B 2 16 D 9 16 18. A type of gene interaction in which product of one gene overrides the effect of another gene is known as A Codominance B Epistasis C Mutation D Linkage

194 Biology Term 3 STPM Chapter 17 Inheritance and Genetic Control 17 19. In brinjal, brown stem is dominant over green stem and purple fruit is dominant over white fruit. What is the proportion of the offspring with brown stem and white fruit if two dihybrids are crossed? A 1 16 C 3 16 B 2 16 D 9 16 20. A plant with cut leaves and round pollens is crossed with another with cut leaves but angular pollens. The following numbers of plants are obtained when the seeds produced are planted later: Phenotypes Number Cut leaves and round pollens 120 Cut leaves and angular pollens 122 Smooth leaves and round pollens 38 Smooth leaves and angular pollens 40 Which of the following are the genotypes of the parents? A First parent is heterozygous for both characters; second parent is heterozygous for leaf character B First parent is heterozygous for the leaf character; second parent is heterozygous for pollen character C First parent is heterozygous for both characters; second parent is heterozygous for pollen character D First parent is homozygous for leaf character; second parent is homozygous for both characters 21. Radish can be elongated (LL), round (ll) or oval (Ll) while its colour can be red (RR), white (rr) or purple (Rr). If a cross is made between white oval radish plant with that of a red oval, which of the following observation is correct? Red elongated Purple elongated Red oval Purple oval Red round Purple round A 1 – 2 – 1 – B – 1 – 2 – 1 C 1 – – 2 1 – D – 2 – 1 – 1 22. In a certain melon, a dominant allele W determines white fruit while that of w determines colour fruit. Another dominant allele Y determines yellow fruit while its recessive determines that of green. What is the phenotypic ratio obtained if a WwYy white fruit plant is crossed with that of a wwyy green one? A 1 white: 2 yellow: 1 green B 1 white: 1 yellow: 2 green C 2 white: 1 yellow: 1 green D 1 white: 1 yellow 23. Red colour kernel in wheat is determined by genotype B1 _B2 _ while white is determined by b1 b1 b2 b2 and brown is determined by B1 _b2 b2 or b1 b1 B2 _. A homozygous red plant is crossed with a white plant. Their offspring is test-crossed with a white plant. What is the phenotypic ratio of the cross? A 1 red: 1 white B 3 brown: 1 white C 1 red: 3 white D 1 red: 2 brown: 1 white 24. A test-cross was performed in which two genes were mapped to be 16.8 cm apart. Which is correct? I A single crossing-over has occurred between the two genes II 83.2% of the offspring are showing the parental phenotypes III 168 of the offspring are showing the recombinant phenotypes IV The COV can be calculated by dividing total number of recombinant phenotypes with total number of parental phenotypes A I and II C II and III B I and IV D III and IV 25. Allele R that determines round tomato is dominant over that of r for oval while allele S that determines smooth fruit is dominant over s for rough. A test cross is carried out on a round smooth plant and produces the following progeny: 130 round smooth; 1200 oval smooth; 1310 round rough; 110 oval rough. Which of the following is the genotype of the round smooth parent?

195 Biology Term 3 STPM Chapter 17 Inheritance and Genetic Control 17 A RS RS C RS rs B RS rS D Rs rS 26. A cross of AB ab × ab ab produces 40% of AB ab , 40% of ab ab , 10% of AB ab and 10% of aB ab . The presence of Ab ab and aB ab in the offspring is because of A linked genes B crossing over C polygenes D reciprocal cross 27. The following table shows map distances between genes along a chromosome. Linked genes AB AC AD BC BD CD Map distance(%) 30 5 10 35 20 15 Which of the following is the correct gene sequence? A BADC B BDAC C BCAD D BACD 28. In Drosophila, eye colour is controlled by a series of alleles found on the X chromosome. Which of the following statement is false? A Males have only one allele for eye colour B Males exhibit only recessive phenotype C Its eye colour is sex-linked D Its eye colour has several phenotypes 29. The red-green colour blindness gene is located in the X chromosome and not in the Y chromosome. A homozygous woman with normal vision is married to a colour blind man. What is the chance that their offspring is colour blind? A 0 B 0.25 C 0.5 D 1.0 30. White eye colour in Drosophila is inherited as sex-linked recessive. What are the results obtained if a white-eyed female Drosophila is crossed with a red-eyed male Drosophila? A All males and females have white eyes. B All males have white eyes and all females have red eyes. C 50% males have white eyes, 50% males have red eyes and all females have red eyes. D 50% males have white eyes, 50% males have red eyes, 50% females have white eyes and 50% females have red eyes. 31. Yellow body in Drosophila is determined by a recessive sex-linked gene. If a cross produces all male progeny with yellow body and all female progeny with grey body, what are the parent's genotypes? Female parent Male parent A XYXY Xy Y B XYXy Xy Y C XYXy XYY D Xy Xy XYY 32. Allele B that controls black hair is dominant over allele b that controls brown hair in humans. What is the chance of individual X in the family below having brown hair? I II III X Key: Black hair male Brown hair female Brown hair male A 0% C 50% B 25% D 100%

196 Biology Term 3 STPM Chapter 17 Inheritance and Genetic Control 17 33. In the pedigree below, individual I2 has blood group O, II1 has blood group B and II2 has blood group AB. Key: 1 I II III 2 1 2 Female Male 1 2 What is the chance of every generation having an individual of blood group B? A 0 C 0.5 B 0.33 D 0.67 34. Alleles M and N are codominant. What are the types of phenotypes produced from MN × NN? A 1 C 3 B 2 D 4 35. The inheritance pattern of red hair in human is shown by a pedigree below. I 1 2 3 9 10 11 19 Brown-haired male Key : Brown-haired female Red-haired male Red-haired female 20 21 12 13 14 15 16 17 18 4 5 6 7 8 II III IV If brown hair is controlled by the dominant allele R and red hair is controlled by the recessive allele r, what are the genotype and phenotype of the individual III-9? A rr, red C Rr, brown B Rr, red D RR, brown 36. Which has no relationship with gene pool? A Phenotypic frequency B Genotypic frequency C Allelic frequency D Gene frequency 37. The frequency of a dominant allele in a population can be maintained from one generation to another if there is A mutation C migration B selection D random mating 38. Which statement is false of HardyWeinberg equation? A The phenotypic ratio is 3:1 B The sum of allelic frequencies is 1 C The sum of genotypic frequencies is 1 D There will be no change in allelic frequencies from one generation to the next 39. In a chicken farm of 100,000 birds, 1000 chickens die of influenza. By assuming that the virus resistance gene is dominant, how many chickens in the farm are carriers of recessive genes? A 1000 C 18,000 B 10,000 D 90,000 40. If 235 of an animal population suffers from a recessive genetic disease and 412 are normal, what percentage of the population is homozygous dominant? A 4 C 36 B 16 D 48 41. The frequency of a genotype of a mutant which is recessive homozygous is 0.09. What is the frequency of the mutant allele? A 0.03 C 0.7 B 0.3 D 0.42 42. Two species of plants have chromosome sets XX and YY respectively. Which genotype is an allotetraploid plant? A XXXX C XYYY B XXYY D YYYY 43. In a population of 400 small mammals, 144 of them are albinos. Albinism is a recessive trait in the population. What is the frequency of the allele for albinism? A 30% C 60% B 40% D 72%

197 Biology Term 3 STPM Chapter 17 Inheritance and Genetic Control 17 44. Cystic fibrosis is a recessive condition affecting one in every 2000 people. What is the frequency of the heterozygous genotype in the population? A 0.0005 C 0.043 B 0.022 D 0.978 45. 36% of a population of 400 mice were found to be resistant to a poison. The resistance is caused by a dominant allele R. What is the number of mice expected to have Rr genotype? A 128 C 192 B 144 D 256 46. A dominant allele Y for yellow coat in rat functions as a lethal gene in homozygous state. If 16% of 1000 foetuses are stillborn, what percentage is expected to be heterozygous in the population? A 4 C 24 B 16 D 48 47. In goats, the normal fur is controlled by dominant allele R, while the non-uniform fur is controlled by recessive allele r. In a population of 1000 goats that mate randomly, it is found that 250 goats have non-uniform fur. What is the frequency of the recessive allele in the population? A 0.25 C 0.75 B 0.50 D 0.87 48. Which is the definition of gene pool? A The relative proportion of the various genotypes present in a population B The relative proportion of the alleles of a gene present in a population C The total number of alleles that will not change over time D The sum total of all the alleles present in a population 49. The table below shows the genotype frequency and ratio in a population. Phenotypes MM MN NN Frequency 0.2 0.5 0.3 Ratio X Y Z What is the frequency of allele M? A X + 1 2 Y B Y + 1 2X C Z + 1 2 Y D Y + 1 2 Z 50. State the function of DNA polymerase I during DNA replication? A It catalyses the synthesis of Okazaki fragments B It catalyses the joining of Okazaki fragments C It catalyses the synthesis of a short RNA primer D It catalyses the replacement of RNA primer by DNA 51. The lactose operon model presented by Jacob and Monod is represented by the diagram below. Operon I P O DNA Which is true when gene I is activated? A The structural genes encode for the synthesis of b-galactosidase, permease and transacetylase. B The structural genes do not encode for the synthesis of b-galactosidase, permease and transacetylase. C Gene P is inactivated while gene O is not. D The whole operon fails to function. 52. The sequence of the template DNA strand is 5’-TTCGAAATG-3’. The corresponding anticodons for the mRNA transcribed are respectively A 3’-UUC-5’, 3’-GAA-5’ and 3’-AUG-5’ B 3’-AAG-5’, 3’-CUU-5’ and 3’-UAC-5’ C 5’-UUC-3’, 5’-GAA-3’ and 5’-AUG-3’ D 5’-AAG-3’, 5’-CUU-3’ and 5’-UAC-3’

198 Biology Term 3 STPM Chapter 17 Inheritance and Genetic Control 17 53. Which about the characteristics of genetic code is correct? I AUG is not the only codon for methionine. II The genetic code is non-overlapping. III Almost all genetic codes degenerate. IV Codon could be ambiguous. A I and III B I and IV C II and III D II and IV 54. The diagram below shows the lactose operon. K M Enzymes L X DNA N Which of the following structure could stop the production of the enzymes? A K C N B M D X 55. Various components and their functions in the lactose operon are shown in the table below. Component Function I Operator II Promoter III Regulatory gene p Site of RNA polymerase attachment q Production of repressor r Site of repressor attachment Which components correspond to their functions? I II III A r q p B q r p C r p q D p r q 56. Of a mutation occurs in the operator gene of lac operon, what will happen to the operon if lactose is present? A Operon will be activated but abnormal enzymes are produced. B Transcription of structural genes occurs to produce mRNA. C Repressor molecule will bind tightly to the operator. D RNA polymerase cannot bind to the promoter gene. 57. Which statement is false of the lactose operon in the presence of lactose? A The transcription of structural genes occurs B The allolactose binds to the repressor C The repressor is not synthesised D The repressor is prevented from binding to the operator 58. A mutation in the lac operon causes a repressor protein that is not able to bind to the promoter. What is the result of such a mutation? A β-galactosidase is not produced at all B β-galactosidase is produced continuously C β-galactosidase is produced continuously in the absence of lactose D β-galactosidase is produced continuously in the presence of lactose 59. Which of the following combination correctly describes operons? Operon Characteristics (a)Lac operon i. example of enzyme induction (b)Trp operon ii. example of enzyme repression iii.repressor produced is active iv.repressor produced is inactive A (a) i, iii (b) ii, iv C (a) i, iv (b) ii, iii B (a) ii, iv (b) i, iii D (a) ii, iii (b) i, iv

199 Biology Term 3 STPM Chapter 17 Inheritance and Genetic Control 17 60. DNA molecules as a genetic material must have all the following features except A the formation of a very stable doublestranded when not transcribed or replicated. B the ability to form complementary base pairs with other DNA nucleotides. C the ability to be decoded into a sequence of amino acids in a protein. D the association of histone proteins with the double helix. 61. If the regulatory gene in E. coli was mutated, A the repressor produced binds with operator gene B RNA polymerase is unable to bind with promoter gene C only b-galactosidase is produced D b-galactosidase, permease and transacetylase are constitutively produced 62. Which statement is true of the structural genes in lactose operon? A lac A gene encodes permease B lac Y gene encodes transacetylase C lac Z gene encodes β-galactosidase D The structural genes are transcribed into three mRNA molecules 63. A substitution of a single nucleotide base pair for another is called A point mutation B missense mutation C nonsense mutation D frameshift mutation 64. What are the genetic abnormalities caused by chromosomal mutation? I Cri-du-chat II Thalassemia III Turner syndrome IV Sickle-cell anaemia A I and II B I and III C II and IV D III and IV 65. Mutation is caused by A disjunction B fertilisation process C loss of one chromosome D exchange of segments of chromosome 66. The base sequence of a gene ABCDEFG undergoes two mutations to become ACDFG. What mutations are these? A Two inversions B Two translocations C Two deletions D Two substitutions 67. Which event causes frameshift mutation? A Addition of a nucleotide before the start codon B Substitution of three nucleotides C Inversion of six nucleotides D Addition of a nucleotide 68. Induced mutation is also known as A radiotherapy B mutagenesis C chemotherapy D genetic engineering 69. If (+) represents the addition of one base and (–) represents the deletion of one base, which mutation results in a frameshift mutation in adjacent bases? A (+)(–) C (+)(+)(+)(–) B (+)(+)(+) D (+)(–)(–)(+) 70. If nondisjuction of chromosome 21 occurs during meiosis II in either one of the two cells, the end products of meiosis are A four daughter cells with one chromosome 21 each. B two daughter cells with two chromosomes 21 and two daughter cells with one chromosome 21. C two daughter cells with two chromosomes 21 and two daughter cells without chromosome 21. D two daughter cells with one chromosome 21, one daughter cell with two chromosomes 21 and one daughter cell without chromosome 21.

