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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.
| 401. |
Which of the following contributed to the success of Mendel?A. Qualitative analysis of dataB. Observation of distinct inherited traitsC. His knowledge of BiologyD. Consideration of one character at one time |
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Answer» Correct Answer - D Mendel took one trait at a time for his experiments, this contributed a lot to his success. |
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| 402. |
Phenotype of an organism is the result ofA. mutations and linkagesB. cytoplasmic effects and nutritionC. environmental changes and sexual dimorphismD. genotype and environmental interactions |
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Answer» Correct Answer - D Phenotype is the observable characteristics or the total appearance of an organism. It is determined by its genes, the relationships between the alleles and by the interaction during development between its genetic constitution (genotype) and the environment. |
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| 403. |
Phenotype of an organism is result ofA. Mutations and linkagesB. Genotype and environment interactionsC. Cytoplasmic effects and nutritionD. Environment changes and sexual dimorphism |
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Answer» Correct Answer - B Phenotype of an organism is the result of genotype and environment interactions. |
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| 404. |
Which is wrong about Mendel ?A. He was born in 1822.B. Mendel presented his work in the form of a paper at Heinzendorf in 1856.C. Mendel carried out his experiments for 7 years.D. Mendel died in 1884. |
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Answer» Correct Answer - B Mendel presented his work in the form of a paper at Heinzendorf in 1865. |
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| 405. |
Father of human genetics isA. CurvierB. BatesonC. MendelD. Garrod |
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Answer» Correct Answer - D Father of human genetics is Garrod. |
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| 406. |
Genes controlling seven traits in Pea studied by Mendel were actually located onA. Seven chromosomesB. Six chromosomesC. Four chromosomesD. Five chromosomes |
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Answer» Correct Answer - C Genes controlling seven traits in Pea stupied by Mendel were actually located on four chromosomes. |
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| 407. |
In pea, wrinkling of seeds is due to nonformation of starch because of the absence of:A. AmylaseB. InvertaseC. Branching enzymeD. Diastase |
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Answer» Correct Answer - C In Pea, wrinkling of seeds is due to non-formation of starch because of the absence of branching enzyme. |
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| 408. |
`50%` sperms are devoid of sex-chromosomes inA. MelandriumB. MothC. GrasshopperD. Bee |
| Answer» Correct Answer - C | |
| 409. |
Select the correct statementA. Franklin stahl coined the term "linkage"B. Punnett square was developed by a British scientistC. Spliceosmoes take part in translationD. Trnsduction was disscovered by a Allman |
| Answer» Correct Answer - B | |
| 410. |
Which of the following in wrongly matched.A. Starch synthesis in pea `" " : " "` Multiple allelesB. ABO blood grouping : Co-dominanceC. `XO type sex determinations : GrasshopperD. T.H. Morgen `" " : " "` Linkage |
| Answer» Correct Answer - A | |
| 411. |
Which one of the following pairs is correctly matchedA. morgan Discovered the process of linkageB. Linus Pauling isolated DNA for the first timeC. Francis Crick Discovered the phenomenon of transformationD. H.Khorana Discovered that a sequence of 3 nucleotides codes for a single amino acid |
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Answer» Correct Answer - A (a) It was Morgan (1910) who clearly proved and defined linkage on the basis of his breeding experiments in fruitfly Drosophila melanogaster. |
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| 412. |
Which of the following pairs is wrongly matched?A. Starch synthesis in pea : Multiple allelesB. ABO blood grouping : Co-domianceC. XO type sex determination : GrasshoperD. T.H. Morgon : Linkage |
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Answer» Correct Answer - A Starch synthesis in pea : incomplete dominance. |
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| 413. |
Select the correct match.A. Ribozyme-Nucleic acidB. `F_(2) xx `Recessive parent -Dihybrid crossC. T.H Morgan - TransductionD. G. Mendel-Transformation |
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Answer» Correct Answer - A `F_(2) xx "Recessive parent"-"Test cross"` T.H. Morgan - Sex chromosome G. Mendel-Father of genetics. |
