NEET Biology Principles of Inheritance and Variation MCQs Set 06

Biology Objective Questions and Answers: Principles of Inheritance and Variation

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Multiple Choice Questions

Question 1: If there are four allelic forms for the gene controlling the ABO blood group, what is the number of possible genotypes?
(a) 6
(b) 10
(c) 12
(d) 14
Show Answer & Explanation

Answer: (b) 10

Explanation:
1. With n alleles the number of genotypes is n(n+1)/2.
2. For n = 4 this is 4 x 5 / 2 = 10.

Teacher's Note:
1. With 3 alleles (ABO) there are 6 genotypes.

Question 2: The shape of seeds depends on starch grain size, so the inheritance of seed shape shows ..... relationship while the inheritance of starch grain size shows ..... .
(a) Dominant-recessive, codominance
(b) Incomplete dominance, codominance
(c) Dominant-recessive, incomplete dominance
(d) Codominance, incomplete dominance
Show Answer & Explanation

Answer: (c) Dominant-recessive, incomplete dominance

Explanation:
1. Round seed shape is dominant over wrinkled, so it shows a dominant-recessive relationship.
2. Starch grain size in the heterozygote is intermediate, so it shows incomplete dominance.

Teacher's Note:
1. The same gene can show different dominance relations at different levels.

Question 3: In any dihybrid cross, the segregation of one pair of characters is independent of the other pair. This is known as:
(a) Law of segregation
(b) Law of purity of gametes
(c) Law of independent assortment
(d) Law of dominance
Show Answer & Explanation

Answer: (c) Law of independent assortment

Explanation:
1. The law of independent assortment states that alleles of different genes assort independently in gamete formation.
2. It gives the 9:3:3:1 ratio in F2.

Teacher's Note:
1. It is true only for genes on different chromosomes.

Question 4: Mendel published his work in 1865, but it remained unrecognised until 1900. Which of the following was not a reason for this?
(a) Communication was not easy
(b) His concept of genes as stable and discrete units was not accepted as an explanation for apparently continuous variations
(c) The use of mathematics to explain a biological phenomenon
(d) The use of the emasculation technique
Show Answer & Explanation

Answer: (d) The use of the emasculation technique

Explanation:
1. Communication was slow in those days, his factors as discrete units were not accepted, and his use of mathematics seemed odd for biology.
2. Emasculation was only a pollination technique and did not delay recognition.

Teacher's Note:
1. Mendel did not provide a physical proof for factors.

Question 5: The parallelism between the behaviour of chromosomes and genes was pointed out by:
(a) de Vries, Correns and Tschermak
(b) Sutton and Boveri
(c) Bateson and Punnett
(d) Landsteiner and de Castello
Show Answer & Explanation

Answer: (b) Sutton and Boveri

Explanation:
1. Walter Sutton and Theodor Boveri noted that chromosomes and Mendelian factors behave alike.
2. De Vries, Correns and Tschermak rediscovered Mendel's work.

Teacher's Note:
1. This led to the chromosomal theory of inheritance.

Question 6: Experimental verification of the chromosomal theory of inheritance was done by:
(a) Tschermak
(b) de Vries
(c) Sutton
(d) Morgan
Show Answer & Explanation

Answer: (d) Morgan

Explanation:
1. Thomas Hunt Morgan and his colleagues worked on Drosophila.
2. They experimentally verified the chromosomal theory.

Teacher's Note:
1. Morgan also discovered linkage and sex-linked inheritance.

Question 7: Who among the following used the frequency of recombination between gene pairs on the same chromosome as a measure of the distance between genes and mapped their position?
(a) Davenport
(b) Sturtevant
(c) Morgan
(d) Nillson
Show Answer & Explanation

Answer: (b) Sturtevant

Explanation:
1. Alfred Sturtevant, a student of Morgan, used recombination frequency as a distance measure.
2. He prepared the first genetic map.

Teacher's Note:
1. Map unit: 1% recombination = 1 centimorgan.

Question 8: If a yellow-bodied, white-eyed Drosophila is crossed with a wild-type brown-bodied, red-eyed Drosophila, what would be the frequency of recombinants in the F1 generation?
(a) 100%
(b) 1.3%
(c) 98.7%
(d) 0%
Show Answer & Explanation

Answer: (d) 0%

Explanation:
1. The cross gives F1 hybrids that are all heterozygous and wild type.
2. Recombinants can only be detected in the next generation (F2 or a test cross), so the F1 has 0% recombinants.