200 Biology Term 3 STPM Chapter 17 Inheritance and Genetic Control 17 Structured Questions 1. In oilseed rape, the plant can be tall or short and the leaves can be green, variegated or yellow with no chlorophyll. Those with no chlorophyll usually die. A cross was made between pure breeding tall green plant and short variegated plant. The progeny produced had a ratio of 1 2 tall green: 1 2 tall variegated. When the F1 tall variegated plants were self-pollinated, the F2 seeds were germinated and the seedlings, after six weeks, had the following numbers: Tall green 182 Short green 64 Tall variegated 358 Short variegated 116 (a) Using suitable symbols, explain how the F1 ratio was obtained. [4] (b) With the help of suitable diagrams and Punnet box, explain how the F2 result was obtained. [6] 2. Pure breeding sweet pea plants with purple flowers and long pollen grains were crossed with pure breeding plants with red flowers and round pollen grains. All the F1 plants had purple flowers and long pollen grains. These F1 plants were test-crossed with plants of red flowers and round pollen grains and produced the following results after the seeds produced were grown. 3998 purple flowers and long pollen grains 403 purple flowers and round pollen grains 397 red flowers and long pollen grains 4002 red flowers and round pollen grains (a) Using suitable symbols, draw a genetic diagram to explain the results of the F1 generation. [3] (b) What is clearly indicated by the test cross data? Calculate the map distance between the genes involved. [3] (c) Diagram the test cross, indicating the percentage of the F1 gametes. [4] 3. A cross between a wild type red-eyed female (R) and a white-eyed male (r) Drosophila is shown in the diagram below. P generation Red-eyed female White-eyed male The eye colour in Drosophila is a sex-linked trait. (a) State the genotype of each of the above Drosophila. [2] (b) Draw a genetic diagram to show the crosses until the F2 generation is produced when the F1 generation is allowed to interbreed. [4] (c) A reciprocal cross of the previous cross was carried out. (i) State the genotype of each parent Drosophila. [2] (ii) Draw a genetic diagram to show the crosses until the F1 generation. [1] (d) What is meant by a sex-linked gene? [1]

201 Biology Term 3 STPM Chapter 17 Inheritance and Genetic Control 17 4. In a population of rats, 64% have agouti coat. Agouti is controlled by a recessive allele y whereas yellow coat is controlled by a dominant allele Y. (a) Calculate the allelic frequency of agouti and that of yellow. [5] (b) Will the allelic frequencies change in the next generation? Explain. [3] (c) Yellow allele is found to be a lethal allele. What is a lethal allele? [2] 5. The base sequence of a gene for a particular polypeptide which is coded by the gene and the genetic code table are shown in the diagram below. Template DNA Start codon T C C A T G A G T Y A T T G G A A T AT A T T 3’ 5’ Stop codon Polypeptide Met P Q R S Genetic code Second base First base U C A G Third base U UUU UUC UUA UUG Phenyl-alanine (Phe) Leucine (Leu) UCU UCC UCA UCG Serine (Ser) UAU UAC UAA UAG Tyrosine (Tyr) Stop Stop UGU UGC UGA UGG Cysteine (Cys) Stop Tryptophan (Trp) U C A G C CUU CUC CUA CUG Leucine (Ser) CCU CCC CCA CCG Proline (Pro) CAU CAC CAA CAG Histidine (His) Glutamine (Gln) CGU CGC CGA CGG Arginine (Arg) U C A G A AUU AUC AUA AUG Isoleucine (Ile) Methionine (Met) GCU GCC GCA GCG Threonine (Thr) AAU AAC AAA AAG Asparagine (Asn) Lycine (Lys) AGU AGC AGA AGG Serine (Ser) Arginine (Arg) U C A G G GUU GUC GUA GUG Valine (Val) GCU GCC GCA GCG Alanine (Ala) GAU GAC GAA GAG Aspartic acid (Asp) Glumatic acid (Glu) GGU GGC GGA GGG Glycine (Gly) U C A G With reference to the above polypeptide, answer the following questions: (a) State the anticodon for amino acid S. [1] (b) Name the amino acids P and Q. [2] (c) If the base in position Y is substituted with another base, what will happen to amino acid R? [2] (d) If the base in position Y is deleted, what will happen to the polypeptide produced? [3] (e) Explain why the effect of deletion of three bases is less serious than that of one base. [2]

202 Biology Term 3 STPM Chapter 17 Inheritance and Genetic Control 17 6. The diagram below shows the production of tryptophan according to the hypothesis proposed by Jacob and Monod. O E P Regulator gene Promoter Operator Structural genes Enzyme E repressor D Enzyme D C Enzyme C B Enzyme B A tryptophan Enzyme A I trpE trpD trp operon trpC trpB trpA (a) What is the type of operon involved? What types of enzymes are produced? [2] (b) Explain the nature of the repressor protein. [2] (c) What is the advantage of having just one promoter for all the structural genes? [2] (d) Why are both repressor proteins and enzyme E known as allosteric proteins? [2] (e) What happens if tryptophan is produced in excess? [2] 7. The inability to taste phenylthiocarbamide is controlled by a single recessive gene. In a population under the Hardy-Weinberg equilibrium, 84% of the population are able to taste PTC. (a) State the Hardy-Weinberg equilibrium. [1] (b) Calculate the allele frequency of the non-tasters. [2] (c) Calculate the frequencies of the genotypes for: (i) Homozygous dominant [2] (ii) Heterozygous [2] (d) Give one factor that affects Hardy-Weinberg equilibrium. [1] 8. The diagram below shows a type of mutation. P Q

203 Biology Term 3 STPM Chapter 17 Inheritance and Genetic Control 17 (a) (i) Name the type of mutation in P. [1] (ii) What is the cause of the mutation in (a)(i)? [1] (b) State the stage of cell division where the mutations in Q occur. [1] (c) (i) State the syndrome produced from the fertilisation between a female gamete which has an extra X chromosome and a normal male gamete. [3] (ii) State three physical characteristics of an individual who has the syndrome in (c)(i). [3] Essay Questions 1. (a) Explain what are polygenes, including their characteristic inheritance. [6] (b) The inheritance of height in humans is assumed to be controlled by polygenes. Tall people with mean height of 180 cm, if married to short people with mean height of 150 cm, would produce children with mean height of 165 cm. Children from such marriages if married to those of the same descendents, would have children that show 6.25% with mean height of 180 cm and the same percentage with mean height of 150 cm. Calculate the contribution of each dominant allele to the overall height. Using suitable symbols and genetic diagram, show how such results are obtained, including the F2 phenotypic ratio. [9] 2. (a) Explain what are sex-linked genes as referred to mammals, including their characteristic ways of inheritance. [7] (b) Calico or tortoiseshell cats are always females. These females are produced as a result of crossing between female black cat and male ginger cat or female ginger cat and male black cat. In the former cross the males produced are black only while in the latter, the males produced are ginger only. Explain the results using suitable symbols and diagrams. [8] 3. (a) Explain the concept of gene pool. [7] (b) Explain the relationship between population genetics and evolution. [8] 4. In tomatoes, alleles for red flower (R) is dominant over white flower (r), and smooth skin (S) is dominant over wrinkled skin (s). A plant that is homozygous for both red flower and smooth skin tomato is crossed with a plant producing white flower and wrinkled skin tomato. A test cross is done on F1 and the progeny produced are given below. Red flower, smooth skin 295 White flower, wrinkled skin 305 Red flower, wrinkled skin 99 White flower, smooth skin 101 Based on the result shown, (a) explain why such ratio is obtained from the test cross, [3] (b) calculate the map distance between the two genes and draw its genetic map, [5] (c) draw the test cross above. [7]

204 Biology Term 3 STPM Chapter 17 Inheritance and Genetic Control 17 5. (a) Radiation and chemicals are agents of mutations. (i) State one example of each mutation agent. [2] (ii) Using your answer in (i), describe two effects of each example on mutation. [4] (b) Explain how autopolyploidy arises in plant. [9] 6. (a) Explain the lactose operon that exists in bacteria E. coli. [6] (b) Explain the effect of the absence and presence of lactose on it. [9] 7. (a) Explain how mutagens can cause mutations. [5] (b) Explain the nature of mutation that brought about sickle cell anaemia and thalassaemia major. [10] 8. What is meant by chromosomal mutation? With the help of suitable examples, explain how the various types of chromosomal mutations occur. [15] Quick Check 1 1. (a) Phenotype is any inheritable trait expressed in a given environment by an individual while genotype is the genetic content of an individual. Phenotype is determined by genotype while genotype is not affected by its phenotype. Phenotype can be affected by the environment while genotype is not so easily affected by the environment, except through mutation. Example of phenotype is albinism and body height while genotype is aa homozygous recessive for an albino person. (b) A gene is a smallest unit of inheritance in a chromosome that controls a trait while an allele is a form of a gene at a particular locus. A gene is part of a DNA with specific sequence of bases while an allele is the same gene that has undergone mutation with different base sequence. A gene determines the synthesis of a protein while an allele determines a nonfunctional protein or another protein with different function. An example of a gene is albino gene with dominant A allele that controls the formation of melanin while recessive a allele that cannot form melanin. (c) A dominant allele expresses itself in either homozygous state or heterozygous state, while a recessive allele only expresses in homozygous state. A dominant allele masks the effect of the recessive allele when in a heterozygous state. A dominant allele controls a functional protein while a recessive allele may control a non-functional gene. A dominant allele can become a recessive allele through mutation while recessive allele seldom become dominant through mutation. An example of dominant allele is A for normal skin colour while recessive allele is a, an allele that is not able to form enzyme for the formation of melanin. Quick Check 2 1. (a) YyRr × yyRr (b) Yyrr × yyRr 2. (a) One locus with dominant-recessive alleles and the other with incomplete dominant or codominant alleles. (b) Both loci are controlled by incomplete dominant or codominant alleles. (c) Both loci are controlled by polygenes. (d) Recessive epistasis, homozygous recessive alleles at one locus inhibit the expression of alleles at another locus. (e) Duplicate dominant genes with cumulative effects, a dominant gene at either locus has the same phenotype but if at both loci, a darker coloration is produced. ANSWERS

205 Biology Term 3 STPM Chapter 17 Inheritance and Genetic Control 17 Quick Check 3 1. Frequency of non-taster = 70 280 = 0.25 Therefore, allelic frequency of non-taster (t) = q = 0.25 = 0.5 So, allelic frequency of taster (T) = p = 1= –q = 1 –0.5 = 0.5 Tasters that produce only taster children have genotype TT. Therefore, frequency of TT = p2 = 0.52 = 0.25 Tasters that produce non-taster children have genotype Tt = 2pq = 2 × 0.5 × 0.5 = 0.25 Quick Check 4 1. Gene frequencies affect each other, depending on whether they are intra or inter loci. Intra locus genes or alleles increase in frequency at the expense of the other. Genes at different loci are independent or may be linked. They may be corelated if both produce advantageous characters. Some are epistatic, affecting negatively on the other gene. 2. A gene frequency tends to increase when it is advantageous or selected for and decrease when selected against in an environment. It may randomly drift if the population is small. 3. It is not necessary so. Most dominant genes control functional proteins; thus, if it is favoured, it will increase faster. Few mutated genes produce proteins with disastrous effects. Then, the allelic frequency will decrease. 4. No. Evolution is the sum changes of genetic frequencies in a certain direction. Quick Check 5 1. It is to control the production of enzymes related to the metabolism of lactose. Energy is saved for not producing the enzymes when not needed. The bacteria can switch to use another nutrient when the environment changes. This allows the bacteria to survive and adapt better. 2. This is because there is another control in the promoter called CRP or CAP site. This site allows the binding of the CRP or CAP protein which is activated by cAMP when there is no glucose. The RNA polymerase can bind better on the operator. When there is glucose, there is no cAMP to activate the CRP or CAP and the polymerase cannot bind to the operator to start transcription, even in the presence of lactose. Quick Check 6 1. It can originate by non-disjunction of Y chromosome during spermatozoan formation in meiosis II. This causes the formation of spermatozoon with YY chromosome and when fertilises a normal ovum, produces an XYY individual. 2. Tetraploid can be sterile because during the formation of gametes by meiosis, the pairing of each group of 4 homologous chromosomes may not be neatly 2 by 2, forming 2 bivalents. More than 2 of them start pairing at different places along the length, resulting in a group of tetravalent or trivalent, plus one with no pairing. Such untidy pairings result in imbalance gametes that cannot survive, so no gametes may be formed. 3. This is done by treating the developing flower buds or shoots with colchicine. Flowers with unreduced pollen is selfed to produce unreduced seed of tetraploid. The seed is germinated to produce tetraploid fertile plant. Developing shoot or branch is either propagated vegetatively or observed if seeds are produced. 4. When a gene is transferred, only part of it and not the whole gene may be transferred, resulting in its malfunction or loss of its function. Besides that, certain gene needs to be expressed in a group in order to produce a certain phenotype. STPM Practice 17 Objective Questions 1. A 2. B 3. D 4. C 5. B 6. D 7. D 8. B 9. A 10. D 11. D 12. B 13. D 14. D 15. D 16. D 17. C 18. B 19. C 20. A 21. B 22. C 23. D 24. A 25. D 26. B 27. B 28. B 29. A 30. B 31. D 32. B 33. B 34. B 35. C 36. D 37. D 38. A 39. C 40. B 41. B 42. B 43. C 44. C 45. A 46. D 47. B 48. B 49. A 50. D 51. C 52. A 53. C 54. B 55. C 56. B 57. C 58. D 59. A 60. A 61. D 62. C 63. A 64. B 65. C 66. C 67. A 68. B 69. C 70. D Structured Questions 1. (a) Let T be the dominant allele for tall and t be the recessive allele for short; CG be the codominant allele for green and CY be the codominant allele for yellow. Therefore, CGCG will have green leaves, CGCY variegated leaves and CYCY yellow leaves and die. P1: tCG tCY TtCGCG TtCGCY – tall green : – tall variegated F1: TCG Gametes: TT CGCG Tall green tt CGCY Short variegated 1 2 1 2 (b) P2: TCY tCG tCG TCY TCG TCY Gametes: Tt CGCY Tall variegated Tt CGCY Tall variegated TCG tCY

