Solutions
Chapter
3
Principles of Inheritance and Variation
SECTION - A
Objective Type Questions 1. Mark the odd one (w.r.t. true breeding line)
(1) Shows the stable trait inheritance (2) Shows expression for few generations only (3) Undergone continuous self-pollination (4) Both (1) & (3)
Sol. Answer (2)
True breeding means Pure lines.
2. Which of the following is not a dominant trait in edible pea?
(1) Axial flower (2) Inflated pod (3) Green seed colour (4) Green pod Sol. Answer (3)
Green seed Recessive Yellow seed Dominant
3. The phenotype of an individual may be affected if the modified allele produces (a) No enzyme at all
(b) The normal/less efficient enzyme (c) A non-functional enzyme
(1) Only (a) is correct (2) (a) and (c) are correct (3) (b) and (c) are correct (4) Only (c) is correct Sol. Answer (2)
Recessive allele may not produce enzyme or may be non functional enzyme.
4. What will be possible blood group in children from the parents with B and AB blood groups?
(1) A, O (2) A, B, AB & O (3) A, B, AB (4) B, O Sol. Answer (3) Phenotype B AB Genotype IBi × IAIB IAi, IBi IAIB . IBIB So A, AB B are possible
5. In garden pea, starch is synthesised effectively in
(1) Heterozygous round seeded plants (2) Homozygous round seeded plants
(3) Wrinkled seeded plants (4) Pure and hybrid round seeded plants
Sol. Answer (2)
Bb Intermediate starch grain BB Large starch grain bb Small starch grain
6. F1 progeny of Mendelian dihybrid cross produces
(1) Two types of pollen grains (2) Four genotypes of gametes
(3) Two types of eggs (4) Four types of pollens only
Sol. Answer (2) F1 AaBb A B AB b Ab a B aB b ab So four gametes
7. When Mendel self hybridised the F1 plants (RrYy), he found that dominant and recessive traits of one character are segregated in a
(1) 9 : 1 ratio (2) 3 : 3 ratio (3) 10 : 6 ratio (4) 3 : 1 ratio
Sol. Answer (4)
8. Mendel published his work on inheritance of characters in 1865 but it remained unrecognised till 1900 because (a) He could not provide any physical proof for the existence of factors
(b) His concept of factors as stable, discrete units that controlled the expression of traits did not find acceptance from the contemporaries
(c) Mendel's approach of using mathematics to explain biological phenomena was totally old (d) Communication was not easy (as it is now)
(1) (a), (b) & (c) are correct (2) (c) & (d) are correct (3) (a), (b) & (d) are correct (4) Only (a) is correct Sol. Answer (3)
9. Which of the following statement for chromosomal theory of inheritance is incorrect?
(1) Pairing and separation of a pair of chromosomes would lead to the segregation of a factor they carried (2) Behaviour of chromosomes is parallel to the behaviour of genes
(3) The two alleles of a gene pair are located on homologous sites on homologous chromosomes (4) Chromosomes as well as genes occur in pairs
Sol. Answer (1)
10. Experimental verification of the chromosomal theory of inheritance was given by
(1) Sutton and Boveri (2) Correns (3) T.H. Morgan (4) Tschermak
11. Fruit flies are one of the best materials for genetic studies because of all, except (1) Ability to grow on simple synthetic medium in the laboratory
(2) Short life span
(3) Production of a large number of progeny in each mating
(4) Presence of few externally visible and identifiable contrasting traits Sol. Answer (4)
12. Generation of non-parental gene combinations is termed as
(1) Linkage (2) Polyploidy (3) Recombination (4) Aneuploidy
Sol. Answer (3)
Non-parental gene combination - Recombination Produce by crossing over
13. Initial clue about the genetic/chromosomal mechanism of sex-determination can be traced back to some of the experiments carried out in
(1) Human beings (2) Birds (3) Insects (4) Plants
Sol. Answer (3)
Henking discovered X-body.
14. In which of the sex determination both male and female have same number of chromosomes?
(1) XY type (2) ZO type (3) XO type (4) Both (1) & (3)
Sol. Answer (1)
A A + X X
A A + X Y Same number of chromosome in and+
15. Two different types of gametes in terms of the sex chromosomes, are produced by
(1) Female fruit fly (2) Male butterfly
(3) Male human and female Drosophila (4) Female birds Sol. Answer (4)
AA + XW A + X A + W
Two type of eggs are produced by + bird because it is heterogametic. 16. Individuals having homomorphic sex chromosomes produce
(1) Only one gamete in complete life span (2) One type of gametes
(3) No gametes (4) Two type of gametes
Sol. Answer (2)
AA + XX A + X
Produce only one type of gamete.
17. Which of the following phenomena leads to variation in DNA?
(1) Linkage, mutation (2) Recombination, linkage
(3) Mutation, recombination (4) Aneuploidy, linkage
Sol. Answer (3)
Mutation Sudden change in genetic Material Recombination Non-parental combination
18. Sickle cell-anaemia disorder arises due to
(1) Duplication of a segment of DNA (2) Substitution in a single base of DNA (3) Deletion of a segment of DNA (4) Duplication in a base pair of RNA Sol. Answer (2)
Substitution of Glutamine T A CAC
G AG GUG Valine C T C
5 3
3 5
19. In pedigree analysis, symbol given for sex unspecified is
(1) (2) (3) (4)
Sol. Answer (1)
20. Cystic fibrosis, Myotonic dystrophy and Thalassemia are
(1) Chromosomal disorders (2) Autosomal recessive disorders
(3) Mendelian disorders (4) Autosomal dominant disorders
Sol. Answer (3)
Mendelian disorder because their inheritance follow Mendelion Inheritance.
