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.
| 101. |
Incomplete dominance is found inA. Neurospora crassaB. Lathyrus odoratusC. Pisum sativumD. Antirrhinum/Mirabilis |
| Answer» Correct Answer - D | |
| 102. |
If both parents are albino, all the offspring shall beA. AlbinoB. Some albino, some normalC. Homozygous normalD. Heterozygous normal |
| Answer» Correct Answer - A | |
| 103. |
Hybrid vigour is due toA. Mixing of traits of male and femaleB. Homozygosity in gametes stateC. Superiority of genesD. Heterozygosity |
| Answer» Correct Answer - D | |
| 104. |
What is incorrect in the following Mendelian traitsA. `{:("Character","Dominant","Recessive"),("Height","Tall","Dwarf"):}`B. `{:("Character","Dominant","Recessive"),("Seed colour","Green","Yellow"):}`C. `{:("Character","Dominant","Recessive"),("Flower position","Axillary","Terminal"):}`D. `{:("Character","Dominant","Recessive"),("Flower colour","Violet","White"):}` |
| Answer» Correct Answer - B | |
| 105. |
A cross between yellow flowered and round fruit ( both dominant ) bearing plant and white flowered elongated fruit ( both recessive ) bearing plant yielded 20 plants in `F_(1)` progeny which formed 960 plants in `F_(2)` generation. The number of plants with yellow flowered and round fruits in `F_(1)` and `F_(2)` generations would beA. 20960B. 20540C. 10180D. 10,60 |
| Answer» Correct Answer - B | |
| 106. |
A couple both carries of sickle cell anaemia planning to get married, wants to know the chances of having anaemic progenyA. 1B. 0.75C. 0.5D. 0.25 |
| Answer» Correct Answer - D | |
| 107. |
A genetic clone isA. Plants produced by asexual meansB. Hybrid produced by sexual meansC. Homozygous plant prouduced by sexual meansD. Heterozygous plant proudced by sexual means |
| Answer» Correct Answer - A | |
| 108. |
A cross between black flowered plant and white flowered plant yielded grey floweredA. CodominanceB. PseudodominanceC. Incomplete dominanceD. Epistasis |
| Answer» Correct Answer - C | |
| 109. |
Medel conduted hybridisation experiments on Garden Pea ofA. 4 yearsB. 5 yearsC. Six yearD. Seven years |
| Answer» Correct Answer - D | |
| 110. |
First generation after a cross isA. First filial generationB. `F_(1)` generationC. Second filial generationD. Both A and B |
| Answer» Correct Answer - D | |
| 111. |
Heterozygous organism for two genes shall beA. RRYYB. RrYYC. RrYyD. RRYy |
| Answer» Correct Answer - C | |
| 112. |
Which is incorrect in Mandelian characters ?A. `{:("Character","Dominant","Recessive"),("Pod colour","Green","Yellow"):}`B. `{:("Character","Dominant","Recessive"),("seed shape","Round ","Wrinkled"):}`C. `{:("Character","Dominant","Recessive"),("Flower position","Terminal","Axillary"):}`D. `{:("Character","Dominant","Recessive"),("Shape of pod ","Full","Constricted"):}` |
| Answer» Correct Answer - C | |
| 113. |
Which pair of features represents polygenic inheritance ?A. Human eye colour and sickle cell anaemiaB. Hair pigment of mouse and tongue rolling in humansC. ABO blood groups in humans and flower colour of Mirabilis jalapaD. Human height and skin colour. |
| Answer» Correct Answer - B | |
| 114. |
Medel studied inheritance of seven pairs of traits in pea which can have 21 possible combinations. If you are told that in one of these combinations. If you are told that in one of these combinations, independent assortment is not observed in later studies, your reaction will beA. Independent assortment principle may be wrongB. Mendel might not have studied all the combinationsC. It is impossibleD. Later sudies may be wrong. |
| Answer» Correct Answer - B | |
| 115. |
The ratio of 9:3:3:1 is due toA. Segregation of charactersB. Independent assortment of genesC. Crossing over of chromosomesD. Homologous pairing between chromosomes. |
