Multiple Choice Questions

Transcription

1 1. (2) 2. (2) 3. (2) Multiple Choice Questions Each parent contributes an allele of the gene. The probability of tallness = 3 4. The probability of pink flower = 2/4 So probability of tall plant and pink flower = (2) 5. (3) 6. (3) 7. (2) 8. (2) I A and I B are codominant. A total of 4 phenotypes and 6 genotypes are possible. 1% crossing over = 1 map unit Birds show ZW ZZ type sex determination. X C Y X X X X C Y X XX C XY all males expected to be normal. 9. (2) The trait is not showing itself in every generation and is not confined to sex chromosomes. 10. (3) The mutation has caused change in - chain of haemoglobin. 11. (3) In case of complete linkage recombinants are absent. 12. (3) It resemble neither of the parents. 13. (1) 14. (4) 15. (4) In insects the XX XO type sex determination in seen. 1

2 16. (2) 17. (1) 18. (4) 19. (4) 20. (2) 21. (4) 22. (4) Holandric genes are Y-linked and are absent in females so will not move in a criss-cross pattern. If b denotes freq of recessive allele and a denotes freq of dominant allele then a + b = 1 (Hardy Weinberg law) b 0.3 then a 0.7 Number of individuals with heterozygous dominant trait in population of 3000 would be 2ab or = 1260 Klinefelter s syndrome causes sterility. Ss Ss 23. (3) (medium) ss ss (small) None showed SS for large starch grain size. Frequency of aa genotype in a cross Aa Aa= 1 4 Frequency of Bb genotype in a cross = 2/4 Frequency of cc genotype in a cross = 1 4 Frequency of aabbcc genotype = (3) 25. (2) 26. (3) 27. (1) 28. (1) Haemophilia shows X-linked or criss-cross inheritance. 2

3 Aa Aa 29. (3) 30. (4) 31. (3) 32 (2) 33. (2) 34. (1) 35. (4) 36. (3) 37. (4) 38. (2) 39. (4) 40. (2) 41. (3) 42. (2) 43. (4) aa AA/Aa Aa/Aa aa AA/Aa No. of Barr bodies = No. of X chromosomes 1 = 1 1 = 0 In case of Mendel s cross 1 : 2 : 1 genotypic ratio. Tall and dwarf 2 phenotypes Rr Yy Rr Yy Heterozygous for first gene frequency=2/4 Homozygous for second gene frequency = 2/4 Both cases together Hence for 1000 seeds Seeds. 4 Child s blood group genotype I A I A Father s genotype I B I B or I B I o or I A I o Mother s could be genotype I A I B or I A I A or I A I o The phenotypic ratio would be 1 Red : 2 Pink:1White due to incomplete dominance. 3

4 44. (1) 45. (3) Probability of having either sex is (2) X h Y X X h X X h Y X h X h X h X h Y X XX h XY 47. (1) 48. (2) 49. (4) 50. (2) 51. (1) 52. (4) 53. (2) 54. (4) 55. (3) 56. (1) 57. (1) Contribution of each dominant allele = 18/6=3ft Contribution of each recessive allele = 6 1ft 6 So in case of Aa bb cc it is = = 8 ft Since parent AB is unable to contribute allele I o. In ratio 9 : 3 : 3 : 1, recombinants are 6 16 I A I o I B I o I A I o I B AB B I o A O 58. (3) 59. (1) 4

5 60. (3) 61. (3) 62. (2) bb Bb B b Bb b bb 63. (4) 64. (4) 65. (1) 66. (2) 67. (4) 68. (2) 69. (1) 70. (2) 71. (1) 72. (3) 73. (4) 74. (4) 1 : 1 or 50% intermediate. Hence out of In case of genotype aa Bb Cc Dd, height of the plant would be Rr Rr 1RR : 2Rr : 1 rr (1) Probability of passing X h gene to sperm = 1/2 Probability of passing PKU gene to sperm = 1/2 Probability of passing both traits = =25% (3) 77. (4) 5

6 78. (4) 79. (4) 80. (2) 81. (4) 82. (2) 83. (4) Ratio X A 3 It is intersex. C 7 3 A B 84. (2) 85. (4) 86. (3) 87. (3) 88. (1) 89. (4) 90. (2) 91. (2) 92. (1) 93. (3) 94. (4) 95. (3) 96. (1) 97. (1) 98. (4) 99. (2) 100. (1) Number of linkage groups = Haploid number of chromosomes. p = 0.6 q = 0.4 Heterozygous individuals = Out of 2000 individuals = = 960 a and b show c and d show XY and XX type sex determination (male) (female) ZZ and (male) ZW type sex determination (female) 6