21. Which of the following trait shows transmission from carrier female to male progeny?
(1) Autosomal dominant (2) X-linked recessive (3) Y-linked recessive (4) X-linked dominant Sol. Answer (2)
1 X linked Recessive gene
X XC × X Y
X YC XY
22. Phenylketonuria is an inborn error of metabolism that is inherited as
(1) Autosomal recessive trait (2) Sex-linked dominant trait
(3) X-linked recessive trait (4) Autosomal dominant trait
Sol. Answer (1)
Genes located on chromosome 12
23. Which of the following abnormalities is due to autosomal dominant mutation?
(1) Colour blindness (2) Thalassemia (3) Myotonic dystrophy (4) Haemophilia Sol. Answer (3)
24. Absence or excess or abnormal arrangement of one or more chromosomes results in
(1) Point mutation (2) Chromosomal disorders
(3) Mendelian disorders (4) Gene mutation
Sol. Answer (2)
25. Mark the odd one w.r.t. syndrome which occur due to failure of segregation of homologous pair of chromosomes during cell division cycle.
(1) Klinefelter's syndrome (2) Down's syndrome (3) Turner's syndrome (4) Thalassemia Sol. Answer (4)
Thalassemia is autosomal recessive disease.
26. Heterozygous round and yellow seeded pea plants were selfed and total 800 seeds are collected. What is the total number of seeds with first dominant and second recessive traits?
(1) 950 (2) 300 (3) 200 (4) 150 Sol. Answer (4) 3 800 150 16
27. Which of the disorder is related with the Karyotype given below?
(1) Turner's syndrome (2) Down's syndrome (3) Myotonic dystrophy (4) Cystic fibrosis Sol. Answer (2)
21-Trisomy
28. Mark the correct match
(1) Turner's syndrome - 45 + XO (2) Phenylketonuria - 44 + XYY (3) Klinefelter's syndrome - 44 + XXY (4) Thalassemia - 44 + YO Sol. Answer (3)
Trisomy of X chromosome 44 + XXY
29. Physical, psychomotor and mental development is retarted in an individual affected with
(1) Down's syndrome (2) Sickle cell-anaemia (3) Turner's syndrome (4) Colour blindness Sol. Answer (1)
Trisomy of 21 chromosome
30. In which of the following disorder's affected individual's possess 47 chromosomes?
(1) Turner's syndrome (2) Klinefelter's syndrome
(3) Down's syndrome (4) Both (2) & (3)
Sol. Answer (4)
X Chromosome Klinefelter Syndrome 2n + 1 47 21 Chromosome Down's Syndrome 2n + 1 47
31. The affected individuals are short statured in disorders like
(1) Turner's syndrome, phenylketonuria (2) Down's syndrome, Turner's syndrome (3) Klinefelter's syndrome, Down's syndrome (4) Turner's syndrome, Klinefelter's syndrome Sol. Answer (2)
32. In which of the following disorder gynaecomastia symptom is seen in individuals?
(1) Down's syndrome (2) Turner's syndrome (3) Klinefelter's syndrome (4) Phenylketonuria Sol. Answer (3)
Development of breast in males.
33. Mark the correct option (w.r.t. monosomy)
(1) Klinefelter's syndrome (2) Down's syndrome (3) Turner's syndrome (4) Haemophilia Sol. Answer (3)
34. Allosomic trisomy condition is seen in
(1) Turner's syndrome (2) Klinefelter's syndrome (3) Down's syndrome (4) Both (2) & (3) Sol. Answer (2)
2n + 1 Trisomy of × Chromosome
35. Which of the following disorder is seen in human female only?
(1) Turner's syndrome (2) Down's syndrome
(3) Haemophilia (4) Klinefelter's syndrome
Sol. Answer (1)
2n – 1 Monosomic of X Chromosome
SECTION - B
Objective Type Questions
1. When a pink flowered Antirrhinum plant is test crossed, then phenotypic ratio in resulting progenies is (1) 1 Red : 1 White (2) 3 Red : 1 White (3) 2 Pink : 1 White (4) 1 Pink : 1 White Sol. Answer (4)
RR rr Rr Red White Pink Rr × Rr
RR : Rr : rr 1 : 2 : 1 Red : Pink : White
2. Heterozygous tall and violet flowered pea plants were selfed and total 512 seeds are collected. What will be total number of seeds for both heterozygous traits?
(1) 128 (2) 256 (3) 384 (4) 64 Sol. Answer (1) Number of individual 512 TtRr 4 16 ⇒ 4 512 128 16
3. Mark the odd one (w.r.t. F2 generation of Mendelian dihybrid cross)
(1) Frequency of TtRR genotype = 12.5% (2) Frequency of ttrr genotype = 6.25% (3) Frequency of TTRR genotype = 6.25% (4) Frequency of ttRr genotype = 25% Sol. Answer (4)
2 1
ttRr 12.5% 16 8
4. Morgan hybridised yellow-bodied, white-eyed females to brown-bodied, red-eyed males and intercrossed their F1 progeny. He observed that
(a) F2 ratio was deviated very significantly from the 9 : 3 : 3 : 1 ratio (b) Both genes did not segregate independently of each other (c) Recombinant types are not obtained in F2 generation (d) Both genes segregate independently of each other Select the correct set of statements :
(1) (a) & (b) only (2) (b) & (c) only (3) (b) & (d) only (4) (c) & (d) only Sol. Answer (1)
y+ w+ yw
Brown body and red eye are linked gene so F2 ratio deviated from 9 : 3 : 3 : 1.
5. ___(A)___ used the frequency of recombination between gene pairs on the ___(B)___ as a measure of the distance between genes and mapped their position on the chromosome.
(A) (B)
(1) Morgan Same chromosome
(2) Sturtevant Different chromosomes
(3) Morgan Different chromosomes
(4) Sturtevant Same chromosome
Sol. Answer (4)
6. While solving the problem of sex determination in large number of insects, it was observed that (1) All eggs lack sex chromosome
(2) Some of the sperms bear the X-chromosome (3) All eggs as well as sperms bear the X-chromosome (4) Some of the eggs bear the X-chromosome
Sol. Answer (2) Insects XX XO A + X + A + X A + O
7. Loss or gain of a segment of DNA results in
(1) Frame-shift muation (2) Point mutation
(3) Polyploidy (4) Chromosomal aberration
8. Which one of the following is a physical factor that induce mutation?
(1) Acridines (2) HNO2 (3) UV-rays (4) Base analogue
Sol. Answer (3)
UV-rays and Non-ionised radiation.