| Answer» Correct Answer - B | |
| 116. |
In Mirabilies jalapa when two `F_(1)` pink flowered plants were corossed with each other, the `F_(2)` generation produced 40 red, 80 pink and 40 white flowering plants. This is a case ofA. Duplicate genesB. Lethal genesC. Incomplete dominanceD. Epistasis |
| Answer» Correct Answer - C | |
| 117. |
`F_(2)` generation is produced as a result ofA. Crossing `F_(1)` individual with dominant individualB. Crossing `F_(1)` individual with recessive individualC. Crossing `F_(1)` individuals amongst themselves.D. All the above |
| Answer» Correct Answer - C | |
| 118. |
A polygenic inheritance in human beings isA. Skin colourB. PhenylketonuriaC. Colour blindnessD. Sickle cell anaemia |
| Answer» Correct Answer - A | |
| 119. |
Ratio `9 : 3 : 3 : 1` is due to:A. Duplicate genesB. Lethal genesC. Monohybrid crossD. Dihybrid cross |
| Answer» Correct Answer - D | |
| 120. |
`F_(2)` generation is produced as a result ofA. Crossing `F_(1)` progeny with one of the parentsB. Selfing the progeny of two individual parentsC. Selfing the parentsD. Recessive cross between individual parents. |
| Answer» Correct Answer - B | |
| 121. |
Hybrid are generally superior to parents due toA. HeterosisB. HomozygosityC. HeterozygosityD. Parents are generally weak. |
| Answer» Correct Answer - A | |
| 122. |
The process of removing stamen from flora buds during hybridisation experiments isA. CappingB. SelfingC. EmasculationD. Crossing |
| Answer» Correct Answer - C | |
| 123. |
"Gametes are never hybrid ". It is a statement of law ofA. DominanceB. SegregationC. Independent assortmentD. Random fetilisation |
| Answer» Correct Answer - B | |
| 124. |
Select the correct statements from the ones given below with respect to dihybrid crossA. Genes far apart on the same chromosomes show very few recombinationsB. Genes loosely linked in the same chromosomes show similar recombinations as the tightly linked onesC. Tightly linked genes on the same chromosome show very few recombinationsD. Tightly linked genes on the same chromosomes show higher recombinations. |
| Answer» Correct Answer - C | |
| 125. |
The ratio of `9 : 7` is due to:A. Supplementary genesB. Lethal genesC. Complementary genesD. Epistatic genes |
| Answer» Correct Answer - C | |
| 126. |
Monohybrid test cross isA. `1:1:1:1`B. `1:1`C. `9:3:4`D. `9:3:3:1` |
| Answer» Correct Answer - B | |
| 127. |
Monohybrid cross involvesA. Individuals of two different parentsB. Individuals different in one traitsC. Individuals different in two traitsD. Individual with parent. |
| Answer» Correct Answer - B | |
| 128. |
A plant of genotype RRTt (red fruits, tall ) is crossed with a plant of genotype rrtt ( white furits, dwarf). What is percentage of tall plants with red fruits in next generation ?A. 1B. 0.75C. 0.25D. 0.5 |
| Answer» Correct Answer - D | |
| 129. |
A person with blood group AB has which of the antigens in RBCs.A. AB. BC. A and BD. None of the above |
| Answer» Correct Answer - C | |
| 130. |
Mendel found that reciprocal crosses yielded indentical results. From that he concludedA. Sex plays a role in deciding dominance of a traitB. There is no dominance of any traitC. There is independent assortment of the traitD. Sex has no influence on the dominance of the trait. |
| Answer» Correct Answer - D | |
| 131. |
Phenotype is influenced byA. EnvironmentB. DevelopmentC. AgeingD. All the above |
| Answer» Correct Answer - D | |
| 132. |
Sexually reproducing organism contribute in their offspringA. All of the geneB. One half of their genesC. One fourth of their genesD. Double the number of genes. |
| Answer» Correct Answer - B | |
| 133. |
In a case of red and white flowered Snapdragon, `F_(1)` plants will have flowersA. RedB. WhiteC. Both A and BD. Pink |
