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MCQOPTIONS
This section includes 5751 Mcqs, each offering curated multiple-choice questions to sharpen your Biology knowledge and support exam preparation. Choose a topic below to get started.
| 1451. |
Down's Syndrome is caused by an extra copy of chromosome number 21. What percent age of off spring produced by an affected mother and a normal father? AIPMT - 2003 |
| A. | 100% |
| B. | 75% |
| C. | 50% |
| D. | 25% |
| E. | cessive |
| Answer» E. cessive | |
| 1452. |
. A male human is heterozygous for autosomal genes A and B and is also hemizygous for haemophilic gene h. What proportion of his sperms will be abh? AIPMT - 2004 |
| A. | 1/8 |
| B. | 1/32 |
| C. | 1/16 |
| D. | 1/4 |
| E. | cessive |
| Answer» B. 1/32 | |
| 1453. |
.The most popularly known blood grouping is the ABO grouping. It is named ABO and not ABC, because "O" in it refers to having AIPMT - 2009 |
| A. | Overdominance of this type on the genes for A and B types |
| B. | One antibody only - either anti - A or anti - B on the RBCs |
| C. | no antigens A and B on RBCs |
| D. | other antigens besides A and B on RBCs |
| E. | cessive |
| Answer» D. other antigens besides A and B on RBCs | |
| 1454. |
What kind of chromosomal aberrations is seen in the given diagram? |
| A. | Deletion |
| B. | Duplication |
| C. | inversion |
| D. | Translocation |
| E. | cessive |
| Answer» D. Translocation | |
| 1455. |
.Which one of the following conditions in human is correctly matched with its chromosomal abnormality / linkage ? AIPMT - 2008 |
| A. | Erythro blastosis foetalis - X - linked |
| B. | Down's syndrome - 44 autosomes +XXY |
| C. | Kline Felter's syndrome - 44 autosomes +XXY |
| D. | Colour blindness - Y - linked. |
| E. | cessive |
| Answer» D. Colour blindness - Y - linked. | |
| 1456. |
.Which one of the following is an example of polygenic inheritance ? AIPMT - 2006 |
| A. | Skin colour in humans |
| B. | Flower colour in Miralibilis jalapa |
| C. | Production of male honey bee |
| D. | Pod shape in garden pea |
| E. | cessive |
| Answer» B. Flower colour in Miralibilis jalapa | |
| 1457. |
Which one of the following symbols and its representation used in human pedigree analysis is correct? AIPMT - 2010 |
| A. | = Mating between relatives |
| B. | = unaffected male |
| C. | = unaffected female |
| D. | = affected male |
| E. | cessive |
| Answer» B. = unaffected male | |
| 1458. |
. The genotype of a plant showing the dominatnt phenotype can be determined by AIPMT - 2010 |
| A. | test cross |
| B. | dihybrid cross |
| C. | pedigree analysis |
| D. | Back Cross |
| E. | cessive |
| Answer» B. dihybrid cross | |
| 1459. |
.Phenotype of an organism is the result of AIPMT - 2006 |
| A. | genotype and environmental interactions |
| B. | mutations and linkages |
| C. | Cytoplasmic effects and nutrition |
| D. | environmental changes and sexual dimorphism |
| E. | cessive |
| Answer» B. mutations and linkages | |
| 1460. |
.How many different gametes will be produced by a plant having the genotype AABbCC? AIPMT - 2006 |
| A. | Two |
| B. | Three |
| C. | Four |
| D. | Nine |
| E. | cessive |
| Answer» B. Three | |
| 1461. |
. A man and a woman, who do not show any apparent signs of a certain inherited disease, have Seven Children (2 daughters and 5 sons). Three of the Sons suffer from the given disease but none of the daughters affected. Which of the following mode of inheritance do you suggest for this disease? AIPMT - 2005 |
| A. | Sex - linked dominant |
| B. | Sex - linked recessive |
| C. | Sex - limited recessive |
| D. | Autosomal dominant |
| E. | cessive |
| Answer» C. Sex - limited recessive | |
| 1462. |