9. In the given pedigree, indicate whether the shaded symbols indicate dominant or recessive allele.
(1) Recessive (2) Codominant
(3) Dominant (4) It can be recessive or dominant both
Sol. Answer (4)
If propositus is Aa
Then it will be dominant pedigree
If is aa then it will be recessive pedigree.
10. In which of the following disorder a single protein that is a part of the cascade of proteins involved in blood clotting is affected?
(1) Thalassemia (2) Sickle-cell anaemia (3) Haemophilia (4) Phenylketonuria Sol. Answer (3)
Haemophilia A Blood clot factor VIIIth is absent
Haemophilia B Blood clot factor IXth is absent
11. Mark the correct statement (w.r.t. sickle cell-anaemia) (1) Homozygous individuals for HbS are apparently unaffected
(2) Heterozygous individuals exhibit sickle-cell trait
(3) Heterozygous individuals are affected as well as carrier (4) Homozygous individuals for HbA show the diseased phenotype
Sol. Answer (2)
HbA HbS Sicklecell trait
12. The defect sickle-cell anaemia is caused by the ______ of glutamic acid by valine at the 6th position of the
______ globin chain of the haemoglobin molecule.
(1) Substitution, (2) Deletion, (3) Duplication, (4) Translocation, Sol. Answer (1)
Substitution of Pyrimidine Purine
DNA CTC CAG RNA GAG GUG AA Glutanic
acid
13. A Y-linked gene is responsible for hypertrichosis (long hair on ears). When an affected man marries a normal woman, what percentage of their daughters would be expected to have hairy ears?
(1) 25% (2) 0% (3) 50% (4) 100%
Sol. Answer (2)
Daughter will not receive Y chromosome from father.
14. A normal woman, whose father had colour blindness, married a normal man. What is the chance of occurrence of colour blindness in the progeny?
(1) 25% (2) 50% (3) 100% (4) 75% Sol. Answer (1) XXC × XY XXX XXC X YC XY X Y XC
25% of progeny will be colour blind.
15. Mr. Stevan is suffering from haemophilia and cystic fibrosis. His father is hetrozygous for cystic fibrosis. The probability of Stevan's sperm having recessive X-linked as well as autosomal allele is
(1) 1 4 (2) 1 16 (3) 1 2 (4) 1 8 Sol. Answer (3) AhAC Xh Y AC Xh 1 2 AC Y 1 2
16. Select incorrect one (w.r.t. reciprocal cross)
(1) To know whether the alleles are present on sex chromosomes or autosomes (2) It is made to eliminate the effect of nuclear traits
(3) Two individuals with contrast genotypes are involved (4) Results are not changed for autosomal traits Sol. Answer (2)
17. The chromosome maps are not accurate maps because
(1) Crossing over frequency is higher than recombination frequency
(2) One crossing over interferes and increases the frequency of nearby crossing over (3) Crossing over frequency decreases towards the ends of chromosome
(4) Heterochromation increases crossing over Sol. Answer (1)
18. In Lathyrus odoratus, hybrid blue flowered and long pollen plant is test crossed with homozygous recessive red flowered and round pollen plant then how many parental types are obtained when genes are present in cis stage in parents?
(1) 50% (2) 43.7% (3) 87.4% (4) 12.6% Sol. Answer (3) BbLl × bb ll BbLlBL BbllBl bbLlbL bbllBl bl 7 1 1 7 14 100 87.4%16
19. Find out the frequency of AabbCcDdee if parents are AabbCCddEe and AabbccDdee
(1) 0.78% (2) 12.5% (3) 25% (4) 50% Sol. Answer (2) AabbCCddEe × AabbCcDdee AabbCcDdee 2 2 2 8 12.5 4 4 4 64 20. In incomplete dominance
(1) Dominant trait is completely expressed in F1 generation (2) Phenotypic and genotypic ratio are different
(3) Two dominant alleles are needed to express the complete dominant trait (4) F1 individuals have the equal traits of both parents
Sol. Answer (3)
21. Progeny with blood group ‘O’ can not be obtained in cross
(1) A×A (2) A×B (3) O×AB (4) B×B
Sol. Answer (3) O × AB ii IAIB
IAi, IBi So no O blood group
22. If a agouti mice (CcAa) is crossed with albino mice (ccAA), then how many albino mice are produced in resulting progeny?
(1) 4 (2) 9 (3) 2 (4) 3
Sol. Answer (3) CcAa × ccAA Agouti Albino
CcAACA CcAaCa ccAAcA ccAaca cA
23. Match the following - (w.r.t. Pedigree analysis)
Column - I Column - II
a. Solid symbol (i) Carrier of sex linked trait
b. Horizontal line (ii) Offspring
between symbols
c. Horizontal line (iii) Trait to be studied
above the symbols
d. Dot in centre (iv) Parents
(1) a(iv), b(iii), c(ii), d(i) (2) a(ii), b(iii), c(iv), d(i) (3) a(iii), b(iv), c(ii), d(i) (4) a(i), b(ii), c(iv), d(iii) Sol. Answer (3)
24. Which of the following parental combination has produced mutant offspring?
(1) Tt × tt = Tt (2) tt × tt = Tt (3) Tt × Tt = tt (4) TT × tt =Tt
Sol. Answer (2) tt × tt = Tt
25. Epistasis and dominance are respectively
(1) Intragenic, Intergenic (2) Non-allelic, Extra-allelic (3) Non-allelic, Interallelic (4) Intergenic, Non-allelic Sol. Answer (3)
26. Which of the following combination seems to have some linkage in character selected by Mendel? (1) Stem height and pod colour (2) Flower colour and flower position (3) Seed shape and seed colour (4) Plant height and pod shape Sol. Answer (4)
Plant Height and Pod shape
27. A diploid organism is heterozygous for five loci and homozygous for 2 loci, how many types of gametes can be produced? (1) 128 (2) 32 (3) 4 (4) 14 Sol. Answer (2) AaBbCcDdEe 5 Heterozygous FFGG Homozygous 2n⇒25 ⇒3
28. Lesch Nyhan disease is an X-linked recessive disorder that causes neurological damage in human beings. A survey of 500 mates from a caucasion population revealed that 20 were effected with this disorder. What is the frequency of the normal allele in this population?