206 Biology Term 3 STPM Chapter 17 Inheritance and Genetic Control 17 F2 : TCG TCY tCG tCY TCG TT CGCG Tall green TT CGCY Tall variegated Tt CGCG Tall green Tt CGCY Tall variegated TCY TT CGCY Tall variegated tt CYCY Dies Tt CGCY Tall variegated tt CYCY Dies tCG Tt CGCG Tall green Tt CGCY Tall variegated tt CGCG Tall green tt CGCY Tall variegated tCY Tt CGCY Tall variegated tt CYCY Dies tt CGCY Short variegated tt CYCY Dies F2 phenotypic ratio is 3 tall green: 6 tall variegated: 1 short green: 2 short variegated 2. (a) Let P be dominant allele for purple flower and p be recessive allele for red flower, and L be dominant allele for long pollen grain and l be recessive allele for round pollen grain P1: F1: Gametes: PL pl PL Purple long pl Red round PL PL pl pl Purple long (b) Linkage, the two genes involved are linked. Map distance = Number of crossing over- type offspring Total offspring produced × 100% = 403 + 397 3 998 + 403 + 397 + 4 002 × 100% = 800 8 800 × 100% = 9.09% (c) P1: G1: F1: PL pl pl Purple long pl Red round PL pl Pl pL PL pl Pl pL pl 45.45% parental types cross-over types 45.45% 4.55% 4.55% 100% pl Purple long (3998) pl Red round (403) pl Purple round (4002) pl Red long (397) 3. (a) Red-eyed female: XRXR White-eyed male: Xr Y (b) P1 : XR XR Red-eyed female Xr Y White-eyed male F1 : XR Xr Red-eyed female XR Y Red-eyed male F2 : XR XR Red-eyed female XR Xr Red-eyed female XR Y Red-eyed male Xr Y White-eyed male 2 : 1 : 1 P2 : XR Xr Red-eyed female XR Y Red-eyed male G1 : XR X Y r G2 : XR Xr X Y R Genotypic ratio : 1 – XR XR 4 1 – XR Xr 4 1 – XR Y 4 1 – Xr Y 4 : : : Phenotypic ratio : 2 – 4 1 – 4 1 – 4 Red-eyed : : female Red-eyed male White-eyed male fi fi (c) (i) Xr Xr , XRY (ii) P1 : Xr Xr White-eyed female XR Y Red-eyed male fi F1 : XR Xr Red-eyed female Xr Y White-eyed male G1 : Xr X Y R (d) The gene involved is located in the X chromosome. 4. (a) • Assume that the population is in equilibrium and Hardy-Weinberg law applies. • Let p be the frequency of Y allele and q be the frequency of y allele. p2 (YY) + 2pq(Yy) + q2 (yy) = (p + q)2 = 1 • Therefore, frequency of agouti (yy) = q2 = 64% = 0.64. • Allelic frequency of agouti (y) = q = 0.64 = 0.8. • Allelic frequency of yellow (Y) = p = 1 – q = 1 – 0.8 = 0.2 (b) • No, if the population isin equilibrium and the population is very large, mating is random, there is no selection, mutation and migration. • Yes, if the population is small, there is no occurence of random mating, selection, mutation and migration. (c) • Lethal gene is a form of a gene that results in death in homozygous or heterozygous states. This is caused by an essential protein that is not produced, or a harmful protein is produced.

207 Biology Term 3 STPM Chapter 17 Inheritance and Genetic Control 17 5. (a) UGA (b) P: Aspartic acid Q: Asparagine (c) There is no change in amino acid R i.e. serine. This is because any substitution will change the DNA code to AGA, AGC or AGT and the mRNA codon will be UCU, UCG or UCA still codes serine. (d) The polypeptide produced will have six amino acids. The first four amino acid sequences are still the same. The 5th amino acid is changed into leucine and 6th amino acid is aparagine. A frame-shift mutation is produced. (e) This is because the genetic code is a triplet code. Three bases deleted will result in a loss of an amino acid. There is no change in the amino acid sequence in the polypeptide. 6. (a) Repressible operon. The types of enzymes are synthetase or ligase. (b) The repressor protein is inactive though it has an active site that can bind to the DNA of the operator. It has an allosteric site that can bind to a tryptophan that activates it to bind to the operator. (c) This will allow the promoter to control the transcription of all the structural genes by one RNA polymerase to produce a polycistronic mRNA. Then, all the five enzymes are synthesised at the same time and rapidly initiate the production of tryptophan. (d) The repressor protein and enzyme E have an additional allosteric site for chemical binding that control the functioning rate. The repressor is activated after tryptophan binding it but enzyme E is inactivated after tryptophan binding. (e) The enzymes are not produced. • Tryptophan acts as a corepressor and binds to the inactive repressor, forming anactive repressor-corepressor complex. • The active complex binds to the promoter, blocking the RNA polymerase from transcribing the five structural genes. • No polycistronic mRNA is produced, so no enzymes are produced for the synthesis of tryptophan. 7. (a) Hardy-Weinberg equilibrium means after one generation of random mating, the gene frequency would not change from generation to generation. (b) The percentage of non-tasters = (100 – 84)% = 16% = 0.16 = 0.4 Allele frequency of non-tasters = q = 0.16 = 0.4 (i) Frequency of homozygous dominant = p2 , p = 1– q = 1 – 0.4 = 0.6 p2 = 0.62 = 0.36 (ii) Frequency of heterozygous = 2 pq = 2 × 0.6 × 0.4 = 0.48 (c) Migration 8. (a) (i) Chromosomal mutation involving changes in number (ii) Non disjunction (b) Meiosis II (c) (i) Klinefelter syndrome (ii) • male with small testes • normally sterile • with well developed breasts Essay Questions 1. (a) • Polygenes are genes that individually contribute little to the phenotype but together, in groups, control quantitative traits. • Polygenes usually exhibit a dominantrecessive relationship, in which only the dominant alleles contribute to the phenotype whereas the recessive alleles do not. • They are duplicate genes. Each dominant allele contributes equally to the overall phenotype. • If the genes are located on different chromosomes, they follow the Mendelian law of inheritance. • Some of the genes may be linked and other may be sex-linked. • A normal quantitative character may be controlled by ten or more such genes. • Such quantitative characters include colour intensity, height, length, weight, volume and character that can be measured such as intelligence. • Quantitative character can be easily influenced by environmental factors, such as food and fertility of the soil. (b) • Because the extreme phenotypes of the F2 are equal to 6.25% (= 1 16 ), they are two polygenes involved. • Therefore, contribution of each allele = tallest – shortest 4 = 180 – 150 4 cm = 7.5 cm Let the 2 dominant alleles be A1 , A2 and recessive alleles be a1 , a2 .

208 Biology Term 3 STPM Chapter 17 Inheritance and Genetic Control 17 The marriage can be summarised as follows: P1: F1: Phenotype: A1 a1 A2 a2 mean height 165 cm P2: Gametes: A1 a1 A2 a2 165 cm A1 a1 A2 a2 165 cm Gametes : A1A1A2A2 180 cm a1a1a2a2 150 cm A1A2 A1A2 A1a2 a1A2 a1a2 A1A2 A1a2 a1A2 a1a2 a1a2 F2 : A1 A1 A1 a2 a1 A2 a1 a2 A1 A2 A1 A1 A2 A2 180 cm A1 A1 A2 a2 172.5 cm A1 A1 A2 a2 172.5 cm A1 a1 A2 a2 165 cm A1 a2 A1 A1 A2 a2 172.5 cm A1 A1 a2 a2 165 cm A1 a1 A2 a2 165 cm A1 a1 a2 a2 157.5 cm a1 A2 A1 A1 A2 a2 172.5 cm A1 A1 A2 a2 165 cm a1 a1 A2 A2 165 cm A1 A1 A2 a2 157.5 cm v A1 a1 A2 a2 165 cm A1 a1 a2 a2 157.5 cm a1 a1 A2 a2 157.5 cm a1 a1 a2 a2 150 cm Summary: Genotypes Ratios Phenotypes Ratios A1 A1 A2 A2 1 16 180 cm 1 16 A1 A1 A2 a2 2 16 172.5 cm A1 a1 A2 A2 123 2 16 172.5 cm 4 16 A1 a1 A2 a2 4 16 165 cm 6 16 A1 A1 a2 a2 1 16 165 cm A1 a1 a2 a2 2 16 157.5 cm a1 a1 A2 A2 1 16 165 cm 4 16 a1 a1 A2 a2 2 16 157.5 cm a1 a1 a2 a2 1 16 150 cm 1 16 So, the phenotypic ratio is 1 : 4 : 6 : 4 : 1 2. (a) • Sex-linked genes are genes found in the sex chromosomes i.e. in the X chromosomes of mammals. • Only females can be a carrier of a recessive allele as the individual has two such sex chromosomes. • If a phenotype is controlled by codominant or incomplete dominant alleles, only the females have a heterozygous phenotype. For example, only female cats are calico (tortoise-shell). • Reciprocal crosses always produce offspring of different phenotypic ratio. For example, a red-eye Drosophila is crossed with a whiteeye male and a white-eye female is crossed with a red-eye male produce different ratios. • If a phenotype is determined by a recessive allele, then it is more frequently found in the males because the males have no heterozygous genotype. • Conversely, if a phenotype is determined by a dominant allele, then it is more frequently found in females because females have heterozygous genotypes. • If a genetic disease or mutant character such as haemophilia is controlled by a recessive allele and if the mother is affected, then all the sons are affected. • Conversely, if the genetic disease or mutant type is controlled by dominant allele and if the father is affected, all the daughters are affected • Sex-linked genes manifest their phenotypes usually in an alternate generation pattern. The phenotype appears in one generation and may disappear in the next generation. (b) Let XB be the allele for black and XG be the allele for ginger Females: XBXB – black, XGXG – ginger and XBXG – calico Males: XBY – black and XGY - ginger P1: Phenotype: Black Ginger F1: XB XG Females calico XB Y Males black Gametes : XB XB XGY XB X Y G P1: Phenotype: Ginger Black F1: XB XG Females calico XGY Males ginger Gametes : XGXG XB Y XG X Y B

209 Biology Term 3 STPM Chapter 17 Inheritance and Genetic Control 17 3. (a) • A gene pool is an aggregate of genes or gametes of a Mendelian population from which the next generation is produced. • It can be considered as the total genetic information possessed by reproductive members in a population of sexually reproducing organisms. • Genes in the pool have dynamic relationships with one another and with the environment around where the organisms live. • Environmental factors such as selection can alter allelic frequencies and cause evolutionary changes in the population. • For example, only one gene or locus is considered. That locus consists of dominant and recessive alleles i.e. A and a alleles. • The frequencies of allele A and a depend on the genotypic frequencies of AA, Aa and aa. So, if the frequency of AA is very high, the frequency of A would be high too. • From the frequencies of the alleles, the frequencies of the genotypes of the next generation can be calculated if we assume that random fertilisation of the gametes occurs. • After one generation of random mating, the population is in equilibrium and the allelic frequency will not change if the population is large. (b) • A population with changing gene pool in every generation brought about by many forces will cause evolution in the long run. • Evolution does not involve individual in its lifetime but the whole population from one generation to the next generation. • The ultimate cause of genetic variation is mutation. It creates a new allele. • The allele may be recessive or dominant. It is inherited from one generation to the next generation through sexual reproduction. • The rate of transmission to the future generation depends on many factors. One of it is environment. • If the phenotype of an individual possesses the allele that has an advantage for survival in that particular environment, the allelic frequency will increase. • For example, a mutation that causes a black bird to become brown has an advantage for blending well in the brown grassland so that predator birds like eagles may not see it. Such allele will spread well in future generations. • This is an example of natural selection proposed in Darwin theory of evolution, the survival of the fittest. • The sum of all the mutations followed by natural selection in a given environment over a period of millions of years will cause the evolution of a new species of organism. • Therefore, evolution is the result of the sum changes of allelic frequencies of many genes involved that causes the chromosome to have genes so different that they are no longer homologous to the original ones. Population genetics determines evolution. 4. (a) • This is due to the two genes involved are linked on the same chromosome. • The alleles do not assort independently in the dihybrid. • The test cross produced two smaller groups due to crossing over. (b) • Map distance is the percent of crossing over = (number of recombinant types / total offspring) ×100 % Number of recombinant is 99 +101 / (295 + 305 + 99 +101) = (200 / 800) × 100 = 25 % R/r R/r S/s S/s 25% (c) F1 test cross : RS rs × rs rs Red, smooth White, wrinkled Gametes : 3/4 RS, 3/4 rs, 1/4 Rs, 1/4 rS Test cross offspring : RS rs Red, smooth (295) parental type, 3/4 rs rs White, wrinkled (301) parental type 3/4 Rs rs Red, wrinkled (99) crossing over, type 1/4 rS rs White, smooth (101) crossing over type 1/4 5. (a) (i)X-ray; Colchicine (ii) X-ray causes the bonds within the base of DNA to change and pair wrongly during replication resulting in point mutation. • Stronger X-ray causes the phosphodiester bonds of DNA to break, fragments of chromosome are lost or translocated resulting in deletion of genes in chromosomal mutation. Colchicine inhibits the formation of spindle fibres causing nondisjunction during nuclear division resulting in aneuploidy mutation of chromosome number. • A more concentrated colchicine causes nondisjunction of all chromosomes during nuclear division resulting in polyploidy mutation of chromosome number. (b) • Autopolyploidy arises when all the chromosomes undergo nondisjunction during mitosis or meiosis.

