InterviewSolution
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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
When a wheat variety of red kernel is crossed with white kerneled wheat, the `F_(2)` ratio would beA. 1:10:4:1B. `9:7`C. 1:2:4:2:4:2:1D. 1:4:6:4:1 |
| Answer» Correct Answer - D | |
| 2. |
A black female is test crossed, producing 6 black offsprings. The Probability that a black heterozygous female would this by chance alonge is approximatelyA. 0.5B. 0.25C. 0.01D. cannot be determined |
| Answer» Correct Answer - A | |
| 3. |
The proginies of test cross can easily be analysed to predict theA. Genotype of the test organismsB. Phenotype of the test organismsC. Ratio of (a) and (b)D. all of the above |
| Answer» Correct Answer - A | |
| 4. |
A 15:1 `F_(2)` ratio of a cross between a wheat variety with red kernels (homozygous for two dominant genes) and another with white kernels showsA. Polygenic inheritanceB. That the two genes are complementaryC. Single factor inheritanceD. That it is a test cross |
| Answer» Correct Answer - A | |
| 5. |
A heterozygous parent producesA. Onel king of gameteB. Two kinds of gametes each having two allelesC. Two kinds of gametes each having one allele with equal proportionD. two kind of gametes |
| Answer» Correct Answer - C | |
| 6. |
Organism of pureline is that which produces individuals ofA. Dominant charactersB. Recessive charactersC. Its own charactersD. intermediate type |
| Answer» Correct Answer - C | |
| 7. |
Indentify the recessive trait among the following given below.A. Coloured lint in cottonB. Rust immunity in wheatC. Round starcny kernal in maizeD. Branched habit in sunflower |
| Answer» Correct Answer - B | |
| 8. |
In a monhybird cross when `F_(1)` is crossed with homozygous dominant parent then which type of offspring will obtainA. Dominant: Recessive 3: 1B. Only recessiveC. Dominant: recessive 1:1D. No recessive |
| Answer» Correct Answer - D | |
| 9. |
In a dihybrid cross of heterozygotes, what proportion of the offsprin will be phenotypically dominant for both traits?A. 3/4B. 3/16C. 1/4D. 9/16 |
| Answer» Correct Answer - D | |
| 10. |
In a typical Mendelian cross, which is a dihybrid cross, one parent is homoxygous for both dominant traits and another parent is homozygous for both recessive traits. In the `F_(2)` -generation, both parental combination and recombinantions appear. The phenotypic ratio of parental combinations to recombinations isA. `10:6`B. `12:4`C. `9:7`D. `15:1` |
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Answer» Correct Answer - A A cross of round yellow seeds (both dominant and greenwrinkled seeds(both recessive) plants produced 9:3:3:1 ratio of plants (phenotypic) in `F_(2)` -generation. The ratio of parental to recombinant is 10:6 here bacause the 9 and 1 are of parental type and 3 and 3 are recombinant type. |
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| 11. |
When two hybrids of rrTt and Rrtt are crossed, the phenotypic ratio of offspring shall be:A. 3 :1B. 9 :3: 3 :1C. 1 : 1D. 1 : 1 : 1 : 1 |
| Answer» Correct Answer - D | |
| 12. |
A human male produces sperms with the genotypes AB, Ab, aB, and ab pertaining to two diallelic characters in equal proportions. What is the corresponding genotype of this person?A. AaBbB. AaBBC. AABbD. AABB |
| Answer» Correct Answer - B | |
| 13. |
Genotypic and phenotypic ratio in monohybrid cross remains the same in case ofA. Sex linked genesB. pseudoallelic genesC. intermediate inheritanceD. dominant and recessive genes |
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Answer» Correct Answer - C Intermediate inheritance is incomplete dominance in which dominant factor of a heteroxygote does not completely mask the expression of recessive allele. In incomplete dominance, genotypic and phenotypic ratio remains the same and it is 1:2:1 |
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| 14. |
Cross between unrelated group of organisms is calledA. HybridB. test crossC. back crossD. heterosis |
| Answer» Correct Answer - A | |
| 15. |
Genotypic ratio of a monohybrid cross isA. 1:2:1B. 3:1C. 9:3:3:1D. 1:1:1 |
| Answer» Correct Answer - A | |
| 16. |
