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.
| 551. |
What will be the sex of the offspring developing from 44 A+xx zygote? |
| Answer» Correct Answer - Female | |
| 552. |
what is a locus |
| Answer» Site of gene is a chromosome | |
| 553. |
Man produced 2 types of sperms. Is it true? |
| Answer» Correct Answer - Yes. | |
| 554. |
How many chromatids are involved in crossing over at one chiasma? |
| Answer» Correct Answer - Two | |
| 555. |
A woman with normal vision, but whose father was colour blind, marries a colour blind man. Suppose that the fourth child of this couple was a boy. Thus boyA. Will be partially colour blind since he is heterozygoustor for the colour blind mutant ableB. Must have normal colour visionC. Must be colour blindD. May be colour blind or may be of normal vision, |
| Answer» Correct Answer - D | |
| 556. |
Morgan hybridised yellow-bodied, white-eyed females to brown-bodied red-eyed males and intercrossed their `F_(1)` progeny. He observed that (a) `F_(2)` ratio was deviated very significantly from the `9:3:3:1` ratio (b) Both genes did not segregate independently of each other (c) Recombinant types are not obtained in `F_(2)` generation (d) Both genes segregate independently of each other Select the correct set of statements :A. (a) & (b) onlyB. (b) & (c ) onlyC. (b) & (d) onlyD. (c ) & (d) only |
| Answer» Correct Answer - A | |
| 557. |
The fruit fly Drosophila melanogaster was found to be very suitable for expermiental verification of chromosomal theory on inheritanc by Morgan and his coleagues becauseA. It reproduces parthenogeneticallyB. A single mating produces two young fliesC. Smaller female is easily recognisable from larger maleD. It completes lige cycle in about two weeks |
| Answer» Correct Answer - D | |
| 558. |
Which of the following animal was selected by morgan for studying linkageA. Apis indicaB. Agrobacterium tumafaciensC. Drosophila melanogasterD. E.Coil |
| Answer» Correct Answer - C | |
| 559. |
Which of the following combination seems to have some linkage in character selected by Mendel ?A. Stern height and pod colourB. Flower colour and flowerC. Flower colour and flower positionD. Seed shape and seed colour |
| Answer» Correct Answer - D | |
| 560. |
Which of the following parental combination has produced mutant offspring ?A. `Tt xx tt = Tt`B. `tt xx tt = Tt`C. `Tt xx Tt = tt`D. `"TT" xx tt = Tt` |
| Answer» Correct Answer - B | |
| 561. |
In a population of 1000 individuals 360 belong to genototype AA ,480 to Aa and the remaining 160 to aa Based on this data ,the frequency of allele A in the population isA. `0.7`B. `0.4`C. `0.5`D. `0.6` |
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Answer» Correct Answer - D In 1000 individuals, person with genotype AA = 360 `therefore` I individual, person with genotype `"AA" = 360//1000 = 0.36` `"AA" = A^(2) = 0.36` `therefore A = sqrt(0.36) = 0.6` |
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| 562. |
Assertion : The honey bee queen copulates only once in her life time. Reason : The honey bee queen can lay fertilized as well as unfertilized eggs.A. If both assertion and reasons are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If the assertion is true but reason is false.D. If both assertion and reason are false. |
| Answer» Correct Answer - A | |
| 563. |
In pea plants, yellow seeds are dominant to green. If a heterozygous yellow seeds plant is crossed wit a green seeded plant, what ratio of yellow and green seeded plants would you expect in `F_(1)` generation :-A. `9:1`B. `1:3`C. `3:1`D. `50:50` |
| Answer» Correct Answer - D | |
| 564. |
Which is dominant, a factor (gene) wrinkeled seeds ? |
| Answer» Correct Answer - Round Seeds | |
| 565. |
In a plant, red fruit (R ) is dominant over yellow fruit (r ) and tallness (T) is dominant over shortness (t). If a plant with RRTt genotype is crossed with a plant that is rrtt :-A. `25%` will be tall with reed fruitB. `50%` will be tall with red fruitC. `75%` will be tall with red fruitD. All the offsprings will be tall with red fruit |
| Answer» Correct Answer - B | |
| 566. |
A gene is said to be dominant, ifA. it is expressed only in heterozygous combinationB. It is expressed only in homozgous combinationC. It is expessed in both homozgous and heterozygous conditionD. it is expressed only in second genertion |
| Answer» Correct Answer - C | |
| 567. |
Hybrid breakdown refers to the condition when offspring are physiological inferior to the following generationA. `F_(1)`B. `F_(2)`C. `P_(1)`D. All of these |
| Answer» Correct Answer - A | |
| 568. |
From a cross Aa BB`xx`aa BB, following genotypic ratio will be obtained in `F_(1)` generationA. 1 Aa BB : 1 aa BBB. 1 Aa BB : 3 aa BBC. 3 Aa BB : 1 aa BBD. All Aa BB : No aa BB |
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Answer» Correct Answer - A (a) `AaBBxxaaBB` Gametes for `F_(1)`=AB,aB and aB, aB After crossing =AaBB,aaBB Ratio= 1:1 |
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| 569. |