Teacher's Note:
1. Parental and recombinant types are compared in the F2.

Question 9: In some insects, half of the sperms possess an X chromosome along with the autosomes while the other half:
(a) Carry only autosomes
(b) Carry a Y chromosome along with the autosomes
(c) Carry only the X chromosome
(d) Carry an X chromosome along with the autosomes
Show Answer & Explanation

Answer: (a) Carry only autosomes

Explanation:
1. In XO insects the males have one X, so meiosis gives two kinds of sperms.
2. Half have an X plus autosomes and half have only autosomes.

Teacher's Note:
1. XO males have no Y chromosome.

Question 10: Female heterogamety can be seen in:
(a) Human beings
(b) Drosophila
(c) Hen
(d) Honey bees
Show Answer & Explanation

Answer: (c) Hen

Explanation:
1. In birds such as the hen, the female has ZW chromosomes and makes two kinds of eggs.
2. Humans and Drosophila show male heterogamety, and honey bees show haplodiploidy.

Teacher's Note:
1. Birds: ZZ males, ZW females.

Question 11: Chromosomal aberrations are commonly observed in:
(a) Germinal cells
(b) Cancer cells
(c) Nail base cells
(d) Gametes
Show Answer & Explanation

Answer: (b) Cancer cells

Explanation:
1. Chromosomal aberrations are commonly seen in cancer cells.
2. Cytogenetic study of these cells is used in cancer diagnosis.

Teacher's Note:
1. Aneuploidy and structural changes are typical of cancer cells.

Question 12: The study of the family history about the inheritance of a particular trait over several generations of a family is called:
(a) Phylogeny
(b) Ontogeny
(c) Pedigree analysis
(d) Cladistics
Show Answer & Explanation

Answer: (c) Pedigree analysis

Explanation:
1. A pedigree chart shows the inheritance of a trait through generations.
2. Phylogeny is evolutionary history and ontogeny is individual development.

Teacher's Note:
1. Pedigree analysis is used for human traits because crossing is not possible.

Question 13: In pedigree analysis, the symbol of a diamond with the number 5 inside represents:
(a) Five offspring of unspecified sex
(b) Five diseased offspring
(c) Five unaffected offspring
(d) Five affected offspring
Show Answer & Explanation

Answer: (a) Five offspring of unspecified sex

Explanation:
1. A diamond stands for an individual of unspecified sex.
2. A number inside indicates how many such offspring there are, so it means five offspring of unspecified sex.

Teacher's Note:
1. Shading tells whether they are affected.

Question 14: Genetic disorders determined by the alteration or mutation of a single gene are known as:
(a) Chromosomal disorders
(b) Mendelian disorders
(c) Non-inheritable disorders
(d) All of the above
Show Answer & Explanation

Answer: (b) Mendelian disorders

Explanation:
1. A single-gene mutation leads to Mendelian disorders.
2. They follow the Mendelian pattern of inheritance in a family.

Teacher's Note:
1. Chromosomal disorders involve whole chromosomes.

Question 15: In sickle cell anaemia, which of the following genotypes will show the disease phenotype?
(a) HbA HbA
(b) HbS HbS
(c) HbS HbA
(d) Both HbA HbA and HbS HbS
Show Answer & Explanation

Answer: (b) HbS HbS

Explanation:
1. Sickle cell anaemia is an autosomal recessive disease.
2. Only the homozygous HbS HbS individual shows the disease; HbS HbA is a carrier.

Teacher's Note:
1. Carrier (HbA HbS) shows sickle cell trait.

Question 16: Polymerisation of the mutant haemoglobin molecule in sickle cell anaemia is due to:
(a) Sulphadrugs
(b) High oxygen
(c) Low oxygen concentration
(d) Plasmodium falciparum
Show Answer & Explanation

Answer: (c) Low oxygen concentration

Explanation:
1. Under low oxygen tension the mutant haemoglobin polymerises.
2. This changes the RBCs to a sickle shape.

Teacher's Note:
1. Sickling is triggered by hypoxia.