Back cross is used for determiningA. Purity of gametsB. mutant genesC. Genotype of one of the parentD. sibling relationship |
| Answer» Correct Answer - C | |
| 17. |
What will be the phenotypic ratio in a situation of complementary gene interction?A. `9:7`B. `15:1`C. `13:3`D. `9:3:4` |
| Answer» Correct Answer - A | |
| 18. |
Complementary genes were demonstratedby Bateson inA. CapsellaB. Lathyrus odoratusC. Summer squashD. Mirabilis jalapa |
| Answer» Correct Answer - B | |
| 19. |
Which of the following gene controls the formationof enxyme which catalyses transformation of chromogen into anthocyanin?A. Gene PB. Gane IC. Gane CD. Gane B |
| Answer» Correct Answer - A | |
| 20. |
Which of the following genotypes of sweet pea plant is related with the production of purple coloured flowersA. CcPpB. CCppC. ccPPD. Ccpp |
| Answer» Correct Answer - A | |
| 21. |
The graphical representation to calculate the probability of all possible genotypes of offspring in a genetic cross is calledA. Pedigree analysisB. KaryotypeC. Punnett squareD. Chromosome map genotype ratio |
| Answer» Correct Answer - C | |
| 22. |
which of the following statements is correct, when we use alphabetical symbols for each gene?A. Capital letter is used for the trait expresed at the `F_(1)` stage and small alphabet for the other traitB. Small letter is used for the trait expressed at the `F_(1)` stage and capital alphabet for the other traitC. Both capital and small letters are used for the both `F_(1)` stage and otherD. Either (a) and (b) |
| Answer» Correct Answer - A | |
| 23. |
The law of dominance is used to explainA. The expression of only one of the parental characters in a monohybrid cross in the `F_(1)`B. The expression of both the parental characters ina monhybrid cross in the `F_(2)`C. The proportion of 3:1 obtained at the `F_(2)`D. All of the above |
| Answer» Correct Answer - D | |
| 24. |
The crossing of a homoxygous tall plant with a dwarf in `F_(2)` would yield plants in the ratio ofA. All homoxygous dwarfB. All heteroxygous tallC. One homoxygous tall, one homoxygous dwarf and two heteroxygous tallD. two tall and two dwarf |
| Answer» Correct Answer - C | |
| 25. |
On crossing two similar hybrids, the percentage of recessive isA. 0.25B. 1C. 0.5D. 0.75 |
| Answer» Correct Answer - A | |
| 26. |
Self-incompatibility in tobacco is an example ofA. Plaeiotropic geneB. Complementary genesC. Multple. AllelismD. Epistasis |
| Answer» Correct Answer - C | |
| 27. |
The genotypes of a plant variety were TtHh, Tthh, ttHhand tthh, were T=tallness and H=hairy stem. Which one of the following crosses would produce progeny giving a phenotypic ratio approximately 1:1:1:1?A. TtHh x TtHhB. TtHh x TthhC. TtHh x ttHhD. TtHh x tthh |
| Answer» Correct Answer - D | |
| 28. |
In wheat, when a green plant was self-fertilised, the progeny had 209 green seedlings and 14 white seedlings. The above result indicates that the parents wereA. heteroxygous for two duplicate allelesB. True-breedingC. heteroxygous for one alleleD. None of the above |
| Answer» Correct Answer - A | |
| 29. |
What is the major advantage of using a Punnett square?A. Show all gametic combinationsB. Show genotypic ratiosC. Show all phenotypic ratiosD. all of the above |
| Answer» Correct Answer - D | |
| 30. |
If a cross is made between AA and aa, the nature of `F_(1)` progeny will beA. Genotypically AA, phenotypically aB. Genotypically Aa, phenotypicallyaC. Genotypically Aa, phenotypically AD. Genotypicaly aa, phenotypically A |
| Answer» Correct Answer - C | |
| 31. |
In first filial progeny of his experiment, Mendel observed that all the `F_(1)` progeny poants wereA. dwarfB. tallC. hybridD. All of these |
| Answer» Correct Answer - B | |
| 32. |
When a dihybrid cross is fit into a Punnett square with 16 boxes, the maximum number of different phenotypes available are:A. 8B. 4C. 2D. 16 |
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Answer» Correct Answer - B The phenotypic ratio in `F_(2)` -generation of a dihybrid cross is 9:3:3:1, therefore, the maximum number of different phenotypes available are four. |
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| 33. |
