principles of inheritance and variation multiple choice questions · 2018-05-30 · principles of...

Principles of Inheritance and Variation

PACE – IIT & MEDICAL: DELHI / BHIWADI / MUMBAI / AKOLA / LUCKNOW / KOLKATA / NASHIK / GOA / RAIBAREILLY 1

Multiple Choice Questions 1. (2) Each parent contributes an allele of the gene. 2. (2) 3. (2)

The probability of tallness = 34

.

The probability of pink flower = 2/4

So probability of tall plant and pink flower = 3 2 64 4 16

4. (2) IA and IB are codominant. A total of 4 phenotypes and 6 genotypes are possible. 5. (3) 1% crossing over = 1 map unit 6. (3) Birds show ZW – ZZ type sex determination. 7. (2) 8. (2)

XC Y X X X

XC Y

X XXC 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.



16. (2) Holandric genes are Y-linked and are absent in females so will not move in a criss-cross pattern. 17. (1) 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 2 × 0.3 × 0.7 × 3000 = 1260 18. (4) Klinefelter’s syndrome causes sterility. 19. (4) 20. (2) 21. (4) 22. (4) Ss ss

Ss ss| |(medium) (small)

None showed SS for large starch grain size. 23. (3)

Frequency of aa genotype in a cross Aa × Aa= 14

Frequency of Bb genotype in a cross = 2/4

Frequency of cc genotype in a cross = 14

Frequency of aaBbcc genotype = 1 2 1 24 4 4 64

24. (3) Haemophilia shows X-linked or criss-cross inheritance. 25. (2) 26. (3) 27. (1) 28. (1)



Aa × Aa

aa AA/Aa Aa/Aa aa AA/Aa

29. (3) No. of Barr bodies = No. of X chromosomes –1 = 1 – 1 = 0 30. (4) 31. (3) In case of Mendel’s cross 1 : 2 : 1 genotypic ratio. Tall and dwarf 2 phenotypes 32 (2) Rr Yy × Rr Yy Heterozygous for first gene frequency=2/4 Homozygous for second gene frequency = 2/4

Both cases together 2 2 44 4 16

Hence for 1000 seeds

11000 2504

Seeds.

33. (2) 34. (1) 35. (4) Child’s blood group genotype IA IA or IA Io Father’s genotype IB IB or IBIo Mother’s could be genotype IA IB or IAIA or IAIo 36. (3) 37. (4) 38. (2) 39. (4) 40. (2) 41. (3) 42. (2) 43. (4) The phenotypic ratio would be 1 Red : 2 Pink:1White due to incomplete dominance.



44. (1) 45. (3)

Probability of having either sex is 12

.

46. (2)

XhY X XhX

Xh YXh XhXh XhY X XXh XY

47. (1) 48. (2) Contribution of each dominant allele = 18/6=3ft

Contribution of each recessive allele = 6 1ft6

So in case of Aa bb cc it is 1 × 3 + 5 × 1 = 3 + 5 = 8 ft 49. (4) Since parent AB is unable to contribute allele Io. 50. (2) 51. (1) 52. (4) 53. (2) 54. (4)

In ratio 9 : 3 : 3 : 1, recombinants are 616

55. (3) 56. (1)

57. (1)

IA Io × IB Io

IA Io IB AB B Io A O

58. (3) 59. (1)



60. (3) 61. (3) 62. (2)

bb × Bb

B b b Bb bb

1 : 1 or 50% intermediate. Hence out of 400 200. 63. (4) In case of genotype aa Bb Cc Dd, height of the plant would be 64. (4) 65. (1) 66. (2) 67. (4) 68. (2) 69. (1) 70. (2) 71. (1) 72. (3) 73. (4)

60 2030 5 358 8

74. (4)

Rr Rr

1RR : 2Rr : 1| rr

1 8000 20004

75. (1) Probability of passing Xh gene to sperm = 1/2 Probability of passing PKU gene to sperm = 1/2

Probability of passing both traits = 1 1 12 2 4 =25%

76. (3) 77. (4)



78. (4)

Ratio X 2 0.6A 3

It is intersex. 79. (4) 80. (2) 81. (4) 82. (2) 83. (4)

A BC

7 3

84. (2) Number of linkage groups = Haploid number of chromosomes. 85. (4) 86. (3) p = 0.6 q = 0.4 Heterozygous individuals = 2 × 0.6 × 0.4 Out of 2000 individuals = 2 × 0.6 × 0.4 × 2000 = 960 87. (3) 88. (1) 89. (4) 90. (2) 91. (2) 92. (1) 93. (3) 94. (4) 95. (3)

a and b show (male)XY and

(female)XX type sex determination

c and d show (male)ZZ and

(female)ZW type sex determination

96. (1) 97. (1) 98. (4) 99. (2) 100. (1)