(1) 9.6 (2) 0.8 (3) 0.096 (4) 96 Sol. Answer (2) 2 20 1 q 0.2 500 25 ⇒ ⇒ ⇒ q = 0.2 p + q = 1 p = 0.8
29. How many types of zygotic combinations are possible between a cross Aa BB Cc Dd × AA bb Cc DD? (1) 32 (2) 128 (3) 64 (4) 16 Sol. Answer (4) Aa BB Cc Dd × AA bb Cc DD Number of gametes 2n 23 8 × 2 = 16
30. In a complimentary gene interaction calculate the number of phenotype and genotype produced in a cross AaBb × aaBB
(1) 1 phenotype, 2genotypes (2) 2 phenotypes, 4 genotypes
(3) 4 phenotypes, 4 genotypes (4) 2 phenotypes, 2 genotypes
Sol. Answer (2) AB
AaBB AaBbAb aaBBaB aaBbab aB
1 1 1 1 Phenotype Phenotype
Genotype
31. Select incorrect statement
(A) Linked genes cause absolute lethality
(B) Persons affected by PKU do not show mental disorder (C) F2 ratio in codominance and incomplete dominance are same (D) Sex of male Drosophila is dependent on Y-chromosome
(1) (A) & (B) (2) (B) & (C) (3) (A), (B) & (D) (4) All of these Sol. Answer (3)
32. In phenylketonuria
(1) Break down of phenylalanine is rapid (2) Accumulation of phenylalanine in body (3) Chromosomal constitution of patient changes
(4) TSD gene situated on chromosome 15 undergoes mutation Sol. Answer (2)
Phenylalanine hydroxylase enzyme is absent.
33. How many types of gametes will be produced by a O Drosophila having following arrangement of two genes (y+ and w+) on X-chromosome?
X Y y+ w+ (1) 2 (2) 4 (3) 1 (4) 8 Sol. Answer (1) y+ w+ Y X
34. If interference is complete or cent percent then the frequency of observed double crossover will be (1) Equal to expected frequency (2) Greater than expected frequency
(3) Lesser than expected frequency (4) Zero
Sol. Answer (4)
35. In F2 generation of a Mendelian dihybrid cross (TTRR × ttrr)
(1) Tall plants and violet flowered plants are obtained in 1 : 1 frequency (2) Ratio of parental and non-parental plants is 1 : 15
(3) Recombinant plants are obtained in 1 : 1 frequency (4) More than one option is correct
Sol. Answer (4)
SECTION - C
Previous Years Questions
1. A colour blind man marries a woman with normal sight who has no history of colour blindness in her family. What is the probability of their grandson being colour blind? [Re-AIPMT-2015]
(1) 0.25 (2) 0.5 (3) 1 (4) Nil
Sol. Answer (2)
Father (Colourblind) Daughter (Carrier)
Grandson [50% Probability (0.5)]
2. The term "Linkage" was coined by [Re-AIPMT-2015]
(1) W. Sutton (2) T.H. Morgan (3) T. Boveri (4) G. Mendel
Sol. Answer (2)
The term "linkage" was coined by T.H. Morgan.
3. A pleiotropic gene : [Re-AIPMT-2015]
(1) Controls multiple traits in an individual. (2) Is expressed only in primitive plants. (3) Is gene evolved during Pliocene
(4) Controls a trait only in combination with another gene Sol. Answer (1)
The gene which controls multiple traits in an individual.
4. In his classic experiments on pea plants, Mendel did not use [Re-AIPMT-2015]
(1) Flower position (2) Seed colour (3) Pod length (4) Seed shape
Sol. Answer (3)
5. A gene showing codominance has [Re-AIPMT-2015] (1) Both alleles independently expressed in the heterozygote.
(2) One allele dominant on the other
(3) Alleles tightly linked on the same chromosome (4) Alleles that are recessive to each other
Sol. Answer (1)
Both alleles are independently expressed in heterozygote during codominance.
6. In the following human pedigree, the filled symbols represent the affected individuals. Identify the type of given pedigree. I II III IV [Re-AIPMT-2015] (1) X-linked dominant (2) Autosomal dominant (3) X-linked recessive (4) Autosomal recessive Sol. Answer (4)
The given pedigree represents inheritance of Autosomal recessive trait.
I II III IV aa aa Aa Aa Aa Aa Aa Aa aa Aa Aa Aa Aa aa Aa aa aa Aa
7. Alleles are [AIPMT-2015]
(1) Heterozygotes (2) Different phenotype
(3) True breeding homozygotes (4) Different molecular forms of a gene Sol. Answer (4)
Alleles are slightly different molecular forms of the same gene.
8. The movement of a gene from one linkage group to another is called [AIPMT-2015]
(1) Crossing over (2) Inversion (3) Duplication (4) Translocation
Sol. Answer (4)
Translocation is illegitimate crossing over between non-homologous chromosome.
9. Multiple alleles are present [AIPMT-2015]
(1) On non-sister chromatids (2) On different chromosomes
(3) At different loci on the same chromosome (4) At the same locus of the chromosome Sol. Answer (4)
10. An abnormal human baby with 'XXX' sex chromosomes was born due to [AIPMT-2015] (1) Fusion of two sperm and one ovum (2) Formation of abnormal sperms in the father (3) Formation of abnormal ova in the mother (4) Fusion of two ova and one sperm
Sol. Answer (3)
Due to non-disjunction of X-chromosomes in mother A + XX (egg) × A + X (sperm)
11. How many pairs of contrasting characters in pea plants were studied by Mendel in his experiments?
[AIPMT-2015]
(1) Seven (2) Five (3) Six (4) Eight
Sol. Answer (1)
7 pairs of contrasting characters in pea plant were studied by Mendel in his experiment.
