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MCQOPTIONS
This section includes 14620 Mcqs, each offering curated multiple-choice questions to sharpen your NEET knowledge and support exam preparation. Choose a topic below to get started.
| 5651. |
In a dihybrid cross 80 plants are obtained. The ratio of homozygous and heterozygous shall be |
| A. | 1.69444444444444 |
| B. | 0.465277777777778 |
| C. | 0.875 |
| D. | 1.28472222222222 |
| Answer» C. 0.875 | |
| 5652. |
In Mendelian monohybrid cross, phenotypic ratio in \[{{F}_{2}}\] is 3:1. How many types of gametes are formed in \[{{F}_{1}}\] generation [Bihar MDAT 1995] |
| A. | Only one type |
| B. | Two types |
| C. | Four types |
| D. | Eight types |
| Answer» C. Four types | |
| 5653. |
From a mating of two dihybrid individuals (AaBb × AaBb), how many of the 16 possible zygote combinations would be expected to be homozygous dominant for gene |
| A. | One |
| B. | Three |
| C. | Four |
| D. | Six |
| Answer» D. Six | |
| 5654. |
In 1900 A.D. three biologists independently discovered Mendel's principles. They are [RPMT 1997; BHU 1989; MP PMT 2002] |
| A. | De Vries, Correns and Tschermak |
| B. | Sutton, Morgan and Bridges |
| C. | Avery, McLeod and McCarthy |
| D. | Bateson, Punnet and Bridges |
| Answer» B. Sutton, Morgan and Bridges | |
| 5655. |
From a mating of two dihybrid individuals (AaBb × AaBb), how many of the 16 possible zygote combinations would be expected to contain only two dominant genes |
| A. | One |
| B. | Three |
| C. | Four |
| D. | Six |
| Answer» E. | |
| 5656. |
The genotypes of a plant variety were TtHh, Tthh, ttHh and tthh, where 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 × TtHh |
| B. | TtHh × Tthh |
| C. | TtHh × ttHh |
| D. | TtHh × tthh |
| Answer» E. | |
| 5657. |
In wheat, when a green plant was self-fertilized, the progeny had 209 green seedlings and 14 white seedings. The above result indicates that |
| A. | The parents were heterozygous for two duplicate alleles |
| B. | The parents were true breeding |
| C. | The parents were heterozygous for one allele |
| D. | None of the above |
| Answer» B. The parents were true breeding | |
| 5658. |
In pea plants, red flowers (R) are dominant over to white flowers (r) and tall plants (T) are dominant over to dwarf plants (t). The table below shows the gametes and the possible offsprings produced in a dihybrid cross. The numbers 1 to 16 represent the genotypes of each individual cross (e.g. 3 = RrTT) RT Rt rT rt RT 1 2 3 4 Rt 5 6 7 8 rT 9 10 11 12 rt 13 14 15 16 If plant 7 is crossed with plant 12, then what proportion of the offsprings produced will be homozygous for both the recessive characters |
| A. | 0.5 |
| B. | 0.375 |
| C. | 0.25 |
| D. | 0.125 |
| Answer» D. 0.125 | |
| 5659. |
In a cross between white-flowered parents, \[{{F}_{1}}\] progeny was uniformly purple-flowered. The \[{{F}_{2}}\] progeny showed a phenotypic ratio 9 : 7 between purple-flowered and white-flowered plants. The above result indicates that the flower colour is controlled by |
| A. | Two alleles |
| B. | One allele |
| C. | Three alleles |
| D. | Four alleles |
| Answer» B. One allele | |
| 5660. |
When experimenting with the factor of tallness in Pisum sativum, Mendel obtained the results 73.97% tall and 26.03% dwarf from one of the cross. This shows that the parents are |
| A. | Hybrid tall and pure dwarf plant |
| B. | Hybrid tall and pure tall plant |
| C. | Hybrid tall and hybird tall plant |
| D. | Pure tall and pure dwarf plant |