.A human male produces sperms with the genotypes AB , Ab , aB , ab pertaining to two diallelic characters in equal proportions. What is the corresponding genotype of this person? AIPMT - 2007 |
| A. | AaBB |
| B. | AABb |
| C. | AABB |
| D. | AaBb |
| E. | cessive |
| Answer» E. cessive | |
| 1463. |
. In the hexaploid wheat, the haploid (n) and basic (X) numbers of chromosomes are AIPMT - 2007 |
| A. | n = 21 and X = 21 |
| B. | n = 21 and X =14 |
| C. | n = 21 and X = 7 |
| D. | n = 7 and X =21 |
| E. | cessive |
| Answer» D. n = 7 and X =21 | |
| 1464. |
Test cross involves AIPMT - 2006 |
| A. | Crossing between two genotypes with dominant trait |
| B. | Crossing between two genotypes with recessive trait |
| C. | Crossing between two F1 hybrids |
| D. | Crossing the F1 hybrid with a double recessive genotype |
| E. | cessive |
| Answer» E. cessive | |
| 1465. |
Study the pedigree chart of a certain family given below and select the correct conclusion which can be drawn for the character |
| A. | The female parent is heterozygous |
| B. | The parent could not have had a normal daughter for this character |
| C. | The trait under study could not be colourblindness |
| D. | The male parent is homozygous dominant |
| E. | cessive |
| Answer» B. The parent could not have had a normal daughter for this character | |
| 1466. |
. Inheritance of skin colour in humans is an example of AIPMT - 2007 |
| A. | Point Mutation |
| B. | Polygenic inheritance |
| C. | Codominance |
| D. | Chromosomal aberrations |
| E. | cessive |
| Answer» C. Codominance | |
| 1467. |
.A woman with 47 chromosomes due to 3 copies of chromosome 21 is characterized by AIPMT - 2005 |
| A. | Super Femaleness |
| B. | Triploidy |
| C. | Turner's Syndrome |
| D. | Down's Syndrome |
| E. | cessive |
| Answer» E. cessive | |
| 1468. |
.Which of the following is not a hereditary disease? AIPMT - 2004 |
| A. | Cystic Fibrosis |
| B. | Thalassaemia |
| C. | Haemophilia |
| D. | Cretinism |
| E. | cessive |
| Answer» E. cessive | |
| 1469. |
.Haemophilia is more commonly seen in human males than in human females because |
| A. | a greater proportion of girls die in infancy AIPMT - 2004 |
| B. | this disease is due to a Y- linked recessive mutation |
| C. | this disease is due to an X - linked recessive mutation |
| D. | this disease is due to an X- linked dominant mutation |
| E. | cessive |
| Answer» D. this disease is due to an X- linked dominant mutation | |
| 1470. |
.ABO blood groups in humans are controlled by the gene I. It has three alleles - IA , IB and i . Since there are 3 different alleles six different genotypes are possible. How many phenotypesOccur? AIPMT - 2010 |
| A. | Three |
| B. | One |
| C. | Four |
| D. | Two |
| E. | cessive |
| Answer» D. Two | |
| 1471. |
. Inorder to find out the different types of gametes produced by a pea plant having the genotype AaBb it should be crossed to a plant with the genotype AIPMT - 2004 |
| A. | AABB |
| B. | AaBb |
| C. | aabb |
| D. | aaBB |
| E. | cessive |
| Answer» D. aaBB | |
| 1472. |
Give the name in following |
| A. | P-terminal bud , q-old flower r-floral bud, s-leaf |
| B. | P- terminal bud, q- floral bud, r- old flower, s- leaf |
| C. | P- old flower, q- terminal bud r- leaf s-floral bud |
| D. | P- leaf, q- floral bud, r- old flower, s- terminal bud |
| Answer» B. P- terminal bud, q- floral bud, r- old flower, s- leaf | |
| 1473. |
. Select the correct statement from the ones given below with respect to dihybrid cross. |
| A. | Tightly linked genes on the same chromosomes show higher recombinations |
| B. | Genes far apart on the same chromosome show very few recombinations |
| C. | Genes loosely linked on the same chrososome show similar recombinations |
| D. | Tightly linked genes on the samechromosome show very few recombinations |
| E. | cessive |
