Tiêu đề 5 principles of inheritance and variation-Entrance questions.pdf ✅

© All rights reserved www.bankofbiology.com 1 MENDEL’S LAWS OF INHERITANCE 1. Select the odd one out A. Violet flower B. Round seed C. Yellow pod D. Tall stem 2. The graphical representation to calculate probability of all genotypes of offspring in a genetic cross, is called A. Pedigree analysis B. Karyotype C. Punnett square D. Chromosome map E. Genotype ratio 3. A pure tall pea plant is crossed with pure dwarf pea plant. The progeny is self-pollinated. The ratio of true breeding tall pea plants to true breeding dwarf pea plants shall be A. 2:1 B. 1:1 C. 3:1 D. 1:2 4. “Gametes are never hybrid”. It is a statement of law of A. Dominance B. Random fertilization C. Independent assortment D. Segregation 5. According to the Law of independent assortment, in a dihybrid cross A. There are 4 genotypes in F2 B. F2 contains 16 phenotypes C. There is a single individual which is homozygous recessive for both the characters D. It is not possible to forecast the different phenotypes 6. How many types of gametes will be produced by individuals of AABbcc genotype? A. 2 B. 4 C. 6 D. 9 7. The Mendelian principle which has always stood true is A. Law of independent assortment B. Law of segregation C. Law of dominance D. All the above 8. Alleles of a gene are found on A. Same chromosome B. Homologous chromosomes C. Non-homologous chromosomes D. Any chromosomes 9. Crossing of an organism with dominant phenotype to a recessive individual is called A. Test cross B. Back cross C. Reciprocal cross D. Dihybrid cross 10. A heterozygous round seeded plant is crossed to recessive wrinkled seeded plant. The progeny would be A. 303 rounded: 301 wrinkled B. 301 rounded: 100 wrinkled C. 20 rounded: 99 wrinkled D. 99 rounded: 301 wrinkled 11. Test cross ratio in a dihybrid cross would be A. 1:1 B. 1:2:1 C. 1:1:1:1 D. 9:3:3:1 12. A pure tall plant is reared in a soil poor in nutrition and reached the size of dwarf plant. If this plant is selfed, the phenotype in F1 generation is likely to be A. All tall plants B. 50% tall and 50% dwarf C. All dwarf D. Data insufficient 13. What should be the composition of gametes? A. GgLl B. Ggl C. Gg D. Gl 14. Percentage of heterozygous individuals obtained from selfing of Rr individual is A. 100 B. 75 C. 50 D. 25 15. In rabbits, the gene for grey fur (G) is dominant over that for black fur (g). in a litter, if 50% rabbits are grey, then the possible parental cross combination is A. GG X Gg B. GG X GG C. gg x gg D. Gg x gg 16. In a population, a multiple allelic gene contains 5 alleles. Then the number of possible genotypes in that population would be A. 5 B. 15 C. 10 D. 18 17. Out of a population of 800 individuals in F2 generation of a cross between yellow round & green wrinkled pea plants, what would be number of yellow and wrinkled seeds? A. 800 B. 400 C. 200 D. 150 NON-MENDELIAN INHERITANCE 18. Two pink flowered snap dragon plants (Rr) are self- pollinated. The probability of the offspring to have white flowers are A. 25% B. 50% C. 75% D. 0% 19. A blue fowl obtained from mating between black and white fowls, is self-crossed. The F2 ratio is A. 1 black: 2 white: 1 blue B. 2 black: 1 white: 1 blue C. 1 black: 2 blue: 1 white D. None of the above 20. ABO blood grouping shows A. Co-dominant genes B. Polygenes C. Dominant-recessive genes D. Both co-dominant & dominant recessive genes 21. A child of blood group O cannot have parents of blood groups A. AB and AB/O B. A and B C. B and B D. O and O 22. Four children belonging to same parents have the blood groups A, B, AB & O. The genotypes of the parents are A. Both parents are homozygous for A group B. One parent is homozygous for A and another parent is homozygous for B C. One parent is heterozygous for A and another parent is heterozygous for B D. Both parents are homozygous for B group 23. In ABO system, blood group genes show multiple allelism. They include A. One dominant allele and 2 recessive alleles www.bankofbiology.com www.bankofbiology.com www.bankofbiology.com www.bankofbiology.com