7 101. (4) AA AAAA Gametes A AA 102. (1) 103. (4) 104. (4) 105. (1) 106. (2) 107. (3) 108. (3) 109. (1) 110. (1) 111. (1) 112. (2) 113. (4) 114. (1) 115. (1) 116. (2) 117. (3) 118. (4) 119. (1) 120. (2) 121. (4) 122. (2) 123. (3) 124. (3) 125. (3) 126. (2) 127. (1) (Progeny) AAA (Triploid) Child has blood group O so genotype I o I o. So both parents have given allele I o. In that case parents are I A I o and I B I o. No. of genetically different sperms = 2 n No. of genetically different eggs = 2 n or 2 3 = 8 or 2 2 =4 (where n is number of heterozygous pairs) 7

8 128. (2) 129. (2) 130. (4) X c Y XX h X c X XX c Y XY X h X h X c X h Y 131. (4) 132. (2) 133. (4) 134. (2) 135. (4) 136. (1) 137. (2) P 4 Q 7 R 138. (3) Bb Bb 1 BB : 2Bb : 1 bb 75% have black coats (3) Gene B and C are closely linked (4) In honey bees males/drones are haploid while females are diploid. ASSERTION AND REASON 1. (3) Seven characters studied by Mendel were present on 4 different chromosomes. 2. (2) 3. (1) 4. (1) 5. (1) 6. (1) 8

9 7. (4) Antlers in male deer are sex-limited traits. 8. (2) 9. (2) 10. (1) 11. (1) 12. (2) 13. (4) In grasshoppers XX XO sex determination is seen. 14. (4) Complete linkage would give a ratio of 1 : 0 : 0 : 1 in test cross. 15. (3) Turner syndrome female has XO condition hence no Barr body is seen. 16. (3) All gene mutations are not lethal. 17. (2) 18. (3) PREVIOUS YEAR QUESTIONS 1. (2) 2. (4) Parents Tt Tt Gametes T or t T or t Progeny 1 TT : 2Tt : 1 tt 25% 3. (3) 4. (3) 5. (4) 6. (4) Heterosis is hybrid vigour. 7. (3) The genotypes of both parents would be Rr. 8. (3) The 64 results obtained would give the ratio 1 : 6 : 15 : 20 : 15 : 6 : 1 9

10 9. (3) 10. (3) 11. (2) 12. (1) 13. (3) 14. (4) Number of phenotypes = 2n + 1 where n is number of genes. 15. (1) Colour of pod, shape of pod and position of pod. 16. (2) It is a test cross ratio 1 : (4) 18. (2) 19. (3) A ratio of 1 : 2 : 1 is obtained. 20. (2) 21. (3) The parents have genotype I A I o and I B I o. 22. (1) In dihybrid ratio 9 : 3 : 3 : 1 obtained the parental to recombinant ratio would be 10 : (3) 24. (1) 25. (3) The (3) statement is explained by law of segregation. 26. (4) Linkage is inversely related to crossing over. 27. (1) 28. (3) 29. (2) 30. (3) T t t t Gametes T or t t Progeny Tt and tt 1 : (3) 10

11 32. (2) 33. (3) 34. (3) 35. (4) 36. (2) 37. (2) 38. (3) 39. (4) 40. (2) R R r r (red) (white) F 1 R r R r F 2 1 R R : 2 R r : 1 r r 25% white flowers. 41. (2) 42. (4) 43. (1) 44. (3) 45. (3) 46. (2) 47. (2) 48. (2) 49. (2) 50. (1) 51. (2) 52. (1) 53. (1) 54. (4) 55. (5) 56. (2) Punnet square boxes = 4 n So if 4 n =16 then n = 2 (where n is number of heterozygous alleles) 11