101. (4)

AA × AAAA

Gametes A AA

(Progeny) AAA (Triploid)

102. (1) 103. (4) Child has blood group O so genotype Io Io. So both parents have given allele Io. In that case

parents are IA Io and IB Io. 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) No. of genetically different sperms = 2n or 22=4 (where n is number of heterozygous pairs) No. of genetically different eggs = 2n or 23 = 8 117. (3) 118. (4) 119. (1) 120. (2) 121. (4) 122. (2) 123. (3) 124. (3) 125. (3) 126. (2) 127. (1)



128. (2) 129. (2) 130. (4)

XcY × XXh

Xc Y

X XXc XYXh XhXc XhY

131. (4) 132. (2) 133. (4) 134. (2) 135. (4) 136. (1) 137. (2)

Q RP

47

138. (3)

Bb × Bb

1 BB : 2Bb : 1 bb

75% have black coats. 139. (3) Gene B and C are closely linked. 140. (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)



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. (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 : 1. 17. (4) 18. (2) 19. (3) A ratio of 1 : 2 : 1 is obtained. 20. (2) 21. (3) The parents have genotype IA Io and IB Io. 22. (1) In dihybrid ratio 9 : 3 : 3 : 1 obtained the parental to recombinant ratio would be 10 : 6. 23. (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 : 1

31. (3)



32. (2) 33. (3) 34. (3) 35. (4) 36. (2) 37. (2) 38. (3) 39. (4) 40. (2)

R R × r r

F1 R r × R r

(red) (white)

F2 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 = 4n (where n is number of heterozygous alleles) So if 4n =16 then n = 2



Different phenotypes possible = 2n In this case 22 = 4 57. (2) 58. (4) 59. (2) Nutrient deficiency will not effect genotype hence tall plant TT. Cross would be TT × tt. 60. (3) 61. (1) 62. (1) 63. (1) The F2 generator obtained from the cross Aa Bb × Aa Bb would be having 9 : 3 : 3 : 1

phenotypic ratio. 64. (1) On selfing Rr Tt we get Rr Tt × Rr Tt

Frequency of Rr from the cross = 24

Frequency of Tt from the cross = 24

Frequency of Rr Tt in progeny = 2 2 4 14 4 16 4

Hence in case of 400 plants it would be 1400 1004

.

65. (1) 66. (1)

Probability of Rr in a cross between Rr and Rr parents is 24

||y probability of Yy is 24

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.



71. (3) 72. (2) 73. (3) 74. (4) 75. (1) It is a test cross. 76. (2) 77. (4) 78. (2) 79. (3) Ratio of 1 : 2 : 1 is obtained phenotypically due to incomplete dominance. 80. (1) 81. (2) 82. (3) 83. (4) 84. (1) 85. (3) 86. (3) 87. (3) 88. (2) 89. (4) Since the genotype shows one heterozygous pair of allele the number of gametes = 2n = 21 = 2. 90. (4) 91. (4) 92. (3) 93. (1) Cytoplasmic (in this case due to mitochondrial genes) inheritance is responsible for male

sterility. 94. (2) Jumping genes/transposons were discovered in maize. 95. (1) 96. (2) 97. (2) 98. (1) 99. (4) 100. (3) The ratio is nearly 1 : 1.



101. (3) 102. (1) 103. (1) 104. (2) 2n = 23 = 8. 105. (1) 106. (1) 107. (4) 108. (2) 109. (1) 110. (1) Psuedoalleles are very closely placed and crossing over is negligible. 111. (2) 112. (1) 113. (2) 114. (4) 115. (3) Total progeny = 1280.

In F2 recombinant frequency is 6

16

Hence 61280 480

16

116. (1) 117. (2) 118. (2) Prokaryotes have naked DNA (without histones) 119. (3) 120. (4) 121. (3) 122. (2) 123. (3) 124. (2) 125. (1) AaBbCc can form 2n = 23 = 8 gametes 4n = 43 = 64 zygotes 126. (1)



AA × AAAA

gametes A AA

Since endosperm would be triploid with only haploid contribution from male the endosperm would be

A + A + AA = AAAA (Tetraploid) 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) F1 progeny = Aa Bb Cc Gametes = 2n = 23 = 8 140. (2) 141. (4) 142. (1) 143. (3) Biological concept of species (Mayr). 144. (2) 145. (1) 146. (1) 147. (3) 148. (3) 149. (3) 150. (1) 151. (4) 152. (4)



153. (4) 154. (2) 155. (1) 156. (4) 157. (2) 158. (5) 159. (4) 160. (3) Man inherits his X chromosome from his mother who must have inherited it from her father or

her mother. 161. (4) 162. (2) 163. (1) 164. (4)

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.

