Previous Year Questions (2016-19) - Heredity and Variation Notes | EduRev

Biology 28 Years NEET/AIPMT Question Papers of Class 12

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The document Previous Year Questions (2016-19) - Heredity and Variation Notes | EduRev is a part of the NEET Course Biology 28 Years NEET/AIPMT Question Papers of Class 12.
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Q.1. In Antirrhinum (Snapdragon), a red flower was crossed with a white flower and in F1 generation all pink flowers were obtained. When pink flowers were soiled, the F2 generation showed white, red and pink flowers. Choose the incorrect statements from the following.    (2019)
(a) Law of segregation does not apply in this experiment.
(b) This experiment docs not follow the Principle of Dominance.
(c) Pink colour in is due to incomplete dominance.
(d) Ratio of FisPrevious Year Questions (2016-19) - Heredity and Variation Notes | EduRev
Ans:
(a)
Law of segregation applies in this case as when pink (lowers obtained in F1, are selfed then red and while (lowers are obtained in F2 which indicates that there is no mixing of gametes.

Q.2. Select the incorrect statement.    (2019)
(a) Human males have one of their sex- chromosome much shorter than other.
(b) Male fruit fly is heterogametic.
(c) In male grasshoppers, 50% of sperms have no sex-chromosome.
(d) In domesticated fowls sex of progeny depends on the type of sperm rather than egg.

Ans: (d)
In birds, e.g., fowls, males have homomorphic sex chromosomes (AA + ZZ.) and females have homoromorphic sex chromosomes (AA + ZW). So, males produce only one type of sperms containing (A + Z) whereas females produce two types of eggs (A + Z and A + W). Therefore, the sex ofthe progeny depends on the type ofthe egg which is fertilised by the sperm.

Q.3. What map unit (Centimorgan) is adopted in the construction of genetic maps?    (2019)
(a) A unit of distance between genes on chromosomes, representing 50% cross over.
(b) A unit of distance between two expressed genes, representing 10% cross over.
(c) A unit of distance between two expressed genes, representing 100% cross over.
(d) A unit of distance between genes on chromosomes, representing 1% cross over.

Ans: (d)
Genetic map is a linear graphic representation of the sequence and relative distance of various genes present in a chromosome. 1% crossing over between two linked genes is known as 1 map unit or centiMorgan (cM).

Q.4. The frequency of recombination between gene present on the same chromosome as a measure of the distance between genes was explained by    (2019)
(a) Sutton Boveri
(b) T.H. Morgan
(c) Gregor J.Mendel
(d) Alfred Sturtevant

Ans: (d)
T.H. Morgan coined the term linkage to describe the physical association of genes on chromosome and term recombination to describe the generation of noil-parental gene combinations. Alfred Sturtevant used the frequency of recombination between gene pairs on the same chromosome as a measure of the distance between genes.

Q.5. What is the genetic disorder in which an individual has an overall masculine development, gynaecomastia, and is sterile?    (2019)
(a) Down’s syndrome
(b) Turner’s syndrome
(c) Klinefelter’s syndrome
(d) Edward syndrome

Ans: (c)
Klinefelter’s syndrome occurs by the union of an abnormal XX egg and a normal Y sperm or a normal X egg and abnormal XY sperm. The individual has 47 chromosomes (44 + XXY). Such persons are sterile males with undeveloped testes, mental retardation, sparse body hair, long limbs and with some female characteristics such as enlarged breast, i.e., gynaecomastia.

Q.6. Which of the following pairs is wrongly matched?    (2018)
(a) Starch synthesis in pea : Multiple alleles
(b) ABO blood grouping : Co-dominance
(c) XO type sex determination: Grasshopper
(d) T.H. Morgan : Linkage

Ans: (a)
The gene for starch synthesis in pea seeds can produce more than one effect which implies it is a pleiotropic gene.

Q.7. Which of the following characteristics represent ‘inheritance of blood groups’ in humans?    (2018)
(i) Dominance
(ii) Co-dominance
(iii) Multiple allele
(iv) Incomplete dominance
(v) Polygenic inheritance
(a) (ii), (iii) and (v)
(b) (i), (ii) and (iii)
(c) (ii), (iv) and (v)
(d) (i), (iii) and (v)

Ans: (b)
ABO blood group system in human beings is an example of dominance, co-dominance and multiple allelism.
IIO IB IO or i - Alleles shows dominant-recessive relationship.
lA IB - Codominance is a phenomenon in which alleles of a gene do not show dominant recessive relationship and express themselves independently when present together.
IIO IB IO or i - More than two alternate forms of a gene present on the same locus are called as multiple alleles and the mode of inheritance in the alleles is called multiple allelism.