12 57. (2) 58. (4) 59. (2) 60. (3) 61. (1) 62. (1) 63. (1) 64. (1) Different phenotypes possible = 2 n In this case 2 2 = 4 Nutrient deficiency will not effect genotype hence tall plant TT. Cross would be TT tt. The F 2 generator obtained from the cross Aa Bb Aa Bb would be having 9 : 3 : 3 : 1 phenotypic ratio. On selfing Rr Tt we get Rr Tt Rr Tt Frequency of Rr from the cross = (1) 66. (1) Frequency of Tt from the cross = 2 4 Frequency of Rr Tt in progeny = Hence in case of 400 plants it would be Probability of Rr in a cross between Rr and Rr parents is 2 4 y probability of Yy is 2 4 So probability of Rr Yy is 2/4*2/4=1/4 67. (1) Yy yy is a test cross. 68. (4) Linkage would reduce crossing over so more parentals would result. 69. (4) 70. (3) It s a test cross. 12

13 71. (3) 72. (2) 73. (3) 74. (4) 75. (1) 76. (2) 77. (4) 78. (2) 79. (3) 80. (1) 81. (2) 82. (3) 83. (4) 84. (1) 85. (3) 86. (3) 87. (3) 88. (2) 89. (4) 90. (4) 91. (4) 92. (3) 93. (1) 94. (2) 95. (1) 96. (2) 97. (2) 98. (1) 99. (4) 100. (3) It is a test cross. Ratio of 1 : 2 : 1 is obtained phenotypically due to incomplete dominance. Since the genotype shows one heterozygous pair of allele the number of gametes = 2 n = 2 1 = 2. Cytoplasmic (in this case due to mitochondrial genes) inheritance is responsible for male sterility. Jumping genes/transposons were discovered in maize. The ratio is nearly 1 : 1. 13

14 101. (3) 102. (1) 103. (1) 104. (2) 105. (1) 106. (1) 107. (4) 108. (2) 109. (1) 110. (1) 111. (2) 112. (1) 113. (2) 114. (4) 115. (3) 116. (1) 117. (2) 118. (2) 119. (3) 120. (4) 121. (3) 122. (2) 123. (3) 124. (2) 125. (1) 126. (1) 2 n = 2 3 = 8. Psuedoalleles are very closely placed and crossing over is negligible. Total progeny = In F 2 recombinant frequency is Hence Prokaryotes have naked DNA (without histones) AaBbCc can form 2 n = 2 3 = 8 gametes 4 n = 4 3 = 64 zygotes 14

15 AA AAAA 127. (1) 128. (1) 129. (3) 130. (3) 131. (3) 132. (1) 133. (4) 134. (2) 135. (1) 136. (3) 137. (3) 138. (2) 139. (2) 140. (2) 141. (4) 142. (1) 143. (3) 144. (2) 145. (1) 146. (1) 147. (3) 148. (3) 149. (3) 150. (1) 151. (4) 152. (4) gametes A AA Since endosperm would be triploid with only haploid contribution from male the endosperm would be A + A + AA = AAAA (Tetraploid) F 1 progeny = Aa Bb Cc Gametes = 2 n = 2 3 = 8 Biological concept of species (Mayr). 15

16 153. (4) 154. (2) 155. (1) 156. (4) 157. (2) 158. (5) 159. (4) 160. (3) 161. (4) 162. (2) 163. (1) 164. (4) 165. (4) 166. (4) 167. (5) 168. (1) 169. (4) 170. (4) 171. (1) 172. (3) 173. (3) 174. (4) 175. (1) 176. (1) 177. (3) 178. (1) 179. (4) 180. (2) Man inherits his X chromosome from his mother who must have inherited it from her father or her mother. Rh mother can give Rh gene to the child but if the father is Rh, then the child would be Rh. This would develop antibodies in the mother leading to development of haemolytic disease in the child. X c X c X XY X Y X c X c X X c Y 16

17 181. (3) Sickle cell anemia is a point mutation (4) 183. (1) First marriage : Husband I o I A Child I o I o So the woman I o I B or I o I o or I o I A Second marriage : Husband I B I B or I o I B 184. (3) 185. (3) 186. (2) 187. (2) 188. (3) 189. (2) 190. (3) 191. (4) 192. (4) Child I A I B So woman I A I A or I A I o Taking both the cases: Woman I A I o X h Y X XX X h Y X XX h XY C is normal but is able to pass the allele for the disease to the next generation so C is a carrier female XX c. While D is a normal male since males in criss cross inheritance are not carriers so XY. Woman I o I o Child I o I o If her claim is right father should be I A I o. The genotypes of the offsprings are I o I o are I A I A / I A I o. The possible genotypes of the parents would be I A I o and I A I o. 17