210 Biology Term 3 STPM Chapter 17 Inheritance and Genetic Control 17 • During mitosis, the centromeres of each chromosome divide but the sister chromatids do not separate as no spindle is formed. • No cytokinesis occur,so only one cell product is formed after mitosis. • If that cell is an initial cell at the shoot apex. • A new branch or a runner of tetraploid is formed as autoploid. • During meiosis, nondisjunction will result in unreduced diploid gametes. • If two unreduced gametes are fertilised, autoplolyploidy tetraploid is formed. • An autoplolyploidy of triploid is formed if a normal haploid gamete fertilises an unreduced diploid gamete. • When nondisjunction occurs during meiosis of tetraploid plants, higher ploidy of pentaploid (5n) and hexaploid (6n) can be produced. 6. (a) • The lactose operon is asshown in the diagram below. Operator lac Z lac Y lac A Regulator gene Structural genes Lac operon Promoter • It consists of regulator gene (I), promoter (P), operator (O) and structural genes that code for enzymes required for metabolising lactose. • The regulator gene is the segment of DNA that can be transcribed into a mRNA molecule and then translated into a protein called repressor. • The repressoris active; it bindsto the operator and blocks the binding of RNA polymerase to stop (switches off) the transcription of the structural genes. • The promoter is the segment of DNA that allows the binding of RNA polymerase so that transription of the structural genes can occur. • A single promoter is used to control the production of three enzymes required for lactose breakdown. • The operator is the segment of DNA that allows the binding of an active repressor. • Once the operator is bound, the repressor blocks the binding of RNA polymerase to the promoter and transcription of the structural genes are not possible. • It switches on when it is not attached with a repressor and production of enzymes can occur. (b) • In the absence of lactose, no enzymes for its metabolism are produced. • The regulatory gene is transcribed into a mRNA that is translated into protein, an active repressor. • The repressor binds to the operator and blocks the RNA polymerase from binding to the promoter. • When the RNA polymerase cannot bind to the promoter, there is no transcription of the enzymes and the enzymes are not produced as shown in the diagram below. I RNA polymerase Bind to operator No transcription of structural genes – no enzymes produced DNA mRNA5� 3� Protein Active repressor lac Z lac Y lac A Regulator gene Promoter Structural genes Operator • Thus, the cell saves energy when it does not need to produce unnecessary enzymes for a non-existent respiratory substrate. • In the presence of lactose, the three enzymes are produced. • The lactose bindsto the repressor, inactivating it and causing it to lose affinity for the operator. • RNA polymerase can bind to the promoter and transcription of the structural genes can start. • Thus, a polycistronic mRNA is produced, which is translated to produce β-galactosidase, permease and transcetylase, as shown in the diagram below. I RNA polymerase RNA polymerase can move across DNA mRNA5� mRNA5� 3� 3� Protein Allolactose Inactive repressor Enzymes are produced β -galactosidase Permease Transacetylase lac Z lac Y lac A • Therefore, the cell only producesthe enzymes to transport and to break down lactose when there is lactose. Lactose is an inducer for production of enzymes. The process of control is called enzyme induction. 7. (a) • Mutagens can cause mutations through radiation mutagens and chemical mutagens. • Radiation such as X-ray and radioactive rays break the DNA molecules directly, causing mutations.

211 Biology Term 3 STPM Chapter 17 Inheritance and Genetic Control 17 • These ionising radiations indirectly cause mutations through the production of free radicals inside the cells and deactivate the DNA or its base sequence in the next round of the cell replication. • Ultra violet light cannot penetrate deep into the tissue but causes mutation in skin cells of higher animals and microorganisms. It affects the double pyrimidine adjacent bases to pair, which is later cut off in the DNA, causing mutations. • An oxidising chemical such as nitrous acid deaminates adenine or cytosine oxidatively, altering the nature of the base and causes it to pair differently during replication and mutation later. • Another example is base analogue such as 5-bromouracil, a thymine analogue that replaces thymine during replication but later may pair with guanine, causing mutations. • Colchicine inhibits the formation of spindle fibre, causing non-disjunction and changes in chromosome numbers. (b) • The mutation that causes sickle cell anaemia is substitution mutation where the 6th code of β gene of haemoglobin is changed. • The substitution of a base of CTT to CAT causes the code which codes glutamic acid to change to code valine. • The β polypeptide formed has a non-polar valine instead of acidic glutamic acid, resulting in the alteration of the quaternary structure of haemoglobin. • If it occurs as a heterozygote, half of the red blood cells formed is sickle-shaped because the abnormal haemoglobin molecules distort the shape. • If it occurs in homozygous state, all the haemoglobin molecules have valine and all the red blood cells formed are sickleshaped. • Such mutation should be considered an incomplete dominant mutation as heterozygote shows the intermediate stage whereas the homozygote gene causes all the red blood cells to be sickle-shaped. • Thalassemia is caused by a deletion of nucleotide mutation which occurrs in β haemoglobin gene. • In a heterozygous condition, it produces thalassemia minor, which can only be detected through a blood test. • In the homozygous state, thalassemia major is resulted, in which the β haemoglobin subunits are not functional. • This impairs the oxygen transport capacity, resulting in severe anaemia. Children with thalassemia major usually cannot live a long life. 8. • Chromosomal mutation involves the change in chromosomal number or its structure. • The change in number may be aneuploid, involving an addition or loss of a few chromosomes but less than a set. • The change in number may involve an increase in the number of sets known as polyploid. • Aneuploids are caused by nondisjunction of usually one chromosome that can occur in meiosis or mitosis. • Usually, it occurs in meiosis in the formation of gametes, with some gametes having one extra chromosome and the others having one less chromosome. • This can happen on the autosome such as chromosome 21, which results in Down syndrome. • Down syndrome is formed from nondisjunction of chromosome 21 in the ovum. The ovum with the extra chromosome is fertilised by a normal spermatozoon, resulting in a baby with three chromosome 21. • Nondisjunction can occur in the X chromosome, resulting a gamete with an extra chromosome X (XX/XY) or lack of X chromosome. • If an XX gamete is fertilised by another X, a triple X syndrome is formed. • IftheXXgamete isfertilised by aYspermatozoon, then Klinefelter syndrome (XXY) is resulted. • A gamete without an X chromosome, when fertilised by another with X chromosome, will result in the formation of Turner (XO) syndrome. • Mutationsinchromosomalstructureoccurduring prophase I of meiosis when the chromosomes break and rejoin. • A segment of the chromosome may be lost, resulting in deletion of dominant alleles and the gametes. The offspring cannot survive. An example is cat cry syndrome. • A segment may be inserted or duplicated. It usually does not bring about harmful effects. An example of insertion or duplication in Drosophila causes the eye to become bar shape. • Translocation of a segment of chromosome from one to another can cause position effect. An example is the transfer of chromosome 21 to chromosome 9, resulting in a syndrome which is almost the same as Down syndrome. • Inversion of a segment chromosome can also occurs. It seems to be common in human but it does not bring about adverse effects.

CHAPTER Bilingual Keywords GENE TECHNOLOGY Concept Map 18 Recombinat DNA: DNA gabungan semula Restriction enzyme: Enzim potong tapak khas Endonuclease: Nuklease potong dalam Nomenclature: Tatanama Palindrome: Ejaan dua hala sama Vector: Vektor/pembawa Cloning: Hasil individu seiras Isolation: Pemencilan Target DNA: DNA sasaran/diperlukan Transduction: Dimasukkan DNA oleh virus Amplification: Tambahan bilangan Transformant: Individu telah diubah Genome: Keseluruhan gen sesuatu organisma cDNA: DNA asal dari mRNA Transgenic: Dipindah dengan gen asing Screening: Penyaringan Amniocentesis: Ujian bendalir janin Gene therapy: Rawatan tambahan gen DNA fingerprinting: Cap jari DNA Forensic science: Sains perundangan Paternity suit: Tuduhan akuan si bapa Defective gene: Gen rosak Gene Technology Nomenclature Recognition site (palindrome) Importance Insertion Sticky or blunt ends generated Ligation Amplification Screening Genomic and cDNA libraries Properties of vectors EcoRI & SmaI Genomic and cDNA cloning Recombination DNA Technology /genetic engineering Restriction Enzyme Vectors Uses in Cloning Human Insulin Cloning Differences between Genomic & cDNA cloning (enzyme involved) Reverse transcription Transformation/ transduction

213 Biology Term 3 STPM Chapter 18 Gene Technology 18 18.1 Recombinant DNA Technology Genetic Engineering 1. Genetic engineering, also known as recombinant DNA technology, can be defined as a technology used in the isolation of gene from one organism and transferring it to another organism of the same species or different species, so that it benefits the organism or humans. 2. Gene is usually isolated from humans or from any plants or animals, by cutting with a specific restriction enzyme. The gene DNA cut will have a specific type of ends. 3. It is than spliced into the DNA of a vector that carries the gene into a recipient cell. The cell may be a bacterial cell, ovum or stem cell, which becomes transgenic. 4. The transgenic cell is cultured as a clone before it is screened and tested. Transgenic organisms can be formed and tested before they are cloned to increase their numbers. 5. Gene banks in the form of gene libraries are kept. The genes stored are analysed, their potential values are tested and used later. 6. Many successes have been achieved to transfer valuable genes from one organism to another forming transgenic organisms. These bring benefits in the field of agriculture, medicine, industry and others. 7. More benefits are on the way, though danger may arise out of indiscriminate uses. Besides this, there are problems and ethics in manipulating genes, especially on humans. 8. However, this field of genetic engineering can open a whole different window in the advances of human race. Restriction Enzymes 1. Restriction enzymes are endonucleases that cut phosphodiester bonds between nucleotides in the DNA at specific sites. 2. They are different from exonucleases like DNAse or RNase that cut phosphodiester bonds one at a time from both ends of the DNA or RNA. 3. Restriction enzymes cut the phosphodiester bonds in both strands of the DNA molecule. The cut is between the same two base sequences when read from 5’ to 3’. The enzyme may cut in a zig-zag pattern leaving “sticky” ends or at cleavage sites leaving blunt ends. Learning Outcomes Students should be able to: (a) explain recombinant DNA technology/genetic engineering; (b) differentiate between genomic and cDNA cloning and genomic and cDNA libraries; (c) describe the vectors used in cloning and their properties; (d) describe the restriction enzyme (EcoRI and SmaI), including its nomenclature, recognition site (palindrome), importance and the types of ends generated; (e) explain reverse transcription, insertion, ligation, transformation/ transduction, amplification and screening; (f) desribe the steps involved in genomic and cDNA clonong, including the enzymes involved, and explain human insulin production in E.coli as an example. Language Check Language Check • Transgenic = containing a gene or genes transferred from another species 2014 2017

214 Biology Term 3 STPM Chapter 18 Gene Technology 18 4. They are naturally found in some bacteria for cutting up invading DNA from viruses and restricting the multiplication of viruses within them. Their own DNA is not cut as its G and C are protected by addition of methyl groups by the enzyme methylase. 5. The restriction site is restricted to four to eight pairs of specific base sequences in the DNA. 6. The short base sequences at the restriction site is a palindrome, which means it is the same when read from 5’ to 3’ in both strands. 7. One of an example of such enzyme is EcoR1 (E-KO-R-1) that cuts a restriction site of GAATTC. (a) It recognises the site and cuts it between base G and A in a ‘staggered’ way so that the DNA ends are ‘sticky’ with single strands of short unpaired bases, as shown in Figure 18.1. – G – CTTAA Cut 5’ – GAATCC – 3’ 3’ – CTTAAG – 5’ Sticky end Cut AATTC – G – Figure 18.1 The action of EcoR1 forming two sticky ends (b) There is a specific nomenclature i.e. the system for naming enzyme. EcoR1 is named so because it is obtained from bacteria E. coli RY 13 and the 1 represents the first from that bacteria strain. 8. Another example is SmaI (S-ma-I) that cuts a restriction site of CCCGGG. (a) The action of SmaI is as shown in Figure 18.2. (b) It is obtained from Serratia marcescens H and it is the first from this bacteria strain. Cut Cut 3’ 5’ G C G C G C C G C G C G 5’ 3’ G C G C G C C Blunt ends G C G C G Figure 18.2 The action of SmaI 9. The importance of a restriction enzyme is to cut two different sources of DNA, one from a donor, for example human DNA with an insulin gene and another from a recipient DNA, for example, from a plasmid. As both of these DNA have the same complementary sticky ends, they can be joined to form a single recombinant DNA. This recombinant DNA formed can be inserted into a bacterial cell to produce human insulin or any human protein for therapeutic purposes. Besides that, recombinant DNA can be stored as gene libraries and used in genetic manipulation. Exam Tips Remember the meaning of restriction enzymes, their importance, how they act (STPM 2000 structured question), two examples and nomenclature of EcoRI and SmaI, together with their palindromic sites.