Person whose father is colourblind marries a lady whose mother is daughter of a colourblind man. Their children will beA. All sons colour blindB. Some sons normal and some daughters colour blindC. All sons and daughters colour colour blindD. All daughter normal |
| Answer» Correct Answer - D | |
| 570. |
A diseased man marries a normal woman.They get three daughters and five sons were normal . The gene of this disease is :A. Sex-influenced diseaseB. Blood group inheritance diseaseC. Sex-linked diseaseD. Sex-limited disease |
| Answer» Correct Answer - C | |
| 571. |
A diseased man marries a normal woman.They get three daughters and five sons were normal . The gene of this disease is :A. Sex-linked recessiveB. Sex-linked dominantC. Autosomal characterD. Sex-limited character |
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Answer» Correct Answer - C A diseased man marries a normal woman. The couple has 3 daughters and 5 sons. The daughters are diseased while the sons are normal. The gene of the disease is autosomal character. |
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| 572. |
Heterozygous tall plant (Tt) is crossed with homozygous dwarf (tt) plant. Then what will be the percentage of dwarf plants in the next generation? |
| Answer» Correct Answer - B | |
| 573. |
From a single ear of corn, a farmer planted 200 kernals which produced 140 tall and 40 dwarf plants. The genotype of these offsprings are most likelyA. TT, Tt and ttB. TT and tt onlyC. TT and Tt onlyD. Tt and tt only |
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Answer» Correct Answer - A TT is homozygous tall plant, Tt is heterozygous tall plant and tt is homozygous dwarf plant. |
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| 574. |
A : Blood group phenotype is controlled by presence or absence of antigens present on surface coating of RBC. R : These antigens are of three types and four in the oligosaccharides rich head regions on glycophorin.A. If both Assertion & Reason are true and the reason is the correct explanation of the assertion, then mark (1)B. If both Assertion & Reason are true and the reason is not the correct explanation of the assertion , then mark (2)C. If Assertion is true statement but Reason is false, then mark (3)D. If both Assertion and Reason are false statements, then mark (4) |
| Answer» Correct Answer - B | |
| 575. |
When red blood corpuscles containing both A and B antigens are mixed with your blood serum, they agglutinate. Hence you blood group is……type.A. ABB. OC. AD. B |
| Answer» Correct Answer - B | |
| 576. |
A male human is heterozygous for autosomal genes A and B. He is also hemizygous for haemophilic gene h. What proportion of sperms will carry abh?A. `1//8`B. `1//32`C. `1//4`D. `1//16` |
| Answer» Correct Answer - A | |
| 577. |
A male human is heterozygous for autosomal genes A and B and is also hemizygous for hemophilic gene h. What proportion of his sperms will be abhA. `1//8`B. `1//32`C. `1//16`D. `1//4` |
| Answer» Correct Answer - A | |
| 578. |
The daughter born to haemophilic father and normal mother could beA. NormalB. CarrierC. HaemophilicD. None |
| Answer» Correct Answer - B | |
| 579. |
If there is complete linkage in `F_(2)` generationA. Parental types and recombinants appear in equal ratioB. Recombinants are less than parental typesC. Recombinants are more than parental typesD. There will be only parental types |
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Answer» Correct Answer - D If there is complete linkage in `F_(2)` generation then there will be only parental types. |
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| 580. |
In Drosophila melanogaster, the genes white and yellow shows……a……recombination and genes white and miniature wing shows…….b…….linkageA. `a to 98.7%, b to 37.2%`B. `a to 98.7%, b to 62.8%`C. `a to 1.3%, b to 37.2%`D. `a to 1.3%, b to 62.8%` |
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Answer» Correct Answer - D In Drosophila melanogaster the genes white and yellow shows 1.3% recombination and genes white and miniature wing shows 62.8% linkage. |
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| 581. |
`overset((A))("_____")` used the frequency of recombination between gene pairs on the `overset((B))("_____")` as a measure of the distance between genes and mapped their position on the chromosome.A. `{:("A","B"),("Morgan","Same chromosome"):}`B. `{:("A","B"),("Sturtevant","Different chromosomes"):}`C. `{:("A","B"),("Morgan","Different chromosomes"):}`D. `{:("A","B"),("Sturtevant","Same chromosome"):}` |
| Answer» Correct Answer - D | |
| 582. |
Find the odd one out w.r.t complete linkageA. `100%` parental combinations in `F_(2)` generationB. `F_(2)` phenotypic ratio is `3:1` in monohybrid crossC. Dihybrid test cross ratio is `1:1` in `F_(2)` generationD. Linked genes tend to separate frequnently |
| Answer» Correct Answer - D | |
| 583. |
