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This set of Biology Multiple Choice Questions & Answers (MCQs) focuses on Heredity and Variation Set 4

Q1 | .Which of the following is not a hereditary disease? AIPMT - 2004
  • Cystic Fibrosis
  • Thalassaemia
  • Haemophilia
  • Cretinism
  • cessive
Q2 | .Haemophilia is more commonly seen in human males than in human females because
  • a greater proportion of girls die in infancy AIPMT - 2004
  • this disease is due to a Y- linked recessive mutation
  • this disease is due to an X - linked recessive mutation
  • this disease is due to an X- linked dominant mutation
  • cessive
Q3 | .A woman with 47 chromosomes due to 3 copies of chromosome 21 is characterized by AIPMT - 2005
  • Super Femaleness
  • Triploidy
  • Turner's Syndrome
  • Down's Syndrome
  • cessive
Q4 | . 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
  • Sex - linked dominant
  • Sex - linked recessive
  • Sex - limited recessive
  • Autosomal dominant
  • cessive
Q5 | .Which one of the following is an example of polygenic inheritance ? AIPMT - 2006
  • Skin colour in humans
  • Flower colour in Miralibilis jalapa
  • Production of male honey bee
  • Pod shape in garden pea
  • cessive
Q6 | .Phenotype of an organism is the result of AIPMT - 2006
  • genotype and environmental interactions
  • mutations and linkages
  • Cytoplasmic effects and nutrition
  • environmental changes and sexual dimorphism
  • cessive
Q7 | .How many different gametes will be produced by a plant having the genotype AABbCC? AIPMT - 2006
  • Two
  • Three
  • Four
  • Nine
  • cessive
Q8 | .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?
  • Round Seeds with yellow cotyledons, and wrinkled seeds with yellow cotyle dons
  • Only round seeds with green cotyledons
  • Only wrinkled seeds with yellow cotyledons
  • Only wrinkled seeds with green cotyledons
  • cessive
Q9 | Test cross involves AIPMT - 2006
  • Crossing between two genotypes with dominant trait
  • Crossing between two genotypes with recessive trait
  • Crossing between two F1 hybrids
  • Crossing the F1 hybrid with a double recessive genotype
  • cessive
Q10 | . If a colourblind woman marries a normal visioned man, their sons will be AIPMT - 2006
  • all colourblind
  • all normal visioned
  • one - half colourblind and one - half normal
  • three - fourths colourblind and one - fourth normal
  • cessive
Q11 | . In the hexaploid wheat, the haploid (n) and basic (X) numbers of chromosomes are AIPMT - 2007
  • n = 21 and X = 21
  • n = 21 and X =14
  • n = 21 and X = 7
  • n = 7 and X =21
  • cessive
Q12 | . Inheritance of skin colour in humans is an example of AIPMT - 2007
  • Point Mutation
  • Polygenic inheritance
  • Codominance
  • Chromosomal aberrations
  • cessive
Q13 | . In pea plants, yellow seeds are dominant to green, If a heterozygous yellow seeded plant isCorssed with a green seeded plants, what ratio of yellow and green seeded plants, would you expect in F1 generation? AIPMT - 2007
  • 9:01
  • 1:03
  • 3:01
  • 50:50:00
  • cessive
Q14 | .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
  • AaBB
  • AABb
  • AABB
  • AaBb
  • cessive
Q15 | .Which one of the following conditions in human is correctly matched with its chromosomal abnormality / linkage ? AIPMT - 2008
  • Erythro blastosis foetalis - X - linked
  • Down's syndrome - 44 autosomes +XXY
  • Kline Felter's syndrome - 44 autosomes +XXY
  • Colour blindness - Y - linked.
  • cessive
Q16 | .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
  • Overdominance of this type on the genes for A and B types
  • One antibody only - either anti - A or anti - B on the RBCs
  • no antigens A and B on RBCs
  • other antigens besides A and B on RBCs
  • cessive
Q17 | . Sickle cell anemia is AIPMT - 2009
  • Caused by substitution of Valine by glutamic acid in the beta globin chain of haemo globin.
  • Caused by a change in a single base pair of DNA
  • Characterized by elongated sickle like RBCS with a nucleus
  • An autosomal dominant trait.
  • cessive
Q18 | .Which one of the following can not be explained on the basis of Mendel's law of dominance? AIPMT - 2009
  • The discrete unit controlling a particular character is called a factor
  • Out of one pair of factors one is dominant, and the other recessive.
  • Alleles do not show any blending and both the characters recover as such in F2 generation.
  • Factors occur in pairs
  • cessive
Q19 | . The genotype of a plant showing the dominatnt phenotype can be determined by AIPMT - 2010
  • test cross
  • dihybrid cross
  • pedigree analysis
  • Back Cross
  • cessive
Q20 | . Select the correct statement from the ones given below with respect to dihybrid cross.
  • Tightly linked genes on the same chromosomes show higher recombinations
  • Genes far apart on the same chromosome show very few recombinations
  • Genes loosely linked on the same chrososome show similar recombinations
  • Tightly linked genes on the samechromosome show very few recombinations
  • cessive
Q21 | .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
  • Three
  • One
  • Four
  • Two
  • cessive
Q22 | Which one of the following symbols and its representation used in human pedigree analysis is correct? AIPMT - 2010
  • #NAME?
  • #NAME?
  • #NAME?
  • #NAME?
  • cessive
Q23 | Study the pedigree chart of a certain family given below and select the correct conclusion which can be drawn for the character
  • The female parent is heterozygous
  • The parent could not have had a normal daughter for this character
  • The trait under study could not be colourblindness
  • The male parent is homozygous dominant
  • cessive
Q24 | Which one of the following conditions correctly describes the manner of determining the sex? AIPMT - 2011
  • Homozygous sex chromosomes (ZZ) determine female sex in birds
  • XO type of sex chromosomes determine male sex in grasshopper
  • XO condition in humans as found in Turner's syndrome, determines female sex
  • Homozygous sex chromosomes (XX) produce male in Drosophila
  • cessive
Q25 | Mutations can be induced with AIPMT - 2011
  • Infra red radiations
  • IAA
  • Ethylene
  • gamma radiations
  • cessive