165. (4) 166. (4)

XcXc X XYX Y

Xc XcX XcY

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)



181. (3) Sickle cell anemia is a point mutation. 182. (4) 183. (1)

First marriage : Husband Io IA

Child Io Io

So the woman Io IB or Io Io or Io IA

Second marriage : Husband IB IB or Io IB

Child IA IB

So woman IA IA or IA Io

Taking both the cases: Woman IA Io

184. (3) 185. (3) 186. (2)

XhY X XXXh Y

X XXh XY

187. (2) C is normal but is able to pass the allele for the disease to the next generation so C is a carrier

female XXc. While D is a normal male since males in criss cross inheritance are not carriers so XY.

188. (3)

Woman Io Io

Child Io Io

If her claim is right father should be IA Io.

189. (2) 190. (3) The genotypes of the offsprings are Io Io are IA IA / IA Io. The possible genotypes of the parents would be IA Io and IA Io. 191. (4) 192. (4)



XcY X XX XcY X XX

XcX X XY

Xc X

X XcX XXY XcX XY

Name of the daughters should be colourblind. 193. (5) 194. (2) 195. (2) 196. (3) 197. (2) Cytoplasmic genes are inherited from cytoplasm of ovum. 198. (4)

Woman : XhX × Man: XY

Children : XX XY XhY

In this case woman must be a carrier XXh and man normal XY. 199. (1) 200. (1) 201. (3)

A

B

a

b

A

b

a

B

cis Trans

202. (1) 203. (3) 204. (3) 205. (2)



Xc Xc

X XcX XcX

XY X XcXc

Y XcY XcY

Progeny

206. (2) 207. (1) 208. (3) 209. (1) 210. (4)

Husband IA Io / IA IA

Wife IB Io / IB IB

IA Io

IB AB B

Io A O

All blood groups possible. 211. (1) 212. (4) 213. (4) 214. (2) Down’s syndrome is trisomy of 21st chromosome. 215. (3) 216. (4)

Isoimmunization is the development of antibodies against Rh antigens. 217. (3) 218. (2) 219. (3) 220. (3) 221. (3) 222. (2) Man could be IA IA or IA Io Woman is IA IB

In first case progeny could be IA IA or IA IB In second case progeny could be IA IAor, IA IB



223. (3) Since it has a single chromosome/nucleoid. 224. (1)

Mother’s genotype IB IB

Father IA IA or IA Io

Progeny IA IB or IB Io

225. (3) 226. (2) 227. (2) 228. (1) Since the gene moved from father to all daughters it showed criss-cross inheritance or sex-linked

dominant nature. 229. (3) 230. (2) 231. (1) 232. (1) 233. (4)

X Y

Xc XXc XcY

XcXc X XY

234. (1) 235. (2) 236. (1) None since all the offspring would be intermediate. 237. (4)

Child A

Father ? Mother B

Mother can be genotype IB IB or IB Io. But since child is A blood group genotype IB IB is not possible. So child gets Io allele from mother. In that case father has contributed IA allele so he may be IA Io or IA IB.

238. (2) 239. (3) 240. (3)



XcY X XX

Xc YX XXc XY

241. (2)

XhY X XhX

Xh YXh XcXh XhYX XXh 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. 245. (2)

XhXh X XY

Xh

X XXh

Y XhY

246. (1) 247. (4)

XhY X XX

Xh YX XXh XY

248. (3) 249. (3) 250. (3) As in this case IA allele is not being given by either of the parents. 251. (1) 252. (2) X-linked disease shows criss cross inheritance. 253. (3) 254. (4)



255. (4)

Aa Aa

AA/Aa Aa/AA Aa/AA aa Aa/AA

Since the parents are heterozygous the above results were obtained. 256. (1) 257. (3) Man can be IB IB or IB Io Woman can be IA IA or IA Io Since child has blood group B he must be having genotype IB Io. 258. (1) 259. (1) One gene controls more than one character it is called pleiotropic gene. 260. (1) 261. (2) Since both parents have to give allele Io to the child the choice AB and O is inappropriate. 262. (2) 263. (2) 264. (4) 265. (1) 266. (2)

267. (2) Boy brother XY sister XcXc

It shows that father was XcY (colourblind) since he passed the gene to daughter but mother was carrier XXc.