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| 414. |
Which of the following cannot be detected in a developing foetus baby amniocentesisA. Down syndromeB. JaundiceC. Klinefelter syndromeD. Sex of the foetus |
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Answer» Correct Answer - B Jaundice cannot be detected in a developing foetus by amniocentesis. |
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| 415. |
A mother with blood group O has a foetus with blood group B.Will there be any problems in the mother or foetus ? If so, specify the problems |
| Answer» No, RBCs of mother with blood group O lack antigenes A and B. Therefore, the plasma factor a of foetus with blood group B will not cause any reaction | |
| 416. |
Identical twins are born when:A. One fertlized egg divided into 2 blastomeres and both separateB. One sperm fertlizes two eggsC. One eggs fertilized with two spermsD. Two eggs are fertilized |
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Answer» Correct Answer - A (a) Identical twins aare formed when one sperm fertilzes one egg to from a single zygote ,they have the same genotype and phenotype and are of same sex. |
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| 417. |
The process of genetic mutations isA. ReversibleB. IrreversibleC. Partially reversibleD. Continuous |
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Answer» Correct Answer - B (b) Murtaiion alter the configurtion and postion tion of nucleotides which is irreversible process except reverse or back mutaion , |
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| 418. |
When AABB and aabb are crossed, in `F_(2)` generations the ratio will beA. `1//16`B. `2//16`C. `8//16`D. `4//16` |
| Answer» Correct Answer - D | |
| 419. |
Among the following characters, which one was not considered by Mendel in his experiment on peaA. Stem- Tall of dwarfB. Trichomes -Gladular or non-glaulatC. Seed- Green or yellowD. Pod -Inflated or Constricted |
| Answer» Correct Answer - B | |
| 420. |
Some genomic representation of skin colour are given below (i) AA bb CC (ii) AA bb cc (iii) AA BB CC (iv) aa bb cc Which of the optin is correct for showing the darkness of colour of the skin in decreasing orderA. `itoivtoiitoiii`B. `iiitoiitoitoiv`C. `iiitoitoiitoiv`D. `itoiiitoiitoiv` |
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Answer» Correct Answer - C (c) The skin shade has to vary from very dark in AABBCC individual to very light in aabbcc individual. |
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| 421. |
How many types of zygotic combinations are possible between a cross Aa BB Cc Dd `xx` AA bb Cc DD ?A. 32B. 128C. 64D. 16 |
| Answer» Correct Answer - D | |
| 422. |
If the cells of an organism heterozygous for two pairs of character via. Aa and Bb undergo meiosis, what will be the genotypes of the gamets producedA. Aa and BbB. AB, aB, Ab and abC. aB and AbD. Ab and ab |
| Answer» Correct Answer - B | |
| 423. |
Change in the number of body parts is calledA. Continous variationB. Discontinous variationC. Meristic variationD. Substantive variation |
| Answer» Correct Answer - C | |
| 424. |
Which is incorrect about thalassemia?A. This blood disease is transmitted from parents to the offspring when both the partners are unaffected carrier for the gene (or heterozygous).B. The defect is due to either mutation or deletion which utimately results in reduced rate of synthesis of one of the globin chains that make up haemoglobin.C. This causes the formation of abnormal haemoglobin molecule resulting into anaemia which is characteristic of the disease.D. Thalassemia differs from sickle cell anaemia in that the former is qualitative problem of synthesising an incorrectly functioning globin while the latter is a quantitative problem of synthesising too few globin molecules. |
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Answer» Correct Answer - D Sickle cell anaemia differs from thalassemia in that the former is a qualitative problem of synthesising an incorrectly functioning globin while the latter is a quantitative problem of synthesising too few globin molecules. |
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| 425. |
Genteic variation in a population arises due toA. Mutations onlyB. Recombination onlyC. Mutations as well as recombinationD. Reproductive isolation and selection |
| Answer» Correct Answer - C | |
| 426. |
Genteic variation in a population arises due toA. Recombination onlyB. Mutaions as well as recombinationC. Reproductive isolation and selectionD. Mutations only |
| Answer» Correct Answer - B | |
| 427. |
Autosomal mutant allele HbS causesA. ThalassemiaB. AlbinismC. Sickle cell anaemiaD. Agammaglobuliema |
| Answer» Correct Answer - C | |
| 428. |
A : Sickel cell anemia occurs due to the point mutation. R : mRNA produced from `Hb(s)` gene has GAG instead of GUG.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 - C | |