12. Fruit colour in squash is an example of [AIPMT-2014]
(1) Recessive epistasis (2) Dominant epistasis
(3) Complementary genes (4) Inhibitory genes
Sol. Answer (2)
Dominant epistasis is the phenomenon of masking or supressing the expression of a gene by a dominant non-allelic gene.
e.g., fruit colour in Cucurbita pepo (Summer squash)
13. A man whose father was colour blind marries a woman who had a colour blind mother and normal father. What percentage of male children of this couple will be colour blind ? [AIPMT-2014]
(1) 25% (2) 0% (3) 50% (4) 75% Sol. Answer (3) X Y+ X+ X Y X+ Xc X+ Xc X+Xc X+Xc X+Y XcY Colourblind male = 50%
14. A human female with Turner's syndrome: [AIPMT-2014]
(1) Has 45 chromosomes with XO (2) Has one additional X chromosome
(3) Exhibits male characters (4) Is able to produce children with normal husband Sol. Answer (1)
Turner's syndrome is caused due to the absence of one of the X chromosomes i.e., 45 with XO (or 44 + XO).
15. In a population of 1000 individuals 360 belong to genotype AA, 480 to Aa and the remaining 160 to aa. Based
on this data, the frequency of allele A in the population is [AIPMT-2014]
Sol. Answer (3)
According to Hardy Weinberg principle. p2 + 2pq + q2 = 1; (p + q)2 = 1
(AA) p2 = 360 out of 1000 individual or p2 = 36 out of 100
q2 = 160 out of 1000 or q2 = 16 out of 100
so q = .16 = .4. As p + q = 1 so, p is 0.6.
16. If both parents are carriers for thalessemia, which is an autosomal recessive disorder, what are the chances
of pregnancy resulting in an affected child? [NEET-2013]
(1) 50% (2) 25% (3) 100% (4) No chance Sol. Answer (2) ATA × ATA AA : AAT : ATAT 1 : 2 : 1 25%
17. The incorrect statement with regard to Haemophilia is [NEET-2013]
(1) It is a recessive disease (2) It is a dominant disease
(3) A single protein involved in the clotting of blood is affected (4) It is a sex-linked disease
Sol. Answer (2)
18. If two persons with 'AB' blood group marry and have sufficiently large number of children, these children could be classified as 'A' blood group : 'AB' blood group : 'B' blood group in 1 : 2 : 1 ratio. Modern technique of protein electrophoresis reveals presence of both 'A' and 'B' type proteins in 'AB' blood group individuals. This
is an example of [NEET-2013]
(1) Incomplete dominance (2) Partial dominance
(3) Complete dominance (4) Codominance
Sol. Answer (4)
Both IA and IB give its expression.
19. Which Mendelian idea is depicted by a cross in which the F1 generation resembles both the parents? [NEET-2013]
(1) Law of dominance (2) Inheritance of one gene
(3) Co-dominance (4) Incomplete dominance
Sol. Answer (3)
20. Which of the following statements is not true of two genes that show 50% recombination fequency ?
[NEET-2013] (1) The genes are tightly linked
(2) The genes show independent assortment
(3) If the genes are present on the same chromosome, they undergo more than one crossovers in every meiosis
(4) The genes may be on different chromosomes Sol. Answer (1)
21. F2 generation in a Mendelian cross showed that both genotypic and phenotypic ratios are same as 1 : 2 : 1.
It represents a case of [AIPMT (Prelims)-2012]
(1) Monohybrid cross with complete dominance (2) Monohybrid cross with incomplete dominance (3) Co-dominance
(4) Dihybrid cross Sol. Answer (2)
22. A normal-visioned man whose father was colour-blind, marries a woman whose father was also colour blind. They have their first child as a daughter. What are the chances that this child would be colour-blind?
[AIPMT (Prelims)-2012] (1) 25% (2) 50% (3) 100% (4) Zero percent Sol. Answer (4) X Y X XC X Y X XC Gametes : X X X C X YC X X X Y X Y XC X
100% daughter = Normal but 50% carrier 50% son – Colourblind
50% son – Normal
23. A test cross is carried out to [AIPMT (Mains)-2012]
(1) Determine the genotype of a plant at F2 (2) Predict whether two traits are linked (3) Asses the number of alleles of a gene
(4) Determine whether two species or varieties will breed successfully Sol. Answer (1)
TT × tt Tt
Cross with recessive parent
24. Represented below is the inheritance pattern of a certain type of traits in humans. Which one of the following
conditions could be an example of this pattern? [AIPMT (Mains)-2012]
Female
Mother FatherMale
Daughter Son
(1) Phenylketonuria (2) Sickle cell anaemia (3) Haemophilia (4) Thalassemia Sol. Answer (3)
25. Which one of the following is a wrong statement regarding mutations? [AIPMT (Mains)-2012] (1) Deletion and insertion of base pairs cause frame-shift mutations
(2) Cancer cells commonly show chromosomal aberrations (3) UV and Gamma rays are mutagens
(4) Change in a single base pair of DNA does not cause mutation Sol. Answer (4)
Point mutation can cause mutation.
26. Which one of the following conditions correctly describes the manner of determining the sex in the given
example? [AIPMT (Prelims)-2011]
(1) Homozygous sex chromosomes (XX) produce male in Drosophila. (2) Homozygous sex chromosomes (ZZ) determine female sex in Birds. (3) XO type of sex chromosomes determine male sex in grasshopper
(4) XO condition in humans as found in Turner Syndrome, determine female sex Sol. Answer (3)
AA – XO – A + X A + O
27. When two unrelated individuals or lines are crossed, the performance of F1 hybrid is often superior of both its
parents. This phenomenon is called [AIPMT (Prelims)-2011]
(1) Metamorphosis (2) Heterosis (3) Transformation (4) Splicing
Sol. Answer (2)
28. Test cross in plants or in Drosophila involves crossing [AIPMT (Mains)-2011] (1) The F1 hybrid with a double recessive genotype (2) Between two genotypes with dominant trait (3) Between two genotypes with recessive trait (4) Between two F1 hybrids
Sol. Answer (1)
29. ABO blood groups in humans are controlled by the gene I. It has three alleles – IA, IB and i. Since there are
three different alleles, six different genotypes are possible. How many phenotypes can occur?