18 X c Y X XX X c Y X XX X c X X XY X c X X X c X Y X c X XX XY Name of the daughters should be colourblind (5) 194. (2) 195. (2) 196. (3) 197. (2) Cytoplasmic genes are inherited from cytoplasm of ovum (4) Woman : X h X Man: XY Children : XX XY X h Y In this case woman must be a carrier XX h and man normal XY (1) 200. (1) 201. (3) A a A a B b b B cis Trans 202. (1) 203. (3) 204. (3) 205. (2) 18

19 XY X X c X c X c Progeny X X c X Y X c Y X c X c X X c Y 206. (2) 207. (1) 208. (3) 209. (1) 210. (4) Husband I A I o / I A I A Wife I B I o / I B I B I A I o I B AB B I o A O All blood groups possible (1) 212. (4) 213. (4) 214. (2) Down s syndrome is trisomy of 21 st chromosome (3) 216. (4) Isoimmunization is the development of antibodies against Rh 217. (3) 218. (2) 219. (3) 220. (3) 221. (3) 222. (2) Man could be I A I A or I A I o Woman is I A I B In first case progeny could be I A I A or I A I B In second case progeny could be I A I A or, I A I B antigens. 19

20 223. (3) 224. (1) 225. (3) 226. (2) 227. (2) 228. (1) 229. (3) 230. (2) 231. (1) 232. (1) 233. (4) 234. (1) 235. (2) 236. (1) 237. (4) Since it has a single chromosome/nucleoid. Mother s genotype I B I B Father I A I A or I A I o Progeny I A I B or I B I o Since the gene moved from father to all daughters it showed criss-cross inheritance or sex-linked dominant nature. X c X c X XY X Y X c XX c X c Y None since all the offspring would be intermediate. Father? Mother B 238. (2) 239. (3) 240. (3) Child A Mother can be genotype I B I B or I B I o. But since child is A blood group genotype I B I B is not possible. So child gets I o allele from mother. In that case father has contributed I A allele so he may be I A I o or I A I B. 20

21 X c Y X XX X c X XX c 241. (2) Y XY X h Y X X h X X h Y X h X c X h X h Y X XX h XY 243. (2) 244. (3) Presence of Barr body suggests XX Presence of F-body suggests Y So syndrome is XXY or Klinefelter s syndrome (2) X h X h X XY X h X XX h Y X h Y 246. (1) 247. (4) X h Y X XX Y X XX h XY X h 248. (3) 249. (3) 250. (3) As in this case I A allele is not being given by either of the parents (1) 252. (2) X-linked disease shows criss cross inheritance (3) 254. (4) 21

22 255. (4) Aa Aa 256. (1) 257. (3) 258. (1) 259. (1) 260. (1) 261. (2) 262. (2) 263. (2) 264. (4) 265. (1) 266. (2) 267. (2) 268. (2) AA/Aa Aa/AA Aa/AA aa Aa/AA Since the parents are heterozygous the above results were obtained. Man can be I B I B or I B I o Woman can be I A I A or I A I o Since child has blood group B he must be having genotype I B I o. One gene controls more than one character it is called pleiotropic gene. Since both parents have to give allele I o to the child the choice AB and O is inappropriate. Boy brother XY sister X c X c It shows that father was X c Y (colourblind) since he passed the gene to daughter but mother was carrier XX c. Woman XX Man XY X X XX Y XY X c X c X X c Y 269. (5) 270. (5) 271. (5) 272. (2) 22

23 I A I B I A A AB I B AB B 273. (2) Husband Wife Rh + Rh Rh Rh Progeny Rh + Rh Rh Rh + Rh Rh Rh 274. (1) Since mother is colourblind she passes the gene to the sons (1) I o I A I o I B I A I o I B AB B I o A O (1) 277. (1) 278. (3) 279. (3) 280. (2) 281. (1) 282. (2) 283. (5) 284. (2) 286. (3) or 25% AB group. In fowls ZW ZZ type sex determination is seen. This is called dosage compensation. X 1 A (female) 23