Q.8. A woman has an X-linked condition on one of her X chromosomes. This chromosome can be inherited by    (2018)
(a) Only daughters
(b) Only sons
(c) Only grandchildren
(d) Both sons and daughters.

Ans: (d)
Woman acts as a carrier when she has the X-linked condition on one of her X-chromosomes. Both son and daughter inherit X-chromosome from mother. Hence, one of the two daughters will be carrier and one of the two sons will be diseased. It can be explained by the given cross :
Previous Year Questions (2016-19) - Heredity and Variation Notes | EduRev
where XC is the X-chromosome carrying the gene for the condition.

Q.9. Thalassemia and sickle cell anaemia are caused due to a problem in globin molecule synthesis. Select the correct statement.    (2017)
(a) Both are due to a quantitative defect in globin chain synthesis.
(b) Thalassemia is due to less synthesis of globin molecules.
(c) Sickle cell anaemia is due to a quantitative problem of globin molecules.
(d) Both are due to a qualitative defect in globin chain synthesis.
Ans: 
(b)
Sickle cell anaemia is caused due to point mutation in which at the 6th position of beta globin chain, glutamic acid is replaced by valine, Huts, it is a qualitative defect in functioning of globin molecules.
Thalassemia is caused due to either mutation or deletion which ultimately results in reduced rate of synthesis of one of the globin chains lltal make up haemoglobin. Hence, it is a quantitative defect in functioning of globin molecules.

Q.10. The genotypes of a husband and wife are IAIB and IAi.
Among the blood types of their children, how many different genotypes and phenotypes arc possible?    (2017)
(a) 3 genotypes; 4 phenotypes
(b) 4 genotypes; 3 phenotypes
(c) 4 genotypes; 4 phenotypes
(d) 3 genotypes; 3 phenotype

Ans: (b)
If the genotypes of husband and wife are IA and IAi respectively, then the probabilities of genotypes and phenotypes among their children can be worked out as:
Previous Year Questions (2016-19) - Heredity and Variation Notes | EduRev
Genotype: IIA     IAi IA      IAIB      IBi
Phenotype: A       A            AB       B
Thus, there are four possible genotypes, viz., II, IAi, IAIand IBi and three possible phenotypes, viz.,A, AB and B among the children.

Q.11. A disease caused by an autosomal primary non disjunction is    (2017)
(a) Klinefelter’s syndrome
(b) Turner’s syndrome
(c) Sickle cell anaemia
(d) Down’s syndrome.

Ans: (d)
Down’s syndrome is an autosomal aneuploidy, caused by the presence of an extrachromosome number 21. Both the chromosomes of the pair 21 pass into a single egg due to non disjunction during oogenesis.

Q.12. Among the following characters, which one was not considered by Mendel in his experiments on pea?    (2017)
(a) Trichomes-Glandular or non-glandular
(b) Seed-Green or yellow
(c) Pod-Inflated or constricted
(d) Stem-Tall or dwarf

Ans: (a)
Mendel considered the following characters of pea in his experiments:


 Character Dominant   Recessive
 1. Seed shape Round (R) Wrinkled (r)
 2. Seed cotyledon colour Yellow (Y) Green (y)
 3.  Flower colour Violet (V) White (v)
 4.  Pod shape Inflated (I) Constricted (i)
 5.  Pod colour Green (G) Yellow (g)
 6.  Flower position Axial (A) Terminal (a)
 7. Stem height Tall (T) Dwarf (t)


Q.13. Which one from those given below is the perf for Mendel’s hybridisation experiments?    (2017)
(a) 1840-1850
(b) 1857-1869

(c) 1870-1877
(d) 1856-1863

Ans: (d)
Mendel carried out hybridisation experiments on garden pea for 7 years from 1856-1863.

Q.14. The mechanism that causes a gene to m< from one linkage group to another is called    (2016)
(a) Inversion
(b) Duplication
(c) Translocation
(d) Crossing-over.

Ans: (c)
Translocation is a chromosomal abnormality caused by rearrangement of parts between non- homologous chromosomes. It may cause a gene to move from one linkage group to another.