[AIPMT (Prelims)-2010]
(1) Three (2) One (3) Four (4) Two
Sol. Answer (3)
30. Which one of the following cannot be explained on the basis of Mendel’s Law of Dominance?
[AIPMT (Prelims)-2010] (1) Factors occur in pairs
(2) The discrete unit controlling a particular character is called a factor (3) Out of one pair of factors one is dominant, and the other recessive
(4) Alleles do not show any blending and both the characters recover as such in F2 generation Sol. Answer (4)
31. The genotype of a plant showing the dominant phenotype can be determined by [AIPMT (Prelims)-2010]
(1) Back cross (2) Test cross (3) Dihybrid cross (4) Pedigree analysis
32. Select the correct statement from the ones given below with respect to dihybrid cross
[AIPMT (Prelims)-2010] (1) Tightly linked genes on the same chromosome show very few recombinations
(2) Tightly linked genes on the same chromosome show higher recombinations (3) Genes far apart on the same chromosome show very few recombinations
(4) Genes loosely linked on the same chromosome show similar recombinations as the tightly linked ones Sol. Answer (1)
Tightly linked gene will show less recombination.
33. A cross in which an organism showing a dominant phenotype is crossed with the recessive parent in order
to know its genotype is called [AIPMT (Mains)-2010]
(1) Monohybrid cross (2) Back cross (3) Test cross (4) Dihybrid cross
Sol. Answer (3)
34. Study the pedigree chart of a certain family given below and select the correct conclusion which can be drawn for the character
[AIPMT (Mains)-2010] (1) The female parent is heterozygous
(2) The parents could not have had a normal daughter for this character (3) The trait under study could not be colourblindness
(4) The male parent is homozygous dominant Sol. Answer (1)
Aa aa Aa aa Aa aa
35. ABO blood grouping is controlled by gene I which has three alleles and show co-dominance. There are six
genotypes. How many phenotypes in all are possible? [AIPMT (Mains)-2010]
(1) Six (2) Three (3) Four (4) Five
Sol. Answer (3)
36. The fruit fly Drosophila melanogaster was found to be very suitable for experimental verification of chromosomal theory of inheritance by Morgan and his colleagues because: [AIPMT (Mains)-2010] (1) It reproduces parthenogenetically
(2) A single mating produces two young flies
(3) Smaller female is easily recognisable from larger male (4) It completes life cycle in about two weeks
37. In Antirrhinum two plants with pink flowers were hybridized. The F1 plants produced red, pink and white flowers in the proportion of 1 red, 2 pink and 1 white. What could be the genotype of the two plants used for hybridization? Red flower colour is determined by RR, and white by rr genes. [AIPMT (Mains)-2010]
(1) rrrr (2) RR (3) Rr (4) rr Sol. Answer (3) Antirrhinum majus RR × rr F1 Rr RR : Rr : rr 1 : 2 : 1 F2
38. Which one of the following symbols and its representation, used in human pedigree analysis is correct? [AIPMT (Prelims)-2010]
(1) = male affected (2) = mating between relatives
(3) = unaffected male (4) = unaffected female
Sol. Answer (2)
39. Point mutation involves [AIPMT (Prelims)-2009]
(1) Deletion (2) Insertion
(3) Change in single base pair (4) Duplication
Sol. Answer (3)
Sickele cell anaemeia
40. Study the pedigree chart given below:
What does it show? [AIPMT (Prelims)-2009]
(1) Inheritance of a condition like phenylketonuria as an autosomal recessive trait (2) The pedigree chart is wrong as this is not possible
(3) Inheritance of a recessive sex-linked disease like haemophilia
(4) Inheritance of a sex-linked inborn error of metabolism like phenylketonuria Sol. Answer (1) Aa Aa aa Aa Aa A– aa A– aa A–
41. Select the incorrect statement from the following: [AIPMT (Prelims)-2009] (1) Galactosemia is an inborn error of metabolism
(2) Small population size results in random genetic drift in a population (3) Baldness is a sex-limited trait
(4) Linkage is an exception to the principle of independent assortment in heredity Sol. Answer (3)
Baldness is sex influenced character.
42. Haploids are more suitable for mutation studies than the diploids. This is because [AIPMT (Prelims)-2008] (1) All mutations, whether dominant or recessive are expressed in haploids
(2) Haploids are reproductively more stable than diploids
(3) Mutagens penetrate in haploids more effectively than diploids (4) Haploids are more abundant in nature than diploids
Sol. Answer (1)
43. Which one of the following conditions in humans is correctly matched with its chromosomal abnormality/
linkage? [AIPMT (Prelims)-2008]
(1) Down syndrome - 44 autosomes + XO (2) Klinefelter syndrome - 44 autosomes + XXY (3) Colour blindness - Y-linked (4) Erythroblastosis foetalis - X-linked
Sol. Answer (2)
44. In the hexaploid wheat, the haploid (n) and basic (x) numbers of chromosomes are [AIPMT (Prelims)-2007] (1) n = 21 and x = 7 (2) n = 7 and x = 21 (3) n = 21 and x = 21 (4) n = 21 and x = 14 Sol. Answer (1)
n = 7 6n = 42 n = 21
x 7 (Monoploid)
45. Inheritance of skin colour in humans is an example of [AIPMT (Prelims)-2007]
(1) Codominance (2) Chromosomal aberration
(3) Point mutation (4) Polygenic inheritance
Sol. Answer (4)
46. A common test to find the genotype of a hybrid is by [AIPMT (Prelims)-2007] (1) Crossing of one F1progeny with male parent (2) Crossing of one F2 progeny with male parent (3) Crossing of one F2 progeny with female parent (4) Studying the sexual behaviour of F1 progenies Sol. Answer (1)
Tt × tt or +
47. Two genes R and Y are located very close on the chromosomal linkage map of maize plant. When RRYY and rryy genotypes are hybridized, the F2 segregation will show [AIPMT (Prelims)-2007] (1) Higher number of the parental types (2) Higher number of the recombinant types
(3) Segregation in the expected 9: 3: 3: 1 ratio (4) Segregation in 3:1 ratio Sol. Answer (1)
48. In pea plants, yellow seeds are dominant to green. If a heterozygous yellow seeded plant is crossed with a green seeded plant, what ratio of yellow and green seeded plants would you expect in F1 generation?