| Answer» E. cessive | |
| 1474. |
. A woman with normal vision, but whose father was colourblind, marries a colourblind man. AIPMT - 2004 Suppose that the fourth child of this couple was a boy. This boy |
| A. | May be colourblind or may be of normal vision |
| B. | Must be colourblind |
| C. | Must have normal colour vision |
| D. | Will be partially colourblind since he is heterozygous for the colourblind mutant allele. |
| E. | cessive |
| Answer» B. Must be colourblind | |
| 1475. |
.Which one of the following can not be explained on the basis of Mendel's law of dominance? AIPMT - 2009 |
| A. | The discrete unit controlling a particular character is called a factor |
| B. | Out of one pair of factors one is dominant, and the other recessive. |
| C. | Alleles do not show any blending and both the characters recover as such in F2 generation. |
| D. | Factors occur in pairs |
| E. | cessive |
| Answer» D. Factors occur in pairs | |
| 1476. |
. Sickle cell anemia is AIPMT - 2009 |
| A. | Caused by substitution of Valine by glutamic acid in the beta globin chain of haemo globin. |
| B. | Caused by a change in a single base pair of DNA |
| C. | Characterized by elongated sickle like RBCS with a nucleus |
| D. | An autosomal dominant trait. |
| E. | cessive |
| Answer» C. Characterized by elongated sickle like RBCS with a nucleus | |
| 1477. |
. If a colourblind woman marries a normal visioned man, their sons will be AIPMT - 2006 |
| A. | all colourblind |
| B. | all normal visioned |
| C. | one - half colourblind and one - half normal |
| D. | three - fourths colourblind and one - fourth normal |
| E. | cessive |
| Answer» B. all normal visioned | |
| 1478. |
.In Mendel's experiments with garden pea, round seed shape (RR) was dominant over wrinkledSeeds (rr), Yellow cotyledon (YY) was dominant over green cotyledon (yy). What are the expected Phenotypes in the F2 generation of the cross RRYY x rryy? |
| A. | Round Seeds with yellow cotyledons, and wrinkled seeds with yellow cotyle dons |
| B. | Only round seeds with green cotyledons |
| C. | Only wrinkled seeds with yellow cotyledons |
| D. | Only wrinkled seeds with green cotyledons |
| E. | cessive |
| Answer» B. Only round seeds with green cotyledons | |
| 1479. |
Labeling the given figure : |
| A. | P- stigma q- style |
| B. | P- anther q- filament |
| C. | P anther q- style |
| D. | P- stigma q- filament |
| Answer» C. P anther q- style | |
| 1480. |
Name the labeled ‘x’ in plant |
| A. | Thorn |
| B. | Hook |
| C. | Prickles |
| D. | Stipules |
| Answer» D. Stipules | |
| 1481. |
Identify this plant modification and Select the correct option |
| A. | Sweet potato – simple tuberous root |
| B. | Dahlia – fasciculated tuberous root |
| C. | Asparagus - simple tuberous root |
| D. | Beet – tap root |
| Answer» C. Asparagus - simple tuberous root | |
| 1482. |
Name of the following aestivation type: |
| A. | Valvate |
| B. | Twisted |
| C. | Imbricate |
| D. | Quincuncial |
| Answer» E. | |
| 1483. |
Match the following with correct combination. |
| A. | : (P)- II, (Q)- I, (R)-IV ,(S)-III |
| B. | : (P)-III, (Q)-IV, (R)–II,(S)- I |
| C. | : (P)- IV ,(Q) – I,(R)-II ,(S)-III |
| D. | : (P)-IV ,(Q)– I, (R)-III,(S)- II |
| Answer» D. : (P)-IV ,(Q)– I, (R)-III,(S)- II | |
| 1484. |
Choose correct option by giving diagram: |
| A. | C- vexillary, D- Quincuncial , E- Imbricate |
| B. | C- vexillary , D- Imbricate , E- Quincuncial, |
| C. | C-Imbricate , D-Quincuncial, E-vexillary |
| D. | C- Imbricate , D- vexillary , E-Quincuncial |
| Answer» D. C- Imbricate , D- vexillary , E-Quincuncial | |
| 1485. |
Match list I with II types of leaves |
| A. | III I IV II |
| B. | I III II IV |
| C. | III IV I II |
| D. | III II I IV |
| Answer» B. I III II IV | |
| 1486. |
