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43. (a) XO type, grasshopper 46. Three alleles are involved which control human (b) ZW type, birds ABO blood group, IA, IB and i. 44. Sex of a child depends on the fusion of the type Mother Father of sperm (‘X’ or ‘Y’ type) with the egg. ii O Group IAIB AB Group Father XY XX Mother Gametes i IA IB Sperms Eggs Progeny IAi IBi XY X X Gametes A group B group Possible blood groups – A Possible blood groups – B Alleles – IA, IB, i XX XX XY XY Offspring 47. Geneticist carries out test cross to know unknown genotype of the dominant trait and weather it Female Male is homozygous or heterozygous. It is done by 45. Male fruit fly and female fowl are heterogametic crossing the unknown genotype with homozygous because they produce two different types of recessive trait. If the progeny is 50% dominant gametes whereas the female fruit fly and male and 50% recessive, then the unknown genotype fowl are homogametic because they produce same is heterozygous, and if all carries dominant trait, type of gametes. the unknown genotype is homozygous. Homozygous Homozygous recessive recessive WW ww ww ? ww WW Ww Ww WW Ww Ww Ww W Dominant Phenotype w Ww Ww (Genotype unknown) ww ww Result All flowers are violet Half of the flowers are violet and half of the flowers are white Interpretation Unknown flower is homozyous dominant Unknown flower is heterozygous 48. Gene responsible for multiple phenotypic IA and i are present only IA expresses and when expression is gene for phenylketonuria. IB and i are present only IB expresses. 51. (a) It is not known as blending because there is Such genes are called Pleiotropic genes. Single gene mutation is the cause for such an effect. no mixing of alleles, both the alleles remain as discrete units which express themselves in 49. Drosophila melanogaster was selected because F2 generation. (i) it can be grown in simple synthetic medium (b) This phenomenon is called incomplete dominance. in the laboratory, 52. Pleiotropism is defined as a phenomenon when (ii) have smaller life cycle just of two weeks, a single gene may produce more than one effect (the multiple effects of a gene) or control several ( iii) large number of progeny is obtained and phenotypes depending on its position. This is due to the interrelationship among various metabolic (iv) sexes are distinguishable. pathways that may contribute towards different 50. ABO blood groups are controlled by the gene phenotypes. I. The plasma membrane of the red blood Example: In phenylketonuria, mutation is in cells has sugar polymers that protrude from the gene which codes the enzyme phenyl-alanine- its surface and the kind of sugar is controlled hydroxylase. The multiple effects of this mutation by the gene. The gene (I) has three alleles result in mental retardation, reduction in hair IA, IB and i. The alleles IA and IB produce a and skin pigmentation. slightly different form of the sugar while allele i doesn’t produce any sugar. IA and IB are completely dominant over i, in other words when 100 Biology–12

53. (i) Case I are having Tt genotype, only one is dwarf with genotype tt. So the genotype ratio is 1:2:1 56. If both the parents are heterozygous for the tall character, then some of the plants will be dwarf in F2 generation as given in the cross below. Tt Tt Parents T t T t Gametes Tt tt Selfing t t Tt Tall Tall + T T TT Tt Tall Tall Tt Tt t Tt tt Small Small Tall Dwarf t 57. Refer answer no. 50. tt tt 58. 50% of the offspring will be heterozygous. Refer Heterozygous answer no. 56. 59. This can be explained in the cross given in answer (ii) The type of cross carried out here is a test cross. It is a cross in which an individual with no. 56. an unknown dominant phenotype is crossed 60. (a) Progeny have characters of both the parents with an individual (parent) homozygous recessive for that trait. which are expressed equally (b) Law of dominance 54. Refer answer no. 40. (c) Progeny will not resemble any of the parent, 55. A monohybrid cross for tall and dwarf characters but will show intermediate feature of both the is given below. parents. 61. (a) The genotype A and C = Tt, B = TT and D = tt. TT tt Parents (b) Phenotype of A, B and C is tall and of D is dwarf. Tall dwarf (c) Phenotypic ratio of progeny is 3 : 1 (d) Genotypic ratio of progeny is 1 : 2 : 1 62. Parents IAi IBi + IA i IB i Gametes TT tt Gametes IA i F1 Generation IB IA IB IBi B blood group Tt F1 Generation All Tall ABgBrolouopd i IA i ii Selfing A Blood O Blood Tt + Group Group T TT Tt 63. Drosophila melanogaster Morgan preferred fruit Tall Tall flies for his experiments because they are easy to handle and culture in lab, life cycle is short, male t Tt tt and female are easily distinguishable. Tall dwarf 64. When two or more non-allelic genes are present Phenotype ratio = Tall : Dwarf more or less closely on the same chromosome 3:1 and their tendency to be inherited together in gametes is called linkage whereas in crossing Genotype ratio = TT : Tt : tt over two homologous chromosomes pair up and 1:2 :1 exchange of genetic material occurs. If two genes are closely located, the crossing over will be low As shown in the cross, there are 3 tall plants and and if two genes are distantly located, crossing one dwarf plant so the phenotypic ratio is 3:1 but over will be more. among tall, one is having TT genotype and two In Drosophila, a yellow-bodied white- eyed female when crossed with brown- bodied red- Principles of Inheritance and Variation  101

sehyeodwmMaelne,dtehlieapnr9og:3e:n3y:1orfaFt2iog. eBnoetrhatyieolnlowwilbl ondoyt • ABO blood group as an example of multiple allelism: The ABO blood grouping is controlled by three alleles, IA, IB and i, hence produces 6 white eye and brown body red eye remain in the same combination as of parents even in aFre2 genotype and 4 phenotype. generation although very few recombinants 72. (a) The ABO blood group is controlled by formed due to crossing over. three alleles, IA, IB and ‘i’ and thus show 65. Strength of linkage between white eye and yellow multiple allelism. In case of presence body is higher; this combination is passed as such of IAIB alleles, the blood group will be to next generations. Recombinants obtained are AB, here both IA and IB will express. very few which indicated that no crossing over occurred between these two genes. This confirms co-dominance. 66. Haemophilia is sex linked recessive trait. Females (b) (i) IA i – A blood group have two X-chromosomes and she will suffer from (ii) i i – O blood group haemophilia if both the X chromosomes carry 73. (a) Law of segregation states that the paired alleles segregate during gamete formation haemophilic trait. and each gamete receives only one of the two 67. The given pedigree represent the criss-cross inheritance. Here the parents pass their trait to alleles. cross TT × tt the grand child of the same sex through offspring of the opposite sex. Here father has been shown to pass the traits to grandson through his daughter. gamete T T tt 68. GAA also codes for glutamic acid so there will (b) Phenotypic ratio of a dihybrid cross is 9:3:3:1 Law proposed on basis of dihybrid cross is law of be no change in the structure of haemoglobin in independent assortment. It states that when two the case of mutation A. In mutation B, the amino acid is changed from glutamic acid to valine thus pairs of alleles for different traits are combined changing the shape of haemoglobin molecule. in a hybrid, segregation of one pair of alleles is 69. Refer answer no. 63 independent of other pair of alleles. 3 Marks Questions In law of segregation two alleles of same trait segregate, while in law of independent assortment 70. The test cross is used to find the genotype of any individual and know whether its dominant two genes pairs of different traits segregate and character will be heterozygous or homozygous. are not dependent on each other for segregation. In this to investigate the trait it is crossed with 74. (a) This mechanism can be explain by considering the law of incomplete dominance. For example: a homozygous recessive parent. There could be two possible cross which are P generation mentioned below: Red (RR) Cross 1 Cross 2 White (rr) Parent with Parent with Gametes R r unknown genotype unknown genotype TTww × ttww Ttww × ttww F1 generation Tall plants Homozygous Tall plants Homozygous with white recessive with white recessive flowers flowers Tw Tw Tw tw All pink (Rr) tw Ttww Ttww tw Ttww ttww Gametes R R Gametes tw Ttww Ttww tw Ttww ttww All tall plants with 50% plants tall with rr RR white flowers white flowers 50% plants dwarf with Rr Rr white flowers F2 generation 71. ABO blood group as an example of dominance: In rr case of the presence of IAi or IBi alleles, the blood group will be A and B respectively, this shows Phenotypic ratio : red : pink : white that allele IA and IB are dominant over i 1: 2: 1 • ABO blood group as an example of co- Genotypic ratio : RR : Rr : rr dominance: In case of presence of IAIB alleles, 1 : 2 :1 the blood group will be AB, here both IA and (b) Polygenic inheritance in exhibited by skin colour of human which is controlled IB will express. This confirms co-dominance. by three genes where dominant alleles 102 Biology–12

have cumulative effect. These dominant transmitted from parents to the offspring. When alleles expresses a part or unit of the trait. Such genes are called polygenes and their both the parents are carrier for the gene or are inheritance is called polygenic inheritance. There genes are A, B, C. heterozygous. If the genotype is AABBCC, the skin colour 79. Colour blindness is a X-linked recessive disorder will darkest and if genotype is aabbcc, the which shows transmission from carrier female skin colour will be lightest. to male progeny and hence, usually males are 75. (a) (i) Phenylketoneuria (ii) Down’s syndrome affected and females remain carriers. If the given (iii) Klinefelter syndrome couple is normal with a colour blind child, their (b) (i) Mental retardation genotypes can be shown in the following cross. (ii) short stature, small head XXC x XY (iii) Gynaecomastia XX XX XXc XcY 76. (a) Possible genotypes of members 4 – X Xh, normal (carrier) (affected) 5 – XhY, 6 – XY; where X is normal and Xh is haemophilic chromosome. T he affected child is male, with colour blindness. (b) 14 15 OR X Y Normal Carrier Xh X The child is male who is colourblind which is a female male X-linked recessive disorder. XXC × XY XhX XhY XX XY XhY haemophilic male child There are 25% chance of male to be haemophilic. XX XX XXC XCY Progeny 77. Haemophilic gene is recessive over normal gene Normale 1 2 3 Female and is sex linked. It is possible only when a Carrier carrier woman marries a hemophilic man. 80. (a) This representation of HbA peptide indicates a normal human, because the glutamic acid in XhX X XXh the sixth position is not substituted by Valine. Father is Mother is Haemophilic carrier (b) The sufferer’s RBCs become elongated and sickle shaped as compared to the normal XhXh affected female biconcave RBCs. XhXh, is homozygous recessive thus expresses (c) Both males and females are likely to suffer from the disease equally, as this is an itself in female autosomal linked recessive trait. 78. Example of an autosomal recessive trait is sickle cell anaemia. In sickle cell anemia in humans the 81. cHhBrBomgoesnoemoefs11cathrrayntdhHe BmAu1taanndt gHeBnAe 2coafu1s6inthg erythrocytes become sickle shaped under oxygen thalassaemia. These mutant genes cause the deficiency. This occurs due to formation of formation of abnormal haemoglobin molecules abnormal haemoglobin-S (Hbs). When two sickle resulting into anaemia. celled heterozygotes marry they may give birth to three types of children–homozygous normal, These mutant gene cause anaemia, jaundice, heterozygous carrier and homozygous sickle hepatoseplenomegaly and bone changes. All the celled in the ratio of 1 : 2 : 1. This can be shown defective alleles kill the feetus resulting in still by the given cross. Sickle cell anaemia can be birth or death soon after delivery. It may also HbA HbS × HbA HbS results in severe haemolytic anaemia and cardiac enlargement. HbAHbA HbAHbS HbAHbS HbSHbS 82. (a) This is an autosomal recessive disease, which is bcIfoencHotrbmosleliessdiicnbkylhesoisnmhgoalzepypegadoi.urTsohfciaoslnlcedhlieatiHnognbe,AtianhnedshRHaBbpCse. (Normal) (Carrier) (Affected) is caused by single base pair substitution Principles of Inheritance and Variation  103

mutation in b-chain of haemoglobin, in which 89. (a) Homozygous dominant: YYRR, Homozygous glutamic acid is substituted by valine. The recessive: yyrr mutant haemoglobin molecule undergoes polymerisation under low oxygen tension (b) Gametes produced by homozygous dominant: causing the change in the shape of the RBC YR and by homozygous recessive: yr from biconcave disc to elongated sickle like structure. (c) YyRr (b) Father-HbA HbS, Mother HbA HbS (d) Gametes produced by F1 offspring: YR, Yr, 83. (a) The trait is recessive yR, yr (b) The trait is autosomal 90. (a) In grasshopper, males have only one X sex (c) Genotypes of parents of I Generation: chromosome and its homologous pair is Aa and Aa absent, and in Drosophila the males have one X and one Y sex chromosome, upon Genotype of third child of II Generation: aa segregation, the gametes formed are of two different types. So it is male heterogamety. Genotype of fourth child of II Generation: Aa (b) In birds, the sex is determined by the female. 84. Refer answer no. 50. Females have one Z and one W sex chromosome. 85. Refer answer no. 73 (a) When gametes are formed, half of them are of 86. IA and IB are completely dominant over i, in other Z type and half are of W type. So they are said to show female heterogamety. Bird males have words when IA and i are present only IA expresses all same type of gametes, Z type only. and when IB and i are present only IB expresses. But when IA and IB are present together they 91. Sex determination in humans is of XY type. both express their own types of sugars: this is Out of 23 pairs of chromosomes present, 22 are because of co-dominance. Hence red blood cells autosomes. A pair of X chromosomes is present have both A and B types of sugars. Since there in females, whereas one X and one Y chromosome are three different alleles, there are six different are present in males. During gamete formation in combinations of these three alleles that are males two types of gametes are produced half of possible, a total of six different genotypes of the the total carry X chromosome and the rest carry human ABO blood types. The below mentioned Y chromosome. When the egg gets fertilized with table shows the genetic basis of blood groups in the sperm carrying X chromosome the zygote human population. develops into female (XX) and with the sperm carrying Y chromosome the zygote develops into Allele Allele Genotype Blood male offspring (XY). from Parent 1 from of types of XX X XY Parents IA Parent 2 offspring offspring XX X Y Gametes IA IA IA IA IA A IB IB IB IA IB AB XX XY XX XY Gametes IB i IA i A In birds, female birds have Z and W chromosomes, whereas males have a pair of Z chromosomes IA IA IB AB besides autosomes. IB IB IB B 92. Haemophilia is a sex-linked recessive trait. Male members become haemophilic even if i IB i B i i ii O x-chromosome (male have only one x-chromosome) carries defective gene and for female members to 87. Since in an individual only two alleles can be haemopholic, both the x-chromosomes should be present, multiple alleles can be found be carrying defective gene, which is of course a only when population studies are made. For rare occurence. example Refer answer no. 85. (b) Non stop bleeding without formation of clot, due 88. Mendel’s work remained unrecognised till 1900 to missing blood clotting factor proteins. because of: 93. (a) (i) His mathematical method to explain biological phenomenon was not acceptable. h (ii) His concept of gene and blending of alleles Genotype of daughter – X X with each other was not acceptable in the Genotype of daughter's husband – XY light of continuous variations occurring in nature. He could not give proof of factors. ( iii) He published his work in a journal which did not have wide circulation. 104 Biology–12

(b) Haemophilia is a sex linked recessive disorder. 102. (a) The conclusions Mendel arrived at on It is expressed if both the X chromosomes carry dominance of traits on the basis of monohybrid the defected gene which causes haemophilia cross are: in females. The inheritance pattern shows that mother was a carrier and the trait was (i) Characters are controlled by discrete expressed in one of her son as males have only units called factors. one X chromosome. A haemophilic son is born to the normal daughter when she marries to a (ii) Factors occur in pairs normal male, this suggests that the daughter (iii) In dissimilar pair of factors, one member is also a carrier of the trait. of the pair dominates the other one. The 94. The pedigree analysis is done in human genetics one who dominates is dominant factor because: and the other is recessive. (b) The recessive allele is unable to express itself • Controlled crosses are not possible in case of due to- humans beings (i) It may produce less efficient enzyme. (ii) It may produce non-functional enzyme. • Traits can be analysed for several generations (iii) It may not produce any enzyme at all. 103. Refer answer no. 73 (b) • Can trace pattern of inheritance so can help 104. In pea plants the colour of flower is controlled by genetic counselors to guide the couples about a single gene with two alleles 'V' and 'v'. In the the possibility of having children with genetic presence of even single dominant allele, the flowers defects like hemophilia. will be violet and if only recessive genes are present in pairs, the colour of the flower will be white. But • The chances of appearance of a particular human skin colour is controlled by three different trait in future generations can be predicted. genes (six alleles). And their combination will make 64 different colours depending upon the number of • To trace whether the trait is dominant or dominant alleles. Each dominant allele will have recessive or sex-linked or not. additive effect on skin colour. 105. Refer answer no. 54. 95. (a) (i) Short statured with small round head (ii) Furrowed tongue 106. (a) T T t t Parents (b) Both Tall Dwarf (c) Klinefelter’s syndrome T T Gamates (d) Male Tt F1 Selfing (e) (i) Sterile female with rudimentary ovaries Tt F2 (ii) Lack of secondary sexual characters T TT Tt t Tt tt (f) Female From above cross it is clear that 50% of tall plants 96. Turner’s syndrome will be heterozygous in F2 generation. Thus Symptoms: females are sterile as ovaries are genotypic ratio is rudimentary, lack of other secondary sexual TT : Tt : tt characters 1:2:1 Karyotype: XO 97. Klinefelter’s syndrome 25% : 50% : 25% Symptoms: overall masculine development, (b) Law of dominance. For explanation refer development of breast, i.e. Gynaecomastia, answer no. 102 (a) individual is sterile 107. It is a dihybrid cross. Karyotype: XXY For example: Mendel selected pea plants with two 98. Down’s syndrome. Symptoms: Short statured with small round different characters, seed shape and seed colour. He crossed plants having round and yellow seeds head, Furrowed tongue, partially open mouth, (dominant traits) with plants having wrinkled broad palm with palm crease, physical, and green seeds (recessive traits). psychomotor and mental development retarded. 99. (a) 1 will have blood group B with genotype IBi (b) 2 will have blood group A with genotype IAi or blood group O with genotype ii, individual 3 will have blood group AB with genotype IAIB, or B with genotype IBi, or O with genotype ii, or it will be A with genotype IAi 100. Refer answer no. 47. 101. Refer answer no. 47. Principles of Inheritance and Variation  105