© All rights reserved www.bankofbiology.com 2 B. Two dominant alleles and 2 recessive alleles C. One dominant allele and 3 recessive alleles D. Two dominant alleles and one recessive allele 24. Multiple phenotype is seen in A. Multiple gene inheritance B. Pleiotropy C. Multiple allelism D. Co-dominance 25. Which of the following is an example for Pleiotropy? A. Starch synthesis in pea B. Phenylketonuria C. Sickle cell anaemia D. All of these 26. Inheritance of human skin colour and human height is A. Mendelian inheritance B. Monogenic inheritance C. Complementary genes D. Polygenic inheritance 27. Which of the following category of genotypes represent intermediate (mulatto) skin colour in human? A. AABBCC, AaBbCc, aabbcc, AabbCC B. AaBBCC, Aabbcc, aabbcc, AabbCc C. AaBbCc, aaBBCc, AabbCC, AABbcc D. aaBBCC, AaBbCc, aabbcc, aabbCC CHROMOSOMAL THEORY, LINKAGE & RECOMBINATION 28. Who proposed Chromosomal theory of inheritance? A. Walter Sutton and Theodore Boveri B. de Vries, Correns & von Tschermak C. Thomas Hunt Morgan D. Alfred Sturtevant 29. Morgan selected Drosophila for genetic study because A. Short life cycle of 2 weeks B. Its cytology is easy to study C. Numerous offspring in each mating D. Inexpensive to breed E. All of these 30. Regarding Morgan’s study in Drosophila, which is false? A. Genes of eye colour, body colour and wing size were located on the X chromosome B. Genes of white eye & yellow body are tightly linked C. Genes of white eye & small wing are loosely linked D. Loosely linked genes show low recombination 31. Mendelian recombination is due to A. Linkage B. Crossing over C. Mutations D. Independent assortment 32. Genes P & Q are both required in dominant state for normal hearing. A deaf couple has all children with normal hearing. The probable genotype for the couple is A. PPqq x ppQQ B. PPqq x PPqq C. PpQq x ppqq D. PPqq x ppQq 33. What is correct about linked genes? A. Segregation occurs but no dominance B. No segregation but independent assortment C. No segregation, no dominance D. Segregation but no independent assortment 34. Percentage of recombination b/w a & b is 9%, a & c 17% and b & c is 26%. The arrangement of genes is A. a-b-c B. a-c-b C. b-c-a D. b-a-c SEX DETERMINATION 35. The nuclear structure observed by Henking in 50% of the insect sperm after spermatogenesis was A. Polar body B. X body C. Y chromosome D. Nucleolus 36. Which one of the following conditions correctly describes the manner of determining the sex in the given example? A. XO type of sex chromosomes determine the male sex in grasshopper B. XO condition in humans as found in Turner’s syndrome, determine female sex C. Homozygous sex chromosomes (XX) produce male in Drosophila D. Homozygous sex chromosomes (ZZ) determine female sex in birds 37. When released from ovary, human egg contains A. One Y Chromosome B. 2 X chromosomes C. One X Chromosome D. XY chromosome 38. In sex determination of birds A. Male is homogametic (ZZ) and female is heterogametic (Z & W) B. Male is heterogametic (Z & W) and female is homogametic (ZZ) C. Male is homogametic (XX) and female is heterogametic (XO) D. Male is heterogametic (XO) and female is homogametic (XX) 39. XX-XY sex determination mechanism is found in A. Human, Drosophila and Grasshopper B. Chimpanzee, Drosophila and Human C. Birds, Drosophila and Grasshopper D. Human, Periplaneta and Grasshopper 40. Which is wrong about sex determination in honeybee? A. Sex determination in honeybee is called as haplodiploid sex determination system. B. Fertilised egg develops as queen or worker. C. Drones are developed by parthenogenesis. D. Males