[AIPMT (Prelims)-2007] (1) 3 : 1 (2) 50 : 50 (3) 9 : 1 (4) 1 : 3 Sol. Answer (2) Yy × yy Yy : yy 50 50
49. A human male produces sperms with genotypes AB, Ab, aB and ab pertaining to two diallelic characters in equal proportions. What is the corresponding genotype of this person? [AIPMT (Prelims)-2007]
(1) AABB (2) AaBb (3) AaBB (4) AABb
Sol. Answer (2) AaBb
AB, Ab, aB, ab
50. Which one of the following is the most suitable, medium for culture of Drosophila melanogaster ?
[AIPMT (Prelims)-2006]
(1) Moist bread (2) Agar agar (3) Ripe banana (4) Cow dung
Sol. Answer (3)
Cytoplasmic or mitochondrial inheritance is from mother side.
51. Phenotype of an organism is the result of [AIPMT (Prelims)-2006]
(1) Mutations and linkages (2) Cytoplasmic effects and nutrition (3) Environmental changes and sexual dimorphism (4) Genotype and environment interactions Sol. Answer (4)
52. In which mode of inheritance do you expect more maternal influence among the offspring ?
[AIPMT (Prelims)-2006]
(1) Autosomal (2) Cytoplasmic (3) Y-linked (4) X-linked
Sol. Answer (2)
53. How many different kinds of gametes will be produced by a plant having the genotype AABbCC?
[AIPMT (Prelims)-2006]
(1) Three (2) Four (3) Nine (4) Two
Sol. Answer (4) AAB bCC 2n 21 2
54. Which one of the following is an example of polygenic inheritance? [AIPMT (Prelims)-2006] (1) Flower colour in Mirabilis jalapa (2) Production of male honey bee
(3) Pod shape in garden pea (4) Skin colour in humans
55. In Mendel’s experiments with garden pea, round seed shape (RR) was dominant over wrinkled seeds (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? [AIPMT (Prelims)-2006]
(1) Only round seeds with green cotyledons (2) Only wrinkled seeds with yellow cotyledons (3) Only wrinkled seeds with green cotyledons
(4) Round seeds with yellow cotyledons and wrinkled seeds with yellow cotyledons Sol. Answer (4)
RRYY × rryy RY ry
RrYy
R Y Round y Round greenyellow
r Y
Y Wrinkled
56. Test cross involves [AIPMT (Prelims)-2006]
(1) Crossing between two genotypes with recessive trait (2) Crossing between two F1 hybrids
(3) Crossing the F1 hybrid with a double recessive genotype (4) Crossing bet
Sol. Answer (3)
57. If a colourblind woman marries a normal visioned man, their sons will be [AIPMT (Prelims)-2006] (1) All normal visioned
(2) One-half colourblind and one-half normal (3) Three-fourths colourblind and one-fourth normal (4) All colourblind
Sol. Answer (4)
58. Cri-du-chat syndrome in humans is caused by the [AIPMT (Prelims)-2006]
(1) Fertilization of an XX egg by a normal Y-bearing sperm (2) Loss of half of the short arm of chromosome 5
(3) Loss of half of the long arm of chromosome 5 (4) Trisomy of 21st chromosome
Sol. Answer (2)
59. 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 are affected. Which of the following mode of inheritance do you suggest for this disease?
[AIPMT (Prelims)-2005] (1) Autosomal dominant (2) Sex-linked dominant (3) Sex-limited recessive (4) Sex-linked recessive Sol. Answer (4)
XXc × XY XCY or XX
60. At a particular locus, frequency of ‘A’ allele is 0.6 and that of ‘a’ is 0.4. What would be the frequency of heterozygotes in a random mating population at equilibrium? [AIPMT (Prelims)-2005]
(1) 0.16 (2) 0.48 (3) 0.36 (4) 0.24
Sol. Answer (2)
61. A woman with normal vision, but whose father was colour blind, marries a colourblind man. Suppose that the
fourth child of this couple was a boy. This boy: [AIPMT (Prelims)-2005]
(1) Must have normal colour vision
(2) Will be partially colourblind since he is heterozygous for the colourblind mutant allele (3) Must be colourblind
(4) May be colourblind or may be of normal vision Sol. Answer (4)
XXC X XCY
XCY or XX
62. Haemophilia is more commonly seen in human males than in human females because
[AIPMT (Prelims)-2005] (1) This disease is due to an X-linked dominant mutation
(2) A greater proportion of girls die in infancy
(3) This disease is due to an X-linked recessive mutation (4) This disease is due to a Y-linked recessive mutation Sol. Answer (3)
63. A woman with 47 chromosomes due to three copies of chromosome 21 is characterized by
[AIPMT (Prelims)-2005]
(1) Down syndrome (2) Triploidy (3) Turner syndrome (4) Super femaleness
Sol. Answer (1)
2n + 1 21 chromosome
64. In order 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 (Prelims)-2005]
(1) aaBB (2) AaBb (3) AABB (4) aabb
Sol. Answer (4) AaBb × aabb Dihybrid test cross
65. Which of the following is not a hereditary disease? [AIPMT (Prelims)-2005]
(1) Cretinism (2) Cystic fibrosis (3) Thalassaemia (4) Haemophilia
Sol. Answer (1)
66. The salivary gland chromosomes in the dipteran larvae, are useful in gene mapping because
[AIPMT (Prelims)-2005] (1) These are much longer in size (2) These are easy to stain
(3) These are fused (4) They have endoreduplicated chromosomes
67. Genetic variation in a population arises due to
(1) Mutations only (2) Recombination only
(3) Mutations as well as recombination (4) Reproductive isolation and selection Sol. Answer (3)
68. Which one is the incorrect statement with regards to the importance of pedigree analysis? (1) It helps to trace the inheritance of a specific trait
(2) It confirms that DNA is the carrier of genetic information
(3) It helps to understand whether the trait in question is dominant or recessive (4) It confirms that the trait is linked to one of the autosome
Sol. Answer (2)
It is used to study inheritance of character.