The result obtained is shown in the following 117. Morgan found that some characters did not figure: follow Mendelian inheritance instead, they were passed from parents to offsprings together. This Round yellow Wrinkled green association was termed as linkage. RR YY rr yy For example: Morgan conducted a cross between RY ry yellow body and white eyed Drosophila with brown body and red eyed Drosophila. He observed that t9h8e.7re%moafinFi2ngge1n.3e%ratwioans was parental type Round yellow and recombinant type. Rr Yy Cross A Selfing RY RY Gametes rY RRYY rY Gametes yw y+ w+ Wild type Ry RrYY RrYY Ry Parental yw ry RRYy rrYY RRYy ry Yellow, white RrYy RrYy RrYy RrYy rrYy RRyy rrYy Rryy Rryy F1 generation yw yw Yellow, white rryy y+ w+ Wild type Phenotypic ratio Round yellow : Round green : Wrinkled yellow : Wrinkled green 9 3 3 1 The phenotypic ratio obtained is 9:3:3:1 for Parental Recombinant round yellow, round green, wrinkled yellow and type (98.7%) type (1.3%) wrinkled green. y+ w+ y+ w 108. Refer answer no. 64. Wild type white 109. Three principles of Mendel's law of inheritance are: yw y w+ (i) Law of dominance: It states that in a dissimilar pair of factors (genes), one member dominates over other and expresses itself yellow, white yellow in heterozygous condition. The other one is called recessive. F2 generation (ii) Law of segregation: It states that gametes y+ w+ y+ w receive only one allele of the two factors randomly. The alleles do not show any yw yw Wild type white blending and both the characters are yw y w+ recovered as such in F2 generation. (iii) Law of Independent Assortment. It states yw yw that when two pairs of traits are combined in yellow, white yellow a hybrid, segregation of one pair of characters is independent of the other pair of characters. Thus Morgan concluded that the genes which 110. (a) Refer answer no. 90 (b) are passed from parents to off- spring's as such (b) Refer answer no. 89. a linked genes. and here recombinants are found 111. Refer answer no. 91 very few. 112. Refer answer no. 75 (b). 1 18. (a) (i) Genotypes of parents will be IAi and IBi, respectively. 113. Refer answer no. 91 114. The defective gene responsible for haemophilia (ii) Individual 'X' will have blood group 'O' with gentotype ii. 'X' may also have blood is located on X-chromosome and futher passes its gp A and B with genotype Ai and Bi, X-chromosome to his daughters not to sons. respectively. 115. Refer answer no. 97. (b) Individual 'Y' may have blood group 'O' or 'A' 116. (a) Refer answer no. 96. (b) Such chromosomal disorders are caused due (c) hInavFe1AgBenbelroaotdiognroounpe. of the male members to failure of segregation. Both genes A & B are 106 Biology–12

dominant and express equally. This shows gametes formed will be of two types, Z and W type co-dominance. I n humans, XX-XY type of sex determination 119. Haemophilia Thalassemia is present. Females have two X chromosomes and males have two different types of sex Single protein involved Defects in the synthesis chromosomes namely, X and Y. This is an example of male heterogamety as two types of in the clotting of blood of globin leading to gametes are formed, one of X type and other of Y type is affected formation of abnormal This can be shown by the cross given below: haemeoglobin (a) Male heterogamety (b) Female heterogamety Sex linked recessive Autosomal recessive disorder disorder Blood does not clot Results in Father Father anaemia X Y Z Z The category of genetic disorder they both come X XX XY X ZZ ZZ under is Mendelian disorder. Mother Mother 120. I n birds, ZZ-ZW type of sex determination occurs. Females have two different types of chromosomes, Y XX XY Y ZW ZW Z and W. Males have two Z chromosomes. This is an example of female heterogamety as the 5 Marks Questions 121. (a) It is either homozygous dominant or heterozygous dominant. (b) A test cross can be carried out to ensure its genotype. Homozygous Homozygous recessive recessive WW ww ww ? ww WW Ww Ww WW Ww Ww Ww W Dominant Phenotype w Ww Ww (Genotype unknown) ww ww Result All flowers are violet Half of the flowers are violet and half of the flowers are white Interpretation Unknown flower is homozyous dominant Unknown flower is heterozygous 122. S.No. Mendelian observation Deviation Explanation and examples (i) One gene controls one trait Pleiotropy Here, single gene effects multiple phenotypic expressions e.g. Disease phenylketonuria, in which a single gene codes for enzyme phenylalanine hydroxylase and the individual suffers from mental retardation, reduction in hair and skin pigmentation. (ii) O n l y o n e o f t h e p a r e n t a l Incomplete Idnontheiws,itthhecFo1ngtreansetriantgiotnraoiftas mono-hybrid cross characters which is dominant dominance does not resemble ampopneoahryibnriFd1 gene ration in a any of the two parents but is intermediate of cross done wit h the two. Example, in snapdragon, cross of red contrasting traits. flower with white flower produces pink flower. (iii) Only one of the parental traits Co-dominance In a monohybrid cross with which is dominant appear in a monohybrid cross with contrasting contrasting characters, phenotype in the Fin1 traits in F1 generation. generation resembles both the parents e.g. ABO blood group, genotype IAIB will have blood group AB. Principles of Inheritance and Variation  107

(iv) All the traits have distinct Polygenic Here, the occurrence of traits is spread across a alternate forms as one gene inheritance gradient. Such traits are controlled by three or controls single phenotype. more genes, e.g. Human skin colour is controlled by three genes A, B and C. 123. (a) Law of dominance states that characters are controlled by discrete units called factors (gene), which occur in pairs and in a dissimilar pair (alleles) of factors one member dominates the other (recessive) in F1 generation. For example when pea plants with round seeds (RR) are crossed with plants with wrinkled seeds (rr), all seeds in F1 generation were found to be round (Rr). When these round seeds were self fertilized, both the round and wrinkled seeds appeared in F2 generation in 3 : 1 ratio. Hence, in F1 generation, the dominant character (round seeds) appeared and the recessive character (wrinkled seeds) got suppressed, which reappeared in F2 generation. Round seed Wrinkled seed Parent (RR) (rr) Progeny Round seed Round seed F1 generation (Rr) (Rr) Selfing of F1 Round seed (Rr) F2 generation Round seed Round seed Round seed Wrinkled seed (RR) (Rr) (Rr) (rr) Round : wrinkled 3 : 1 ratio (b) Mendelian Monohybrid Incomplete dominance Co-dominance Cross F1 All progeny is similar to All progeny is intermediate of All progeny shows features of dominant parent the parents both the parents equally F2 Both parental traits reappear Shows both parental and Shows both parental and co- intermediate trait dominant trait 124. Human blood is a good example of multiple allelism as a single character is controlled by more than two alleles. Since there are three different alleles, six genotype and four phenotype are produced. Human blood also exhibits a good example of co-dominance as the genes IA and IB express themselves equally and produce AB blood group. 125. A monohybrid cross is a cross in which an organism differs only by a single trait. For example a cross of Pisum sativum seeds of different colour i.e yellow and green. (a) Parents YY x yy yellow green Yy yellow Selfing Yy Y YY Yy Yellow Yellow y Yy yy Yellow green F2 – Phenotypic ratio = 3:1 Genotypic ratio = 1:2:1 108 Biology–12

(b) Law of Dominance: In heterozygous condition, Phenotypic ratio one member of the pair is dominant over other. Tall yellow : Tall green : Dwarf yellow : Dwarf green Law of Segregation: Factor or an allele of a pair segregates from each other so that a 9 :3: 3 : 1 gamete receives only one of the two factors. 128. A single gene with two alleles ‘B’ or ‘b’ controls (c) Phenotypic ratio of F2 in monohybrid cross is the size of the starch grains and the seed shape. 3:1 whereas in a dihybrid cross it is 9:3:3:1. 126. The pattern of inheritance of ABO blood groups in As shown in the above cross, the heterozygous humans is determined by three mechanisms, i.e. multiple allelism, co-dominance and dominance. individuals are of intermediate size, thus deviating from Mendelian law of dominance. Human ABO blood system comprises of four However, it has been observed that the trait of blood groups A, B, AB and O which is controlled seed shape follows Law of Dominance and the by single gene I. This ‘I’ gene has three alleles IA, heterozygotic individuals are round in shape IB and i which explains multiple allelism. which is a dominant trait. IA and IB are completely dominant over i, showing dominance. When IA and IB are present together they both express themselves and produce AB blood group. This phenomenon is known as co-dominance. The blood groups and their genotype are given below in a table. Allele Allele Genotype Blood from from of types of Parent 1 Parent 2 offspring offspring IA IA IA IA A IA IA IB IA IB AB IB IB i IA i A IB IA IA IB AB 129. (a) Colour blindness is a sex-linked disorder. T h e d e f e c t i n c o l o u r b l i n d n e s s i s IB IB IB B present on X chromosome and a person is unable to distinguish red and green colour. i IB i B i i ii O (b) Males have only one X-chromosome. If the gene for colour blindness is present on that 127. (i) Parents TTyy ttYY chromosome, it will always express. Tall plant Dwarf plant green seeds yellow seeds In females, two X-chromosomes are present. The female will be affected if the defected F1 generation Tt Yy gene is present on both the X-chromosomes. Tall plant If only one chromosome carries gene for colour blindness, it will not express as it is recessive. yellow seeds Carrier woman Normal man The phenotype of F1 generation will be tall and yellow and its genotype will be TtYy Parents X XC XY (ii) F2 generation Selfing Gametes X XC XY TY tY Ova Sperms Ty ty TY TTYY TTYy TtYY TtYy Offsprings Tall & yellow Tall & yellow Tall & yellow Tall & yellow X Y TTYy TTyy TtYy Ttyy Ty Tall & yellow Tall & green Tall & yellow Tall & green X XX XY Normal girl Normal boy tY TtYY TtYy ttYY ttYy X XC XC Y Dwarf & yellow Dwarf & yellow Tall & yellow Tall & yellow XC Carrier girl Colour blind boy ty TtYy Ttyy ttYy ttyy 1 Normal Girl : 1 Carrier girl Tall & yellow Tall & green Dwarf & yellow Dwarf & yellow 1 Normal Boy : 1 Colour blind boy Principles of Inheritance and Variation  109

130. (a) Polygenic Pleiotropy 133. (a) Law of Independent Assortment states that inheritance when two pairs of traits are combined in a l One gene shows hybrid, segregation of one pair of characters l More than two multiple effects. is independent of the other pair of character. alleles control a single character E.g. In (b) Let the genotypes of two heterozygous parents Drosophila, be RrYy. E.g. human white eye skin colour is mutation leads to × controlled by depigmentation in three genes A, B many other parts Parents Round yellow Round yellow and C of the body. RrYy RrYy RY RY Gametes rY RRYY rY Gametes (b) Dominance Co- Incomplete Ry RrYY RrYY Ry dominance dominance ry RRYy rrYY RRYy ry Out of Both the Here a third the two genes which phenotype RrYy RrYy RrYy RrYy F1 generation rrYy RRyy rrYy contrasting control a is produced pairs of character which is a alleles, equally combination Rryy Rryy only one express of the rryy expresses in themselves phenotypes Phenotypic ratio the presence of both Round yellow : Round green : Wrinkled yellow : Wrinkled green of another. alleles. 9 3 3 1 131. (a) genotype of W : IAi, X : ii, phenotype of Y : B, It is clear from the above cross that factors Z : AB responsible for shape of seed assort independent (b) Refer answer no. 123 (b). of factors responsible for colour of the seed. (b) ‘Y’ exhibits dominance while ‘Z’ exhibits co- 134. Thalassaemia are categorised as Mendelian dominance pattern of inheritance. disorders because this disorder occurs due 132. (a) This can be explained by the following cross in to single gene mutations in chromosome 11 which parents are heterozygous for terminal, & 16 and the inheritance follow Mendelian and violet flower character. pattern. Thalassemia is an autosomal-linked TtVv ttVv recessive blood disorder and symptoms include, Axial Violet terminal violet Parents in case of Thalassemia minor, mild anaemia, tv TV tV Tv tV tv Gametes low haemoglobin level and in case of major thalassaemia, the child up to 6-9 months old tV tv develop serve anaemia, skeletal deformities, TtVV TtVv jaundice and fatigue. TV Axial Axial Violet Sickle cell anaemia is an autosomal recessive Violet trait where RBCs become sickle shaped under ttVV ttVv F1 Generation low oxygen concentration. It occurs due to tV Terminal terminal replacement of glutamic acid by valine at sixth Violet violet position in beta haemoglobin chain. TtVv Ttvv 135. (a) Chromosomal disorder is caused due to the Tv Axial Axial absence or excess of one or more chromosome(s) white Violet which causes abnormal arrangement of ttVv ttvv chromosomes whereas Mendelian disorder is tv Terminal terminal caused due to alteration or mutation of one Violet white Axial violet : Axial white gene. 3 :1 (b) Klinefelter’s and Turner’s syndrome. (b) This cross is based on Law of Independent (c) Turner’s syndrome: Females are sterile as ovaries are rudimentary, lack of other Assortment. This law states that when two secondary sexual characters. pairs of traits are combined in a hybrid, Klinefelter’s syndrome: Overall masculine segregation of one pair of characters is development, development of breast, i.e. independent of the other pair of character. Gynaecomastia, individuals are sterile. 110 Biology–12

136. (a) Refer answer no. 73 (b) 128 (a) Phenotypic ratio: Red : Pink : White 137. (a) Mendelian Polygenic Pleiotropy 1 : 2 : 1 Inheritance Inheritance Genotypic ratio: One gene Refer answer Refer answer RR : Rr : rr controls 128 (a) 128 (a) 1 : 2 : 1 single In pea plant there is no blending of clours. It follows Mendelian inheritance. Phenotypic character and follows ratio and genotypic ratio are 3 : 1 and 1 : 2 : 1 Mendelian respectively. Law of dominance & segregation is laws e.g. T followed. & t controls IsnhoAwnntiirnrhteirnmumed(iSatneacpodlroaugro. nTsh)ethpearFe1ngteanl ecoralotuiorns height of pea plant reappear irnatFio2 generation. The phenotypic and genotypic in this case is 1 : 2 : 1. This follows (b) Refer answer no. 73 (b). 138. (a) Refer answer no. 62 (b) Refer answer no. 85 law of incomplete dominance. 139. Monohybrid cross of pea plant for flower colour: 140. (a) Refer answer no. 70 (b) Refer answer no. 70 WW X ww Parents Violet flowers white flowers 141. (a) Refer answer no. 85 W w Gametes (b) (i) Yes, if the woman is homozygous for blood group 'A' IAIA X ii 'A' blood group 'O' blood group F1 Ww + ii All violet flowers IA IAi IAi Selfing +Ww IA IAi IAi W WW Ww all children will be of blood group 'A' Violet Violet (ii) Yes, if the woman is heterozygous for blood group 'A' w Ww ww Violet White IAi X ii Phenotypic ratio: 'A' blood 'O' blood Violet : White group group 3 : 1 + ii IA IAi IAi Genotypic ratio: WW : Ww : ww 1 : 2 : 1 i ii ii Monohybrid cross for Antirrhinum flower for colour. 50% of children will be of blood group 'A' and RR X rr Parents 50% will be of blood group 'O'. Red flowers white flowers 142. (a) Refer answer no. 137 R r Gametes (b) (i) This is a monohybrid cross. (ii) Refer answer no. 110 Law of dominance and law of segregation. F1 143. For cross: refer answer 56. Rr For explanation refer answer no. 72 (a) All pink flowers 144. (a) Refer answer no. 51 Selfing (b) Refer answer no. 70. +R r 145. Refer answer no. 126. 146. (a) Refer answer 122. R RR Rr (b) In an individual only two alleles are present. Red Pink To study more than two alleles for a gene one r Rr rr has to study a population. Pink While Principles of Inheritance and Variation  111

147. (a) Flower colour. 148. Gg X gg Parents Ww X ww Parents Green pods Yellow pods (White) Gametes (Violet) Ww w + Gg + Ww F1 generation g Gg gg w Ww ww g Gg gg w Ww ww 50% of green pods (Gg) 50% flowers violet (Ww) 50% of yellow pods (gg) (b) Refer answer no. 145 (b) 50% flowers white (ww) 149. Refer answer no. 137 150. Refer answer no. 137 (b) Cross is test cross. It helps to determine the genotype of parent (heterozygous or homozygous) (c) Refer answer no. 137. 151. Homozygous Homozygous recessive recessive WW ww ww ? ww WW Ww Ww WW Ww Ww Ww W Dominant Phenotype w Ww Ww (Genotype unknown) ww ww Result All flowers are violet Half of the flowers are violet and Interpretation Unknown flower half of the flowers are white is homozyous dominant Unknown flower is heterozygous (a) In this case one of the parent is homozygous dominant and other is homozygous recessive. (b) In this case one of the parent is heterozygous dominant and other parent is homozygous recessive. 152. (a) Incomplete dominance (b) Parents Red (RR) Red (rr) Gametes R r F1 generation All pink (Rr) Gametes RR Gametes r RR r Rr Rr F2 generation rr (c) Refer answer no. 137 Phenotypic ratio : red : pink : white 1 : 2 :1 112 Biology–12 Genotypic ratio : RR : Rr : rr 1 : 2 :1

153. TTRR X ttrr (Parents) Tw TTWw TTww TtWw Ttww tall violet tall white tall violet tall white Tall Round Dwarf wrinkled TR tr Gametes tW TtWW TtWw ttWW ttWw tall v tall dwarf dwarf violet violet violet violet F1 tw TtWw Ttww ttWw ttww tall tall dwarf dwarf TtRr violet violet white Tall Round white Selfing Tall : Tall : Dwarf : Dwarf + TR Tr tR tr violet white violet white 9: 3 :3: 1 TR TTRR TTRr TtRR TtRr Case II: It the plant is homozygous for colour of tall tall tall tall flower and height, then upon selfing the result round round round round will be. Tr TTRr TTrr TtRr Ttrr TTWW X TTWW tall tall tall tall Tall violet tall violet round wrinkled round wrinkled Gametes (unknown genotype) TW tR TtRR TtRr ttRR ttRr tall tall dwarf dwarf (Selfing) round round round round TW × TW = TTWW tr TtRr Ttrr ttRr ttrr All plants will be tall and have violet flowers tall tall dwarf dwarf Case III: if the plant in homozygous for height round wrinkled round round and heterozygous for flower colour. Phenotypic ratio TTWw X TTWw tall violet Tw Tall : Tall : Dwarf : Dwarf (unknown genotype) Gametes round wrinkled round wrinkled (Selfing) TW 9: 3 :3: 1 TW Tw TWTTWW TTWw Such results are obtained because tall and Tw TTWw TTww round are dominant over dwarf and wrinkle respectively. 154. (a) Cross – Refer answer no. 151 Tall violet : Tall white (i) Genotype – TtRr. 3:1 Phenotype – tall round. Case IV: If the plant is heterozygous for height and homozygous for flower clour. (ii) Tr, Tr, tR, tr (iii) Phenotype & its ratio Refer answer no. 151. TtWW X TtWW For explanation of Law of independent assortment refer answer no. 72. TW tW Gametes (b) Refer answer no. 108 (Selfing) 155. Case I: TW tW If the plant is heterozygous for colour of flower and height then upon selfing the result will be TW TTWW TtWW TtWw X Tt Ww Tall violet Tall violet Tall violet tall violet tW TtWW ttWW (unknown genotype) Tall violet Dwarf violet TW Tw tW tw Tall violet : Dwarf violet Gametes will be similar for both parents (Selfing) TW Tw tW tw 3: 1 TW TTWW TTWw TtWW TtWw 156. Case I: If plant is homozygous for both seed tall violet tall violet tall violet tall violet shape and colour. Principles of Inheritance and Variation  113