268. (2)

Woman XX Man XYX Xc

X XX XcXY XY XcY

269. (5) 270. (5) 271. (5) 272. (2)



IA IB

IA A ABIB AB B

273. (2)

Husband Wife

Rh+ Rh– Rh– Rh–

Rh+ Rh–

Rh– Rh+ Rh– Rh– Rh–Progeny

274. (1) Since mother is colourblind she passes the gene to the sons. 275. (1)

Io IA × Io IB

IA Io

IB AB B

Io A O

14

or 25% AB group.

276. (1) 277. (1) 278. (3) 279. (3) 280. (2) 281. (1) 282. (2) 283. (5) In fowls ZW – ZZ type sex determination is seen. 284. (2) This is called dosage compensation. 286. (3)

X 1A (female)



X 0.5A (male)

XA between 1 and 0.5 (intersex)

X 1A (super female)

X 0.5A (super male)

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



311. (4) 312. (2)

2N – 2 Nullisomic

a pair of chromosomes is missing. 313. (1) 314. (2) ‘Cri du chat’ is a case of deletion in short arm of chromosome 5. 315. (1) 316. (3) Both colour blindness and haemophilia are cases of sex-linked inheritance. 317. (3) 318. (4) Sickle cell anaemia is an autosomal recessive disorder. 319. (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. 327. (3) 328. (4) 329. (3) 330. (1) 331. (3) Colchicine is metaphasic poison and causes non-disjunction. 332. (3) Histones are rich in basic proteins lysine and arginine. 333. (1) 334. (4) 335. (1) 336. (3) 337. (1) 338. (3) 339. (4)



340. (2) 341. (3) 342. (1) 343. (3) 344. (3) 345. (4) These are called polytene chromosomes. 346. (1) 347. (2) Down’s syndrome is trisomy of 21st chromosome. 348. (4) 349. (3) 350. (3) 351. (4) Wheat is a hexaploid having 42 chromosomes. Haploid number would be half of total number.

So it is 42 212 and basic number is 7 as it is a hexaploid and 42 7

6 .

352. (4) 353. (4) 354. (2) The patient lacks the enzyme to metabolise amino acid phenylalanine. 355. (2) Balbiani rings are swollen regions of polytene chromosomes where transcription rates are high. 356. (3) This is called non disjunction. 357. (2) 358. (3) 359. (1) Cretinism is a thyroid malfunction. 360. (2) 361. (4) In case of wheat 2N = 42 So monosomic 2N – 1 = 41

Haploid 2N 42 212 2

Nullisomic 2N – 2 = 40



Trisomic 2N + 1 = 43 362. (3) Since both genes of haemophilia and colour blindness are present on X-chromosome/X linked. 363. (3) 364. (1) 365. (3) 366. (3) Lampbrush chromosomes are seen in oocytes of amphibians. Since they appear in diplotene

stage of prophase I they are seen as homologous pairs. 367. (2) 368. (2) 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. 369. (2) Position of centromeres leads to metacentric, acrocentric, telocentric shapes. 370. (1) 371. (4) A point mutation causes changes in chain of haemoglobin.

372. (5) 373. (4) 374. (4) 375. (2) 376. (1) 377. (4) 378. (4) 379. (2) Albinism is due to lack of melanin. 380. (1) 382. (4) 383. (4) Endoreduplication is due to DNA replication producing large size chromosomes. 384. (5) 385. (1) 386. (5) 387. (5) 388. (3) 389. (2)



390. (3) Polyethylene glycol helps in fusion of protoplasm during somatic hybridization. 391. (3) 392. (3) 394. (4) Point mutation causes substitution of a single nucleotide in this case. 395. (3) Gamma radiations are deeply penetrating rays causing mutations and used in mutational

breeding. 396. (3) 397. (1) 398. (1) 399. (2) 400. (2) 401. (4) 402. (1) 403. (4) 404. (4) Haploids have only one set of genes and all are expressed. 405. (2) 406. (3) 407. (2) 408. (1) 409. (3) Substitution of purine by purine or pyrimidine by pyrimidine is called transition. 410. (1) 411. (1) 412. (3) 413. (3) 414. (1) 415. (1) 416. (4) 417. (3) 418. (3) 419. (3) 420. (1)



2 360p1000

then 6p 0.610

and p is the frequency of A in population.

421. (3) 422. (3)

XcY X XcX(man) (woman)

Xc Y

Xc XcXc XcY

X XXc XY

50% of male children will be colourblind. 423. (3) 424. (4) Recessive epistasis gives a ratio of 9 : 3 : 4 due to gene interaction.

principles of inheritance and variation multiple choice questions · 2018-05-30 · principles of...

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