24 287. (3) 288. (5) 289. (1) 290. (1) 291. (3) 292. (1) 293. (3) 294. (3) 295. (3) 296. (1) 297. (2) 298. (2) 299. (3) 300. (2) 301. (2) 302. (2) 303. (1) 304. (3) 305. (1) 306. (2) 307. (1) 308. (3) 309. (1) 310. (2) X 0.5 A (male) X between 1 and 0.5 (intersex) A X 1 (super female) A X 0.5 (super male) A Percentage of having child of either sex is 50% each time. Number of Barr bodies = No of X chromosomes 1. So in this case = 4 1 = 3 Since in males only one X chromosome is present. Barr body is heterochromatised extra X chromosomes, so seen in females. 24

25 311. (4) 312. (2) 2N 2 Nullisomic a pair of chromosomes is missing (1) 314. (2) Cri du chat is a case of deletion in short arm of chromosome (1) 316. (3) Both colour blindness and haemophilia are cases of sex-linked inheritance (3) 318. (4) Sickle cell anaemia is an autosomal recessive disorder (1) 320. (4) 321. (5) 322. (4) 323. (2) 324. (5) 325. (4) 326. (3) Histone proteins play an important role in DNA packing in a eukaryotic chromosome (3) 328. (4) 329. (3) 330. (1) 331. (3) Colchicine is metaphasic poison and causes non-disjunction (3) Histones are rich in basic proteins lysine and arginine (1) 334. (4) 335. (1) 336. (3) 337. (1) 338. (3) 339. (4) 25

26 340. (2) 341. (3) 342. (1) 343. (3) 344. (3) 345. (4) 346. (1) 347. (2) 348. (4) 349. (3) 350. (3) 351. (4) 352. (4) 353. (4) 354. (2) 355. (2) 356. (3) 357. (2) 358. (3) 359. (1) 360. (2) 361. (4) These are called polytene chromosomes. Down s syndrome is trisomy of 21 st chromosome. Wheat is a hexaploid having 42 chromosomes. Haploid number would be half of total number. So it is and basic number is 7 as it is a hexaploid and The patient lacks the enzyme to metabolise amino acid phenylalanine. Balbiani rings are swollen regions of polytene chromosomes where transcription rates are high. This is called non disjunction. Cretinism is a thyroid malfunction. In case of wheat 2N = 42 So monosomic 2N 1 = 41 Haploid 2N Nullisomic 2N 2 = 40 26

27 362. (3) 363. (3) 364. (1) 365. (3) 366. (3) 367. (2) 368. (2) 369. (2) 370. (1) 371. (4) 372. (5) 373. (4) 374. (4) 375. (2) 376. (1) 377. (4) 378. (4) 379. (2) 380. (1) 382. (4) 383. (4) 384. (5) 385. (1) 386. (5) 387. (5) 388. (3) 389. (2) Trisomic 2N + 1 = 43 Since both genes of haemophilia and colour blindness are present on X-chromosome/X linked. Lampbrush chromosomes are seen in oocytes of amphibians. Since they appear in diplotene stage of prophase I they are seen as homologous pairs. Torsion during synapsis may lead to deletion of a part of one chromosome and its attachment to its homologue causing duplication of genes in it. Position of centromeres leads to metacentric, acrocentric, telocentric shapes. A point mutation causes changes in chain of haemoglobin. Albinism is due to lack of melanin. Endoreduplication is due to DNA replication producing large size chromosomes. 27

28 390. (3) 391. (3) 392. (3) 394. (4) 395. (3) 396. (3) 397. (1) 398. (1) 399. (2) 400. (2) 401. (4) 402. (1) 403. (4) 404. (4) 405. (2) 406. (3) 407. (2) 408. (1) 409. (3) 410. (1) 411. (1) 412. (3) 413. (3) 414. (1) 415. (1) 416. (4) 417. (3) 418. (3) 419. (3) 420. (1) Polyethylene glycol helps in fusion of protoplasm during somatic hybridization. Point mutation causes substitution of a single nucleotide in this case. Gamma radiations are deeply penetrating rays causing mutations and used in mutational breeding. Haploids have only one set of genes and all are expressed. Substitution of purine by purine or pyrimidine by pyrimidine is called transition. 28

29 360 p 2 then (3) 422. (3) 6 p 0.6 and p is the frequency of A in population. 10 X c Y X X c X (man) (woman) X c Y X c X c X c X c Y X XX c XY 50% of male children will be colourblind (3) 424. (4) Recessive epistasis gives a ratio of 9 : 3 : 4 due to gene interaction. 29