YYRR X yyrr YyRR X yyrr yellow round green wrinkled Yellow round Green wrinkled (unknown genotype) (Homozygous Recessive) (unknown genotype) (recessive homozygous) YR yr Gametes YR yR yr Gametes F1 YyRr YR yR All yellow round Case II: It plant is heterozygous for both seed yr YyRn yyRr shape and colour yellow green YyRr X yyrr Green wrinkled round round (Homozygous Recessive) Phenotypic ratio Yellow round Yellow round : Green round (unknown genotype) 1: 1 Conclusion: These crosses are called test cross. The unknown genotype is crossed with YR Yr yR yr yr Gametes homozygous recessive parents. The phenotypic ratios obtained in above four cases will tell us about the genotype of the parent plant. YR Yr yR yr 157. For cross, refer answer no. 107 and for Mendel's law, refer answer no. 72 (b). yr YyRr Yyrr yyRr yyrr yellow yellow green green 158. (a) Refer answer no. 154 case II. round wrinkled round wrinkled (b) Refer answer no. 154 case II. Phenotypic ratio (c) This cross is known as test cross. It is conducted to find the unknown genotype of Yellow : Yellow : Green : Green the plant. round wrinkled round wrinkled 159. (a) Refer answer no. 154 case II. 1 : 1 :1: 1 (b) Refer answer no. 156 (c) 160. (a) Refer answer no. (for definition) 72. For example: Round yellow Wrinkled green Case III: If plant in homozygous for seed colour RR YY rr yy & heterozygous for seed shape. RY ry YYRr X yyrr Round yellow yellow round Homozyous Rr Yy (unknown genotype) recessive Selfing YR Yr yr Gametes RY RY Gametes rY RRYY rY Gametes Ry RrYY RrYY Ry YR Yr ry RRYy rrYY RRYy ry yr YyRr Yyrr RrYy RrYy RrYy RrYy yellow yellow round wrinkled rrYy RRyy rrYy Phenotypic ratio Yellow round : Yellow wrinkled Rryy Rryy rryy 1: 1 Phenotypic ratio Case IV: If plant is heterozygous for seed colour Round yellow : Round green : Wrinkled yellow : Wrinkled green and homozygous for seed shape. 9 3 3 1 (b) Refer answer no. 64 114 Biology–12

161. Refer answer no. 158 (a) GA Ga gA ga 162. Parents GGRR ggrr + Green pod Round seeds X Yellow pods GA GGAA GGAA GgAA GgAa wrinkled seeds Green Green Green Green Axial Axial Axial Axial Gametes GR gr F1 GgRr Ga GGAa GGaa GgAa Ggaa Gametes Green pod Green Green Green Green Round seeds Axial Terminal Axial Terminal gA GgAA GgAa ggAA ggAa Green Green Yellow Yellow Axial Axial Axial Axial GR Gr gR gr ga GgAa Ggaa ggAa ggaa Green Green Yellow Yellow GR Gr gR gr Axial Terminal Axial Terminal GR GGRR GGRr GgRR GgRr Phenotypic ratio of F2 generation Green Green Green Green Green : Green : Yellow : Yellow pod Axial Terminal Axial Terminal Round pod pod pod seeds Round Round Round 9: 3 :3: 1 seeds seeds seeds Gr GGRr GGrr GgRr Ggrr (b) The Mendelian Principle which can be derived Green Green Green Green from such a cross is \"Law of Independent pod pod pod pod Assortment\". It states that when two pairs Round Wrinkled Round Wrinkled of alleles for different traits are combined in seeds seed seeds seed a hybrid, segregation of one pair of alleles is independent of other pair of alleles. gR GgRR GgRr ggRR ggRr Green Green Yellow Yellow 164. (a) FFGG ffgg Parents pod Round pod pod pod inflated and constricted yellow seeds Round Round Round seeds seeds seeds green pods pods gR GgRR GgRr ggRR ggrr FG fg Gametes Green Green Yellow Green pod FfGg F1 Round pod pod pod inflated and seeds Round Round Wrinkled green pods seeds seeds seed Selfing Phenotypic ratio F2 generation Green : Green : Yellow : Yellow Pods Pods Pods Pods Round Round Seeds Wrinkled seeds Wrinkled seeds seeds 9: 3 :3: 1 FG Fg fG fg + FfGg 163. (a) GGAA ggaa FG FFGG FFGg FfGG Inflated Green pods yellow pods Parents Green axial flowers terminal flower Inflated Inflated Inflated Green Green Green GA ga Gametes Fg FFGg FFgg FfGg Ffgg Inflated Inflated Inflated Inflated Yyellow Green Yellow Green fG FfGG FfGfg ffGG ffGg GgAa F1 Inflated Inflated Constricted Constricted Green pods axial flowers Green Green Green Green Selfing fg FfGg Ffgg ffGg ffgg F2 generation Inflated Inflated Constricted Constricted Green Yellow Green Yellow Principles of Inheritance and Variation  115

Phenotypic ratio of F2 generation occur due to gene mutations and the inheritance follows Mendelian pattern. Inflated : Inflated : Constricted : Constricted For Thalassemia symptoms refer answer answer green yellow green yellow no. 132 9: 3: 3 : 1 For Colour blindness refer answer no. 127 (a) (b) Refer answer no. 127 (b) (b) Refer answer no. 161 (b) 169. Refer answer no. 132 165. (a) AAVV X aavv Parents 170. (a) Thalassemia is a genetic defect in which Axial Violet Terminal White the globulin chains of haemoglobin are less than the actual required. Like SEA, it is an Flowers Flowers autosomal hereditary disease. Symptoms of thalassaemia include anaemia caused due to GA ga Gametes defective or abnormal form of haemoglobin, fatigue and shortness of breath. Fertilization AaVv (b) Since this disease is an autosomal recessive Axial Violet disease and can manifest only in homozygous recessive state. i.e. when an individual has Selfing both the alleles defective. This condition can be achieved only when one of the defective + AV Av aV av allele comes from father. Thus mother should not be blamed for developing a child suffering AV AAVV AAVv AaVV AaVv from thalassaemia. Axial Axial Axial Axial Violet Violet Violet Violet (c) Knowledge and awareness. Av AAVv AAvv AaVv Aavv 171. (a) Refer answer no. 98 & 36 Axial Axial Axial Axial (b) Thalassaemia and haemophilia are categorised Violet White Violet White as Mendelian disorders because these aV AaVV AaVv aaVV aaVv disorders occur due to single gene mutations Axial Axial Terminal Terminal and the inheritance follow Mendelian pattern. Violet Violet violet Violet Thalassemia is an autosomal linked recessive blood disorder and symptoms include, in av AaVv Aavv aaVv aavv case of Thalassemia minor, mild anaemia, low haemoglobin level and in case of major Axial Axial Terminal Terminal thalassaemia, the child up to 6-9 months old develop serve anaemia, skeletal deformities, Violet White Violet White jaundice and fatigue. Phenotypic ratio Aoaf VFv1:(hAeltlraoxzyiaglouviso)let flowers Haemophilia is a sex-linked recessive disorder Genotype of F1 = and the symptoms include prolonged clotting (b) Phenotypic ratio of F2 generation time and continuous bleeding, even in a minor injury. Axial : Axial : Terminal : Terminal Sickle cell anaemia is an autosomal recessive Violet White Violet White trait where RBCs become sickle shaped under low oxygen concentration. It occurs due to 9 :3: 3 : 1 replacement of glutamic acid by valine at sixth position in beta haemoglobin chain. (c) Refer answer no. 161 (b) 172. Refer answer no. 169 (b) 166. (a) TtYy X ttyy Parents 173. (a) Genotype of parents Gen I = Aa, Aa. Tall, yellow seeds dwarf, green seeds Gen II = Son – Aa Gametes Affected daughter – aa TY tY Ty ty ty normal daughter–Aa F1 generation (b) aa. + TY tY Ty ty (c) It is an autosomal trait. Only daughters are ty TtYy ttYy Ttyy ttyy affected, but fathers are not affected. Tall Dwarf Tall Dwarf Green Green Yellow Yellow 167. Refer answer no. 158 (a) & 161 (b) 168. (a) Colour blindness and thalassemia are categorised as Mendelian disorders because these 116 Biology–12

174. (a) Traits that are generally controlled by three AA aa or more genes, the phenotype reflects the (Axial flower) (Terminal flower) contributory effect of each allelele. e.g. Human skin colour, controlled by three AA aa genes A, B and C. In multiple allelism more than two alleles govern the same character, e.g. Human blood group, controlled by three different alleles (IA, IB, i) Aa Aa Aa Aa (b) In pleiotropy a single gene can show multiple phenotypic expressions, in phenyl ketonuria (50% plants are with Axial Flower single mutated gene expresses mental and 50% plants with terminal flower) retardation, reduction in hair and skin (b) Law of segregation states that the allelic pairs pigmentation. segregate during gamete formation without 175. (a) If the plant is homozygous for the dominant trait the cross will show dominant trait (all losing their identity. plants with axial flowers) in F1 generation. 176. Thomas Hunt Morgan worked on Drosophila melanogaster, the fruit fly. He observed that AA aa some genes remain in physical association as (Terminal flower) (Axial flower) they are passed from parents to their offspring due to which the parental combination was AA aa obtained in higher percentage than non-parental recombinants. Few genes which lie close to each other are tightly linked whereas those which are located at a distance are loosely linked. Tightly Aa Aa Aa Aa linked genes show low recombination whereas (All plants with Axial Flower) loosely linked genes show higher recombination. If the plant is heterozygous for dominant trait The type of linkage, tight or loose, helped him the F1 generation will show ratio of 1 : 1 of to measure the distance between the genes and axial and terminal flowers. mapping their location on the chromosome. Assertion & Reason Type Questions For question numbers 1-9: Two statements are given 5. Assertion (A): Test cross is a back cross. one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer to these questions from Reason (R): In test cross, individual is crossed the codes (a), (b), (c) and (d) as given below. with recessive parent. (a) Both Assertion and Reason are true and Reason 6. Assertion (A): Quantitative inheritance is called is the correct explanation of Assertion. polygenic inheritance. (b) Both Assertion and Reason are true but Reason Reason (R): Several genes control the expression is NOT the correct explanation of Assertion. of a trait. (c) Assertion is true but Reason is false. 7. Assertion (A): In birds, the chromosome (d) Assertion is false and Reason is also false. composition of the egg determines the sex. 1. Assertion (A): Mendel was successful in his Reason (R): Female birds are heterogametic. hybridisation experiments. Reason (R): Garden pea proved to be an ideal 8. Assertion (A): Y chromosome causes maleness. experimental material. Reason (R): If the number of X chromosome is 2. Assertion (A): The principle of segregation given more than one, femaleness dominates. by Mendel is the principle of purity of gametes. 9. Assertion (A): There ie expression of only one Reason (R): Gametes are pure for a character. gene of the parental character in a Mendelian 3. Assertion (A): A gene may have several Monohybrid cross in F1 generation. allelomorphs. Reason (R): In a dissimilar pair of factors one Reason (R): Wild form can mutate in more than member of the pair dominates the other. one ways. [CBSE 2022] 4. Assertion (A): Complementary genes are non- 1. (b) Answers 5. (a) allelic genes. 6. (a) Reason (R): Complementary genes interact to 2. (a) 3. (a) 4. (b) 7. (b) 8. (c) 9. (a) produce a completely new trait. Principles of Inheritance and Variation  117

Case Based Questions 1. Turner’s Syndrome is an example of monosomy. will continue to bleed even from a minor cut It is formed by the union of an allosome free egg since he or she does not possess the natural phenomenon of blood clotting due to absence and a normal ‘X’ containing sperm or a normal of anti-haemophilic globulin or factor VIII and plasma thromboplastin factor IX essential for egg and an allosome free sperm. The individual it. As a result of continuous bleeding the patient may die of blood loss. Colour blindness is another has 2n = 45 chromosomes (44 + XO) instead of type of sex linked trait in which the eye fails 46. Such individuals are sterile females who have to distinguish red and green colours. Vision is however, not affected and the colour blind can, rudimentary ovaries, under developed breasts, lead a normal life, reading, writing and driving (distinguishing traffic lights by their position). small uterus, short stature, webbed neck and (i) If a haemophilic man marries a woman whose abnormal intelligence. They may not menstruate father was haemophilic and mother was normal then which of the following holds true or ovulate. This disorder can be treated by giving to their progenies? female sex hormone to the women from the age (a) Of the total number of daughters, 50% daughters are carrier and 50% are of puberty to make them develop breasts and haemophilic. have menstruation. This makes them feel more (b) All the daughters are haemophilic normal. (c) All sons are haemophilic and all daughter are normal. (i) Number of Barr body present in a female with Turner’s syndrome is (d) All sons are normal and all daughters are carrier. (a) 0 (b) 1 (ii) A man whose father was colourblind and (c) 2 (d) <2 mother was normal marries a woman whose father a haemophilic and mother was normal. (ii) Turner’s syndrome is an example of Which of the following is true for their progenies? (a) aneuploidy (b) euploidy (a) 25% female progenies carry the gene for (c) polyploidy both haemophilia and colourblindness. (d) autosomal abnormality (b) 25% male progenies carry only the gene of colourblindess. (iii) Turner’s syndrome is a/an (c) 25% female progenies carry only the gene (a) autosomal recessive Mendelian disorder of colourblindness. (b) autosomal dominant Mendelian disorder (d) 25% male progenies and 25% female progenies carry the gene of haemophilia. (c) sex linked Mendelian disorder ( iii) Which of the following statement is incorrect (d) chromosomal disorder regarding haemophilia? (iv) Which of the following statements regarding (a) It is a dominant. Turner’s syndrome is incorrect? (b) A single protein involved in clotting of (a) It is a case of monosomy of chromosomes. blood is affected. (b) The suffering individual is a sterile female (c) It is recessive disease. having one ‘X’ chromosome missing in the (d) None of these cells (iv) Anup is having colourblindness and is (c) The problem is due to an extra chromosome married to Soni who is normal. What is the chance that their son will have the disease? (d) The individual are of short stature. (a) 100% (v) Assertion: Turner’s syndrome is caused due to absence of anyone of the X and Y sex (b) 50% chromosome. (c) 25% Reason: Individuals suffering from Turner’s (d) 0% syndrome show masculine as well as feminine development. (a) Both assertion and reason are true and reason is the correct explanation of assertion. (b) Both assertion and reason are true but reason is not the correct explanation of assertion. (c) Assertion is true but reason is false. (d) Both assertion and reason are false. 2. Haemophilia is a sex linked disease which is also known as bleeder’s disease as the patient 118 Biology–12

(v) Refer to the given cross (ii) To determine the genotype of a purple Parents XCY × XXh flowered plant, Prashant crossed this plant with a white flowered plant. This cross represents a Gametes XC Y X Xh Progeny (a) test cross (b) dihybrid cross (1) (2) (3) (4) (c) reciprocal cross (d) trihybrid cross Select the correct option regarding 1, 2, 3 (iii) In white flowered plant, allele is expressed in and 4 (a) heterozygous condition only (a) 1. Colourblind carrier female (b) homozygous condition only 2. Colourblind haemophilic female (c) F3 generation (d) both homozygous and heterozygous 3. Normal male condition 4. Haemophilic male (iv) The character, i.e., purple colour of the flowers (b) 1. Colourblind people that appeared in the first filial generation is 2. Haemophilic female called 3. Normal male (a) recessive character. 4. Haemophilic male (b) dominant character. (c) 1. Colourblind female (c) holandric character. 2. Colourblind and haemophilic female (d) lethal character. 3. Normal male (v) Assertion: A geneticist crossed two plants and he obtained 50% purple flowered plants 4. Normal female and 50% white flowered plants. (d) 1. Colourblind female Reason: Purple coloured flower plant might 2. Normal female be heterozygous. 3. Normal male (a) Both assertion and reason are true and reason is the correct explanation of 4. Haemophilic female 3. Prashant wanted to find the genotype of a pea assertion. plant bearing purple coloured flowers in his kitchen garden. For this, he crossed purple (b) Both assertion and reason are true but flowered plant with white flowered plant. As a reason is not the correct explanation of result, all plants which were produced had purple flower only. Upon selfing these plants, 75 purple assertion. flower plants and 25 white flower plants were produced. Now, he can determine the genotype (c) Assertion is true but reason is false. of a purple flowered plant by crossing it with a white flowered plant. (d) Both assertion and reason are false. Ans. 1. (i) (a) (ii) (a) (iii) (d) (iv) (c) (v) (d) (i) Which of the following cannot be derived from 2. (i) (a) (ii) (d) the crosses done by Prashant? (iii) (a) (iv) (d) (a) Mendel’s law of segregation (v) (a) (b) Mendel’s law of dominance 3. (i) (c) (ii) (a) (c) Mendel’s law of independent assortment (iii) (b) (iv) (b) (d) Both (a) and (c). (v) (a) qqq Principles of Inheritance and Variation  119