| 429. |
In sickle cell anaemia, the sequence of amino acid from first to seventh position of `beta`-chain of haemoglobin S (HbS) isA. His, Leu, Thr, Pro, Glu, Val, ValB. Val, His, Leu, Thr, Pro, Glu, GluC. Glu, His, Leu, Pro, Val, Glu, GluD. Val, His, Leu, Thr, Pro, Val, Glu |
| Answer» Correct Answer - D | |
| 430. |
Sickle cell anaemia is controlled byA. Single alleleB. Single pair of alleleC. Multiple alleleD. Polygenes |
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Answer» Correct Answer - B Sickle cell anaemia is controlled by single pair of allele `(Hb^(A) " and " Hb^(S))`. |
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| 431. |
State True or False. (1) Garden pea has seven characters only. (2) Flowers of Pisum sativum naturally show cross pollination. (3) A true breeding line shows the stable trait inheritance. (4) Mendel applied statisfical methods and mathematical logic for analysing the results. |
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Answer» (1) False (2) False (3) True (4) True |
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| 432. |
Select the incorrect statement regarding pedigree analysisA. Solid symbol shows the unaffected individualB. It is useful for genetic counsellorsC. Proband is the person from which case history startsD. It is an analysis of traits in a several generations of a family |
| Answer» Correct Answer - A | |
| 433. |
Which one of the following symbols and its representation, used in human pedigree analysis is correctA. B. C. D. |
| Answer» Correct Answer - B | |
| 434. |
Read the following statement s and choose the correct option In p[henylketonuria the affected person does not secrete the enzyme to con vert pheylalnine to typrosine (II)Possibility of male becoming haemophiliac is extremely rare (III) Sickle cell anaemia is caused by the substitution of glutamic acid by valine at fifth position of beta chain of haemoglobin (IV) Myotonic dystrophy is an autosomla dominant traitA. I and II alone are wrongB. II and III alone are wrongC. II alone are wrongD. III alone is wrong |
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Answer» Correct Answer - B Possibility of female becoming haemophilic is extremely rate. Sickle cell anaemia is caused by the substitution of glutamic acid by valine at sixth position of beta chain of haemoglobin. |
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| 435. |
If a tall plant is crossed with a dwarf plant and obtained progeny is half tall and half dwarf plants. Then the genotype of progony will beA. `TTxxtt`B. `Ttxxtt`C. `TTxxTt`D. `TtxxTt` |
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Answer» Correct Answer - B (b) It is the test cross. |
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| 436. |
If a agouti mice (CcAa) is crossed with albino mice (ccAA), then how many albino mice are produced resulting progeny ?A. 4B. 9C. 2D. 3 |
| Answer» Correct Answer - C | |
| 437. |
transition type of mutation is caused when :A. GC is replaced by TAB. CG is repleaced by GCC. AT is replaced by CGD. At is replaced by GC |
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Answer» Correct Answer - D (d) in transition anitrogen base is replaced by another of its type i.e one puiine is replecred by another purine` (AharrG)` while one pyrimidine by another pyimidine `(Charr TorU)` |
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| 438. |
Euplodiy is best explanined byA. Exact muliple ofa haploid set of chromosomes B. One chromosome less than the haploid set of chromosomesC. One chromosome more than the haploid set of chromosomesD. One chromosome more than the diploid set of chromosomes |
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Answer» Correct Answer - A (a) the term euploidy `(eu="good"+"ploud"="multiple )` is applied to organisms with chromosame numberser number that are multiplles of same basic number `e.g,x,2x,3x,4x5x,6x,` etc. |
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| 439. |
The best method to improve the genetic quality of mankind isA. Marriage restrictionsB. SterilizationsC. Control of immigrationsD. Sexual separation of defective |
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Answer» Correct Answer - D (d) The defective or undesirable persons (haemophila colourblindness etc. and epilspsis , feeble idiocy etc.) should not be pemitted to reproduce so that the uneanted gens aarae gradully eliminted from the gane pool of human population. |
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| 440. |
Genes which code for a pair of contrasting traits are known asA. CistronB. AlleleC. ExonD. Intron |
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Answer» Correct Answer - B Genes which code for a pair of contrasting traits are known as alleles or allelomorphs, i.e. they are slightly different forms of the same gene. |
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| 441. |