Q.15. If a colour-blind man marries a woman who is homozygous for normal colour vision, probability of their son being colour-blind    (2016)
(a) 0
(b) 0.5
(c) 0.75
(d) 1

Ans: (a)
Genotype o f colour blind man -  X°Y
Genotype of women homozygous XX
for normal woman
Previous Year Questions (2016-19) - Heredity and Variation Notes | EduRev
Hence, there is zero (0) probability of their son to be colour-blind.

Q.16. A cell at telophase stage is observed t student in a plant brought from the field, tells his teacher that this cell is not like c cells at telophase stage. There is no form; of cell plate and thus the cell is containing more number of chromosomes as compared to other dividing cells. This would result in    (2016)
(a) Somaclonal variation
(b) Polyteny
(c) Aneuploidy
(d) polyploidy

Ans: (d)
Polyploidy is the phenomenon of occurrence of more than two sets of chromosomes in the nucleus of a cell. Polyploidy is more common in plants. Polyploidy arises as a result of total non-disjunction of chromosomes during mitosis or meiosis.

Q.17. Pick out the correct statements.    (2016)
(1) Haemophilia is a sex-linked recessive Disease.
(2) Down’s syndrome is due to anetiploidy
(3) Phenylketonuria is an autosomal recessive gene disorder.
(4) Sickle cell anaemia is an X-Iinked recessive gene disorder.
(a) (1), (3) and (4) are correct.
(b) (1), (2) and (3) are correct.
(c) (1) and (4) are correct.
(d) (2) and (4) are correct.

Ans: (b)
Sickle-cell anaemia is an autosomal recessive genetic disorder. It can be transmitted from parents to the offspring when both the partners are carriers of the gene (or heterozygous).

Q.18. A tall true breeding garden pea plant is crossed with a dwarf true breeding garden pea plant When the F1 plants were selfed the resulting genotypes in the ratio of    (2016)
(a) 3 : 1 : : Tall; Dwarf
(b) 3 : 1 Dwarf : Tall
(c) 1 : 2 : 1 : : Tall homozygous : Tall heterozygous : Dwarf
(d) 1 : 2 : 1 : : Tall heterozygous : Tall homozygous : Dwarf.
Ans: (c)
When a tall true breeding garden pea plant is crossed with a dwarf true breeding garden pea plant and the F1 plants were soiled the resulting genotypes were in the ratio of 1 : 2 : 1 i.e„ Tall homozygous: Tall heterozygous : Dwarf
It can be illustrated as given below:
Previous Year Questions (2016-19) - Heredity and Variation Notes | EduRev
Phenotypic ratio : 3 : 1 : : Tall: Dwarf
Genotypic ratio -1 :2:1:: TT : T t : tt

Q.19. Match the terms in column I with their description m column II and choose the correct option.    (2016)

 Column I Column II
 A. Dominance (i) Many genes govern a single character
 B. Codominance (ii) In a heterozygous organism only one allele expresses itself
 C. Pleiotropy (iii) In a heterozygous organism both alleles express themselves fully
 D. Polygenic (iv) A single gene inheritance influences many characters

 A
 B
 C
 D
 (a)
 (iv)
 (i) 
(ii) 
(iii)
 (b)
 (iv)
 (iii)
 (i)
 (ii)
 (c)
 (ii)
 (i)
 (iv)
 (iii)
 (d)
 (ii)
 (iii)
 (iv) 
 (i)

Ans: (d)

Q.20. In a test cross involving F1 dihybrid flies, more parental-type offspring were produced than the recombinant-type offspring. This indicates    (2016)
(a) The two genes are linked and present on the same chromosome
(b) Both of the characters are controlled by more than one gene
(c) The two genes are located on two different chromosomes
(d) Chromosomes failed to separate during meiosis
Ans:
(a)
If in a dihybrid test cross more parental combinations appear as compared to the recombinants in F2 generation, then it is indicative of involvement of linkage. Linkage is the tendency of two different genes on the same chromosome to remain together during the separation of homologous chromosomes at meiosis. During complete linkage no recombinants are formed whereas in incomplete linkage few recombinants are produced along with parental combinations.

Q.21. Which of the following most appropriately describes haemophilia?    (2016)
(a) Chromosomal disorder
(b) Dominant gene disorder
(c) Recessive gene disorder
(d) X-linked recessive gene disorder

Ans: (d)
Haemophilia is a sex-linked disease. It occurs due to the presence of a recessive sex linked gene h, which is carried by X-chromosome.

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