5 Molecular Basis of Inheritance Topics Covered 5.2 The Search for Genetic Material 5.4 Replication 5.1 The DNA 5.6 Genetic Code 5.3 RNA World 5.8 Regulation of Gene Expression 5.5 Transcription 5.10 Rice Genome Project: First Crop Genome 5.7 Translation 5.9 Human Genome Project 5.11 DNA Fingerprinting C hapter map Conclusively Hershey Lac Operon Single strand DNA is the and Chase Regulation of mRNA, VNTR as genetic material experiment Gene Expression probe tRNA, rRNA RNA DNA Human Genome DNA is the Fingerprinting Project genetic material MOLECULAR BASIS OF INHERITANCE DNA Replication Transcription Genetic Code Translation Searching for 61 codons, genetic material Double helix Transcription nucleosides, unit Promotor, 3 stop codons, Bacterial nucleotides, Structural gene, degenerate, transformation polarity Terminator universal AUG Griffith’s Spicing, Cistron, initiator codon experiment capping, Exon, Intron tailing It took over hundred years after Mendel's laws to establish DNA as a genetic material. DNA and RNA are the two types of nucleic acids present in all living cells. DNA acts as a genetic material in most of the organisms. RNA present as a only genetic material in some viruses. Most of the RNA act as messenger and it also acts as adapter, structural and catalytic molecule. Topic 1. The DNA • DNA is long polymer of deoxyribonucleotides. Number of base pairs as the number of nucleotides determine the length of this polymer. Its length varies in different organisms but is fixed for any particular species. Example: Length of DNA of bacteriophage F × 174 is 5368 nucleotides, Escherichia coli is 4.6 × 106 bp and of humans it is 3.3 × 109 bp. Structure of Polynucleotide Chain • A nucleotide in a DNA has three components a nitrogenous base, a deoxy-ribose pentose sugar and a phosphate group. 120

• Nitrogenous bases are of two types-Purines (Adenine and Guanine), and Pyrimidines (Cytosine, Uracil, Thymine). Uracil is present only in RNA, thymine is present only in DNA whereas cytosine is common in both. Phospho-di-ester bond de-oxy-ribose sugar 5' phosphate H H H 3' hydroxyl H PC PC PC OH PC H H H H G C AT N-glycosidic linkage Purines nitrogenous base Pyrimidines • Nitrogenous base + pentose sugar N - glycosidic linkage (iv) Two chains are coiled in right-handed fashion. nucleoside, e.g. deoxyadenosine, deoxyguanosine. The pitch (single turn) of helix is 3.4 nm. 5' OH of Nucleoside + Phosphate Phosphoester linkage (v) The plane of one base pair stacks over another, Nucleotide and presence of hydrogen bonds provides stability to helical structure. • N u c l e o t i d e + N u c l e o t i d e 3'–5' phosphodiester linkage Dinucleotide. These two are joined by phosphodiester Francis Crick proposed central dogma which bond. says that the genetic information flows in the following manner: • The free –OH group is the 3' end of DNA and the free phosphate group is 5' end of DNA. DNA → RNA → protein. • The backbone of DNA is formed of sugar and phosphate where the nitrogenous bases are linked to sugar at an angle. The genetic flow of information in some viruses is in reverse direction, i.e. RNA → DNA which is • RNA has a similar structure except few differences called reverse transcription. from DNA Packaging of DNA helix – The sugar is ribose • Length of the DNA is taken as the total number – It has an additional -OH group at its 2' position of base pairs multiplied by distance between two consecutive base pairs. The distance between two – Instead of thymine, it has uracil. consecutive base pairs is 0.34 nm (0.34 × 10–9 m). Therefore human DNA length is 6.6 × 109 bp × 0.34 – Thymine is 5'-methyl uracil. × 10–9 m/bp = 2.2 metres. • In 1896, Friedrich Meischer identified DNA as an acidic substance present in nucleus and called this substance as nuclein. • In 1953, James Watson and Francis Crick analysed • In Prokaryotes, in spite of no defined nucleus, DNA the X-ray diffraction data of Maurice Wilkins and is held in a defined region called nucleoid by some Rosalind Franklin, and observations of Erwin positively charged proteins. Chargaff and proposed the Double Helix model for DNA. • Erwin Chargaff proposed two rules: • In Eukaryotes, there is a set of eight positively charged basic proteins called histones octamer on A + G = T + C = 1 which negatively charged DNA (200 bp) is wrapped. This resulting structure is called nucleosome. A = T and G = C Repeating unit of nucleosome is called chromatin. Salient features of double helix structure of DNA DNA H1 histone (i) DNA is made of two polynucleotide chains Histone arranged in double helix, where sugar and octamer phosphate form the backbone. The bases project inside. Core of histone molecules (ii) The two chains are anit-parallel to each other. Structure of a nucleosome (iii) The nitrogenous bases are joined to each other by hydrogen bonds. A forms two hydrogen bonds with T and C forms three hydrogen bonds with G. Purines are always opposite to pyrimidines and vice versa. Molecular Basis of Inheritance  121

Chromatin fibre showing nucleosomes Nucleosome and Chromatin • Chromatin fibers are further condensed during metaphase of cell division by additional set of proteins called Non-histone Chromosomal (NHC) proteins to form chromosome. • In nucleus, some region of chromatin is lightly stained, loosely packed and is transcriptionally active, this region is called euchromatin. Some regions are darkly stained, densely packed and transcriptionally inactive, such regions are called heterochromatin. Topic 2. The Search for Genetic Material DNA only transformed the non-virulent strain into Transforming Principle virulent one. In 1928, Frederick Griffith did a series of experiments The Genetic Material is DNA • Alfred Hershey and Martha Chase in 1952 conducted with Streptococcus pneumoniae, a bacterium responsible for pneumonia. some experiments with bacteria and bacteriophages. • He observed two strains of bacteria, smooth shiny • They grew bacteriophages in a medium that colony called S type which was virulent and had contained radioactive phosphorous and radioactive capsule, the other R type rough colony is non sulphur, separately. virulent and without capsule. The experiment was conducted in the following way: • Bacteriophages became radioactive after few generations. The one grown on radioactive S strain → Inject into mice → Mice die phosphorus had radioactive DNA as phosphorus is the component of DNA. R strain → Inject into mice → Mice live S strain (heat killed) → Inject into mice → S strain (heat killed) Mice live • The one grown on radioactive sulphur had radioactive protein shell as sulphur is the component of protein. + → Inject into mice → Bacteriophages Bacteriophages grown on a grown on a R strain (live) Mice die medium containing medium containing • The bacteria isolated from this dead mice were radioactive Sulphur radioactive living S-type. Phosphorous • Griffith concluded that there was some transforming • The labelled bacteriophages from both media were principle (genetic material) which was transferred allowed to infect E. coli. to R-strain from heat killed S-strain which made R-strain turned virulent. But he could not establish • In both the cases viral coats were removed from the the biochemical nature of this genetic material. bacteria by agitating them in a blender. Biochemical Characterisation of Transforming • The virus particles were separated from the bacteria by spinning them in a centrifuge. No Principle radioactivity was detected in E. coli but was detected in supernatant in case where bacteriophages were • Oswald, MacLeod and McCarty worked to determine labelled with radioactive sulphur. No radioactivity the transforming principle of Griffith's experiment. was detected in supernatant but was detected in They purified proteins, DNA and RNA from heat E. coli where bacteriophages were labelled with killed S-cells to see which out of the three could radioactive phosphorus. transform R-cells into S-cells. They established that 122 Biology–12

• This indicated that DNA was the genetic material Properties of Genetic Material (DNA versus RNA) that was passed from virus to bacteria. • A molecule that can act as a genetic material must The Hershey-Chase Experiment fulfil the following criteria: (i) It should be able to generate its replica. (ii) It should be chemically and structurally stable. (iii) It should provide the scope for slow changes i.e. mutations that are required for evolution. (iv) It should be able to express itself in the form of ‘Mendelian characters’. • Protein does not replicate itself, RNA has -OH group at 2 ' position, this makes RNA very reactive and thus unstable. RNA is also known to be catalytic and hence is reactive. Protein and RNA do not fulfil one or more criterion of genetic material. • DNA is structurally more stable, it can replicate, mutate slowly and express itself in Mendelian way. Being double stranded and presence of thymine provides additional stability to DNA. Topic 3. RNA World • Since RNA was unstable and very reactive, DNA evolved from RNA with few chemical modifications • RNA was the first genetic material. Many life which made DNA more stable than RNA. processes such as metabolism, translation, splicing etc. evolved around RNA. • RNA also acts as catalyst in living systems. Topic 4. Replication Semi-conservative mode of replication–experimental proof: Messelson and Stahl proved that DNA replicates in semi-conservative manner. • They took E. coli and cultured them iCnsmCel ddieunmsictoyngtraaindiinengt15cNenHt4rCiflufgoartmioannfyrogmentehraetnioonrsm. 1a5lNDwNaAs.incorporated into new DNA which separated by • They then transferred these cells with medium containing 14NH4Cl. • The sample was taken after 20 minutes and then after 40 minutes. And both were separated by CsCl density gradient to measure the density of DNA. • After 20 minutes, the hybrid DNA obtained were of intermediate density. • After 40 minutes, equal amount of hybrid DNA and light DNA (normal DNA) were obtained. Generation I Generation II 15N-DNA 14N-DNA 14N-DNA 15N-DNA 15N-DNA 20 min 40 min 14N-DNA 14N-DNA Gravitational force 14N14N 14N15N 14N14N 14N15N Light Hybrid Light Hybrid Separation of DNA by Centrifugation Messelson and Stahl’s Experiment Molecular Basis of Inheritance  123

Later, similar experiments were conducted by Taylor • The major enzyme used for DNA replication is DNA- and the colleagues in 1958 on vicia faba using dependent DNA polymerase. radioactive thymidine. This also proved that DNA replicates in semi conservative manner. • Deoxyribonucleoside triphosphates act as substrates as well as provide energy for polymerization reaction The Machinery and the Enzymes through terminal phosphates just like ATP. 5' 3' • Replication starts at a specific region called origin of replication. The two strands of DNA unwind and Template form a small opening in helix called as replication fork. DNA polymerase adds new complementary Continuous 5' Discontinuous synthesis bases in one direction only, i.e. 5'–3'. The strand synthesis 3' (lagging strand/Okazaki with polarity 3'–5' has continuous replication. On (leading strand) the other strand with polarity 5'–3', fragments of fragments) newly synthesized strand are formed called okazaki fragments, which are later joined by DNA ligase. 3' Newly 5' This is called discontinuous synthesis. 5' synthesised 3' • In eukaryotic cells, replication occurs in S-phase strands of cell cycle. Failure in cell division after DNA replication results in polyploidy. Replicating Fork Topic 5. Transcription • The segment of DNA coding for RNA or a protein is termed as gene and if it codes only for a polypeptide • Copying of information from one strand of DNA to and is called cistron. RNA is called transcription where uracil is added instead of thymine in RNA. • In prokaryotes, the structural gene is polycistronic, i.e. it codes for more than one protein. • B o t h s t r a n d s o f D N A a r e n o t t r a n s c r i b e d simultaneously because they will form two RNA • In eukaryotes, the structural gene is monocistronic, with different sequence, and both will form i.e. it codes for only one polypeptide. As the different proteins. The two RNA strands will be structural gene is interrupted so it is called split complementary to each other, so they can form gene. Portions of structural gene which are double stranded RNA which can prevent translation. expressed are called exons, and other parts which are not expressed are called introns. Transcription Unit • Transcription unit of DNA is defined by three regions Types of RNA and the process of Transcription • In prokaryotes, three major types of RNA, are Promoter: It provides binding site for RNA polymerase. needed to synthesise a protein in a cell. These RNA are: Structural gene: Portion of DNA which is transcribed into RNA. (i) mRNA (messenger RNA), Terminator: P dissociates RNA polymerase (ii) tRNA (transfer RNA), enzyme and marks the end of transcription. (iii) rRNA (ribosomal RNA). Schematic structure of a transcription unit • DNA has two strands, and DNA dependent RNA T h e r e i s s i n g l e D N A - d e p e n d e n t R N A polymerase that catalyses transcription of all polymerase catalyses the polymerization in only one types of RNA. direction, i.e. 5'–3', so the strand which has polarity of 3'-5', called template strand is transcribed. The • Transcription involves three different stages other strand is called coding strand. • Some regulatory sequences which do not code for any (i) Initiation: RNA polymerase binds with sigma RNA or protein are situated upstream (towards 5' factor and then it binds to promoter to initiate w.r.t. coding sequence) or downstream (towards 3' transcription. w.r.t. coding sequence) the structural gene. (ii) Elongation: Sigma factor dissociates and RNA 124 Biology–12 polymerase uses nucleoside triphosphate as substrate and polymerise RNA depending on the template. Complementary bases are added during polymerization.

(iii) Termination: Rho factor binds with RNA RNA polymerase III transcribes tRNA, 5S rRNA, polymerase and it dissociates from the template and snRNAs (small nuclear RNAs). thus completing the process of termination. • The hnRNA has both exons and introns. The splicing occurs by spliceosomes which removes introns and join exons. Process of transcription in Bacteria Process of Transcription in Eukaryotes • Since there is no separation of nucleus and cytoplasm • After transcription, RNA should move to cytoplasm in prokaryotes, the translation starts simultaneously for translation. Cytoplasm contains RNA degrading as soon as RNA begins to polymerize. enzymes. To protect RNA from degradation, an additional process capping and tailing occurs. In • In eukaryotes, transcription occurs inside the capping, methyl guanosine triphosphate is added nucleus. Apart from initiation, elongation and to 5' end of hnRNA. In tailing, 200-300 adenylate termination, there are some additional processes residues are added at 3' end. Fully processed hnRNA and complexities in eukaryotes. There are three is now called functional mRNA. different RNA polymerases, one separate for each type of RNA. • The split gene arrangement is an ancient feature of genome which represent the dominance of RNA RNA polymerase I transcribes rRNAs (28S, 18S, world. and 5.8S). RNA polymerase II transcribes heterogeneous nuclear RNA (hnRNA). Topic 6. Genetic Code • Discovery of Servo Ochoa enzyme, a polynucleotide phosphorylase also helped synthesizing RNA in • According to physicist, George Gamow, the template independent manner. genetic code is a set of three-letter combinations of nucleotides (triplet) called codons, each of which • Finally the checker board for genetic code was codes for a specific amino acid or stop signal. As prepared as given below. there are only four bases and if only one base codes for an amino acid, they will code for only 4 amino acids. If the code for an amino acid is doublet, there will be 16 possible combinations so they will code for only 16 amino acids. But if the code is triplet, 64 combinations are possible . This will cover all twenty amino acids. • Dr Har Gobind Khorana synthesized synthetic RNA molecules by using three repeating units with defined combinations of bases, homopolymers and copolymers. Khorana was the first scientist to chemically synthesize oligonucleotides. He also identified stop codons. • Marshall Nirenberg developed cell free system to synthesize proteins by adding poly uracil RNA to the system. This proved that genetic code is a triplet. Molecular Basis of Inheritance  125

Salient Features of Genetic Code • Change of base at only one position causes point (i) The codon is triplet. 61 codons code for amino mutation. Example is sickle cell anaemia in which the A is replaced by the T in DNA. As a result, amino acids and 3 codons UAA, UGA, UAG act as stop acid glutamine is replaced by valine at 6th position. codons which does not code for any amino acid. • Insertion or deletion of one or two base pairs causes (ii) One codon codes for only one amino acid, hence, change in reading frame from the point of insertion it is unambiguous and specific. or deletion. This is called frame shift mutation. Due to this the entire sequence of amino acids is (iii) Some amino acids are coded by more than one changed from the point where insertion or deletion codon, hence the code is degenerative. has happened. (iv) The code is read in mRNA in a contiguous • If three base pairs are inserted or deleted, there is fashion. There are no punctuations. no change in the sequence ahead of mutation. This further proves that codon is triplet. (v) The code is nearly universal: for example, from bacteria to human UUU would code for tRNA - the Adapter Molecule Phenylalanine (Phe). Some exceptions to this rule • tRNA is also called soluble RNA. The molecule is have been found in mitochondrial codons, and in some protozoa. clover leaf like (within cell it looks like inverted L) and has an anticodon loop that has bases complementary (vi) AUG has dual functions. It codes for Methionine to triplet codon. It has an amino acid acceptor end (met), and it also acts as initiator codon. and amino acid binds to it. For each amino acid, there is a specific tRNA. There are no tRNAs for Mutations and Genetic code stop codons. • Any change in genes is called mutation. Topic 7. Translation energetically to form peptide bond between two amino acids. Ribosome moves from codon to codon • Translation is the process by which a protein is along mRNA and amino acids are added one by one synthesized from the information contained in a thus forming a polypeptide chain. Ribosome (23S molecule of messenger RNA (mRNA) . But before an rRNA in bacteria called ribozyme) acts as catalyst amino acid is added to the protein chain, it is linked for the formation of peptide bond. to its cognate tRNA in the presence of ATP. This is called charging of tRNA or aminoacylation of tRNA. • At the end, release factor binds to stop codon, thus terminating translation and releasing complete • The smaller subunit of ribosome joins first with polypeptide from the ribosome. mRNA at start codon followed by the larger subunit. Larger subunit has two sites for subsequent amino • In addition to start codon, codes for polypeptide and acids to bind. The start codon is recognised by stop codon, a mRNA has additional sequences which initiator tRNA. are not translated but are required for efficient translation process. These are called untranslated • The charged tRNA now binds with appropriate codon regions. These are present at both 5' and 3' ends, in mRNA forming complementary base pairs with before start codon and after stop codon. tRNA anticodon. Two tRNAs carrying amino acids are placed so close to each other that it is favoured Topic 8. Regulation of Gene Expression • In eukaryotes the regulation of gene expression can • In prokaryotes, the gene expression is controlled occur at various levels. during the initiation of transcription. The activity of RNA polymerase in turn is regulated by interaction transcriptional level (formation of primary of some accessory proteins which affects its ability transcript), to recognize start sites. processing level (regulation of splicing), • These regulatory proteins act both positively, i.e. they act as activators, and negatively, i.e. repressors. transport of mRNA from nucleus to the cytoplasm, • The accessibility of promoter region of prokaryotic translational level DNA is in many cases regulated by interaction of proteins with a sequence called operator. These • The genes express to perform a particular function operators lie adjacent to promoter in most of the and depend upon metabolic, physiological or environmental conditions that regulate their expression. 126 Biology–12

operons. Each operon has its specific operator and specific repressor.Example: lac operator is present only in lac operon. The Lac operon • When lactose is present, it binds with repressor and changes its properties. Repressor then becomes • Lac operon in bacteria was first described by unable to bind with operator. RNA polymerase Francois Jacob and Jacque Monod. binds with promoter and transcription of all three structural genes occurs. Such a regulation is called • Lac operon consists of one regulatory gene ‘i’ which negative regulation. refers to inhibitor and three structural genes ‘z’, ‘y’ and ‘a’. • ‘i’ gene codes for repressor, ‘z’ codes for beta- galactosidase (b-gal) which hydrolyses lactose into galactose and glucose, ‘y’ codes for permease which increases permeability of cell to b-gal, ‘a’ gene codes for transacetylase. All these are required for the metabolism of lactose, so all genes are required to function together. • Lactose acts as substrate for enzyme b-gal and also acts as inducer. It is transported to cell through cell membrane due to action of permease. • Repressor is translated from ‘i’ gene. This repressor binds to the operator region and prevents RNA polymerase from transcribing operon. Topic 9. Human Genome Project Sequence annotation, whole genome was sequenced and the functions were assigned to Human genome project was launched in 1990. It is different regions later. called a mega project because it was launched for 13 years, involving many countries such as US, UK, • For sequencing DNA was isolated and cut into Japan, France, Germany, China and some others. fragments → each fragment was cloned into This project involved huge cost which was estimated suitable host (BAC or YAC) → cloned fragment to be 9 billion US dollars, huge data was supposed to was amplified → the sequencing was done using be generated from a single sequence. The retrieval automated DNA sequencers which worked on the and analysis of this huge data was again an issue. principle of method developed by Frederick Sanger This lead to the development of a new field of biology → different fragments were aligned and arranged called bioinformatics. on the basis of overlapping regions → the data was analysed by special computer based programme Major Goals of HGP → sequences were annotated and assigned to each • Identify all the approximately 20,000–25,000 genes chromosome and data was stored. in human DNA • Genetic and physical mapping was done using • Determine the sequences of the 3 billion chemical information on polymorphism or restriction endonuclease recognition sites and repetitive DNA base pairs that make up human DNA sequence called microsatellite. • Store this information in the databases • Improve tools for data analysis • Last chromosome to be sequenced was chromosome-1 • Develop transfer related technologies to other and was completed in May 2006. sectors, such as industries Salient Features of Human Genome • To address the ethical, legal, and social issues (ELSI) • The human genome contains 3164.7 million (3.1647 that may arise from the project. billions) nucleotide bases. Methodology • The average gene consists of 3000 bases, largest • For deciphering human genome sequence, two gene is dystrophin with 2.4 million bases. approaches were involved: • The total number of genes is estimated at 30,000. Expressed sequence tags (ESTs) the genes from RNA were identified. Molecular Basis of Inheritance  127