are diploid and females are haploid. MUTATION, PEDIGREE ANALYSIS AND GENETIC DISORDERS 41. When a mutation is limited to the substitution of one nucleotide for another, it is called A. Translocation B. Point mutation C. Base inversion D. Frame shift mutation 42. Loss (deletion) or gain (insertion/ duplication) of DNA segment cause A. Chromosomal aberration which are not seen in cancer cells B. Chromosomal aberration which do not lead to frameshift mutation C. Frame shift mutation which is not present in cancer cells www.bankofbiology.com www.bankofbiology.com www.bankofbiology.com
© All rights reserved www.bankofbiology.com 3 D. Frame shift mutation which is present in cancer cells 43. In pedigree analysis, the symbol gives which of the following information? A. It is the mating between relatives and male is not affected B. It is the mating between relatives and female is not affected C. Male is homozygous affected and female is recessive D. Male is heterozygous normal and female is homozygous normal 44. Which one of the patterns of inheritance is consistent with this pedigree? A. Autosomal recessive B. Autosomal dominant C. Y-linkage D. Sex-linked recessive 45. Odd one out A. Haemophilia B. Phenylketonuria C. Mongolism D. Cystic fibrosis 46. Haemophiliac man marries a carrier woman. % of daughters becoming haemophiliac shall be A. 0% B. 50% C. 75% D. 100% 47. In sickle cell anaemia, the glutamic acid is replaced by A. Proline B. Alanine C. Serine D. Valine 48. Sickle cell anaemia is due to A. Mutation of CTC to CAC on gene in chro. 11 B. Change of GAG to GUG on the mRNA C. Change of GUG to GAG on the mRNA D. Both A & B 49. Both husband and wife have normal vision though their fathers were colourblind. The probability of their daughter becoming colourblind is A. 0% B. 25% C. 50% D. 75% 50. Male child with blood group AB is colourblind. His parents could be A. Father normal vision with blood group A, mother colourblind with group O B. Father colourblind with group O, mother colourblind with group AB C. Father normal vision with blood group A, mother colourblind with group B D. Father colourblind with group O, mother normal vision with group O. 51. Albinism and phenylketonuria are disorders due to A. Recessive Autosomal genes B. Dominant Autosomal genes C. Dominant sex genes D. Recessive sex genes 52. Which is not true about Phenylketonuria? A. It causes mental retardation B. Phenyl alanine hydroxylase absent C. Phenylalanine is converted to tyrosine D. Phenyl pyruvic acid is excreted through urine 53.  Thalassemia is due to the mutation of A. HBA1 & HBA2 genes on chromosome 16 B. HBB gene on chromosome 11 C. HBB gene on chromosome 16 D. HbA gene on chromosome 11 into HbS 54. Marriage between close relatives should be avoided because it induces more A. Recessive alleles to come together B. Mutations C. Multiple births D. Blood group abnormalities 55. Haploid chromosome number is 12. What is the number in a Monosomic? A. 23 B. 22 C. 25 D. 26 56. Which one is not a feature of Down’s syndrome? A. Broad flat face B. Furrowed big tongue C. Gynaecomastia D. Partially open mouth 57. Ovum producing Klinefelter’s syndrome shall have chromosome number A. 21 B. 22 C. 23 D. 24 58. The chromosomal condition in Turner’s syndrome is A. 21 trisomy + XY B. 44 autosomes + XXY C. 44 autosomes + XYY D. 44 autosomes + XO 59. Which one of the following is responsible for mental abnormalities in humans? A. XXX and XY B. XX and XXX C. XO and XXX D. XX and XO >> Answer Key: Click Here 👉 Want Other Chapters? Click Here 👉 NEET/ Entrance Basic Level (T-10) Online Test Series 👉 NEET/ Entrance Online Test Series (Chapter-wise) 👉 NEET/ Entrance Random 20 Test Series 👉 NEET/ Entrance Mock Test Series (90 questions) 👉 NEET/ Entrance Questions Chapter-wise with answers (PDF) www.bankofbiology.com www.bankofbiology.com