Give the answer of following questions if in a test cross `AaBb xx aabb, 87.4%` of the progeny are like parents. (i) Are the genes linked? (ii) Is there any crossing over between the genes ? |
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Answer» (i) As parental types are `87.4%` (or `gt 50%`) and non-parental types are `12.6%` ( or `lt 50%`), it means genes are linked. (ii) Yes, Crossing over lead to recombination (generation of non-parental gene combinations). |
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| 584. |
A condition where an individual heterozygous for two pairs of linked genes ( AaBb ) possesses the two dominant genes on one homologous chromosomes pair and two recessive on the other it is said to beA. Cis-arrangmentB. Trans-arrangementC. Partly cis partly transD. More than one option is correct |
| Answer» Correct Answer - A | |
| 585. |
Assertion: Number of chromosomes in one genome is equal to the number of linkage groups. Reason : Linkage groups give important information about the location of genes in the chromosomes.A. If both assertion and reasons are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If the assertion is true but reason is false.D. If both assertion and reason are false. |
| Answer» Correct Answer - B | |
| 586. |
What will be the number of linkage groups in maize if it has 10 pairs of chromosomes or What will be the number of linkage groups in a cell having 2n=20A. 5B. 10C. zeroD. 20 |
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Answer» Correct Answer - B (b) In maize n=10, hence linkage groups =10 |
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| 587. |
How many linkage groups are present in human male ?A. 24B. 23C. 46D. 22 |
| Answer» Correct Answer - A | |
| 588. |
When a cluster of genes shows linkage behaviour theyA. Do not show a chromosome mapB. Show recombination during melosisC. Do not show independent assortmentD. Induce cell division |
| Answer» Correct Answer - C | |
| 589. |
In human the inhertiance of sex linkage takes place throughA. AutosomeB. Y-chroosomeC. X-chromosomeD. Both (b) and (c ) |
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Answer» Correct Answer - D Sex linkage is the transmission of characters and their determining genes alongwith sex determining genes which are found on the sex chromosomes . Y chromosome of male carries a few genes but X chromosome which is common to male and female carries a number of genes. |
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| 590. |
Which one of the following is a genetically transmitted characterA. Colour blindnessB. HydrocephalusC. HemophiliaD. all of these |
| Answer» Correct Answer - D | |
| 591. |
In the above given pedigree, assume that no outsider marrying in, carry a disease. Write the genotypes of II and IIIA. All `X^(d)Y`B. `X^(D)Y" and "X^(D)X^(d)`C. `X^(d)XX^(d)Y" and "X^(d)Y^(D)`D. `X^(d)X^(d)" and "X^(d)Y` |
| Answer» Correct Answer - C | |
| 592. |
Karyotype isA. Chromosome complement which at specific for each species of living organismB. All organism posseing same type of chromosomesC. Division of necleusD. None of these |
| Answer» Correct Answer - A | |
| 593. |
What is the chromosome number of plasmodiumA. 18B. 14C. 10D. 9 |
| Answer» Correct Answer - B | |
| 594. |
A child receivesA. 25% genes form his fatherB. 50% genes form his fatherC. 75% genes form his fatherD. 100% genes form his father |
| Answer» Correct Answer - B | |
| 595. |
Genes are in the form of :A. Sequence of nucleotideB. Base pairC. Proportion of base pairD. None of these |
| Answer» Correct Answer - A | |
| 596. |
Telemere repetitive DNA sequences control the function of eukarote chromosomes because theyA. Act as repliconsB. Are RNA transciption initiatorC. Help chromosome pairingD. Prevent chromosome loss |
| Answer» Correct Answer - D | |
| 597. |
The most likely reason for the development of resistence against pesticides in insects damaging a crop isA. Random mutationsB. Genetic recombinationC. Directed mutationsD. Acquired heritable changes |
| Answer» Correct Answer - A | |
| 598. |
Represented below is the inheritance pattern of certain type of traits in humans. Which one of the followings conditions could be an example of the pattern ? A. PhenylketonuriaB. Sickle cell anaemiaC. HaemophiliaD. Thalasemia |
| Answer» Correct Answer - C | |
| 599. |
Identical twins areA. HeterozygousB. HomozygousC. MonozygoticD. Dizygotic |
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Answer» Correct Answer - C (c ) Identical twins are product when one fertilled egg dividesinto two blastomeres and both give rise to divdes in to young ones. |
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| 600. |
Fraaternal twins are produced whenA. A fertiilzed egg divided in to twoB. An egg is fertilzd by two set spermsC. A divvided egg has two set of chromosesD. Two eggs are fertized simultaneously |
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Answer» Correct Answer - D (D) they are also known as dizygotic twins. |
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