215 Biology Term 3 STPM Chapter 18 Gene Technology 18 Restriction enzyme cuts the insulin gene. The fragment with sticky ends is produced. Restriction enzyme cuts the plasmid. The fragment with sticky ends is produced. DNA fragments produced by the restrition enzyme. Can be joined to form a single recombinant DNA. Cut Plasmid Sticky ends Cut Cut Insulin gene Sticky ends Cut plasmid Recombinant DNA + Insulin gene Figure 18.3 The use of a restriction enzyme Vectors 1. A vector or cloning vector is a transfer agent for a gene or DNA from one organism to another. 2. There are two types of vectors frequently used in genetic engineering i.e. plasmid and virus. 3. Plasmids are small rings of DNA found in many types of bacteria. (a) Each plasmid may contain around 1,000 to 10,000 bases or three to ten genes. (b) It replicates at the same time or independently when the main DNA of the bacterial cell replicates. (c) Genes found inside the plasmid express themselves just like normal genes and may confer extra characters, such as resistance to antibiotic ampicillin. (d) A bacterial cell may contain one to several hundreds of plasmids. (e) A plasmid can be extracted and genes are inserted into it and put back into the bacterial cell. (f) It is natural for bacterial cells to take in plasmids, though some cells need Ca2+ to help in the uptake. 2. Viruses are simple parasitic organisms that infect cells to reproduce. (a) They can be used to transfer genes into cells that they attack. (b) Their harmful genes must be removed before they can be used. (c) When they attack cells, their DNA or DNA formed from RNA that they brought in can be inserted into the normal chromosome of the host cell. 2017

216 Biology Term 3 STPM Chapter 18 Gene Technology 18 (d) They are very useful as different types can bring specific genes to different types of eukaryotic cells that plasmids cannot do. (e) One of the viruses is phage λ, a bacteriophage that also brings genes to a bacteria cell. (f) Another example is a retrovirus that can transfer genes into stem cells. (g) Viruses are better vectors than plasmids as larger DNA or more genes can be inserted into the cell. 5. Vectors must have the following properties to make them effective agents. (a) They must be easy to maintain and stored so that they can be used when required. (b) They can easily enter the cell so that the genes can be brought into the cells. (c) They must not be harmful or cause side effects to the host cell. (d) They usually have antibiotic resistant genes and restriction sites for the screening of transformed bacteria. (e) Genes brought into the cells by them should be able to express themselves. 6. Vectors are now commercially produced by scientific companies. They are sold to universities and research laboratories for genetic engineering. The vectors have different markers (genes) for screening purpose. They also have spesific restriction sites for insertion of genes. Cloning 1. Cloning is a technique used to produce clones i.e. a group of individuals, cells or DNA with identical genetic makeup. 2. An example of cloning is to produce transgenic bacteria that can form insulin. 3. There are seven steps in the cloning process, as shown in Figure 18.4. (a) Isolation of target DNA and vector DNA/reverse transcription (i) There are two ways to obtain insulin gene. One is to cut the gene from the cell that contains it and another way is to get it from cDNA. (ii) Cultured human cells or β-cells from the islets of Langerhans are used. DNA is isolated by ultracentrifugation after the cells are homogenised. However, the insulin gene contains introns that must be removed before insulin can be formed. (iii) cDNA of the insulin gene can be obtained by extracting the mRNA from the active β pancreatic cell. Reverse transcriptase is then added to the mRNA to use it as Exam Tips Remember the vectors characteristics of cloning vectors and the characteristics of bacterial host cell. (STPM 2015 essay question) Islets of Langerhans are the regions of the pancreas that contain its endocrine cells. Info Bio

217 Biology Term 3 STPM Chapter 18 Gene Technology 18 Insulin gene mRNA of insulin gene Insulin gene Ampicillin resistance gene (ampR ) lac Z gene lac Z gene split non-functional lac Z gene Restriction site Plasmid vector Restriction cDNA (a) Isolation of target DNA and vector DNA (b) Restriction (c) Insertion (d) Ligation (e) Transformation (f) Amplification (g) Screening + Ampicillin + X-gal Insulin gene Probe comfirms insulin gene + ligase + or or Recombinant plasmid E.coli E. coli with insulin gene Figure 18.4 Cloning of the human insulin gene in bacteria (iv) A cloning vector is obtained either by using plasmid or phage λ. Both vectors are specially developed for genetic engineering. (v) Each plasmid used has an ampicillin resistance gene and β-galactosidase (lac Z) gene. (b) Restriction of target DNA and vector DNA by restriction endonuclease or enzyme (i) Both target and vector DNA are treated with the same restriction enzyme, such as EcoR1. (ii) This is to ensure that both target DNA and vector DNA have the same complementary sticky ends and can join to form recombinant DNA. (iii) The circular plasmid is cut open in the middle of the β-galactosidase gene that has a restriction site for EcoR1. template to form single DNA strands. Then, the RNA is digested with RNase and the single DNA strands are made into double-stranded DNA by adding DNA polymerase. Exam Tips Student should be able to explain the production of protein from mRNA using genetic engineering technique. (STPM 2015 essay question)

218 Biology Term 3 STPM Chapter 18 Gene Technology 18 (iv) The insulin cDNA is added to both restriction sites at both ends before it is cut. (c) Insertion of target DNA into vector DNA (i) The two DNA are mixed to get the two ends of the target gene joined by hydrogen bonds to the open plasmid. (ii) Usually, millions of copies of each DNA type are mixed so that the chance of successful joining is high. (d) Ligation of target DNA into vector DNA by DNA ligase (i) An enzyme DNA ligase is added so that rings of recombinant DNA are formed, as shown in figure 18.3. (ii) The ligase ensures phosphodiester bonds are formed between the backbones of the two types of DNA molecules. (e) Transformation or transduction of recombinant DNA into host cells (i) The recombinant DNA in the form of plasmid or phage λ is added to a culture flask of E. coli. (ii) Calcium chloride is added, followed by a short heat shock that helps to open pores in the plasma membrane to allow the plasmid to enter. (iii) The bacteria that receives the plasmid are transformed, whereas the bacteria infected with phage λ are transduced. (f) Amplification (i) This is a process where the number of bacteria transformed or transduced are allowed to increase naturally as bacteria divide every 30 minutes. (ii) The bacteria are diluted to a low concentration and plated. When they develop into a colony or clone, they do not overlap or mix with each other. (iii) The plating medium contains ampicillin and a type of sugar called X-gal. (g) Screening for transformants (i) Finally the bacteria which are transformed and are able to take up the insulin gene are selected. (ii) The first screening is done by the antibiotic ampicillin put in the medium, which kills all bacteria that did not take up the plasmid. (iii) The second screening is done by the X-gal in the medium. Any bacterium that has not taken a foreign DNA can produce β-galactosidase, which digests the X-gal to blue coloration. Bacteria clones that have taken up foreign DNA produce white coloration as the enzyme is non-functional and they are selected for mass production. Exam Tips Remember the seven steps in cloning E.coli that can produce human insulin (STPM 1996 and 1998 essay questions, 2001 structured question).

219 Biology Term 3 STPM Chapter 18 Gene Technology 18 Quick Check 1 1. What are cloning vectors? 2. What is the advantage of viruses over plasmids in gene transfer? 3. State the properties of vectors. Genomic and cDNA Libraries Cloning of genomic and cDNA libraries 1. Cloning or construction of genetic libraries is a process of storing DNA fragments especially genes of particular interests in the form of plasmids. Such plasmids are usually inserted inside bacteria especially the laboratory strain of E. coli which are allowed to divide naturally to form clones. Thus, each clone will only have a particular gene of interest e.g. human insulin gene. 2. Cloning of a genomic library is the storing of all the genes of a species of organism, in the form of bacterial clones. The best clones would only have one particular gene. Thus, any time such a gene is required to be studied or transferred into another organism can be extracted from the bacteria. A genomic library enables researchers to have the whole set of genes at his disposal to manipulate. 3. The cloning of genomic library is summarised in the Figure 18.5. Thousands of genomic DNA fragments Genomic library DNA fragments inserted into plasmid using ligase Recombinant Introduction of DNA molecules plasmid into bacteria Human DNA Cleave with restriction nuclease Figure 18.5 The cloning of human genomic library 4. The steps in cloning of genomic library are as follows: (a) Extract cells of interest by homogenisation of the tissue followed by selective digestion of unwanted large molecules such as proteins and RNA. The long DNA left can be easily extracted just by stirring with a wire. (b) Treat with one or more restriction enzymes. The DNA may be cut into thousands of fragments each containing no gene, one gene or more than one gene. Thus, repeated cutting with different restriction enzymes enables the ideal gene to

220 Biology Term 3 STPM Chapter 18 Gene Technology 18 be obtained i.e. each fragment contain one gene or group of related genes with regulator and promoter genes so that the protein under the gene control can be synthesised in vitro under the right conditions in a test tube or in vivo in the cells of another organism. (c) The DNA fragments are inserted into laboratory developed plasmids (cloning vectors). Such small rings of DNA have at least two specific markers or genes such as antibiotic resistant genes and β-galactosidase (lac z) gene with their regulator and promoter genes. Such genes are used to select the bacteria that have taken in the required gene. The plasmid also has a restriction site in the middle of one of such specific marker gene as shown Figure 18.6. (d) The plasmid with the foreign gene inserted formed is added with ligase to form a recombinant DNA. Normally the cut human DNA fragments are just mixed with cut plasmids. The plasmids are inserted into the laboratory strain of E. coli developed especially for cloning purpose. The plasmid with the inserted gene can be easily taken up by the bacteria. Electric shock and salts like calcium chloride may help the bacteria to take up the plasmid. (e) Screening or selection of the right bacteria that have taken the right gene. The bacteria are plated into Petri dishes with the specific antibiotic or chemicals like lactose. Bacteria with the right gene will survive in the antibiotic. The human gene taken up can be tested to see if the protein under its control can be produced. Such selected bacteria are allowed to form clones, to be labelled and stored. 5. The cloning of cDNA follows different steps and are summarised in Figure 18.7. Culture Cells Extract RNA RNA cDNA Insert into Bacterial Plasmid Isolate plasmids Grow RNA Reverse Transcriptase RNAase H + PCR Double-stranded cDNA Generate competent bacteria Figure 18.7 The cloning of cDNA Ter Ori Plasmid for use in cloning AmpR LacZ Polycloning site Figure 18.6 A plasmid used for cloning

221 Biology Term 3 STPM Chapter 18 Gene Technology 18 (a) Extraction of the mRNA of interest. This mRNA is produced using a gene as template before a protein is synthesised (Figure 18.8). Eukaryotic cells have genes containing introns that are removed before the exons are spliced into proper mRNA before a required protein is produced. Promoter Intron 1 Exon 1 Gene (DNA) Primary transcript (RNA) Mature transcript (mRNA) Protein Transcription Splicing Protein synthesis Intron 2 Exon 2 Intron 3 Exon 3 Exon 4 Figure 18.8 Eukaryotic genes have introns (b) Synthesis of cDNA by PCR (Polymerase Chain Reaction). AAAAAAAAA 3’ AAAAAAAAA 3’ TTTTTTTTTT AAAAAAAAAGGGGG 3’ TTTTTTTTTT TTTTTTTTTT TTTTTTTTTT AAAAAAAAAA TTTTTTTTTT mRNA mRNA:DNA hybrid poly G trailing mRNA hydrolysis double-stranded DNA after 1st PCR cycle Add PCR primers PCR amplification AAAAAA AAAA Insert into rector 5’ 5’ 3’ 5’ 3’ GGGGG 3’ GGGGG 3’ GGGGG 3’ GGGGG 3’ RE1CCCCC RE1CCCCC RE1CCCCC RE1CCCCC TTTT-RE2 TTTT-RE2 5’ Figure 18.9 Synthesis of cDNA from mRNA

222 Biology Term 3 STPM Chapter 18 Gene Technology 18 The culture cells or tissue are homogenised and selectively digested with proteinase and DNase leaving RNA including rRNA, tRNA and mRNA intact. mRNA differs from other RNA as they have poly-A tails. So, only mRNA with poly-A tails are converted into cDNA. (c) The double stranded cDNA are inserted in specially prepared plasmids. Such plasmids with the necessary markers are cut with restriction enzyme (RE) that cuts the restriction site forming blunt ends like Sma1.The plasmids are then added with the right linkers so that the cDNA can be inserted and ‘sealed’ with ligase to form recombinant plasmid or DNA. (d) Screening of clones. The recombinant plasmids are inserted into laboratory-developed bacteria to form clones and the process of clone screening is done to identify each clone and their genes. The clone is also tested for the presence of the functional gene with the right protein. The clone is labelled and stored as cDNA library. 6. The differences between the two cloning processes are summarised in Figure 18.10. Cell DNA Restriction enzyme digestion Ligation Bacteria Restriction enzyme digestion Vector cDNA Library Genomic Library mRNA cDNA Restriction enzyme digestion Ligation Restriction enzyme digestion Ligation of linkers Bacteria Figure 18.10 Cloning of genomic and cDNA libraries Summary Cloning of genomic library Cloning of cDNA library 1. Extraction of all cell DNA. Extraction of all cell mRNA. 2. The DNA are cut into fragments using restriction enzyme. The mRNA are used to form cDNA using reverse transcriptase and DNA polymerase. 3. The cut fragments have ‘sticky’ ends and can be inserted into plasmids cut with the same restriction enzyme. Linkers have to be added to cDNA or ‘sticky’ ends have to be added before the cDNA can be cut. 4. Millions of clones may be produced but the clones with functional genes may be small. Different cells have different mRNA and clones with functional genes are produced. 5. The cloning is storing fragments of DNA in bacterial clones. The cloning is storing functional genes with no introns. 6. It is possible to produce clones with one gene of the species each. Only enables the production of clones with genes that can be transcribed into mRNA. 7. Cloning of both prokaryotic and eukaryotic genomic libraries may be done. Cloning of eukaryotic cDNA is usually done to produce functional protein. 8. The purpose of cloning genomic library is to store DNA fragments of an organism. The purpose of cloning cDNA is to store DNA that encodes a specific protein.