• Almost all (99.9 per cent) nucleotide bases are occur in humans. This information promises to exactly the same in all people. revolutionize the processes of finding chromosomal locations for diseases-associated sequences and • The functions are unknown for over 50 per cent of tracing human history. the discovered genes. Applications and Future Challenges • Less than 2 per cent of the genome codes for proteins. • The knowledge of DNA sequence will help us to • Repeated sequences make up very large portion of understand biological system. the human genome. • This will open new fields of science all over the • Repetitive sequences are stretches of DNA sequences world. that are repeated many times, sometimes hundred • This may enable us to use totally new approach in to thousand times which have no direct coding functions, but shed light on chromosome structure, biological research. dynamics and evolution. • New technology and the whole sequence will help us • Chromosome 1 has most genes (2968), and the Y has to study various functions of genes in a particular the fewest (231). tissue or the development of tumours. • This will also help us to understand the working of • Scientists have identified about 1.4 million locations all the genes as interconnected networks in living where single base DNA differences (SNPs–single beings. nucleotide polymorphism, pronounced as ‘snips’) Topic 10. Rice Genome Project: First Crop Genome • The International Rice Genome Sequencing world over in understanding the domestication and Project (IRGSP) was started in 1997 in Singapore evolution of rice. International Symposium. • The information obtained is being used to develop • 200 million US dollars was the total estimated cost. new superior strains of rice. • Eleven countries participated and the major Salient Features of Rice Genome contributions were from Japan, Korea, China, United Kingdom and United States. Brazil joined • The Rice Genome is the smallest of major cereal the last and worked on chromosome 12. crops. • From India, Nagendra Singh of Indian Agricultural • Size of rice genome is 389 Mb. About 37,544 non- Research Institute and Akhilesh Tyagi of University transposable elements related protein coding of Delhi worked on a part of chromosome 11. Sanger’s sequences and 80,127 polymorphic sites have been method was applied to sequence the rice genome. identified. • The milestone achievement was to sequence • About 35% of rice genome is transposons. centromere on chromosome 8. This sequence was found to be complex and highly repetitive. • Single-nucleotide polymorphism (SNP) frequency varies from 0.53 to 0.78%. • The outcome of the project has enabled an easy • Chromosomes 11 and 12 have the lowest gene density research on rice breeding and is helping the scientists compared to the rest of the rice chromosomes. Topic 11. DNA Fingerprinting • The repetitive sequences show high degree of polymorphism, the variation arises at genetic level • DNA fingerprinting technique was developed by Alec due to mutations. If an inheritable mutation is Jeffery. This enabled us to quickly compare the DNA observed in a population at high frequency, it is sequences of two different individuals. referred to as DNA polymorphism. As it is present in non-coding region, such mutations are not expressed • In this technique, some specific regions of DNA called but accumulate in an individual with time. repetitive DNA are identified. These sequences do not code for any protein. They are separated from • This high degree of polymorphism is called Variable bulk genomic ' DNA in the form of different peaks Number Tandem Repeats (VNTR). These are during density gradient centrifugation. These small mini-satellites with large number of repeats. This peaks are called satellite DNA. Depending on the variability in the number of tendem repeats is base composition (A-T rich or C-G rich), length of exploited in the DNA fingerprinting. segment, and number of repetitive units, satellite DNA is classified into further categories called micro-satellite, mini-satellite etc. 128 Biology–12

• DNA fingerprinting involved following steps: Isolation of DNA; digestion of DNA by restriction endonucleases; separation of DNA fragments by electrophoresis; transferring (blotting) of separated DNA fragments to synthetic membranes such as nitrocellulose or nylon; hybridisation using labelled VNTR probe; detection of hybridised DNA fragments by autoradiography. • The bands obtained give a characteristic pattern for an individual which differs from others. • DNA isolated from single cell is amplified using PCR and is subjected to DNA analysis. • DNA fingerprinting is used in forensic science to identify criminals, rapists, solving paternity issues. It also helps us to determine genetic diversity in a given population and establish evolutionary relationship among organisms. Paternal chromosome Maternal chromosome Chromosome 7 Chromosome 7 Chromosome 2 Chromosome 2 Chromosome 16 Chromosome 16 DNA from individual A DNA from individual B Number of short tandem repeats C AB Number of short tandem repeats 0 11 Chromosome 7 12 Chromosome 2 11 10 Chromosome 16 9 8 7 6 5 4 3 2 1 DNA from crime scene (C) Amplified repeats, separated by size on a gel, give a DNA fingerprint Schematic representation of DNA fingerprinting EXERCISE I. Multiple Choice Questions 3. Both deoxyribose and ribose belong to a class of 1. In a DNA strand the nucleotides are linked sugars called: together by: (a) trioses (b) hexoses (a) glycosidic bonds (c) pentoses (d) polysaccharides (b) phosphodiester bonds (c) peptide bonds 4. The fact that a purine base always paired through (d) hydrogen bonds hydrogen bonds with a pyrimidine base leads to, 2. A nucleoside differs from a nucleotide. It lacks in the DNA double helix: the: (a) the antiparallel nature (a) base (b) sugar (b) the semiconservative nature (c) phosphate group (d) hydroxyl group (c) uniform width throughout DNA (d) uniform length in all DNA 5. The net electric charge on DNA and histones is: (a) both positive (b) both negative Molecular Basis of Inheritance  129

(c) negative and positive, respectively 14. Discontinuous synthesis of DNA occurs in one strand, because: (d) zero (a) DNA molecule being synthesised is very long 6. The promoter site and the terminator site for (b) DNA dependent DNA polymearse catalyses transcription are located at: polymerisation only in one direction (5' → 3') (c) it is a more efficient process (a) 3' (downstream) end and 5' (upstream) end, respectively of the transcription unit (d) DNA ligase has to have a role (b) 5' (upstream) end and 3' (downstream) end, respectively of the transcription unit 15. Which of the following steps in transcription is catalysed by RNA polymerase? (c) the 5' (upstream) end (a) Initiation (b) Elongation (d) the 3' (downstream) end (c) Termination (d) All of the above 7. Which of the following statements is the most appropriate for sickle cell anaemia? 16. Given below are the observations drawn in HGP. Select the option that shows the correct (a) It cannot be treated with iron supplements observations.[CBSE 2022] (b) It is a molecular disease (i) The human genome contains 3164.7 billion (c) It confers resistance to acquiring malaria base pairs. (d) All of the above (ii) The average gene consists of 3000 bases. 8. One of the following is true with respect to AUG (iii) Less than 2% of the genome codes for proteins. (a) It codes for methionine only (iv) Chromosome one has most genes (2698). (a) (i) and (ii) (b) (ii) and (iii) (b) It is also an initiation codon (c) (iii) and (iv) (d) (i) and (iii) (c) It codes for methionine in both prokaryotes 17. The phosphoester linkage in the formation of a and eukaryotes nucleotide involves the bonding between (d) All of the above [CBSE 2022] 9. The first genetic material could be: (a) Phosphate group and OH of 3'C of a nucleoside (a) protein (b) carbohydrates (b) Phosphate group and OH of 5'C of a nucleoside (c) DNA (d) RNA (c) Phosphate group and H of 3'C of a nucleoside 10. With regard to mature mRNA in eukaryotes: (d) Phosphate group and H of 5'C of a nucleoside (a) exons and introns do not appear in the mature RNA 18. The switching ‘on’ and ‘off’ of the lac operon in prokaryotes is regulated by [CBSE 2022] (b) exons appear but introns do not appear in the mature RNA (a) Glucose (b) Galactose (c) introns appear but exons do not appear in the (c) Lactose (d) Fructose mature RNA 19. For ‘in-vitro’ DNA replication, which one of the following substrates need to be added along with (d) both exons and introns appear in the mature the necessary enzymes, the DNA template and RNA specific conditions? [CBSE 2022] 11. The human chromosome with the highest and (a) Ribonucleotide triphosphate least number of genes in them are respectively: (b) Deoxyribonucleoside triphosphate (a) Chromosome 21 and Y (c) Deoxyribonucleotide triphosphate (d) Ribonucleoside triphosphate (b) Chromosome 1 and X 20. Which one of the following factor will associate (c) Chromosome 1 and Y transiently with RNA polymerase to terminate (d) Chromosome X and Y transcription in prokaryotes? [CBSE 2022] 12. Who amongst the following scientists had no (a) sigma factor (b) RHO factor contribution in the development of the double (c) delta factor (d) theta factor helix model for the structure of DNA? 21. Choose the correct pair of codon with its (a) Rosalind Franklin (b) Maurice Wilkins corresponding amino acid from the following list: (c) Erwin Chargaff (d) Meselson and Stahl [CBSE 2022] 13. DNA is a polymer of nucleotides which are linked (a) UAG : Glycine (b) AUG : Arginine to each other by 3'-5' phosphodiester bond. To (c) UUU : Phenylalanine (d) UGA : Methionine prevent polymerisation of nucleotides, which of the following modifications would you choose? 22. During elongation process of translation, the peptide bond formation between amino acids is (a) Replace purine with pyrimidines catalysed by [CBSE 2022] (b) Remove/Replace 3' OH group in deoxy ribose (a) ribosomal RNA (c) Remove/Replace 2' OH group with some other (b) protein in small subunit of ribosome group in deoxy ribose (c) protein in large subunit of ribosome (d) Both ‘B’ and ‘C’ (d) transfer RNA 130 Biology–12

23. A region of coding strand of DNA has the 2. The DNA molecule takes a complete turn after following nucleotide sequence: every _____ base pairs. 5' – TGCGCCA – 3' 3. Genes that shuffle from one location to another The sequence of bases on mRNA transcribed by are called ____ . this DNA stand would be: [CBSE 2022] 4. _____ has the shape of clover-leaf. 5. ______ is the technique by which the three- (a) 3' – ACGCGGT – 5' (b) 5' – ACGCGGT – 3' dimensional structures of macromolecules can (c) 5' – UGCGCCA – 3' (d) 3' – UGCGCCA – 5' be studied. 24. A DNA molecule is 160 base pairs long. It has 20% III. True or False adenine. How many cytosine bases are present 1. RNA also serves catalytic functions. 2. Genetic code is ambiguous. in this DNA molecule? [CBSE 2022] 3. Phosphodiester linkage is present in DNA. 4. Histones are rich in lysines and arginines. (a) 192 (b) 96 5. DNA is basic in nature. (c) 64 (d) 42 1 Mark Questions 1. Why is RNA more reactive in comparison to DNA? 25. A template strand in a bacterial DNA has the following base sequence: [Delhi 2015 C] 2. Mention the polarity of the DNA strands a–b and 5' – TTTAACGAGG – 3' What would be the RNA sequence transcribed c–d shown in the replicating fork given below. from this template DNA? [CBSE 2022] bd (a) 5' – AAATTGCTCC – 3' (b) 3' – AAATTGCTCC – 5' (c) 3' – AAAUUGCUCC – 3' (d) 5' – CCUCGUUAAA – 3' 26. t R N A h a s a n _ _ _ _ _ _ _ t h a t h a s b a s e s complementary to the codon. Its actual structure is a compact molecule which looks like _______ . Select the option that has correct choices for the two ‘blanks’. [CBSE 2022] (a) amino acid acceptor end, clover-leaf (b) anticodon loop, clover-leaf a c [All India 2008] (c) amino acid acceptor end, inverted L 3. When and at what end does the ‘tailing’ of hnRNA (d) anticodon loop, inverted L take place?  [All India 2009] 27. Which type of RNA is correctly paired with its 4. Why hnRNA is required to undergo splicing? function?[CBSE 2022] [Delhi 2009] (a) small nuclear RNA : Processes rRNA 5. Mention one difference to distinguish an exon (b) transfer RNA : attaches to amino acid from an intron. [Foreign 2016] (c) ribosomal RNA : involved in transcription 6. Write the function of RNA polymerase II. (d) micro RNA : involved in translation  [Foreign 2015 (C)] 28. The figure given below has labellings (i), (ii) and 7. Give an example of a codon having dual function. (iii), which two labellings in the given figure are [Delhi 2016] components of a nucleosome? Select the correct 8. Mention the role of the codons AUG and UGA option:[CBSE 2022] during protein synthesis. [Delhi 2011] (i) 9. How does a degenerate code differ from an (ii) unambiguous one? [Foreign 2015 C] 10. Mention the contribution of genetic maps in human genome project. [All India 2011] (iii) 11. Name the transcriptionally active region of (a) (i) – HI histone, (ii) – DNA chromatin in a nucleus. [Delhi 2015] (b) (i) – DNA, (ii) – Histone Octamer (c) (ii) – DNA, (iii) – HI Histone 12. Name the negatively charged and positively (d) (ii) – Histone octamer, (iii) – DNA charged components of a nucleosome. II. Fill in the Blanks [Delhi 2015 C] 1. The formation of a peptide bond is catalysed by 13. Name the specific components and the linkages the enzyme _____ . between them that form deoxyguanosine. [All India 2013 C] Molecular Basis of Inheritance  131

14. Name the two basic amino acids that provide 27. Name the types of synthesis 'a' and 'b' occurring positive charges to histone proteins. in the replication fork of DNA as shown below: [Delhi 2016] [Delhi 2012 C] 15. If the base adenine constitutes 31 percent of an 5' 3' isolated DNA fragment, then what is the expected a synthesis b synthesis percentage of the base cytosine in it? [Delhi 2011 C] 5' 3' 16. Name the positively charged proteins around which the negatively charged DNA usually 28. What is a cistron? [All India 2015] wound?[Delhi 2010 C] 17. How is the length of DNA usually calculated? 29. Name the enzyme that transcribes hnRNA in [All India 2009 C] eukaryotes.[Delhi 2015 C] 18. Name the components 'a' and 'b' in the nucleotide with a purine given below: 30. Which one out of Rho factor and Sigma factor act as initiation factor during transcription in a H prokaryote?[Delhi 2013 C] aC 31. Which one of an intron and an exon is the reminiscent of antiquity? [All India 2013 C] bH     [Delhi 2008] 32. Mention the two additional processing which hnRNA needs to undergo after splicing so as to 19. Mention the carbon positions to which the become functional.[Delhi 2009] nitrogenous base and the phosphate molecule are respectively linked in the nucleotide given below: 33. At which ends do capping and tailing of hnRNA occur respectively? [Foreign 2009] H 34. Name the parts 'A' and 'B' of the transcription given below: [Delhi 2008] Phosphate C 3' A 3' S H B base B   [All India 2008 C] 35. Write the two specific codons that a translational H unit of mRNA is flanked by one on either sides. 20. HPohwospdhaotee s DNA polymorphism arise in a [All India 2015 C] population?[Delhi 2014] 36. Which one of the two sub-units of ribosome 21. Name the sourcCe of energy for the replication of encounters an mRNA?[Delhi 2013 C] DNA.S [Delhi 2015 C] 37. Write the scientific importance of single 22. Why is it not poHssible for an alien DNA to become nucleotide polymorphism identified in human part of chbraosemosomB e anywhere along its length genome.[Foreign 2014] and replicate normally? [All India 2014] 38. Mention any two ways in which single nucleotide 23. What will happen if DNA replication is not polymorphism (SNPs) identified in human followed by cell division in a eukaryotic cell? genome can bring revolutionary change in [All India 2014 C] biological and medical sciences. [All India 2011 C] 24. Name the enzyme and state its property that 39. How is repetitive/satellite DNA separated is responsible for continuous and discontinuous from bulk genomic DNA for various genetic replication of the two strands of a DNA molecule. [Delhi 2013] experiments?[Delhi 2014] 40. W r i t e t h e d u a l p u r p o s e s e r v e d b y Deoxyribonucleoside triphosphates in 25. Name the enzyme that joins the small fragment of DNA of a lagging strand during DNA replication. polymerisation. [CBSE 2018] [Delhi 2013 C] 2 Marks Questions 26. Name the enzyme involved in the continuous 41. (a) Draw a neat labelled diagram of a nucleosome. (b) Mention what enables histones to acquire a replication of DNA strand. Mention polarity of positive charge.  [All India 2012] templet strand. [All India 2010] 132 Biology–12