| Answer» D. Pure tall and pure dwarf plant | |
| 5661. |
Some pure black mice were mistakenly mixed with hybrid black mice. The quickest way to discover whether one individual black mouse is pure is |
| A. | To cross it with pure black mouse and all the offsprings must be black |
| B. | To cross it with a brown mouse and all the offsprings must be black |
| C. | To cross it with a known hybrid black mouse and all the offsprings must be black or brown mice in equal numbers |
| D. | To cross it with a pure brown mouse and all the offsprings must be black or brown in equal numbers |
| Answer» E. | |
| 5662. |
A black hybrid mouse is crossed with a pure brown mouse to produce a total of 24 young ones. Which of the following result is most likely to be correct |
| A. | 13 black males and 11 black females |
| B. | 11 brown males and 13 brown females |
| C. | 8 black males, 10 black females, 3 brown males, 3 brown females |
| D. | 5 black males, 6 brown males, 7 black females, 6 brown females |
| Answer» E. | |
| 5663. |
Who has putforth Mendel's conclusions in the form of laws |
| A. | Bateson |
| B. | Correns |
| C. | Punnet |
| D. | Johanssen |
| Answer» C. Punnet | |
| 5664. |
If a homozygous red-flowered plant is crossed with a homozygous white-flowered plant, the offspring would be [MP PMT 1994, 97; AIIMS 1999, 2002] |
| A. | Half red-flowered |
| B. | Half white-flowered |
| C. | All red-flowered |
| D. | Half pink-flowered |
| Answer» D. Half pink-flowered | |
| 5665. |
In sweet peas, genes C and P are necessary for colour in flowers. The flowers are white in the absence of either or both the genes. What will be the percentage of coloured flowers in the offspring of the cross Cc pp × cc Pp [CMC Vellore 1994] |
| A. | 1 |
| B. | 0.75 |
| C. | 0.25 |
| D. | 0.5 |
| Answer» D. 0.5 | |
| 5666. |
If the cells of an organism heterozygous for two pairs of characters viz. Aa and Bb undergo meiosis, what will be the genotypes of the gametes produced [JIPMER 1994] |
| A. | Aa and Bb |
| B. | AB, aB, Ab and ab |
| C. | aB and Ab |
| D. | Ab and ab |
| Answer» C. aB and Ab | |
| 5667. |
If enough crosses are made between male flies of the genotype 'Aa' and the female flies of the genotype 'aa' to produce about 1000 offsprings. Which one of the following is the most likely distribution of genotypes in the offsprings |
| A. | 250 Aa : 750 aa |
| B. | 750 Aa : 250 aa |
| C. | 243 AA : 517 Aa : 240 aa |
| D. | 481 Aa : 519 aa |
| Answer» E. | |
| 5668. |
From a cross Aa BB × aa BB, following genotypic ratio will be obtained in \[{{F}_{1}}\] generation [CBSE PMT 1990] |
| A. | 1 Aa BB : 1 aa BB |
| B. | 1 Aa BB : 3 aa BB |
| C. | 3 Aa BB : 1 aa BB |
| D. | All Aa BB : No aa BB |
| Answer» B. 1 Aa BB : 3 aa BB | |
| 5669. |
From a single ear of corn, a farmer planted 200 kernels which produced 140 tall and 40 dwarf plants. The genotype of these offsprings are most likely [NCERT 1981; DPMT 1993] |
| A. | TT, Tt and tt |
| B. | TT and tt only |
| C. | TT and Tt only |
| D. | Tt and tt only |
| Answer» B. TT and tt only | |
| 5670. |
If a plant heterozygous for tallness is selfed, the \[{{F}_{2}}\] generation has both tall and dwarf plants. This proves the principle of [CPMT 1984; DPMT 1990] |
| A. | Dominance |
| B. | Segregation |
| C. | Independent assortment |
| D. | Incomplete dominance |
| Answer» C. Independent assortment | |
| 5671. |