223 Biology Term 3 STPM Chapter 18 Gene Technology 18 7. The differences between the genomic library and cDNA library are summarised in Figure 18.11. DNA fragments mRNA DNA cloning Reverse transcription and DNA cloning Genomic DNA clones cDNA clones Gene 1 Gene 2 Gene 3 Gene 1 Gene 2 Gene 3 Exon Exon intron intron Restriction enzyme digestion Transcription intron Chromosomal DNA intron Exon Exon Exon Exon Figure 18.11 The differences between genomic and cDNA libraries Summary Genomic library cDNA library 1. May contain more fragments of gene than the gene itself. Contains complete functional gene that is reverse transcribed from mRNA. 2. May contain gene together with the unnecessary introns. Contains gene with no introns. 3. May contain no gene that encodes a protein. Contains gene that encodes a protein or a polypeptide. 4. Contains gene that needs the control system to enable it to produce functional protein. Contains gene that needs only the bacterial operon system spliced into the plasmid to produce the required protein. 5. May contain the complete genomic genes of a species. May only contain genes that are active in a cell i.e. mRNA can be produced. 6. Contains gene that may not be functional in another eukaryotic cell. Contain gene that may be functional in another eukaryotic cell as the plasmid contains the regulator gene. 7. Usually contain regulatory DNA sequences such as enhancers and promoter sequences Usually contain no regulatory DNA sequences such as enhancers and promoter sequences 8. May store all the gene fragments of an organism. May store all the possible genes that can be transcribed into mRNA that can be translated into protein. Exam Tips Student should be able to differentiate cDNA and genomic DNA libraries. (STPM 2015 essay question)

224 Biology Term 3 STPM Chapter 18 Gene Technology 18 STPM PRACTICE 18 Objective Questions A I only B I and II C I, II and III D I, II, III and IV 6. Which of the following sequence of DNA is a palindrome? A –GAATTG– C –GGATCC– –CTTAAC– –CCTAGG– B –CAATTC– D –CCATCC– –GTTAAG– –GGTAGG– 7. Which combination of components and their functions is correct? Component Function I Plasmid X Joins DNA fragments II DNA ligase Y Cuts both donor and vector DNAs III Restriction enzyme Z Carries gene of interest into the host cell I II III A X Y Z B X Z Y C Z X Y D Z Y X 8. Which about genomic and cDNA cloning are correct? I Using DNA as the starting materials II Using reverse transcriptase III Using DNA polymerase IV Using DNA ligase A I and II B I and III C II and IV D III and IV 1. What enzyme is used to cut DNA in genetic engineering? A DNA ligase B Restriction endonuclease C DNA polymerase D Exonuclease 2. Which combination is true of restriction enzymes? I It restricts gene transcription II It is found in all eukaryotic cells III It acts on palindromic sequences IV It is sensitive to changes in temperature and pH A I and II C II and IV B I and III D III and IV 3. Which statement is true of restriction enzymes? A They are named according to their roles. B They cut double-stranded DNA to produce blunt ends. C They cut double-stranded DNA at specific base sequences. D They are used in genetic engineering to join two DNA fragments. 4. Which statement is true of plasmid? I They are DNA molecules II They carry a small number of genes III Examples include EcoR1 and SmaI IV They replicate any time A I and II C II and III B I and III D I, II and IV 5. What is the role of restriction enzyme in cloning of bacteria? I To cut the vector DNA II To cut the DNA of a receiver bacterial cell III To cut the DNA of a bacterial cell that is already cloned with foreign genes IV To cut DNA ligase

225 Biology Term 3 STPM Chapter 18 Gene Technology 18 9. Which of the following is true of cutting plasmid for the formation of recombinant DNA? A Two different hydrolases B The same type of DNase C The same type of endonuclease that cuts a donor gene D A common DNA ligase is used 10. Which combination is true of the similarities of genomic cloning and cDNA cloning? I Uses DNA ligase II Uses reverse transcriptase III Uses restriction endonuclease IV Uses DNA as the starting materials A I and II B I and III C II and IV D III and IV 11. The production of cloned DNA is shown in the diagram below: Source DNA Plasmid Amplification Screening Cloned DNA Transformation The reason for carry out screening is to rid bacteria that I have not transformed II have not undergone amplification III are transformed with nonrecombinant plasmids A I and II B I and III C II and III D I, II and III 12. Which of the following is true of X, Y and Z in the cloning process shown below? Human cell Y Z E. coli X X Y Z A DNA ligase EcoRI Recombinant DNA B EcoRI DNA ligase Recombinant DNA C Recombinant DNA EcoRI DNA ligase D Recombinant DNA DNA ligase EcoRI 13. The diagram below shows the cloning of a gene. DNA Gene Plasmid Transformation P Q R

226 Biology Term 3 STPM Chapter 18 Gene Technology 18 What are P, Q and R? P Q R A Bacteria Endonuclease DNA ligase B DNA ligase Bacteria Endonuclease C Endonuclease DNA ligase Bacteria D Bacteria DNA ligase Endonuclease 14. Which of the following is the unpaired nucleotides produced by restriction enzymes at one end of DNA fragment A Blunt end B Sticky end C Restriction fragment D Single strand template 15. Name the process of selecting the transformed bacterial clone. A Filtering B Screening C Cloning D Ligating 16. A radioactive short sequence of single stranded DNA used to locate a gene is called A vector B probe C palindrome D phage 17. The techniques gained in recombinant DNA technology can be used to I obtain the genotypic ratio II identify dominant allele III join DNA fragments from different organisms IV form transgenic cells A I and II B I and III C III and IV D II, III and IV 18. Which of the following is the role of reverse transcriptase in the process of cloning bacteria for insulin production? A RNA to RNA B RNA to DNA C DNA to RNA D DNA to DNA 19. Which of the following is used to produce recombinant DNA? A DNA polymerase and ligase B DNA polymerase and restrictive enzymes C RNA polymerase and restrictive enzymes D Restrictive enzymes and ligase 20. Which of the following is not a transgenic organism? A The herbicide-resistant maize B The cloned sheep named Dolly C The bacteria that produce human insulin D The sheep that produces human protein in milk 21. Which step is not required for the cloning process of gene that is isolated from a bacteria? A Ligation B Insertion C Reverse transcription D Cutting by restriction enzyme 22. Which combination is true of restriction enzymes? I Join two DNA fragments II A type of endonuclease III Use to cut DNA and RNA IV Naturally found in some bacteria A I and III B I and IV C II and III D II and IV

227 Biology Term 3 STPM Chapter 18 Gene Technology 18 Structured Questions 1. The diagram below shows the reactions that have taken place during the formation of a recombinant DNA. Plasmid Recombinant DNA cDNA cDNA cDNA mRNA X Y mRNA (a) Name enzymes X and Y; and state the purpose for adding such enzymes. [4] (b) What is the role of plasmid in the cloning of a gene? [1] (c) What are the steps taken after the recombinant DNA is formed so that a transgenic bacterial clone is produced? [2] (d) State three industrial processes that can make use of the recombinant DNA. [3] 2. The diagram below shows the action of a restriction enzyme on a short sequence of DNA. – G – CTTAA Cut 5’ – GAATCC – 3’ 3’ – CTTAAG – 5’ Sticky end Cut AATTC – G – (a) What is a restriction enzyme? [2] (b) State the example of the enzyme as shown in the diagram and explain how it got its name. [3] (c) Explain how a restriction enzyme reacts. [5] Essay Questions 1. (a) Describe the production of protein from mRNA using genetic engineering technology. [1] (b) What are the differences between cDNA and genetic DNA libraries? [4]

228 Biology Term 3 STPM Chapter 18 Gene Technology 18 Quick Check 1 1. • They are agents that transfer foreign genes into cells. • A foreign gene is isolated and inserted into the DNA of vectors. • The DNA is then transferred into another cell and the gene is able to produce a specific desired protein. • Examples are plasmids and viruses. 2. • Plasmids can only transfer genesto certain strains of E.coli and not all different types of bacteria. • Every type of cell can be attacked by a certain type of virus and so can be used to transfer genes into the target cell. • Besides that, virus can transfer a longer piece of DNA than plasmid. 3. • They must be easy to maintain and stored so that they can be used when required. • They can enter the cell easily so that the genes can be brought into the cells. • They must not be harmful or cause side effects to the host cell. • Genes brought into the cells by them should be able to express themselves. STPM Practice 18 Objective Questions 1. B 2. D 3. C 4. D 5. B 6. C 7. C 8. A 9. C 10. B 11. B 12. B 13. C 14. B 15. B 16. B 17. C 18. B 19. D 20. B 21. C 22. D Structured Questions 1. (a) X is reverse transcriptase. Y is RNase. X is added to produce DNA from mRNA. Y is added to digest the mRNA. (b) Its role is to transfer the donor gene to a receiving cell. (c) The recombinant DNA is mixed with bacteria with calcium chloride and heated so that the DNA enters the bacteria. The bacteria are screened so that only the bacteria that have taken up the gene are allowed to be cloned. (d) It is used in the production of human insulin and other human proteins. It is used in the production of bacteria that can decompose oil slicks and other industrial toxins. It is used in the production of bacteria that can fix nitrogen so that no nitrogenous fertiliser is used in farming. 2. (a) A restriction enzyme is an endonuclease that cuts DNA at specific sites. It cuts a palindrome site of six base pairs and leaves the ends ‘sticky’ with unpaired bases. (b) It is EcoR1. It gets its name as it is obtained from E.coli RY 13 strain and the 1 represents the first such enzyme obtained from this strain. (c) It can recognise a specific palindromic site. The site consists of usually 6 pairs of bases. It cuts in the middle, forming blunt ends. Others cut in a zigzag fashion, at different sites in the double strands. It leaves behind unpaired bases at the ends, making the ends ‘sticky’. Essay Questions 1. (a) • mRNA has to be converted into single stranded cDNA using reverse transcriptase with required deoxyribonucleotides. • The mRNA has to be digested by RNase. • The single stranded cDNA is then converted into double stranded using DNA polymerase with required deoxyribonucleotides. • The two ends of the cDNA have to be added with ‘sticky ends’ or special restriction site. • This enables the cDNA to be inserted into a plasmid vector with complementary ‘sticky ends’ or cut by the same restriction endonuclease. • Commercial developed plasmid vector which has antibiotic resistant gene and special restriction site in the middle of a market, β-galactosidase is used. • Millions copies of each cDNA and plasmid vector are made by polymerase chain reactions. • The cDNA and the plasmids cut at the restriction site to form recombinant DNA followed by ligation with DNA ligase. • Such recombinant DNA is inserted into commercial developed bacterial host cells by mixing the two with the help of ions for easy entry of plasmids into the cells. • The host cells are plated into special agar medium with antibiotic and X-gal. • Bacterial colony with blue coloration is selected to be cultured and tested for the production of the protein. • Those bacteria that can produce the protein required are then cultured and tested for the production of the protein. (b) • cDNA library has a particular functional gene while genomic library has cut fragments of genomic DNA. • cDNA library has gene with no introns while genomic library has gene together with the unnecessary introns. • cDNA library has gene that encodes a protein or polypeptide while that of genomic may not encode a protein. • cDNAlibrary has prokaryotic operon system while that of genomic may have eukaryotic system. • cDNA library can only use to produce one type of protein while genomic library can be used to produce many types of proteins. ANSWERS

CHAPTER Bilingual Keywords BIOTECHNOLOGY Concept Map 19 Tillage: Pemugaran Fermentation: Penapaian Herbicide resistant plants: Tumbuhan rintang herbisid Transgenic: Transgenik Genetically modified: Diubah suai genetik Plasmid: Plasmid Restriction enzyme: Enzim pembatas Oncogenes: Onkogen Pollination: Pendebungaan Cross-pollination: Pendebungaan silang Inbred: Biak baka dalam Gene modification: Modifikasi gen Immunisation: Imunisasi Electrophoresis: Elektroforesis Roles Biotechnology Applications of Biotechnology Food and beverage production Agriculture Roles of biotechnology in our life Definition of biotechnology Fermentation & vitamin-enriched eggs Hybrid rice Herbicide resistant plants Transgenic fish Health Genetic screening Diagnostic kits Oil-decomposing bacteria Medicine Human growth hormone Gene therapy Forensic DNA fingerprinting