Question 17: Regarding phenylketonuria, which of the following statements is wrong?
(a) Phenylalanine cannot be converted into tyrosine
(b) Phenylalanine is converted into phenylpyruvate and derivatives
(c) Phenylpyruvate is deposited in the heart, liver and kidney
(d) This is an inborn error of metabolism
Show Answer & Explanation

Answer: (c) Phenylpyruvate is deposited in the heart, liver and kidney

Explanation:
1. The enzyme phenylalanine hydroxylase is absent, so phenylalanine cannot become tyrosine.
2. It is instead converted to phenylpyruvic acid and derivatives, which accumulate in the brain causing mental retardation, not in the heart, liver or kidney.

Teacher's Note:
1. PKU is an autosomal recessive inborn error of metabolism.

Question 18: Chromosomal disorders arise due to:
(a) Absence of one or more chromosomes
(b) Excess of one or more chromosomes
(c) Abnormal arrangement of one or more chromosomes
(d) All of the above
Show Answer & Explanation

Answer: (d) All of the above

Explanation:
1. Aneuploidy is a gain or loss of chromosomes.
2. Structural abnormalities change chromosome arrangement, so all are causes.

Teacher's Note:
1. Failure of segregation causes aneuploidy.

Question 19: Which of the following symptoms is not associated with Down's syndrome?
(a) Flat back of the head
(b) Many loops on the finger tips
(c) Big and wrinkled tongue
(d) Congenital liver diseases
Show Answer & Explanation

Answer: (d) Congenital liver diseases

Explanation:
1. Down's syndrome (trisomy 21) features a flat back of the head, many loops on the fingertips and a furrowed tongue.
2. Congenital heart disease occurs, but liver disease is not a feature.

Teacher's Note:
1. Short stature and a palm crease are also seen.

Question 20: Match the following:
A. Down's syndrome; B. Klinefelter's syndrome; C. Turner's syndrome; D. Phenylketonuria
I. 44 + XY; II. 45 + XY; III. 44 + XO; IV. 44 + XXY
(a) A-I, B-IV, C-III, D-I
(b) A-II, B-IV, C-III, D-I
(c) A-I, B-II, C-III, D-IV
(d) A-I, B-II, C-IV, D-III
Show Answer & Explanation

Answer: (b) A-II, B-IV, C-III, D-I

Explanation:
1. Down's syndrome has trisomy of chromosome 21, so 45 autosomes + XY (II).
2. Klinefelter is 44 + XXY (IV), Turner is 44 + XO (III) and phenylketonuria is a gene disorder with a normal 44 + XY (I).

Teacher's Note:
1. Down's: 47 chromosomes; Klinefelter: 47; Turner: 45.

Question 21: Retarded physical, psychomotor and mental development are consequences observed in:
(a) Down's syndrome
(b) Klinefelter's syndrome
(c) Turner's syndrome
(d) Lesch-Nyhan syndrome
Show Answer & Explanation

Answer: (a) Down's syndrome

Explanation:
1. Down's syndrome causes retarded physical, psychomotor and mental development.
2. Klinefelter's and Turner's syndromes mainly affect sexual development.

Teacher's Note:
1. Trisomy 21 is the commonest autosomal aneuploidy.

Question 22: Gynaecomastia can be seen in:
(a) Down's syndrome
(b) Klinefelter's syndrome
(c) Turner's syndrome
(d) Edward's syndrome
Show Answer & Explanation

Answer: (b) Klinefelter's syndrome

Explanation:
1. Klinefelter's males (44 + XXY) develop enlarged breasts.
2. They are sterile with some feminine characters.

Teacher's Note:
1. Gynaecomastia = development of breasts in males.

Question 23: Match the following:
A. Haemophilia; B. Down's syndrome; C. Klinefelter's syndrome; D. Turner's syndrome
i. Broad palm with characteristic palm crease; ii. Delayed clotting of blood; iii. Some feminine character; iv. Rudimentary ovaries
(a) A-i, B-ii, C-iii, D-iv
(b) A-iii, B-ii, C-i, D-iv
(c) A-ii, B-i, C-iv, D-iii
(d) A-ii, B-i, C-iii, D-iv
Show Answer & Explanation

Answer: (d) A-ii, B-i, C-iii, D-iv

Explanation:
1. Haemophilia causes delayed clotting of blood (ii).
2. Down's syndrome has a broad palm with a palm crease (i), Klinefelter's males show some feminine characters (iii) and Turner's females have rudimentary ovaries (iv).

Teacher's Note:
1. Turner females are sterile with rudimentary ovaries.

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