Normal maize has starchy seeds which remain smooth when dry. A mutant form has sugary seeds which go crinckled when dry. When a mutant was crossed with a normal plant, an F1 was produced which had smooth seeds. What would be the relative ratios of the different seed types, if the F1 was allowed to self [AIIMS 1993] |
| A. | 1 smooth : 3 sugary |
| B. | 3 smooth : 1 sugary |
| C. | 1 smooth : 1 sugary |
| D. | All sugary |
| Answer» C. 1 smooth : 1 sugary | |
| 5672. |
When the tall plants with red flowers were crossed with dwarf plants having white flowers, Mendel found the ratio of progeny as [DPMT 1993] |
| A. | 0.0430671296296296 |
| B. | 0.125694444444444 |
| C. | 9 : 3 : 3 : 1 |
| D. | 1 : 4 : 6 : 4 : 1 |
| Answer» D. 1 : 4 : 6 : 4 : 1 | |
| 5673. |
In sweet pea plants the presence of dominant C and P genes is essential for development of purple colour. The ratio of plants producing flowers of different colours in the progeny of the cross Cc Pp × Cc pp will be [AFMC 1993] |
| A. | 2 white and 6 purple coloured flowers |
| B. | 2 purple and 6 white coloured flowers |
| C. | 3 white and 5 purple coloured flowers |
| D. | 3 purple and 5 white coloured flowers |
| Answer» E. | |
| 5674. |
In genetics, the use of chequer board was done by |
| A. | Mendel |
| B. | Correns |
| C. | Punnet |
| D. | Darwin |
| Answer» D. Darwin | |
| 5675. |
Mendelism is related with [BVP 2003] |
| A. | Heredity in living beings |
| B. | Meiosis during sexual reproduction |
| C. | Mutation in living organisms |
| D. | None of the above |
| Answer» B. Meiosis during sexual reproduction | |
| 5676. |
A linear tetrad of 4 cells lying in an axial row is formed during the development of |
| A. | Embryo sac |
| B. | Pollen grains |
| C. | Ovary |
| D. | Ovule |
| Answer» B. Pollen grains | |
| 5677. |
In which of the following types of embryo sac the megaspore nuclei do not fuse to form a triploid nucleus at chalazal end |
| A. | Fritillaria type |
| B. | Plumbagella type |
| C. | Adoxa type |
| D. | None of the above |
| Answer» D. None of the above | |
| 5678. |
A Drusa type of embryo sac is |
| A. | 16 celled and 16 nucleate |
| B. | 15 celled and 15 nucleate |
| C. | 15 celled and 16 nucleate. |
| D. | 14 celled and 15 nucleate |
| Answer» D. 14 celled and 15 nucleate | |
| 5679. |
If the number of chromosomes in root cells is 14, what will be the number of chromosomes in synergids cells of an ovule of that parent |
| A. | 7 |
| B. | 14 |
| C. | 21 |
| D. | Incomplete information |
| Answer» B. 14 | |
| 5680. |
Through which cell of the embryo sac, does the pollen tube enter the embryo sac [CBSE PMT 2005] |
| A. | Egg cell |
| B. | Central cell |
| C. | Persistant synergid |
| D. | Degenerated synergid |
| Answer» E. | |
| 5681. |
Ovule with funiculus lying close to micropyle is known as [Pb. PMT 2004] |
| A. | Anatropous |
| B. | Campylotropous |
| C. | Atropous |
| D. | Cytokinin |
| Answer» B. Campylotropous | |
| 5682. |
Female gametophyte of angiosperm is generally [MHCET 2004] |
| A. | 7 celled and 7 nucleate |
| B. | 8 celled and 8 nucleate |
| C. | 7 celled and 8 nucleate |
| D. | 8 celled and 7 nucleate |
| Answer» D. 8 celled and 7 nucleate | |
| 5683. |
An ovule which becomes curved so that the nucellus and embryo sac lie at right angles to the funicle is [CBSE PMT 2004] |
| A. | Anatropous |
| B. | Orthotropous |
| C. | Hemitropous |
| D. | Campylotropous |
| Answer» D. Campylotropous | |
| 5684. |
In angiosperms embryosac is developed from [MP PMT 2003] |
| A. | Megaspore mother cell |
| B. | Secondary nucleus |
| C. | Endothecium |
| D. | Microspore mother cell |