OR A B H H How do histones acquire positive charge? [Delhi 2011] PC PC 42. Discuss the role the enzyme DNA ligase plays H H OH G U during DNA replication. [Delhi 2016] 43. Draw a neat labelled sketch of a replicating fork (a) Name the type of sugar guanine base is of DNA.  [Delhi 2012] attached to. OR (b) Name the linkage connecting the two Draw a labelled schematic sketch of replication nucleotides. fork of DNA. Explain the role of the enzymes (c) Identify the 3' end of the dinucleotide. Give a involved in DNA replication.  [Delhi 2009] reason for your answer. [Delhi 2010 C] 44. Explain the dual function of AUG codon. Give the 52. Study the given portion of double stranded sequence of bases it is transcribed from and its polynucleotide chain carefully. Identify a, b, c anticodon.  [All India 2009] and the 5' end of the chain. [All India 2009] 45. How is the translation of mRNA terminated? Explain. H d H 46. i po z ya PC Transcription PC H OH H mRNA lac mRNA Translation b X Enzymes a Inducer Hc OH H (a) Name the molecule ‘X’ synthesized by ‘i’ gene. How does this molecule get inactivated? CP CP (b) Which one of the structural genes codes for b-galactosidase? H H (c) When will the transcription of this gene stop? [All India 2009] 53. (a) A DNA B C Protein Look at the m –RNA 47. Study the figure given below and answer the questions: above sequence and mention the events A, B and C. i po zya (b) What does central dogma state in molecular biology? How does it differ in some viruses? mRNA lac mRNA [Delhi 2009 C] Enzymes 54. State the central dogma in molecular biology. Who proposed it? Is it universally applicable? Explain.[All India 2008 C] Inactive repressor 55. Make a labelled diagram of an RNA showing its 3'–5' polarity. [All India 2010 C] (a) How does the repressor molecule get inactivated? 56. State the functions of the following in a prokaryote: (b) When does the transcription of lac mRNA stop? (a) tRNA (b) rRNA (c) Name the enzyme transcribed by the gene ‘Z’.  [All India 2010 C] [Delhi 2009] 57. Following are the features of genetic codes. What 48. Mention two applications of DNA polymorphism. does each one indicate? Stop codon; Unambiguous [Foreign 2016] codon; Degenerate codon; Universal codon. 49. Write the full form of VNTR. How is VNTR [All India 2016] different from ‘Probe’? [All India 2011] 58. (a) Name the scientist who suggested that the genetic code should be made of a combination 50. Draw a schematic diagram of a part of double of three nucleotides. stranded dinucleotide DNA chain having all the (b) Explain the basis on which he arrived at this four nitrogenous bases and showing the correct conclusion.[Delhi 2014] polarity.[Delhi 2012] 51. Answer the following based on the dinucleotide 59. One of the salient features of the genetic code shown below. is that it is nearly universal from bacteria to Molecular Basis of Inheritance  133

humans. Mention two exceptions to this rule. 77. Genetic code is specific and nearly universal. Why are some codes said to be degenerate? Justify.[Delhi 2008 C] [Foreign 2014] 60. Genetic codes can be universal and degenerate. 78. How would lac operon operate in E. coli growing Write about them, giving one example of each. in a culture medium where lactose is present as [All India 2013] 61. Explain the structure of a tRNA and state why source of sugar? [All India 2014 C] it is known as an adaptor molecule. 79. i p o x y a [All India 2012 C] 62. Name the category of codons UGA belongs to. Given above is a schematic representation of the lac operon in E. coli. What is the significant role Mention another codon of the same category. of 'i' gene in switching on or off the operon? Explain their role in protein synthesis. [All India 2013 C] [Foreign 2009] 63. What is aminoacylation? State its significance. 80. Differentiate between the genetic codes given below: [All India 2017] [All India 2016] 64. State the functions of ribozyme and release factor (a) Unambiguous and Universal in protein synthesis respectively. (b) Degenerate and Initiator [All India 2015 C] 81. Although a prokaryotic cell has no defined 65. Where does peptide bond formation occur in a nucleus, yet DNA is not scattered throughout the bacterial ribosome and how? [Foreign 2014] cell. Explain. [CBSE 2018] 66. Explain aminoacylation of tRNA. [All India 2013, Delhi 2014 C] 3 Marks Questions 67. Mention the role of ribosomes in peptide-bond 82. (i) What is the given diagram representing? formation. How does ATP facilitate it? (ii) Name the parts a, b, c and d. [All India 2010] (iii) In the eukaryotes the DNA molecules are organized within the nucleus. How is the 68. Which human chromosome has (a) maximum DNA molecule organized in a bacterial cell number of genes, and which one has (b) fewest in absence of a nucleus? [All India 2009] genes?[Foreign 2014] bc 69. Expand 'BAC' and 'YAC'. Explain how they are used in sequencing human genome. a [Delhi 2011 C] d 70. What is Central Dogma? Who proposed it? [All India 2015 C] 83. List the salient features of double helix structure 71. Why is DNA considered a better genetic material? of DNA.  [All India 2012] [All India 2013 C] 84. (a) Draw the structure of the initiator tRNA 72. State the dual role of deoxyribonucleoside adaptor molecule. triphosphates during DNA replication. (b) Why is tRNA called an adaptor molecule? [Delhi 2011]  [All India 2008] 73. Compare the roles of the enzymes DNA 85. Unambiguous, universal and degenerate are polymerase and DNA ligase in the replication. some of the terms used for the genetic code. [All India 2008 C] Explain the salient features of each one of them. [All India 2011] 74. State the difference between the structural genes in a transcription unit of prokaryotes and 86. The base sequence in one of the strands of DNA eukaryotes.[All India 2014] is TAGCATGAT 75. A template strand is given below. Write down (i) Give the base sequence of its comp-lementary the corresponding coding strand and the mRNA strand. strand that can be formed, along with their polarity. (ii) How are these base pairs held together in a DNA molecule? 3' ATGCATGCATGCATGCATGCATGC5' (iii) Explain the base complementarity rules. [Foreign 2014] Name the scientist who framed this rule. 76. (a) Differentiate between 'unambiguous' and [Delhi 2011] degenerate' codons. 87. Describe the packaging of DNA helix in a prokaryotic cell and an eukaryotic nucleus. (b) Write two functions of the codon AUG. [Foreign 2016] [Delhi 2008 C] 134 Biology–12

88. (a) How are the following formed and involved in 99. (a) A DNA segment has a total of 1,500 DNA packaging in a nucleus of a cell? nucleotides, out of which 410 are Guanine containing nucleotides. How many pyrimidine (i) Histone octamer (ii) Nucleosome bases this segment possesses? (iii) Chromatin (b) Draw a diagrammatic sketch of a portion of (b) Differentiate between euchromatin and DNA segment to support your answer. heterochromatin.  [Delhi 2016] 89. (a) Why did Hershey and Chase use radioactive  [Delhi 2015] sulfur and radioactive phosphorus in their 100. (a) A DNA segment has a total of 2,000 experiment?  nucleotides, out of which 520 are adenine containing nucleotides. How many purine (b) Write the conclusion they arrived at and how? bases this DNA segment possesses?  [Foreign 2016] 90. It is established that RNA is the first genetic (b) Draw a diagrammatic sketch of a portion of material. Explain giving three reasons. DNA segment to support your answer. [Delhi 2012]  [Delhi 2015] 91. How was a heavy isotope of nitrogen used to provide 101. Draw a labelled diagram of a nucleosome. Where is it found in a cell?  [Foreign 2014] experimental evidence to semiconservative mode 102. Describe the structure of a RNA poly-nucleotide of DNA-replication?  [Foreign 2015 C] chain having four different types of nucleotides. [Delhi 2013] OR Describe the experiment that helped demonstrate 103. With the help of a schematic diagram, explain the location and the role of the following in a the semiconservative mode of DNA replication. transcription unit: [Delhi 2016] 92. (a) Differentiate between a template strand and Promoter, Structural gene, Terminator. coding strand of DNA. [Delhi 2014 C] (b) Name the source of energy for the replication 104. (a) Construct a complete transcription unit with of DNA.  [Delhi 2015 C] promotor and terminator on the basis of the 93. (i) Name the enzyme that catalyses the hypothetical template strand given below: transcription of hnRNA. A TGCATGCATAC (ii) Why does the hnRNA need to undergo (b) Write the RNA strand transcribed from changes? List the changes hnRNA undergoes and where in the cell such changes take place? the above transcription unit along with its [All India 2011] polarity.[Delhi 2012] 94. (a) Name the enzyme responsible for the 105. D e s c r i b e w i t h t h e h e l p o f a s c h e m a t i c transcription of tRNA and the amino acid the representation the structure of a transcription initiator tRNA gets linked with. unit.  [All India 2012 C] (b) Explain the role of initiator tRNA in initiation 106. Monocistronic structural genes in eukaryotes have interrupted coding sequences. Explain. How of protein synthesis. [Delhi 2012] are they different in prokaryotes? [Delhi 2011 C] 95. How are the structural genes activated in the lac operon in E. coli?  [All India 2012] 107. Differentiate between the following: 96. A considerable amount of lactose is added to the (a) Promoter and terminator in a trans-cription growth medium of E. coli. How is the lac operon switched on in the bacteria? Mention the state of unit. the operon when lactose is digested? (b) Exon and intron in an unprocessed eukaryotic mRNA.  [All India 2011 C] [All India 2010 C] 108. Describe the initiation process of tran-scription in bacteria.  [Delhi 2010] 97. What are satellite DNA in a genome? Explain their role in DNA fingerprinting. [All India 2009] 109. Describe the elongation process of transcription in bacteria.  [All India 2010] 98. (a) A DNA segment has a total of 1000 nucleotides, out of which 240 are adenine containing 110. Given below is a part of the template strand of a nucleotides. How many pyrimidine bases this structural gene: DNA segment possesses?   TAC  CAT  TAG  GAT (b) Draw a diagrammatic sketch of a portion of (a) Write its transcribed mRNA strand with its DNA segment to support your answer. polarity. [Delhi 2015] (b) Explain the mechanism involved in initiation of transcription of this strand. [Delhi 2008] Molecular Basis of Inheritance  135

111. Explain the role of DNA-dependent RNA (b) Mention the DNA sequence coding for serine polymerase in transcription.  [Delhi 2008 C] and the anticodon of tRNA for the same amino acid. 112. (a) Name the scientist who postulated the presence of an adapter molecule that can (c) Why are some untranslated sequence of bases seen in mRNA coding for a polypeptide? assist in protein synthesis. Where exactly are they present on mRNA? [Foreign 2009] (b) Describe its structure with the help of a diagram. Mention its role in protein synthesis. [Foreign 2014] 119. Explain the role of regulatory gene in a lac 113. (a) Name the enzyme responsible for the operon. Why is regulation of lac operon called transcription of tRNA and the amino acid the negative regulation?  [Delhi 2010 C] initiator tRNA gets linked with. 120. Explain the process of translation. (b) Explain the role of initiator tRNA in initiation [All Inidia 2014 C] of protein synthesis. [Delhi 2012] 121. (a) What do 'Y' and 'B' stand for in 'YAC' and 114. (a) Name the scientist who called tRNA as an 'BAC' used in Human Genome Project (HGP). adaptor molecule. Mention their role in the project. (b) Draw a clover leaf structure of tRNA showing (b) Write the percentage of the total human the following: genome that codes for proteins and the percentage of discovered genes whose (i) Tyrosine attached to its amino acid site. functions are known as observed during HGP. (ii) Anticodon for this amino acid in its correct (c) Expand 'SNPs' identified by scientists in site (codon for tyrosine is UAC). (c) What does the actual structure of tRNA look HGP. [All India 2016 C] like?  [All India 2011] 122. In human genome which one of the chromosomes 115. (a) How many codons code for amino acids and has the most genes and which has the fewest? how many do not? [All India 2009] (b) Explain the following giving one example of 123. The length of a DNA molecule in a typical each: mammalian cell is calculated to be approximately 2.2 meters. How is the packaging of this long (i) unambiguous and specific codon (ii) degenerate codon molecule done to accommodate it within the (iii) universal codon nucleus of the cell?  [Delhi 2008] (iv) initiator codon [All India 2010 C] 124. Why is DNA a better genetic material when 116. (a) Draw the structure of the initiator tRNA compared to RNA?  [Delhi 2015 C] adaptor molecule. 125. Frederich Griffith claimed that R-strain (b) Why is tRNA called an adaptor molecule? streptococcus pneumoniae had been transformed [All India 2008] by heat-killed S-strain bacteria. Explain the 117. How do the initiation and termination of findings.  [All India 2012 C] translation process occur in bacteria? Where 126. In a series of experiments with Strep-tococcus are untranslated regions located in an mRNA? and mice F. Griffith concluded that R-strain bacteria had been transf-ormed. Explain. Mention their role.  [All India 2010 C] 118. [All India 2010] 127. Explain the role of 35S and 32P in the experiments conducted by Hershey and Chase. SER  [All India 2010 C] 128. List any four properties of a molecule to be able to act as a genetic material.  [All India 2008 C] c 129. (a) Why did Meselson and Stahl use 14N and 15N isotopes in the sources of nitrogen present a a' in the culture medium in their experiment? GAGCUG AG UUA A Explain. (a) Identify the polarity from a to a' in the (b) Write conclusion drawn by them from the above diagram and mention how many more experiment.  [Delhi 2012 C] amino acids are expected to be added to this polypeptide chain. 130. Why do you see two different types of replicating strands in the given DNA replication fork? Explain. Name these strands. 136 Biology–12

3' 5' 137. (a) D r a w a s c h e m a t i c r e p r e s e n t a t i o n o f transcription unit showing the polarity of both the strands. Label the promoter gene 3' and the template strand. 5' (b) Mention the condition when template strand becomes coding strand. (c) Give the function of the promotor gene. 5' 3' 5' 3'  [Delhi 2016] [All India 2009 C] 138. (a) Draw schematic representation of the 131. Answer the following questions based on Meselson and Stahl's experiment: transcription unit of DNA showing its (a) Write the name of the chemical substance polarity. Label terminator and coding strand used as a source of nitrogen in the experiment in it. by them. (b) Explain the role of promoter gene in this unit. (b) Why did the scientists synthesis the light and [Delhi 2008 C] the heavy DNA molecules in the organism 139. Correlate the codons of mRNA strand with amino used in the experiment? acids of a polypeptide translated. (c) How did the scientists make it possible to 5' AUGACCUUUCAUUCGUGUAA 3' distinguish the heavy DNA molecule from the → mRNA light DNA molecule? Explain. Met-Thr-Phe-His-Ser-Cys. (d) Write the conclusion the scientists arrived at → Translated polypeptide after completing the experiment?  Infer any 3 properties of genetic code with [All India 2011] examples from the above information.  132. (a) Identify the polarity at A and B respectively in the figure given below. [Delhi 2010 C] 140. (a) Why is tRNA called an adaptor? (b) Draw and label a secondary structure of Template DNA tRNA. How does the actual structure of tRNA (parental strands) look like?  [All India 2010 C] 141. Identify giving reasons, the salient features of genetic code by studying the following nucleotide AB sequence of mRNA strand and the polypeptide Newly synthesised strands (b) Explain the mechanism the figure represents. translated from it. [All India 2011 C] AUG UUU UCU UUU UUU UCU UAG 133. Explain the mechanism of DNA replication in Met - Phe - Ser - Phe - Phe - Ser[Delhi 2009 C] bacteria. Why is DNA replication said to be 142. One of the codons on mRNA is AUG. Draw the structure of tRNA adaptor molecule for this semiconservative?  [All India 2010 C] 134. (a) Why does DNA replication occur in small codon. Explain the uniqueness of this tRNA? replication forks and not in its entire length? [Delhi 2008] (b) Why is DNA replication continuous and 143. How are the structural genes inactivated in lac operon in E. coli? Explain.  [Delhi 2012] discontinuous in a replication fork? (c) Explain the importance of 'origin of replication' 144. Given below is a schematic representation of lac in a replication fork. [All India 2009 C] operon: 135. (a) What are the transcriptional pro-ducts of RNA 1i po z ya polymerase III? (b) Differentiate between 'Capping' and' 'Tailing'. (c) Expand hnRNA.  [All India 2014 C] 136. (a) Draw a schematic representation of the structure of a transcription unit and show the following in it: (a) Identify i and p. (i) Direction in which the transcription (b) Name the 'inducer' for this operon and explain occurs its role.  [Foreign 2011] (ii) Polarity of the two strands involved (iii) Template strand 145. i po z ya (iv) Terminator gene (b) Mention the function of promoter gene in transcription.  [All India 2009] M Molecular Basis of Inheritance  137

(a) Name the molecule 'M' that binds with the 153. Name a technique to establish the paternity of a operator. new-born baby. Describe the procedure that you (b) Mention the consequences of such binding. would follow. [All India 2008 C] (c) What will prevent the binding of the molecule 154. (a) List the two methodologies which were 'M' with the operator gene? Mention the event involved in human genome project. Mention that follows.  [Foreign 2009] how they were used. 146. Following the collision of two trains a large (b) Expand ‘YAC’ and mention what was it used for.[All India 2017] number of passengers are killed. A majority of 155. (a) Expand VNTR and describe its role in DNA them are beyond recognition. Authorities want to hand over the dead to their relatives. Name a fingerprinting. modern scientific method and write the procedure (b) List any two applications of DNA fingerprinting that would help in the identification of kinship. technique. [CBSE 2018] [Delhi 2015] 5 Marks Questions 147. The following is the flow chart highlighting the steps 156. Answer the following questions based on Hershey in DNA fingerprinting technique. Identify a, b, c, d, and Chases’s experiments: e and f. (a) Name the kind of virus they worked with and why? Isolation of DNA from blood cells (b) Why did they use two types of culture media ↓ to grow viruses in? Explain. Cutting of DNA by 'a' (c) What was the need for using a blender and ↓ later a centrifuge during their experiments? Separation of DNA fragments by elec- (d) State the conclusion drawn by them after the trophoresis using 'b' experiments.  [Delhi 2016] ↓ 157. List the criteria a molecule that can act as genetic material must fulfill. Which one of the criteria are Transfer (blotting) of fragments to 'c' gel ↓ best fulfilled by DNA or by RNA thus making one DNA splits into single strand of them a better genetic material than the other? ↓ Explain.[Delhi 2016] Introduction of labelled 'd' probe 158. Describe Frederick Griffith’s experiment on Streptococcus pneumoniae. Discuss the ↓ conclusion he arrived at.  [All India 2012] 'e' of single strands with 'd' 159. (a) Describe the experiment which demo-nstrated ↓ the existence of “transforming principle”. Detection of banding pattern by 'f ' (b) How was the biochemical nature of this “transforming principle” determined by [Delhi 2010 C] Avery, MacLeod and McCarty?  148. Explain the significance of satellite DNA in DNA fingerprinting technique.  [All India 2015] [Foreign 2015 C] 149. A burglar in a huff forgot to wipe off his bloodstains 160. How did Alfred Hershey and Martha Chase from the place of crime where he was involved in conclusively establish that DNA is the genetic a theft and fight. Name the technique which can material? Explain.  [Foreign 2015 C] help in identifying the burglar from the blood- 161. How did Griffith prove transforming principle in Genetics. Explain the procedure. [Delhi 2015 C] stains. Describe the technique. [All India 2013] 150. In a maternity clinic, for some reason, the 162. (a) Explain the experiment performed by Griffith authorities are not able to hand over the two new- on Streptococcus pneumoniae. What did he born babies to their respective parents. Name and conclude from this experiment? describe the technique you would suggest to sort (b) Name the three scientists who followed up Griffith’s experiments. out the matter.  [All India 2013] 151. (a) Explain DNA polymorphism on the basis of genetic mapping of human genome. (c) What did they conclude and how? [Delhi 2009] (b) State the role of VNTR in DNA fingerprinting. 163. How did Hershey and Chase prove that DNA is the hereditary material? [All India 2013] 152. \"A very small sample of tissue or even a drop of Explain their experiment with suitable diagrams. blood can help determine pate-rnity\". Provide [All India 2009] a scientific explanation to substantiate the 164. What is semi-conservative DNA replication? How was it experimentally proved and by whom? statement.[All India 2015] [All India 2008] 138 Biology–12