230 Biology Term 3 STPM Chapter 19 Biotechnology 19 19.1 Roles of Biotechnology Biotechnology 1. Biotechnology  is the use of living systems and organisms to develop or make useful products. It is a science and technology which is based on the use of cells or part of cells for the production of knowledge, products and services. 2. Modern biotechnology includes the use of gene technology and other methods in molecular biology to produce knowledge and useful products for application in research, medicine, agriculture and industry. 3. Biotechnology was a science discovered in  South San Francisco, California  in the 1970s. Major biotech companies like Genentech and Amgen were two of the first biotech companies. 4. Biotechnology has been used by mankind in agriculture, food production and medicine for thousands of years. The term itself is largely believed to have been coined in 1919 by Hungarian engineer, Karl Ereky. 5. In the late 20th and early 21st century, biotechnology has expanded to include new and diverse sciences such as genomics, recombinant gene  technologies, applied  immunology, and development of pharmaceutical therapies and diagnostic tests. Roles of Biotechnology in Daily Lives Green biotechnology (agriculture) 1. Improving crop yield and farm animal by selective breeding such as super hybrid rice and cows with high milk production. 2. Using recombinant DNA, tissue culture and stem cell technologies to further improve yield, disease resistant, drought resistant, salt tolerant crops and farm animals. 3. Reducing or eliminating the use of tillage to control weeds which significantly enriches soil and reduces erosion. 4. Allowing conservation tillage that enables farmers to make fewer passes over the field, reducing fuel use and soil erosion and contributing to fewer greenhouse gas emissions. 5. Developing crops to meet bio-fuel demands which provide cleaner, environmentally friendly sources of renewable energy. 6. Increasing yields to produce more food on less land, protecting valuable green spaces and the wildlife within. Summary Green biotechnology 1. Improve yield 2. Conservation 3. Renewable energy 4. Protect green spaces 5. Better seed development 6. More efficient nitrogen (less fertilisers) 7. Biosensor 8. Utilising natural medicines Learning Outcomes Students should be able to: (a) define biotechnology; (b) outline the roles of biotechnology in daily lives. 2016 2014

231 Biology Term 3 STPM Chapter 19 Biotechnology 19 7. Developing seeds that grow better in drought conditions or in soils with excessive water or in high salinity soils which would not normally allow healthy plant growth. 8. Improving nitrogen use efficiency in plants to reduce applications of fertiliser. 9. Utilising biosensor for sensing concentration of both inorganic and organic chemicals in the soil or atmosphere. 10. Utilising rainforest as in Malaysia, with its estimated 15,000 flowering plant species and 185,000 animal species, and the practice of traditional and herbal medicines from Malay, Chinese and Indian cultures. White biotechnology (industry) 1. Utilising microorganisms in fermentation industries in the production of alcoholic beverages, food seasonings, cheese, food and food supplements. 2. Utilising gene manipulated microorganisms to produce enzymes and special chemicals for a wide range of industrial functions like conversion of sugar from one less sweet sucrose to a sweeter fructose. 3. Producing industrial enzymes that includes the food enzymesrhamnosidase, pectinases, proteases, tannase; and enzymes used in biomass conversion- cellulases, beta-glucosidases, rhamnosidase and beta-galactosidase. 4. Producing bioactives and food additives such as secondary metabolites, microbial food colorants and organic acids. 5. Utilising bacterial expression system for production of enzyme degraded carbohydrates, and functional foods. 6. Utilising enzymatic hydrolysis of agriculture and forest biomass for ethanol production from a variety of raw materials such as hemp, citrus and wood wastes. 7. Producing bio-diesel using micro-algae, yeast or mixed cultures for the processing of the biomass, selection and modification and biotransformation of the algal oils by lipases and of solvents for the production of bio-fuels. 8. Utilising microorganisms in pollution control and generation of natural gases and bio-fuels that curtail waste generation and use up less energy and water. 9. Characterising novel nano-materials such as nano-particles, nano-fibres, nano-tubes, nano-porous medium, and nanocomposite possess large surface area that can increase the enzyme loading and facilitate reaction kinetics, thus improving the bio-catalytic efficiency for industrial applications for enzyme immobilization food processing and bioenergy generation. Summary White biotechnology 1. Fermentation 2. Produce enzymes 3. Produce bioactives 4. Produce bio-diesel 5. Pollution control 6. Nano-materials

232 Biology Term 3 STPM Chapter 19 Biotechnology 19 Red biotechnology (medicals) 2016 1. Producing human therapeutic proteins including hormones like insulin and enzymes from genetic engineering. 2. Searching for new medicines and drugs as in pharmacogenomics in which different individuals with the same disease often respond differently to a drug treatment because of differences in gene expression. 3. Understanding oncogenes i.e. genes which produce proteins that may function as transcription factors and receptors for hormones and growth factors, as well as serve as enzymes involved in a wide variety of ways to change growth cell properties that cause cancer including tumour suppressor genes that regulate oncogenes. 4. Concocting a personalised medicine for each individual with special gene like BRCA1 or 2 that increases risk of developing breast cancer. 5. Improving techniques for drug delivery and understanding factors that influence drug effectiveness such as its solubility, breakdown, and elimination and inventing microspheres i.e. tiny particles that can be filled with drugs for delivery. 6. Developing nanotechnology and nano-medicine with nanosensors that can monitor blood pressure, hormone concentrations, unblock arteries, detect and eliminate cancer cells. 7. Developing artificial blood such as Hemopure that is made from the haemoglobin of cattle. 8. Developing vaccines that may be useful against conditions such as Alzheimer’s disease or drug addiction. 9. Using antibodies especially monoclonal antibodies in some types of therapies e.g. cancer therapy. 10. Developing gene therapy with viral and non-viral vectors for gene delivery for treating cystic fibrosis and other genetic disorders. The latest gene therapy is using transgenic T-lymphocyte to kill cancer cells. 11. Culturing cells and tissues such as foetal neurone that can be used to replace or repair defective tissues and organs including Parkinson's and other serious diseases. 12. Using xeno-transplantation i.e. transfer between species such as ‘knockout’ pig that lacks a gene called GGTA1 to human and researching more on histocompatibility complex genes to prevent rejection of transplants. 13. Developing cellular therapeutics using encapsulated cells (‘biocapsules’) to deliver important biological molecules e.g. for Type 1 diabetes therapy. Summary Red biotechnology 1. Human therapeutic proteins 2. Pharmacogenomics 3. Oncogenes 4. Personalise medicine 5. Improve drug delivery 6. Nanotechnology 7. Artificial blood 8. Vaccines 9. Monoclonal antibodies 10. Gene therapy 11. Culture cells 12. Xeno-transplantation 13. Cellular therapeutics 14. Tissue engineering 15. Stem cell technologies

233 Biology Term 3 STPM Chapter 19 Biotechnology 19 14. Developing tissue engineering technology for skin grafts and replacement of tissues and organs by growing them in culture. 15. Developing stem cell technologies (Figure 19.1) from both embryo and adult for therapeutic purposes. Summary Blue biotechnology 1. Aquaculture 2. Extract new source of food 3. Enchance seafood safety 4. Drug discovery 5. Nano-technology 6. Cosmetic discovery 7. Bio-processing 8. Bio-polymers 9. Bio-fuels 10. Agriculture 11. Expand biological and geochemical knowledge Egg Sperm Fertilisation Zygote Cell division Blastocyst Liver cells Macrophages Adipocytes Pancreatic Blood cells cells Smooth Neurons muscle cells Other cell types Isolate inner cell mass Differentiation into specific cell and tissue types Grow cells from inner cell mass in culture Cultured human embryonic stem cells Days 5–7 Trophoblast Inner cell mass: source of embryonic stem cells Figure 19.1 Understanding stem cell technology Blue biotechnology (marine and aquatic applications) 1. Increasing the food supply through aquaculture of conventional fishery resources and potential of non-conventional aquatic resources especially planktons and bacteria. 2. Discovering and extracting new sources of food supplements such as essential polyunsaturated fatty acids, amino acids, vitamins, enzymes, cofactors and minerals. 3. Enhancing seafood safety and quality using marine biological system. 4. Discovering and producing of drugs to cure diseases especially cardiovascular diseases; and drugs that have antiviral and anticancer functions. 5. Discovering biological system especially nano-technology using marine microbes to diagnose diseases, improve and monitor human health. 6. Discovering and producing of cosmetics to maintain and to rejuvenate human skin.

234 Biology Term 3 STPM Chapter 19 Biotechnology 19 7. Discovering bio-processing system and new enzymes for industrial process. 8. Discovering new chemicals or bio-polymers for industrial and daily usage. 9. Discovering new source of bio-fuels and bio-processing of new bio-fuels. 10. Developing new types and resources of industrial materials. 11. Discovering new chemicals for agricultural and animal farming industries. 12. Restoring and protecting both marine ecosystems and land environments especially using marine microbial remediation. 13. Expand knowledge of biological and geochemical processes in the world ocean. Agricultural biotechnology Healthcare biotechnology Industrial biotechnology Traditional complementary medicines New agriculture Value-added products Vaccines Bio materials Diagnostics and biomedical instruments Enzymes Green chemistry Biofuel Microbe technologies Clinical trials Contract research organisation Tropical diseases Food supply High-yield crops Nutraceutical Natural substances Bioinformatics Contrast manufacturing CMO Figure 19.2 Traditional roles of biotechnology Microorganisms Process design Plants Proteins Arsenic Hydroponics Activated sludge Wastewater treatment Drinking water purification Biofilm studies Figure 19.3 More modern roles of microorganisms

235 Biology Term 3 STPM Chapter 19 Biotechnology 19 19.2 Applications of Biotechnology Application of Biotechnology in Food and Beverage Production Fermentation 1. Fermentation  in  food production is  typically the conversion of carbohydrates to alcohols and carbon dioxide or organic acids by the action of micro-organisms for the production of different forms of food and food supplements. 2. Fermentation of grain flour or dough is used to make bread. In different countries, different types of dough are baked in different forms and tastes. Grains are also fermented into different alcoholic beverages including rice wine, beer from barley and hops, and whisky from wheat. Cooked grains can be made into vinegar if the content of fermentation is stirred to allow aerobic respiration. 3. Fermentation of milk from different animals is used to make different forms of cheese, yogurt, cultured and traditional fermented milk. The fermentation of cheese takes several weeks when different types of bacteria added to produce lactic acid to coagulate the milk protein. The cheese can be of different textures, colours and tastes. Yogurt is fermented milk with bacteria to produce lactic acid and other organic acid for producing different tastes. Traditional fermented milk depends on which animal the milk comes from and the number of days of fermentation. 4. Fermentation of fruits typically produces different types of wine including the red and white wine from grapes, apple made into cider and pear made into perry. Besides that, fruits can be fermented into vinegar as well. 5. Fermentation of beans especially soy bean produces soy sauce and soybean paste. Cooked soybean left in the sun and stirred periodically for a month produces soy sauce. Other bean-based products include black soybean, stinky tofu, Japanese sticky bean and Beijing mung bean milk. Other types of seeds include cocoa and coffee fermented for beverages. 6. Fermentation of tea leaves has produced different types of tea of different aroma from different countries. Examples are the Yunnan fully fermented tea, i.e. tea leaves added with water and made into hard solid blocks, red teas fermented for about a week including the common Ceylonese tea. 7. Fermentation of different types of vegetables is made into processed vegetables. E.g. Korean kim chi and salted kale. Different countries also produce different types of pickles. Learning Outcomes Students should be able to: (a) describe the application of biotechnology in food and beverages production (fermentation and vitamin-enriched eggs); (b) describe the application of biotechnology in agriculture (hybrid rice, herbicide resistant plants and transgenic fish); (c) describe the application of biotechnology in medicine (human growth hormone, human insulin and gene therapy) and forensic (DNA finger printing); (d) describe the application of biotechnology in public health (genetic screening, diagnostic kits and oildecomposing bacteria) Summary 1. Fermentation: (a) Grains – bread, alcoholic beverages, vinegar (b) Milk – cheese, yogurt, cultured milk (c) Fruits – wine, cider, perry, vinegar (d) Beans – soy sauce, tofu, sticky bean, cocoa, coffee (e) Tea leaves – various kind of teas (f) vegetables – kimchi, pickles (g) Honey – mead (h) Meat – fish sauce, shrimp paste, salami, pepperoni 2. Vitamin-enriched eggs: (a) rich in Omega-3 fatty acids, vitamin E, ostensibly

236 Biology Term 3 STPM Chapter 19 Biotechnology 19 8. Besides, honey can be fermented into mead. Fish can be fermented into fish sauce, shrimp can be fermented into shrimp paste and meat can be fermented into salami and pepperoni. 9. Fermentation of food is the controlled action of microorganisms to alter the texture of food, to preserve the food by the acids and alcohols produced, and to produce the characteristic flavours and aromas. Changes produced by fermentation is shown in Table 19.1. Table 19.1 Changes by fermentation in food Change Description Texture Food is softened as the result of complex changes in proteins and carbohydrates Nutritional value Microorganisms improve digestibility by hydrolisis of polymeric compounds, mainly polysaccharides and proteins; secrete e.g. vitamins Enrichment with Protein, essential amino acids, essential fatty acids Flavour Sugars are fermented to acids, which reduce sweetness and increase acidity, in some cases bitterness is reduced by enzymatic activity Aroma The production of volatile compounds: amines, fatty acids, aldehydes, esters and ketones Colour Proteolytic activity, degradation of chlorophyll and enzymatic browning may produce brown pigments 10. The application of biotechnology involves the selection of microorganism types. Brewer's yeast and Baker’s yeast are the same species of Saccharomyces cerevisiae but they are of different strains. The alcohol content, substrate preparation, fermentation environment and final fermentation products are different. Vitamin-enriched eggs 1. Advances in biotechnology on nutrition of chickens and other poultry enable farms to produce eggs with vitamin-enriched eggs. Though designer eggs are produced with enriched vitamins, the aim is to produce eggs with vitamin content that can at least satisfy the 50% daily requirement index (DRI). 2. It is possible to increase the vitamin A, D, E, K and assorted B complexes content by increasing the vitamin content of the poultry feed 3 to 10 times more supplemental vitamins. 3. However, the vitamin-enriched eggs usually cannot achieve the 50% DRI on any of the vitamin enriched by taking one egg per day. 4. Eggs are enriched in omega-3 fatty acids, which benefit human nutrition. Most of these eggs are also enriched in vitamin E, ostensibly as an additional antioxidant to protect the long-chain polyunsaturated fatty acid.