| Answer» B. Secondary nucleus | |
| 5685. |
Synergids of the polygonum type embryo sac are [MHCET 2002] |
| A. | Haploid |
| B. | Diploid |
| C. | Triploid |
| D. | Polyploid |
| Answer» B. Diploid | |
| 5686. |
For the formation of tetrasporic embryosac, how many megaspore mother cells are required [BHU 2003] |
| A. | 1 |
| B. | 2 |
| C. | 3 |
| D. | 4 |
| Answer» B. 2 | |
| 5687. |
The haploid cell which divides by mitosis to form embryosac is [RPMT 2002] |
| A. | Megaspore mother cell |
| B. | Microspore mother cell |
| C. | Functional megaspore |
| D. | Non-functional megaspore |
| Answer» D. Non-functional megaspore | |
| 5688. |
The point of attachment of funicle with chalazal end is called [MP PMT 1996, 2001; MHCET 2003] |
| A. | Placenta |
| B. | Integument |
| C. | Nucellus |
| D. | Hilum |
| Answer» E. | |
| 5689. |
In which type whole of the megaspore mother cell takes part in the formation of the female gametophyte |
| A. | Monosporic 8 nucleate |
| B. | Monosporic 4 nucleate |
| C. | Bisporic |
| D. | Tetrasporic |
| Answer» E. | |
| 5690. |
If diploid chromosome number in a flowering plant is 12, then which one of the following will have only 6 chromosomes [CPMT 1984, 86] |
| A. | Endosperm |
| B. | Leaf cells |
| C. | Cotyledons |
| D. | Synergids |
| Answer» E. | |
| 5691. |
Mature Polygonum type embryo sac has got [BHU 1999; MP PMT 2000] |
| A. | Seven cells and eight nuclei |
| B. | Seven nuclei and eight cells |
| C. | Eight cells and eight nuclei |
| D. | Seven cells and seven nuclei |
| Answer» B. Seven nuclei and eight cells | |
| 5692. |
The formation of embryo sac is called [CPMT 1998] |
| A. | Megasporogenesis |
| B. | Megagametogenesis |
| C. | Micro gametogenesis |
| D. | None of these |
| Answer» C. Micro gametogenesis | |
| 5693. |
Ovule of Capsella is [RPMT 1995] |
| A. | Orthotropus |
| B. | Anatropus |
| C. | Campylotropus |
| D. | Amphitropus |
| Answer» D. Amphitropus | |
| 5694. |
Female gametophyte of angiospermic plants is represented by [MP PMT 1994, 2000] |
| A. | Oospore |
| B. | Egg |
| C. | Carpel |
| D. | Pollen grain |
| Answer» C. Carpel | |
| 5695. |
In an embryo sac of a typical angiosperm, there are [MP PMT 1993] |
| A. | Egg, synergids and antipodals |
| B. | Egg, synergids, polar nuclei and antipodals |
| C. | Egg, synergids, central cell and polar nuclei |
| D. | Egg, synergids and secondary cell |
| Answer» C. Egg, synergids, central cell and polar nuclei | |
| 5696. |
The ovule in pea are [DPMT 1990; MP PMT 1996] |
| A. | Anatropous |
| B. | Hemianatropous |
| C. | Campylotropous |
| D. | Amphitropous |
| Answer» D. Amphitropous | |
| 5697. |
An orthotropous ovule is one in which micropyle and chalaza are [CPMT 1982; AIIMS 1992; CBSE PMT 1994] |
| A. | In straight line of funiculus |
| B. | Parallel to funiculus |
| C. | At right angles to funiculus |
| D. | Oblique to funiculus |
| Answer» B. Parallel to funiculus | |
| 5698. |
Filiform apparatus is found in which part of angiosperms [CPMT 1993, 95; CPMT 1999] |
| A. | Sperm |
| B. | Antipodal |
| C. | Egg |
| D. | Synergid |
| Answer» E. | |
| 5699. |
Horizontally oriented ovules are called |
| A. | Hemitropous |
| B. | Campylotropous |
| C. | Circinotropous |
| D. | Atropous |
| Answer» B. Campylotropous | |
| 5700. |
Collar like outgrowth arising from the base of ovule and forming a sort of third integument is known as [CPMT 1984; JIPMER 2002] |
| A. | Coma |
| B. | Caruncle |
| C. | Aril |
| D. | Operculum |
| Answer» D. Operculum | |