In Mendelian dihybrid cross how many are recombinants?A. `37.2%`B. `62.8%`C. `37.5%`D. `62.5%` |
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Answer» Correct Answer - C In dihybrid cross, among the 16 progenies, 6 are recombinants then percentage is `= (6)/(16) xx 100 = 37.5` |
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| 442. |
Alleles areA. Similar forms of different geneB. Slightly different forms of the different geneC. Similar forms of the same geneD. Slightly different forms of the same gene |
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Answer» Correct Answer - D Alleles are slightly different forms of the same gene. |
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| 443. |
The proportion of 3:1 at the `F_(2)` generation is explained by theA. Law of DominanceB. Law of SegregationC. Law of Independent assortmentD. Test cross |
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Answer» Correct Answer - A The law of dominance is used to explain the expression of only one of the parental characters in a monohybrid cross in the `F_(1)` and the expression of both in the `F_(2)`. It also explains the proportion of 3:1 obtained at the `F_(2)`. |
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| 444. |
In Punnett square, the possible gametes are written on two sides, usually theA. Top row and left columnsB. Top row and right columnsC. Bottom row and right columnsD. Bottom row and left columns |
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Answer» Correct Answer - A In Punnett square, the possible gametes are written on two sides, usually the top row and left columns. |
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| 445. |
A pea plant parent having violet coloured flowers with unknown genotype was a plant having white coloured flowers in the progeny 50% of the flowers were violet and 50% were white. The genotype constitution of the parent having violet coloured flower was:A. HomozygousB. MerzygousC. HeterozygousD. Hemizygous |
| Answer» Correct Answer - C | |
| 446. |
"Both the characters in a monohybrid cross are recovered as such in the `F_(2)` generation though one of these in not seen at the `F_(1)` stage". This interpretation is based on theA. First Law of MendelB. Second Law of MendelC. Second set of generalisationsD. Incomplete dominance |
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Answer» Correct Answer - b "Both the characters in a monohybrid cross are recovered as such in the `F_(2)` generation though one of these in not seen at the `F_(1)` stage". This interpretation is based on the Second Law of Mendel/Law of Segregation. |
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| 447. |
Punnett square was developed byA. British Zoologist, Reginald C. PunnettB. German Botanist, Reginald C. PunnettC. Stanford Geneticist Reginald C. PunnettD. British Geneticist Reginald C. Punnett |
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Answer» Correct Answer - D The production of gametes by the parents, the formation of the zygotes, the `F_(1) " and " F_(2)` plants can be understood from a diagram called Punnett Square. It was developed by a British geneticist, Reginald C. Punnett. |
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| 448. |
Select one word for the statement. a. If `F_(1)` resembled both the parents b. If `F_(1)` did not resemble either of the two parents and was in between the two c. If `F_(1)` resembled either of the two parentsA. c-dominance, b-co dominance, a - incomplete dominanceB. a-dominance, c-co dominance, c-incomplete dominanceC. b-dominance, a-co dominance, c-incomplete dominanceD. c-dominance, a-co dominance, b-incomplete dominance |
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Answer» Correct Answer - D If `F_(1)` resembled both the parents -Co-dominance. If `F_(2)` did not resemble either of the two parents and was in between the two - Incomplete dominance. If `F_(1)` resembled either of the two parents - Dominance. |
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| 449. |
A person may have one gene for normal heamoglobin and one gene for sickle cell haemoglobin. This heterozygous condition is calledA. GenomeB. AnaemiaC. Gene traitD. Sickle cell trait |
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Answer» Correct Answer - D Sickle cell trait is a heteozygone condition in which an organism may have one gene for normal haemoglobin and the other gene for sicke cell haemoglobine. |
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| 450. |
If a diploid cell is treated with colchicine, then it becomesA. TetraploidB. DiploidC. TriploidD. Monoploid |
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Answer» Correct Answer - A Colchicine inhibts the cell divison or mitosis, but dulpication of chromosomes is continue, as a result diploid becomes tetraploid. |
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