| Answer» Correct Answer - D | |
| 134. |
A heterozygous round seeded (Rr) plant is crossed to recessive wrinkled (rr) seeded plant. The progeny would be:A. 303 rounded : 301 wrinkledB. 301 rounded : 100 wrinkledC. 20 round :99 wrinkledD. 99 rounded : 301 wrinkled. |
| Answer» Correct Answer - A | |
| 135. |
Two pink flowered Snapdragon plants (Rr) are self pollinated. Probability of offspring having white flower isA. 0.25B. 0.5C. 0.75D. `2.5%` |
| Answer» Correct Answer - A | |
| 136. |
When heterozygous red flowered plant is crossed with white flowered plant the progeny will show a ratio ofA. 350 red : 350 whiteB. 450 red : 250 whiteC. 380 red: 220 whiteD. None of the above |
| Answer» Correct Answer - A | |
| 137. |
Dominant epistasis ratio isA. `9:7`B. `9:3:3:1`C. `9:3:4`D. `12:3:1` |
| Answer» Correct Answer - D | |
| 138. |
The factors which represent the constrasting pairs of characters are calledA. Dominant and recessiveB. AllelesC. Homologous pairsD. Deteminants |
| Answer» Correct Answer - B | |
| 139. |
The term test cross refers to a cross between:A. The crossing of `F_(1)` individual with homozygous recessiveB. Crossing and `F_(1)` individual with either of the two parentsC. Crossing and `F_(1)` individual with another `F_(1)` individualD. Crossing `F_(1)` individual with that of `F_(2)` |
| Answer» Correct Answer - A | |
| 140. |
In a case of incomplete dominance, true breeding red flowerd plant (RR) is crossed with true breeding white flowered plant (rr). What would be correct `F_(2)` ratio ?A. 1 red : 2 pink : 1 whiteB. 3 red : 1 whiteC. 1 red : 1 pink : 2 whiteD. 2 red : 1 pink : 1 white |
| Answer» Correct Answer - A | |
| 141. |
A dwarf Pea plant with wrinkled seeds ( ttrr) produces seeds of typeA. 2B. 4C. 1D. 9 |
| Answer» Correct Answer - C | |
| 142. |
Mendel selected Pea as material for his experiments becauseA. It is a annual plant with comparatively short life cycleB. The flowers are self-pollinatedC. The number of seeds produced is quite largeD. All the above |
| Answer» Correct Answer - D | |
| 143. |
Multiple alleles are presentA. At different loci in the same chromosomeB. In different chromosomeC. At the same locus in one type of chromosomesD. None of the above |
| Answer» Correct Answer - C | |
| 144. |
In the first step of monohybrid cross experiment, Mendel selected Pea plants which wereA. Pure tall as male and pure dwarf as femaleB. Pure tall as female and pure dwarf as maleC. Heterozygous tall as male and pure dwarf as femaleD. Heterozygous tall as female and pure dwarf as male. |
| Answer» Correct Answer - B | |
| 145. |
Out of seven characters in Pea plant studied by Mendel, the number of flower based character was 1A. 1B. 3C. 4D. 2 |
| Answer» Correct Answer - D | |
| 146. |
Which trait was not incorporated by Mendel for his experiments?A. Colour of Pea seedB. Colour of Pea flowerC. Colour of Pea plant/Size of Pea seedD. Coloured of pea pod |
| Answer» Correct Answer - C | |
| 147. |
Mendel conducted hybirdisation experiments onA. Pigeon PeaB. Garden PeaC. Wild PeaD. Sweet Pea. |
| Answer» Correct Answer - B | |
| 148. |
Match the columns and choose the correct option A. a-iv,b-iii, c-`i`,d-iiB. a-ii,b-`i`,c-iv,d-iiiC. a-ii,b-iii,c-iv, d-`i`D. a-iv,b-i,c-ii,d-iii |
| Answer» Correct Answer - C | |
| 149. |
When both the parents are of blood group AB, children would be of blood groupA. A,B, AB and OB. A,B and ABC. A and BD. A,AB and O |
| Answer» Correct Answer - B | |
| 150. |
Among the following character which one was not cosidered by Mendel in his experiments on PeaA. Stem-Tall or dwarfB. Trichomes-Glandular or nonglandularC. seed-Green or YellowD. Pod- Inflated or constricted |
| Answer» Correct Answer - B | |