Name in given floral diagram: |
| A. | P-Calyx, q-Corolla, r-Androecium, s- Gynoecium, t- Mother axis |
| B. | P-Calyx, q-Androecium, r- Gynoecium, s-Corolla, t-Mother axis |
| C. | P- Corolla, q- Calyx, r-Androecium, s- Gynoecium, t- Mother axis |
| D. | P- Corolla , q- Calyx ,r- Gynoecium , s Androecium -t- mother axis |
| Answer» C. P- Corolla, q- Calyx, r-Androecium, s- Gynoecium, t- Mother axis | |
| 1487. |
Name the labeling part of given diagram: |
| A. | P – Endosperm q- embryo |
| B. | P-seed coat q- coleoptile |
| C. | P- Endosperm q- cotyledon |
| D. | P- seed coat q –embryo |
| Answer» B. P-seed coat q- coleoptile | |
| 1488. |
Identify the inflorescence |
| A. | Raceme |
| B. | Spike |
| C. | Helicoid |
| D. | Scorpioid |
| Answer» D. Scorpioid | |
| 1489. |
Choose the correct option by given placentation q P r |
| A. | P-free central-Dianthus, q- parietal-Tomato, r-Marginal -Bean |
| B. | P- parietal-Tomato, q- Marginal -sunflower ,r- free central- Bean |
| C. | P-parietal-Argemone, q- free central- Bean, r- Marginal -sunflower |
| D. | P-free central-Dianthus, q- parietal-Argemone, r-Marginal –Bean |
| Answer» E. | |
| 1490. |
Which plant is this and live in _______ habitat. |
| A. | Opuntia , ever green |
| B. | Muehlenbevkia, dry |
| C. | Dioscorea , thorn forest |
| D. | Agave , desert |
| Answer» C. Dioscorea , thorn forest | |
| 1491. |
Choose correct option according to given leaf: |
| A. | Moringa – multipinnate compound leaf |
| B. | Balanites- Bifoliate compound leaf |
| C. | Caesalpinia - bipinnate compound leaf |
| D. | Aegle – multifoliate PX |
| Answer» D. Aegle – multifoliate PX | |
| 1492. |
Name the labeled flower part. |
| A. | P-peduncle, q-ovary r-stigma, s-calyx, t-thalamus |
| B. | P-corolla, q-anther, r-stigma, s-calyx, t- peduncle |
| C. | P-petals, q-style, r-stigma, s-stamen, t- ovary |
| D. | P- corolla, q- anther , r-style, s-calyx t- thalamus |
| Answer» E. | |
| 1493. |
Select the correct pair |
| A. | (a)- (r)- IV, |
| B. | -(s)-III |
| C. | - (p)-II, |
| D. | - (q) -I |
| Answer» C. - (p)-II, | |
| 1494. |
Name the following part of seed: |
| A. | . p-seed, q-endocarp, r-mesocarp, s-exocarp |
| B. | . p-endocarp, q- seed, r- exocarp , s-mesocarp |
| C. | . p- seed, q-endocarp, r-mesocarp, s- exocarp |
| D. | . p-endocarp, q- seed, r- exocarp , s-mesocartp |
| Answer» D. . p-endocarp, q- seed, r- exocarp , s-mesocartp | |
| 1495. |
Choose the correct option by given diagram : |
| A. | Scorpioid - Heliotropium |
| B. | Scorpioid - Hamelia |
| C. | Spike - Achyranthus |
| D. | Spike – musa |
| Answer» B. Scorpioid - Hamelia | |
| 1496. |
Labeling ‘p’ in root section |
| A. | Velamen tissue |
| B. | Meristemaic tissus |
| C. | Growth tissue |
| D. | Fleshy tissue |
| Answer» B. Meristemaic tissus | |
| 1497. |
Labeling the following diagram: |
| A. | . p-leaf q. –stem .r. - fruit s- flower |
| B. | . p- flower q- stem r- leaf s- fruit |
| C. | . p- leaf q-stem r- flower, s- fruit |
| D. | . p- flower q- leaf r- stem s- fruit |
| Answer» D. . p- flower q- leaf r- stem s- fruit | |
| 1498. |
____________ is wraped by cerebrum ? |
| A. | Thalamu |
| B. | Hypothalamus |
| C. | Cerebellar hemisphere |
| D. | Mid- brain |
| Answer» B. Hypothalamus | |
| 1499. |
In a dihybrid cross between AABB and aabb the ratio of AABB, AABb, aaBb, aabb in F2 generation is |
| A. | 9 : 3 : 3: 1 |
| B. | 1: 1: 1: 1 |
| C. | 1: 2: 2: 1 |
| D. | 1 : 1 : 2 : 2 |
| E. | cessive |
| Answer» D. 1 : 1 : 2 : 2 | |
| 1500. |
What is correct in reference with nerve impulse ? |
| A. | Self-induced and unidirectional |
| B. | Selt-induced and bidirectional |
| C. | Electric potential in the nerve by region increase |
| D. | Ion channel get closed in this region. |
| Answer» B. Selt-induced and bidirectional | |