237 Biology Term 3 STPM Chapter 19 Biotechnology 19 Application of Biotechnology in Agriculture Hybrid rice 1. A hybrid rice variety, also referred to as the F1 , is the direct product of crossing two genetically different varieties of parents. In hybrids, the positive qualities of both parents are combined resulting in a phenomenon called “hybrid vigour” or “heterosis”, where young seedlings grow very fast and the mature plant has better reproductive characteristics. These factors result in higher yields than ordinary rice, also called inbreds. 2. The hybrid, on the other hand, is a product of two parents. Hence, new seeds must always be produced for cross-pollination for the following season. Seeds harvested from the hybrid plant cannot be used for replanting because hybrid vigour is lost resulting in lower yield and non-uniform plant height. These seeds are of variable sizes, may be discoloured, partially filled, may have slightly opened hull, generally lighter in weight and much lower yield even compared with inbreds. 3. An example is a new hybrid of “Super rice”, IR8. This kind of shorter rice has a stronger stem and can be plant closer. Its average yield is 3 times higher than normal rice. 4. Traditional varieties are all inbred rice. Normally, each rice flower contains both male and female organs. This allows the plant to perpetuate or reproduce itself through self-pollination or inbreeding, hence the term inbred. This means the grains harvested from inbred rice can be used as seeds for the next planting season. 5. Traditional varieties are tall and lodge easily. These are lowyielding and can be grown only once a year since they take 5 – 6 months to mature. Most are photoperiodic as flowering is affected by day length. They have low tiller number, are often susceptible to insect pests and diseases, and the grains may shatter easily. Many, however, are aromatic and have good taste or eating quality.   6. High-yielding varieties (modern rice varieties) are short or semi-dwarf; mature in about four months or less; and are non-photoperiodic. Thus a second or third crop is possible; high yielding; resistant to insect pests and diseases; responsive to nitrogenous fertiliser; heavy-tillers; and well-suited to irrigated areas. 7. Super hybrid rice is a variety designed for the future. Its target yield of 12 tons/ha will be achieved through a dramatic transformation of the rice plant architecture; less tillers; more grains per panicle; heavier grain weight per panicle; longer and extensive root system; more efficient use of water and nutrients; and greater resistance to insect pests and diseases. However, this is still being developed and improved. Figure 19.4 Wild rice to first generation hybrid rice, and finally super hybrid rice VIDEO Hybrid Rice

238 Biology Term 3 STPM Chapter 19 Biotechnology 19 8. There are three hybrid rice varieties approved by the National Seed Industry Council in Philippines which are the Magat, Mestizo and Panay rice; one released nationally, and two approved for location-specific release. Twenty-one promising hybrids are now under evaluation in the National Cooperative Tests. 9. The significant shifts from farmers’ usual practices include: (a) use of new seeds every planting season, that is, they cannot use their harvest as seeds; (b) use of only 20 kg seeds per hectare (farmers use from 80 to 120 kg of seeds per hectare); (c) application of farm wastes and organic materials in the seedbed to make the soil loose and friable. This is to facilitate seedling pulling while keeping roots of seedlings intact; (d) sparse seeding of 50 g/m2  or a total of 20 kg in 400 m2  seedbed; and (e) transplanting of 1-2 seedlings per group instead of 3-5 seedlings. 10. Besides selective genetic breeding to improve hybrid rice, the future of the hybrid rice may incorporate gene modifications to further improve the plants both quantitatively and qualitatively. The qualitative characters may include colour, texture, aroma, vitamin, mineral and roughage content. Herbicide resistant plants 1. Herbicide-resistant crops were first produced by methods of traditional breeding. 2017 2. They have been grown commercially since 1984, when the first triazine-resistant (resistant to triazine herbicide) oilseed rape was introduced in Canadia. This resistant oilseed rape, developed by methods of traditional breeding from Brassica rapa L. had been backcrossed into a commercial variety of oilseed rape. 3. There are two common herbicides that herbicide resistant crops are resistant to. These are glyphosate and glufosinate for soybean, maize, rapeseed, and cotton. 4. Crops have been genetically modified to be  resistant to nonselective herbicides. These transgenic crops contain genes that enable them to degrade the active ingredient in the herbicide, rendering it harmless.  Usually, Agrobacterium with T1 plasmid is used. T1 plasmids are commercially produced for genetic engineering. The plasmids can be spliced with foreign gene and enter Agrobacterium. The transformed bacteria can transfer the T1 plasmid into cultured crop cells. The cells are stimulated to form tissue and platlets to form stock of transgenic crops. The proceduce is like gene cloning with the help of antibiotic for screening. Info Bio China grows hybrid rice in 17 million ha out of its total 33 million ha rice area. India grows hybrid rice in 150,000 ha while Vietnam cultivates 400,000 ha to hybrid rice. Hybrid rice technology helped China increase its rice production from 140 million tons in 1978 to 188 million tons in 1990. It accounts for 66% of China’s total production and 20 percent of the world’s total rice supply. Furthermore, hybrid rice technology has saved China more than 2 millions hectares of agricultural lands now being used of other purposes while preserving the environment. At least 17 other countries have initiated hybrid rice programs including Malaysia. Figure 19.5 Development of crop variety × Variety B Resistant to diseases but low yielding Variety A susceptible to diseases but high yielding F1 hybrid F1 population Pedigree nurseries F3 –F7 Observational nursery Replicated yield trials New variety Select for resistance to diseases and high yield potential Screen for diseases t t t t t t

239 Biology Term 3 STPM Chapter 19 Biotechnology 19 5. The use of herbicide resistant crops enables the use of ‘nonselective’ herbicides to remove all weeds in a single, quick application. This means less spraying, less traffic on the field, and lower operating costs. 6. Herbicide resistant crops also facilitate low or no tillage cultural practices which are considered to be more sustainable as low disturbances to the soil and to reduce soil erosion. Farmers can manage weeds without turning to some of the more environmentally suspect types of herbicides like ‘broad-spectrum’. 7. Another example is HR-rice varieties, which are becoming commercially available to improve control of the weed flora associated with this crop, especially of red rice and other weedy rice species. The second was to provide an alternative tool for weed management that have already evolved resistance to particular herbicides, especially grasses such as Echinochloa spp. Herbicide resistant rice, furthermore, allows for the substitution of some of the currently used herbicides by others less detrimental to the environment. 8. Critics claim that in some cases, the use of herbicide resistant crops  can  lead to an increase  in herbicide use, promote  the development of herbicide resistant weeds, and  damage biodiversity on the farm. Extensive ecological impact assessments have been addressing these issues. Studies in the United Kingdom have shown that different herbicides and different  herbicide application practices can affect the amount of wild plants on the farm. In comparison with conventional cropping systems, weed and animal populations were negatively affected by herbicide tolerant sugar beet and rapeseed, but biodiversity was increased with the use of herbicide tolerant maize. 9. When a specific herbicide is sprayed at high rates it imposes a high selection pressure on the weed flora. In few years this may cause shifts in weed composition towards species that naturally tolerate the herbicide and other herbicides may be needed to control these weeds that are herbicide tolerant. Furthermore, it may become difficult to control non-herbicide resistant crops in subsequent years. 10. Gene-flow from crops to other crops or related species is another route to the development of resistant weed populations in the field. Once the resistance gene is present in non-herbicide resistant crop or related weed species then it is expected that the same weed control practices by consistent sprayings with herbicides having the same mode of action will lead to a rapid build-up of herbicide resistant weeds and other crops. Info Bio Glyphosate is a broadspectrum herbicide with no residual effect, and it is rapidly degraded in the soil. Although, glyphosate can eventually enter the water table, leading to health risks due to toxicity. Info Bio ‘Broad-spectrum’, or nonselective herbicides are effective at killing a wide range of weeds. The problem is they can also kill valuable crops. Therefore, broad-spectrum herbicides are only useful before seedlings emerge or in special cases like fruit orchards, vineyards, and tree nurseries. Info Bio Glyphosate-resistant soybean 1. It simplifies weed control to the use of a single herbicide and with a more flexible timing than that required for conventional herbicides. 2. The number of herbicide applications in soybeans is estimated to have dropped by 12% from 1995-1999. 3. The American Soybean Association states that glyphosate-resistant soybean protects the environment through changes in tillage practices and herbicide application, and by improved weed control. 4. Farmers are producing cleaner crops containing fewer non-grain materials contaminated from weeds. 5. Argentina adopted the highest rate of soybean, reaching more than 90% in seven seasons. Cost reductions at about US$20 per hectar was estimated as a result from more effective weed management.

240 Biology Term 3 STPM Chapter 19 Biotechnology 19 Transgenic fish 1. Transgenic fish are genetically modified fish in which a foreign gene is transferred into them using genetic engineering techniques. The aim is to introduce a new trait to the fish which do not occur naturally in the species.  2. Some transgenic fish have promoters driving an over-production of “all fish”  growth hormone. This resulted in dramatic growth enhancement in several species, including salmonids, carps and tilapia. For example, promoter had been extracted from ocean pout fish and growth hormone gene extracted from Chinook salmon to transfer into the eggs of north Atlantic salmon. This produced a dramatic increase in size of the transgenic north Atlantic salmon as shown in Figure 19.6. Promoter gene Ocean trout fish Growth hormone gene Chinook salmon Salmon Eggs Fry 1 2 7 4 6 5 3 Transgenic North Atlantic Salmon Figure 19.5 Gene modification of salmon. 3. Transgenic fish  have been developed as potential food by increasing the speed of development and potentially reduce fishing pressure on wild stocks. This has resulted in dramatic growth enhancement in several species, including salmon, trout and tilapia. 4. Salmon belong to the  Salmonidae  family which also includes salmon and trout. Although the smallest species is just 13 centimetres long as an adult, most are much larger, and the largest can reach up to 2 metres. A genetically modified Atlantic salmon (AquAdvantage salmon) grows faster and is bigger than the wild-type Atlantic salmon from which it was derived.  5. Tilapia is the  common name  for several species of cichlid fish from the tilapine cichlid tribe. Tilapia is one of the most important fish in Malaysian fish farming. This is because of their large size, rapid growth, and palatability. Tilapias are very vulnerable to cold temperatures, and thus survive well with temperatures above 16 °C. Thus, attempts to produce transgenic tilapias that can grow well in cold water. Info Bio AquaBounty, the leading company in transgenic fish for the food industry, claims that their transgenic AquAdvantage salmon can mature in half the time it takes non-transgenic salmon and achieves twice the size. AquaBounty has applied for regulatory approval to market their GM salmon in the US but the application is still pending. INFO Transgenic Fish

241 Biology Term 3 STPM Chapter 19 Biotechnology 19 6. Zebra fish (Figure 19.6) are one of the most commonly modified aquarium fish as they have optically clear chorions (shells), rapidly develop, and the 1-cell embryo is easy to see and microinject with transgenic DNA.  They have the capability of regenerating their organ tissues, and are used to understand heart tissue repair and regeneration in efforts to study and discover cures for cardiovascular diseases. 7. Transgenic zebra fish were created to detect aquatic pollution. The laboratory that originated the GloFish had originally developed them to change colour in the presence of pollutants, to be used as environmental sensors. 8. The regulation of genetic engineering and transgenic fish concerns the approaches taken by governments to assess and manage the risks associated with the development and release of genetically modified crops. There are differences in the regulation of genetically modified organisms between countries, with some of the most marked differences occurring between the USA and Europe. For example, a fish not intended for food use is generally not reviewed by authorities responsible for food safety. 9. Critics have objected to use of genetic engineering on several grounds, including ethical concerns, ecological concerns especially gene flow, and economic concerns raised by the fact that genetically modification techniques and genetically modified organisms are subject to intellectual property law. 10. There are controversies over using transgenic fish as food is safe, whether it would exacerbate or cause fish allergies, whether it should be labelled, and whether transgenic fish and organisms are needed to address the world’s food needs. These controversies have led to litigation, international trade disputes, and protests, and to restrictive regulation of commercial products in most countries. 11. Table 19.2 shows the risks and benefits of transgenic fish. Table 19.2 The risks and benefits of transgenic fish Risks Benefits • Transgenic fish are deleterious to health of consumer • Transgenic fish show no advantage over nontransgenics • Transgenic fish, on release or escape, become undesirable alien species • Transgenic fish, on release or escape, interbreed with wild individuals resulting in permanent presence of transgene in wild populations • Transgenic fish, on release or escape, die out quickly but adversely affect ecological balance • Transgenic fish lead to lower-cost, faster growing strains available for aquaculture • Transgenic fish lead to availibility of disease resistant strains • Transgenic fish bring strains of fish with other desirable traits-cold tolerance, freeze resistance or salinity tolerance • Reversibly sterile strains of transgenic fish become available allowing sate use in aquaculture • Transgenic fish offer fish farmers possible exploitation of cheaper and 'greener' diets • Transgenic fish are used for production of valuable pharmaeutical proteins Figure 19.6 Normal zebra fish Info Bio Zebra fish are popular aquarium fish, commonly sold as Zebra danio, and have been very vital as model organisms in research. They derive their name from the uniform horizontal stripes along the side of the body bilaterally. Males bear gold stripes within the blue stripes, while females bear silver stripes within the blue stripes. Zebra fish can mature up to 6.4 cm in the wild, but usually it is rare for them to mature beyond 6 cm in captivity. Info Bio The GloFish is a patented brand of genetically modified fluorescent zebra fish with bright red, green, and orange fluorescent colour. Although not originally developed for the ornamental fish trade, it became the first genetically modified animal to become publicly available as a pet when it was introduced for sale in 2003. They were quickly banned for sale in California.