OR 173. (a) Write the conclusion drawn by Griffith at the end of his experiment with Streptococcus State the aim and describe Messelson and Stahl’s pneumoniae. experiment.  [Delhi 2012] (b) How did O. Avery, C. Macleod and M. 165. A McCarty prove that DNA was the genetic 3' 5' material? Explain. [All India 2013] 5' 3' 174. How were the following proved exper-imentally? B Explain. (a) Biochemical nature of the 'transforming (a) Identify strands ‘A’ and ‘B’ in the diagram of principle' of Griffith's experiment. transcription unit given above and write the (b) DNA is the genetic material. basis on which you identified them.  [All India 2010 C] 175. (a) How did Griffith explain the trans-formation of (b) State the functions of Sigma factor and R strain (non-virulent) bacteria into S strain Rho factor in the transcription process in a (virulent)? bacterium. (b) Explain how Macleod, McCarty and Avery determined the biochemical nature of the (c) Write the functions of RNA polymerase-I and molecule responsible for transforming R RNA polymerase-III in eukaryotes. strain bacteria into S strain bacteria. [Foreign 2016] 166. Explain the process of transcription in Eukaryotes.  [All India 2009] [Foreign 2015 C] 176. (a) Describe the process of transcription in bacteria.[All India 2014 C] 167. Study the mRNA segment given below which (b) Explain the processing the hnRNA needs to is complete to be translated into a polypeptide undergo before becoming functional mRNA of chain. eukaryotes.  [All India 2016] a UACGAG AGAUUU b 177. Explain the process of transcription in prokaryotes. How is the process different in (i) Write the codons for ‘a’ and ‘b’. eukaryotes?  [All India 2012, 2015] (ii) What do they code for? 178. (a) What is a genetic code? (iii) How is peptide bond formed between two (b) Explain the following: amino acids in the ribosome? Degenerate code; Unambiguous code; Initiator code.  [Delhi 2014 C] 168. Explain the role of lactose as an inducer in a lac 179. Name the major types of RNAs and explain operon.  [Delhi 2016] their role in the process of protein synthesis in a 169. (a) Explain the chemical structure of a single prokaryote.  [Foreign 2014] stranded polynucleotide chain. 180. Explain the process of translation. (b) Describe the salient features of the double [Delhi 2014 C] helix structure of DNA molecule. 181. Explain the role of lactose as an inducer in a lac [All India 2012 C] operon.  [Delhi 2016] 170. (a) Mention the contributions of the following 182. Sketch a schematic diagram of lac operon in scientists: switched on position. How is the operon switched (i) Maurice Wilkins and Rosalind Franklin off? Explain. [All India 2015 C] 183. (a) Differentiate between repetitive and satellite (ii) Erwin Chargaff DNA. (b) Draw a double-stranded dinucleotide chain (b) How can satellite DNA be isolated? Explain. with all the four nitrogen bases. Label (c) List two forensic applications of DNA the polarity and the components of the fingerprinting.  [Delhi 2012 C] dinucleotide.  [All India 2011 C] 184. How did Hershey and Chase establish that DNA 171. (a) Describe the experiment which demon-strated is transferred from virus to bacteria? [Delhi 2015] the existence of \"transforming principle\". 185. Describe the Hershey and Chase experiment. (b) How was the biochemical nature of this Write the conclusion drawn by the scientist after \"transforming principle\" determined by their experiment.  [All India 2014] Avery, Macleod and McCarty?[Foreign 2015] 186. Name the scientists who proved experimentally 172. (a) Describe the various steps of Griffith's that DNA is the genetic material. Describe their experiments that led to the conclusion of the experiment.  [Delhi 2012] 187. (a) Name the stage in cell cycle where DNA Transforming Principle. replication occur. (b) H o w d i d t h e c h e m i c a l n a t u r e o f t h e (b) Explain the mechanism of DNA replication. Transforming Principle get established? Highlight the role of enzymes in the process.  [All India 2014, 2013 C] Molecular Basis of Inheritance  139

(c) W h y i s D N A r e p l i c a t i o n s a i d t o b e 197. W h o p r o p o s e d t h a t D N A r e p l i c a t i o n i s semiconservative?  [All India 2016] semiconservative? How was it experimentally proved by Meselson and Stahl? 188. Describe Meselson and Stahl's experiment that was carried in 1958 on E. coli. Write the conclusion [All India 2008 C, 2008, 2009 C] they arrived at after the experiment. 198. Describe the Hershey and Chase experiment. Write [All India 2016] the conclusion drawn by the scientist after their 189. (a) Draw a labelled diagram of a replicating experiment.[All India 2014] fork\" showing the polarity. Why does DNA replication occur within such 'forks'? 199. i p o z y a Given above is the schematic repres-entation (b) Name two enzymes involved in the process of DNA replication, along with their properties. of lac operon of E. coli. Explain the functioning [All India 2015] of this operon when lactose is provided in the growth medium of the bacteria. [Delhi 2013 C] 190. Explain the process of DNA replication with the help of a replicating fork.[Delhi 2015 C] 200. (a) State the arrangement of different genes that in bacteria is referred to as 'operon'. 191. (a) Explain the process of DNA replication with (b) Draw a schematic labelled illustration of lac the help of a schematic diagram. operon in 'switched on' state. (b) In which phase of the cell cycle does replication (c) Describe the role of lactose in lac-operon. occur in eukaryotes? What would happen [All India 2011] if cell-division is not followed after DNA 201. (a) Who proposed the concept of lac operon? replication.  [Delhi 2014] (b) Draw a labelled schematic repre-sentation of a lac operon. 192. Describe Meselson and Stahl's experiment and write the conclusion they arrived at. (c) Explain how does this operon get switched [Delhi 2014 C] 'on' and 'off'. [Delhi 2008 C] 3' 5' 193. 202. Name and describe the technique that will help in solving a case of paternity dispute over the custody of a child by two different families. [All India 2010 C] 5' 3' 203. (a) Describe the structure and function of a t-RNA molecule. Why is it referred to as an adapter molecule? (a) Identify the structure shown above. (b) Explain the process of splicing of hn-RNA in a eukaryotic cell. [All India 2017] (b) Redraw the structure as a replicating fork and label the parts. 204. Write the different components of a lac-operon in E. coli. Explain its expression while in an ‘open’ (c) Write the source of energy for this replication state.[All India 2017] and list the enzymes involved in this process. (d) Mention the difference in the synthesis based 205. (a) What is an operon? on the polarity of the two template strands. [All India 2013 C] (b) Explain how a polycistronic structural gene 194. H o w d i d M e s e l s o n a n d S t a h l e x p e r i - is regulated by a common promoter and a mentally prove that DNA replication is semi- combination of regulatory genes ina lac- conservative? Explain. [All India 2013 C] operon.[All India 2017] 195. State the aim and describe Meselson and Stahl's 206. (a) Absence of lactose in the culture medium affects the expression of a lac-operon in experiment.[Delhi 2012 C] 196. (a) What did Meselson and Stahl observe when E.coli. Why and how? Explain. (i) They cultured E. coli in a medium (b) Write any two ways in which the gene caonndtaceinnitnrgifu15gNedH4thCel for a few generations content? expression is regulated in eukaryotes. (ii) They transferred one such bacterium [All India 2017] tcoulttuhreednoforrmtawlomgeendeiruamtioonfs?NH4Cl and 207. (a) State the ‘Central dogma’ as proposed by Francis Crick. Are there any exceptions to (b) What did Meselson and Stahl conclude from it? Support your answer with a reason and this experiment? Explain with the help of an example. diagrams. (b) Explain how the biochemical charac-terisation (nature) of “Transforming Principle’ was (c) Which is the first genetic material? Give determined, which was not defined from reasons in support of your answer. Griffith’s experiments. [CBSE 2018] [Delhi 2009, 2007] 140 Biology–12

Answers I. Multiple Choice Questions 15. A = 31%, that means, T = 31%, rest are C and G which equals to be 38%. Therefore, percentage of 1. (b) 2. (c) 3. (c) 4. (c) 5. (c) 6. (b) 7. (d) 8. (d) 9. (d) 10. (b) C will be 38/2 = 19% 11. (c) 12. (d) 13. (b) 14. (b) 15. (d) 16. (b) 17. (b) 18. (c) 19. (b) 20. (b) 16. Histones 21. (c) 22. (a) 23. (c) 24. (b) 25. (d) 17. It is calculated by multiplying the total number of 26. (d) 27. (b) 28. (d) base pairs with distance between two consecutive base pairs II. Fill in the Blanks 18. a- phosphate, b- pyrimidine 19. Phosphate is linked to 5th carbon and nitrogenous 1. peptidyl transferase 2. 10 base is joined to 1st carbon. 3. jumping genes 4. tRNA 5. X-ray analysis 20. Polymorphism (arises due to mutations). 21. Deoxyribonucleoside triphosphates provide III. True or False energy in addition to acting as substrates. 22. This DNA must be linked to ori/origin of 1. True 2. False 3. True 4. True 5. False replication site to start replication. 1 Mark Questions 23. It will result in a chromosomal anomaly called 1. Presence of 2'-OH in sugar, and uracil in RNA polyploidy. makes it reactive. 24. The enzyme is DNA dependent DNA polymerase, 2. Polarity of a–b is 3'–5' and of c–d is 5'–3' it uses a DNA template to catalyse the 3. Tailing occurs during processing of hnRNA into polymerisation of deoxyribonucleotides and can functional mRNA. It is added at 3' end of hnRNA. do this only in one direction, that is 5'-3'. 25. DNA ligase. 4. The hnRNA is non-functional, and undergos 26. DNA dependent DNA polymerase, the polarity splicing to remove introns and exons join in a of template strand will be 3’-5’ defined manner. 27. 'a' Continuous, 'b' discontinuous. 5. Exon is a part of hnRNA which is expressed and 28. Segment of DNA coding for a polypeptide forms functional mRNA whereas introns are the 29. RNA polymerase II intervening sequence of nucleotides which do not 30. Sigma factor appear in processed mRNA. 31. Introns 6. RNA polymerase II transcribes precursor of 32. Capping and tailing mRNA i.e. hnRNA. 33. Capping occurs at 5’ end and tailing occurs at 3’ 7. AUG end. 8. AUG codes for methionine and also acts as an 34. A: promoter, B: coding strand initiation codon. 35. Initiator codon ‘AUG’ and any one of the three UGA does not code for any amino acid, it acts as stop codon, UAA, UAG and UGA. stop codon. 36. Smaller subunit 37. Single nucleotide polymorphism helps in finding 9. Degenerate code codes one amino acid by more chromosomal locations for disease-associated than one codon. Whereas unambiguous code means each codon codes for one amino acid. sequences and tracing human history. 10. The HGP involved several stages of which the 38. Refer Answer no. 37 first stage was to construct physical and genetic 39. Density gradient centrifugation. maps of human chromosomes. Genetic mapping 40. Two functions of deoxyribonucloside triphosphates helped in determining the relative positions of in polymerisation are markers and genes on chromosome. • Acts as substrate 11. Euchromatin • Provides energy 12. Negatively charged: DNA 41. (a) DNA H1 histone Positively charged: Histones 13. Guanine is linked to the pentose sugar through Histone a N-glycosidic linkage and form a nucleoside, octamer deoxyadenosine. Core of histone 14. Lysine and arginine molecules Molecular Basis of Inheritance  141

(b) Histones are rich in basic amino acids, lysine (c) 3' end of di-nucleotide is towards–OH end; because –OH is attached on 3' carbon of De- and arginine, which are positively charged oxyribose sugar. due to their side chains. 52. A- triple hydrogen bond between C and G, b-purine, c-sugar, d-phosphate group, 5’ end is 42. During replication, one strand is synthesized in towards phosphate group of each strand. a discontinuous manner forming DNA fragments 53. A- replication; B- transcription; formed called Okazaki fragments which are C- translation Central dogma is the flow of genetic information joined with the help of DNA ligase to from a from DNA → RNA → protein. In virus, the flow continuous stand. of information takes place from RNA → DNA due to reverse transcription. 43. 5' 3' 54. Central dogma is the flow of genetic information Template from DNA → RNA → protein. Francis Crick proposed it. It is not universally accepted as in Continuous 5' Discontinuous synthesis virus, the flow of information takes place from RNA → DNA due to reverse transcription. synthesis 3' (lagging strand/Okazaki (leading strand) fragments) Newly 55. 5' 3' synthesised 5' Phosphate A 5' strands 3' DNA polymerase catalyses the poly-merisation G of nucleotides, Ligase joins the fragments of discontinuous synthesis. U 44. AUG codes for Methionine and also acts as C initiation codon. 3' The sequence of bases from which it is transcribed 56. Function of tRNA in prokaryote: it carries amino is TAC. Its anticodon on tRNA is UAC. acid for translation 45. When ribosome encounters a stop codon (UAG/ Function of rRNA in prokaryote: it forms UGA/UAA) ribosomes. 46. (a) ‘i’ gene synthesizes repressor. This molecule 57. UAA, UGA, UAG are called as stop codons as gets inactivated when lactose (inducer) binds they do not code for any amino acids. with it. Unambiguous codon: One codon codes for only one amino acid and is specific. (b) Z gene Degenerate codon: Some amino acids are coded (c) In the absence of inducer lactose and when by more than one codon. repressor binds with the operator. Universal codon: The code is nearly universal for example, from bacteria to human UUU would 47. (a) When lactose acts as inducer and binds to code for Phenylalanine (phe). Some exceptions repressor. to this rule have been found in mitochondrial codons, and in some protozoans. (b) In the absence of inducer lactose and when repressor binds with the operator. 58. (a) George Gamow (b) There are only four bases A, T, C and G. If (c) b-galactosidase only one base codes for amino acid, then only 4 48. Genetic mapping and DNA fingerprinting. amino acids are coded . If the code is doublet, 49. VNTR–Variable Number of Tandem Repeats. there will be 16 possible combinations so they will code for only 16 amino acids. But if the Probe is a radioactive labelled single stranded code is triplet, 64 combinations are possible. This will cover all twenty amino acids. DNA which is used to hybridise DNA fragments. 59. The code is nearly universal: for example, 50. H H H H from bacteria to human UUU would code for PC PC Phenylalanine (phe). Some exceptions to this rule PC PC have been found in mitochondrial codons, and in H H H H some protozoans. Some amino acids are coded by H H H hydrogen HO C P bonds CP CP H H H H C P 5' 51. (a) deoxyribose H (b) 3'-5' phosphodiester linkage 142 Biology–12

more than one codon, e.g. CUU and CUC both 70. The genetic information flows from DNA → RNA code for Leucine. → protein. This is central dogma and is proposed 60. Refer Answer no. 58 by Francis Crick. 61. The tRNA is also called soluble RNA. The molecule is clover leaf like (within cell it looks 71. DNA is structurally more stable, it can replicate, like inverted L) and has an anticodon loop that mutate slowly and express itself in Mendelian has bases complementary to code. It has an amino way. Being double stranded and presence of acid acceptor end and amino acid binds to it. For thymine provides additional stability to DNA. each amino acid, there is a specific tRNA. There are no tRNAs for stop codons. They are called 72. They act as substrates and also provide energy adaptor molecules because they serve as a link for polymerization reaction. between the mRNA and amino acid sequence of proteins. 73. DNA polymerase enzyme polymerizes new strand 62. It is a stop codon. Other codons of same category of DNA in 5’-3’ direction on DNA template both are UAA and UAG. They are responsible for during continuous as well as discontinuous termination of translation hence called stop synthesis whereas DNA ligase joins two codon. fragments of DNA in discontinuous synthesis. Ser 74. Structural genes in prokaryotes are polycistronic whereas in eukaryotes they are monocistronic. tRNA Introns are present in eukaryotes but are absent in prokaryotes. UC A Anticodon 75. Coding strand: 5’ – TACGTACGTACGTACGTACGTACG- 3’ 63. When an amino acid is linked to its cognate tRNA in the presence of ATP is called charging of tRNA mRNA strand: or aminoacylation of tRNA. If two such charged tRNAs are brought close enough, the formation 5’-UACGUACGUACGUACGUACGUACG -3’ of peptide bond between them would be favoured energetically. 76. (a) Refer answer no. 56. (b) Refer answer no. 08 64. • 23S rRNA in bacteria is the enzyme- ribozyme which acts as a catalyst for the formation of 77. Genetic code will code for same amino acid in all peptide bond during protein synthesis. living organisms ranging from bacteria to human. e.g. UUU codes for phenylalanine in bacteria, • Release factor binds to the stop codon, dog, plants and even humans. terminating translation and releasing the polypeptide from the ribosome. 78. When lactose is present, it binds with repressor and changes its properties. Repressor is then 65. It occurs with the help of 23S rRNA. This unable to bind with the operator. RNA polymerase molecule acts as an enzyme-molecule which acts binds with promoter and transcription of all three as a catalyst for the formation of peptide bond during protein synthesis. structural genes occurs. 66. When an amino acid is linked to its cognate tRNA p i po z y a In presence in the presence of ATP is called charging of tRNA of inducer or aminoacylation of tRNA. Transcription 67. There are two sites in the larger subunit of ribosome where subsequent amino acids bind Repressor mRNA lac mRNA and are placed close enough to each other for the formation of a peptide bond. ATP is required for Translation linking of amino acid with its cognate tRNA . Inducer b-galactosidase Permease Transacetylase 68. Chromosome 1 has maximum genes and chromosome Y has fewest genes. (Inactive repressor) 69. Bacterial Artificial Chromosome, Yeast Artificial 79. ‘i’ gene transcribes the mRNA for repressor Chromosome. They are used as vectors for cloning protein, the repressor protein when present the DNA fragments in human genome project. binds to the operator region of lac operon and does not allow RNA polymerase to transcribe structural genes. Thus the lac operon is switched off by repressor that is why it is called negative regulation. But when lactose is present, the repressor protein is unable to bind to operator, the polymerase enzyme transcribes the structural genes. Thus the lac operon is switched on. Molecular Basis of Inheritance  143