If Aabb x aaBB, then phenotypic ratio of its progeny (`F_(2)` -generation) will beA. 9:3:3:1B. 1:2:1C. 1:1:1:1D. 4:1 |
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Answer» Correct Answer - A 9:3:3:1 is the phenotypic ratio of `F_(2)` -generation when the parents are having Aabb x aaBB alleles. |
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| 34. |
When a cluster of genes shows linkage behaviour theyA. Do not show chromosome mapB. INDUCE CELL DIVISIONC. Do not show independent assormentD. Show recombination during meiosis |
| Answer» Correct Answer - C | |
| 35. |
Alternative form of genes for a particular charcteristic are calledA. homologous chromosomesB. alleleC. linked genesD. genotypes |
| Answer» Correct Answer - B | |
| 36. |
Pure line is connected with development ofA. HomozygosityB. HeterozygosityC. Homozygosity and self assortmentD. heterozygosity and linkage |
| Answer» Correct Answer - C | |
| 37. |
Identify the pair that does not match.A. 1:2:2:4:1:2:1:2:1 - Dihybrid genotypic ratioB. Inheritance of ABO blood group - Multiple allelismC. One gene pair masks the effect of another pair - EpistasisD. Incomplete dominance - Reported by Mendel |
| Answer» Correct Answer - D | |
| 38. |
Allelomorphic pair impliesA. A pair of contrasting charactersB. a pair of non-contrasting charactersC. any two charactersD. sex-linked characters |
| Answer» Correct Answer - A | |
| 39. |
The proportion of the plants that Mendel found on self-pollination of `F_(1)` plants, wasA. 1/4th tall and 3/4 dwarfB. 1/4 th dwarf and 3/4th tallC. 1/4th dwarf, 1/2th hybrid and 1/4 tallD. 1/2 dwarf, 1/2th hybrid and 1/4th tall |
| Answer» Correct Answer - C | |
| 40. |
Mendel found certain traits not to assort independently. It is due toA. DominanceB. LinkageC. Crossing overD. Amitosis |
| Answer» Correct Answer - B | |
| 41. |
Assertion: Mendel worked on garden pea (Pisumsativum). Reason : Garden pea belongs to family-Malvaceae.A. Both Assertion and Reason ae true and the Reason is a correct explanation of the AssertionB. Both Assertion and Reason are true but the Reason is not the correct explanation of the AssertionC. Assertion is true, but the Reason is falseD. Assertion is false, but the Reason is true |
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Answer» Correct Answer - C Mendel worked on garden pea (Pisun satium) as plant material for his expreiments. Garden pea is a member of family- Papilonacease. Pea plant has following advantages i.Well- defined characters ii. Predominantly self-fertilisation iii. Easy hybridisation. |
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| 42. |
The unit of inheritance isA. AlleleB. geneC. factorD. allelomorph |
| Answer» Correct Answer - B | |
| 43. |
The term allele was given byA. BatesonB. MorganC. MendelD. correns |
| Answer» Correct Answer - A | |
| 44. |
Which of the following term represent a pair of contrasting character?A. HomozygousB. PhenotypeC. HeterozygousD. |
| Answer» Correct Answer - A | |
| 45. |
Multiple phenotype is seen inA. PleiotropyB. Incomplete dominanceC. Multiple allelismD. polygenic inheritance |
| Answer» Correct Answer - A | |
| 46. |
Phenotype isA. The genetic make up of an individualB. The same for parent and offspringC. The account of physiological activitesD. The appearance of an individual |
| Answer» Correct Answer - D | |
| 47. |
Diseases caused by pleiotropic genes areA. SyndrmesB. Reversible by diet therapyC. Reversible by gene therapyD. Extremely rare |
| Answer» Correct Answer - A | |
| 48. |
When a gene pair contains two different genes controlling different traits of a character, it isA. HomozygousB. HeterozygousC. MonozygousD. None of the above |
| Answer» Correct Answer - B | |
| 49. |
All genes located on the same chromosome:-A. Form different groups depending upon their relative distanceB. Form one linkage groupC. Will not form any linkage groupD. Form interactive groups that affect the phenotype |
| Answer» Correct Answer - B | |
| 50. |
When two or more non-allelic gene pairs affect the same character in the same way, this is calledA. Polygenic inheritanceB. PleiotropyC. Total penetranceD. Additive expressivity |
| Answer» Correct Answer - A | |