80. (a) U n a m b i g u - Universal: Genetic code/ organisms, e.g. UUU codes for phenylalanine ous: One codon codons are (nearly) same in all organisms from bacteria to humans. codes for only for all organism/from one amino acid. bacteria to human • Genetic code is degenerate as some amino acids are coded by more than one codon, e.g. (b) Degenerate: Initiator: Start codon / UUU and UUC both code for phenylalanine. More than one AUG codon coding 86. (i) Sequence of complimentary strand will be for the same ATCGTACTA. amino acid (ii) The base pairs are held together by hydrogen 81. It happens because DNA being negatively bonds, between A and T by double bond and C charged is held in position in nucleoid region by and G by triple bond. positively charged proteins. (iii) A = T and C = G. This as given by Watson and 82. (i) Structure of Nucleosome Crick. Chargaff stated A+G=C+T= 1. (ii) a'-Histone octamer, 'b'-DNA, c-H1–histone 87. In Prokaryotes, in spite of no defined nucleus, DNA and d-core of histone molecule is held in a defined region called nucleoid by some positively charged proteins. (iii) In bacteria there is no nucleus but DNA is organized in nucleoid region in large loops In Eukaryotes, there is a set of eight positively held by proteins. charged basic proteins called histone octamer on which negatively charged DNA (200 bp) 83. (i) DNA made of two polynucleotide chains is wrapped. The resulting structure is called arranged in a double helix, where sugar and nucleosome. phosphate forms the backbone. The bases project inside. 88. (a) (i) Eight molecules of positively charged basic proteins called histones are organized (ii) The two chains are anti-parallel to each other. into histone octamer. (iii) The nitrogenous bases are joined to each other (ii) Negatively charged DNA wrapped around by hydrogen bonds. A forms two hydrogen positively charged histone octamer forms bonds with T and C forms three hydrogen a structural unit called nucleosome. bonds with G. Purines are always opposite to pyrimidines and vice versa. (iii) It is a beads-on-string structure made up of repeating nucleosomes. (iv) Two chains are coiled in right-handed fashion. The pitch (single turn) of helix is 3.4 nm. In (b) Euchromatin Heterochromatin each turn, 10 base pairs are present, thus, distance between two consecutive bp is 0.34 It is a region in nu- It is a region in nucle- nm. cleus where chroma- us which are darkly tin is lightly stained, stained, densely packed (v) The plane of one base pair stacks over loosely packed and and transcriptionally another, and presence of hydrogen bonds is transcriptionally inactive. provides stability to helical structure. active 84. (a) Met tRNA 89. (a) Hershey and Chase used radioactive sulphur to label protein coat of virus as sulphur is UAC the component of protein, and radioactive phosphorus was used to label the DNA as (b) tRNA is called as an adaptor molecule because phosphorus is the component of DNA. it acts as a carrier of amino acid, type of amino acid depends upon the anticodon loop that the (b) When they allow bacteriophages to infect tRNA has. E. coli bacteria, separately with radioactive sulphur and radioactive phosphorous. 85. • Unambiguous code means that one codon is Bacteria which were infected with viruses particular for one amino acid only. e.g. AUG having radioactive DNA were found to always codes for Methionine. contain radioactive DNA later and bacteria which were infected with viruses having • Genetic code is a universal codon and its radioactive protein coat were not found to corresponding amino acid are the same in all contain radioactivity. 144 Biology–12

On observation they concluded that DNA is the genetic material. Bacteriophage Radioactive (32P) labelled DNA Radioactive (35S) labelled protein capsule Infection Blending Centrifugation No radioactivity (35S) Radioactive (35P) detected in cell detected in cell + + Radioactivity (35S) No radioactivity detected in supernatant detected in supernatant The Hershey–Chase Experiment 90. The first Genetic material is RNA. Evidences to 91. (a) (i) Heavy isotope of nitrogen was fully support this are: incorporated into DNA strands and after the density gradient centrifugation in CsCl, the • Processes like metabolism, translation, DNA strands were found near the lower end splicing evolved around RNA. of test tube in heavy density. • RNA acts as catalyst in living systems. (ii) The light isotope of DNA was found to be the part of DNA, the DNA after centrifugation • In some viruses it is the hereditary material was found of hybrid and light density. and is unstable and hence would have mutated to lead to evolution. T hey concluded that DNA replicates semi conservatively. Generation I Generation II 15N-DNA 14N-DNA 14N-DNA 15N-DNA 15N-DNA 20 min 40 min 14N-DNA 14N-DNA Gravitational force 14N14N 14N15N 14N14N 14N15N Light Hybrid Light Hybrid SeparaSetipoanratoiofnDoNf DANbAybCy Cenenttrriiffuuggaatitoinon 92. (a) Template strand Coding strand • It codes for protein molecule, i.e. it is • No transcription occurs hence does not code for transcribed and later translated. protein • Polarity is 5' - 3' • Polarity is 3 '- 5' (b) S ource of energy for replication of DNA is 93. (i) RNA polymerase II deoxyribonucleoside triphosphates. (ii) Because it has non-functional introns. The hnRNA undergoes splicing to removed introns Molecular Basis of Inheritance  145

and exons join. Capping by addition of methyl 97. DNA sequences which are repeated many times guanosine triphosphate to 5’ end and tailing and show a high degree of polymorph-ism, and by addition of poly A tail at 3’ end takes place form separate peaks from bulk of DNA in a in the nucleus. genome during density gradient centrifugation, are called as satellite DNA. Satellite DNA forms 94. (a) The enzyme responsible for transcription of major portion of DNA. tRNA is RNA polymerase III. The initiator tRNA gets linked with methionine. DNA from every tissue from an individual, shows the same degree of polymorphism and is heritable, (b) The tRNA when charged with amino acid hence very useful in DNA fingerprinting. This methionine reaches the smaller subunit of polymorphic DNA forms specific pattern which ribosome, with its anticodon UAC recognizing is specific for each individual and thus helps in the codon AUG on mRNA and binds by identification. forming complementary base pairs, this initiates protein synthesis. The methionine 98. (a) Total nucleotides are = 1000 is the first amino acid in every protein chain. Adenine = 240 95. Thymine = 240 p i po z y a In presence so, C + G = 1000 – (240 + 240) = 520 of inducer C = G, Transcription therefore C = 520/2 = 260 Repressor mRNA lac mRNA Pyrimidines are C and T, Translation Total Pyrimidines are b-galactosidase Permease Transacetylase Inducer C + T = 240 + 260 = 500 (Inactive repressor) (b) H H H PC PC When lactose in added to the medium, it enters H PC cell due to the activity of permease enzyme. H H H Lactose acts an inducer. It inactivates repressor PC and allows RNA polymerase to access promoter. H 96. When lactose in added to the medium, it enters cell hydrogen due to the activity of permease enzyme. Lactose acts an inducer. It inactivates repressor and bonds allows RNA polymerase to access promoter. RNA polymerase enzyme transcribes the structural H H H H genes present in lac operon. When lactose is HO C P digested, there is no inducer, the repressor binds CP CP C P 5' to the operator. The RNA polymerase access is H H H H blocked and the transcription is stopped. Thus operon is in switch off condition. 99. (a) Total nucleotides = 1500 Guanine = 410 cytosine = 410 So, A + T = 1500 – (410 + 410) = 680 A = T = 680/2 = 340 Pyrimidines are C and T, Thus, total pyrimidines will be p i po z y a In absence C + T = 340 + 410 = 750 of inducer (b) H Repressor mRNA Repressor binds to the operator H H H region (O) and prevents RNA PC PC PC polymerase from transcribing H PC the operon H H H Repressor hydrogen pi y a In presence bonds of inducer po z Transcription H H H HO C P CP CP H H H H C P 5' Repressor mRNA lac mRNA H Translation 100. (a) Total nucleotides = 2000 Adenine = 520 b-galactosidase Permease Transacetylase Inducer Thymine = 520 So, C + G = 2000 – (520 + 520) = 960 (Inactive repressor) C = G = 960/2 = 480 Purines are G and A, 146 Biology–12

Thus, total purines will be Transcription start site 3' Promoter Structural gene Template strand Terminator 5' A + G = 480 + 520 = 1000 (b) H H H H 5' Coding strand 3' PC PC PC PC H H H H hydrogen Schematic structure of a transcription unit bonds 104. (a) The transcription unit. H ATG C A T G C A T AC 5' C P 5' 3' H H HO C P H H Promoter Terminator CP CP H H H 5' TAC G T A C G T A TG 3' DNA H1 histone (b) The mRNA is 5' UAC GUA CGU AUG 3' 101. 105. Refer answer no. 100 106. It is said so because in eukaryotes split genes Histone are present. Portion of genes which are present octamer in mRNA and are translated later are called Core of histone exons. The introns are removed from hnRNA molecules and are not present in mRNA. In prokaryotes Nucleosome is found in nucleus in the cell. there are no such exons or introns. The genes are 102. 5' polycistronic. Phosphate A 107. (a) Promoter: the regulatory sequence which is present upstream in transcription unit and is G responsible for initiation of transcription. U Terminator: it is a regulatory sequence which is present downstream in transcription C unit and is responsible for termination of 3' transcription. RNA is typically single stranded and is made of (b) Exon: the portion which remains in mRNA ribonucleotides that are linked by phosphodiester bonds. A ribonucleotide in the RNA chain Introns: the portion which are removed due contains ribose (the pentose sugar), one of the four nitrogenous bases (A, U, G, and C), and to splicing and are not present in mRNA. a phosphate group. The ribose sugar has –OH group at 2’ position. 108. Initiation: When RNA polymerase binds to promoter, it initiates transcription. RNA 103. Transcription unit of DNA is defined by three polymerase binds with sigma factor and then it regions binds to promoter. (i) Promoter: it provides binding site for RNA 109. Elongation: sigma factor dissociates and RNA polymerase polymerase uses nucleoside triphosphate as (ii) Structural gene: portion of DNA which is substrate and polymerize RNA depending on transcribed into RNA the template. Complementary bases are added (iii) Terminator: it dissociates RNA polymerase during polymerization. enzyme and marks the end of transcription. 110. (a) 5’-AUG GUA AUC CUA-3’ • DNA has two strands, and DNA dependent (b) Initiation: When RNA polymerase binds to RNA polymerase catalyse the polymerization promoter, it initiates transcription. RNA in only one direction, i.e. 5’-3’, so the strand polymerase binds with sigma factor and then which has polarity of 3’-5’ is called template it binds to promoter. strand and is transcribed. The other strand is called coding strand. 111. DNA dependent RNA polymerase in transcription binds to the promoter region of transcription unit. • There are some regulatory sequences which do not code for any RNA or protein situated It helps in polymerization of mRNA. upstream (towards 5’ w.r.t. coding sequence) or downstream (towards 3’ w.r.t. coding 112. (a) Francis Crick sequence) the structural gene. (b) The tRNA is also called soluble RNA. The molecule is clover leaf like (within cell it looks like inverted L) and has an anticodon loop that has bases complementary to code. It has an amino acid acceptor end where amino acid binds. For each amino acid, there is a specific Molecular Basis of Inheritance  147

tRNA and is carried by it during translation depends upon the anitcodon loop the tRNA process. There are no tRNA for stop codons, so has. they terminate the translation process. They are also called adapter molecules because 117. Initiation: the ribosome binds to the mRNA at they serve as a link between the mRNA and the start codon (AUG) that is recognised only by amino acid sequence of proteins. the initiator tRNA. Tyr Termination: release factor binds to the stop codon, terminating translation and releasing the A UG polypeptide from the ribosome. U A C mRNA Untranslated regions are located before start 3' codon and after stop codon. They are required for effective translation. 113. (a) The enzyme responsible for transcription of tRNA is RNA polymerase III, and methionine 118. (a) Polarity of a to a’ is 5'-3’ , no more amino acid is the amino acid that gets linked with the will be added as the last codon on mRNA is initiator tRNA stop codon which does not code for any amino acid. (b) Initiator tRNA with anticodon UCA carries amino acid methionine. It then binds with (b) DNA sequence coding for serine will be TCA codon (AUG) present on the mRNA and helps and anticodon for the same on tRNA will be in the initiation of protein synthesis. TCA 114. (a) Francis Crick (c) Untranslated regions are required for (b) Refer answer 109 (b) effective translation and are located before start codon and after stop codon. (c) The actual structure looks like an inverted 'L' 119. ‘i’ gene transcribes the mRNA for repressor 115. (a) 61 codons code for amino acids and 3 codons protein, the repressor protein when present binds do not code for any amino acid to the operator region of lac operon and does not allow RNA polymerase to transcribe structural (b) (i) One codon codes for only one amino acid, genes. Thus the lac operon is switched off by hence, it is unambiguous and specific. binding of repressor on the operator that is why it is called negative regulation. But when lactose is (ii) Some amino acids are coded by more than present, the repressor protein is unable to bind to one codon, hence the code is degenerate. operator, the polymerase enzyme transcribes the structural genes. Thus the lac operon is switch (iii) The code is nearly universal: for example, on. from bacteria to human UUU would code for Phenylalanine (phe). Some 120. Translation is the process of polymerisation of exceptions to this rule have been found amino acids to form a polypeptide. in mitochondrial codons, and in some protozoans. The different phases of translation are: (iv) AUG has dual functions. It codes for (i) Activation of amino acids. Methionine (met), and it also act as initiator codon. (ii) Initiation of polypeptide synthesis. 116. (a) Met (iii) Elongation of polypeptide chain. tRNA (iv) Termination of polypeptide chain. 121. (a) B stands for Bacterial, Y stands for yeast. U A C Anticodon They are used as vectors for cloning DNA (b) It is called as adaptor molecule because it acts fragments. as carrier of amino acid, type of AA it carries (b) Less than 2 per cent of the genome codes for proteins. The functions are unknown for over 50 per cent of discovered genes and the rest are known. (c) Single Nucleotide Polymorphism. 122. Chromosome 1 has most genes and chromosome Y has fewest genes. 123. In Eukaryotes, there is a set of eight positively charged basic proteins called histones octamer on which negatively charged DNA (200 bp) is wrapped. This resulting structure is called nucleosome. Repeating unit of nucleosome is called chromatin. 148 Biology–12

124. Refer answer no. 70. density gradient centrifugation as 15N is not 125. In 1928, Frederick Griffith did a series of a radioactive isotope, and it can be separated from 14N only based on densities. experiments with Streptococcus pneumoniae, a bacterium responsible for pneumonia. (b) The conclusion drawn from the experiment was that the DNA replicates in semi S strain → Inject into mice → Mice die conservative manner. R strain → Inject into mice → Mice live S strain (heat-killed) → Inject into mice → Mice 130. There are two different types of replicating strand in the given DNA replication fork because the live DNA polymerase enzyme functions only in one direction i.e. 5’-3’. On the template strand with 3’- S strain (heat-killed) + R strain (live) → Inject 5’ polarity, the DNA polymerase can continuously into mice → Mice live polymerize the new strand and is thus called as continuous strand. On the DNA template • He observed two strains of bacteria, smooth shiny with polarity of 5’-3’, the DNA polymerase will colony called S type which was virulent and had replicate the new strand in fragments thus called capsule, the other R type rough colony is non discontinuous strands. The fragments are later virulent and without capsule. The experiment joined by DNA ligase enzyme. was conducted in the following way- 131. (a) The substance used as a source of nitrogen is • S-type living bacteria were isolated from this ammonium chloride. dead mice. (b) To distinguish between original and replicated Griffith concluded that there is some transforming DNA, density gradient centrifugation is used principle (genetic material) which was transferred as 15N is not a radioactive isotope, and it can to R-strain from heat killed S-strain and R-strain be separated from 14N based on densities became virulent. But he could not establish the only. biochemical nature of that genetic material. (c) The heavy DNA and the light DNA molecule 126. Refer answer no. 122. was distinguished by density gradient 127. Hershey and Chase used radioactive sulphur centrifugation to label protein coat of virus as sulphur is the (d) Conclusion was that DNA replicates in semi component of protein, and radioactive phosphorus conservative manner. was used to label the DNA as phosphorus is the component of DNA. 132. (a) Polarity of A: 5’-3’, polarity of B: 5’-3’. Bacteriophages with radioactive sulphur and (b) Replication starts at a specific region called radioactive phosphorous were allowed to infect origin of replication. The two strands of DNA E. coli bacteria separately. Bacteria which were unwind and form a small opening in helix infected with viruses having radioactive DNA called as replication fork. DNA polymerase were found to contain radioactive DNA later on adds new complementary bases in one and bacteria which were infected with viruses direction only, i.e. 5’-3’. The strand with having radioactive protein coat were not found polarity 3’-5’ has continuous replication. On to contain radioactivity. the other strand with polarity 5’-3’, fragments of newly synthesized strand are formed which The conclusion they arrived on was that DNA is are joined later by DNA ligase. This is called the genetic material. discontinuous synthesis. 128. (i) It should be able to generate its replica 133. Replication in bacteria: (ii) It should chemically and structurally be Helicase enzyme separates two strands of DNA stable. to form a replication fork. (iii) It should provide the scope for slow changes The major enzyme used for DNA replication is i.e. mutations that are required for evolution. DNA dependent DNA polymerase. (iv) It should be able to express itself in the form Deoxyribonucleoside triphosphates act as of 'Mendelian Characters’. substrates as well as provide energy for polymerization reaction. 129. (a) Meselson and Stahl used 14N and 15N isotopes in the source of nitrogen present Replication starts at a specific region called in the culture medium in their experiment origin of replication. The two strands of DNA as nitrogen is the component of DNA and unwind and form a small opening in helix called different isotopes are used to distinguish between original and replicated DNA by Molecular Basis of Inheritance  149