69. In our society women are blamed for producing female children. Choose the correct answer for the sex-determination in humans
(1) Due to some defect in the women (2) Due to some defect like aspermia in man
(3) Due to the genetic make up of the particular sperm which fertilizes the egg (4) Due to the genetic make up of the egg
Sol. Answer (3)
Next generation sex is determined by male. 70. Down’s syndrome in humans is due to
(1) Two ‘Y’ chromosomes (2) Three ‘X’ chromosomes
(3) Three copies of chromosome 21 (4) Monosomy
Sol. Answer (3)
2n + 1 21 Chromosome
71. The variation/difference in the offsprings of a species from their parents constitutes an important component of
(1) Genetics (2) Speciation (3) Species fixation (4) Heredity
Sol. Answer (1)
72. If two pea plants having red (dominant) coloured flowers with unknown genotypes are crossed, 75% of the flowers are red and 25% are white. The genotypic constitution of the parents having red coloured flowers will be
(1) Both homozygous (2) One homozygous and other heterozygous
(3) Both heterozygous (4) Both hemizygous
Sol. Answer (3)
Rr × Rr RR : Rr : rr Red Red White
73. Walter Sutton is famous for his contribution to
(1) Genetic engineering (2) Totipotency
(3) Quantitative genetics (4) Chromosomal theory of inheritance
74. A polygenic trait is controlled by 3 genes A, B and C. In a cross AaBbCc × AaBbCc, the phenotypic ratio of the offsprings was observed as
1 : 6 : x : 20 : x : 6 : 1 What is the possible value of x?
(1) 3 (2) 9 (3) 15 (4) 25
Sol. Answer (3)
Polygenic inheritance – 3 genes 1 : 6 : 15 : 20 : 15 : 6 : 1
75. The chromosome constitution 2n – 2 of an organism represents
(1) Monosomic (2) Nullisomic (3) Haploid (4) Trisomic
Sol. Answer (2)
Nullisomic 2n – 2
76. Mendel's principle of segregation means that the germ cells always receive
(1) One pair of alleles (2) One quarter of the genes
(3) One of the paired alleles (4) Any pair of alleles
Sol. Answer (3)
77. Absence of one sex chromosome causes
(1) Turner's syndrome (2) Klinefelter's syndrome
(3) Down's syndrome (4) Tay-Sach's syndrome
Sol. Answer (1) AA – XO
Absence of X chromosome. 78. Chimera is produced due to
(1) Somatic mutations (2) Reverse Mutations
(3) Lethal mutations (4) Pleiotropic mutations
Sol. Answer (1)
Mutation in different tissue.
79. Haploids are more suitable for mutation studies than the diploids. This is because (1) All mutations, whether dominant or recessive are expressed in haploids (2) Haploids are reproductively more stable than diploids
(3) Mutagens penetrate in haploids more effectively than diploids (4) Haploids are more abundant in nature than diploids
Sol. Answer (1)
80. Which one of the following conditions in humans is correctly matched with its chromosomal abnormality/ linkage?
(1) Down’s syndrome – 44 autosomes + XO (2) Klinefelter’s syndrome – 44 autosomes + XXY (3) Colour blindness – Y-linked (4) Erythroblastosis foetalis – X-linked
Sol. Answer (2)
Trisomy of X-Chromosome AA + X X Y
81. The genes, which remain confined to differential region of Y-chromosome, are
(1) Autosomal genes (2) Holandric genes
(3) Completely sex-linked genes (4) Mutant genes
Sol. Answer (2)
82. The colour blindness is more likely to occur in males than in females because (1) The Y-chromosome of males have the genes for distinguishing colours (2) Genes for characters are located on the X-chromosomes
(3) The trait is dominant in males and recessive in females (4) None of these
Sol. Answer (2)
X Linked character and carry one X So it will give its expression.
83. Albinism is a congenital disorder resulting from the lack of the enzyme
(1) Tyrosinase (2) Xanthine oxidase (3) Catalase (4) Fructokinase
Sol. Answer (1)
Loss of pigment caused by Tyrosinase.
84. An abnormal human male phenotype involving an extra Y-chromosome (XYY) is a case of
(1) Edward’s syndrome (2) Jacob syndrome (3) Intersex (4) Down’s syndrome Sol. Answer (2)
Trisomy of Y-Chromosome
85. The phenomenon, in which an allele of one gene suppresses the activity of an allele of another gene, is known as
(1) Epistasis (2) Dominance (3) Suppression (4) Inactivation
Sol. Answer (1)
86. Barr body in mammals represents
(1) All the heterochromatin in male and female cells (2) The Y-chromosome in somatic cells of male (3) All the heterochromatin in female cells
(4) One of the two X-chromosomes in somatic cells of females Sol. Answer (4)
One of X-chromosome change into heterochromatin in +.
87. When two dominant independently assorting genes react with each other producing effect jointly they are called
(1) Collaborative genes (2) Complementary genes
(3) Duplicate genes (4) Supplementary genes
Sol. Answer (2)
88. A genetically diseased father (male) marries with a normal female and gives birth to 3 carrier girls and 5 normal sons. It may be which type of genetical disease?
(1) Sex-influenced disease (2) Blood group inheritance disease
(3) Sex-linked disease (4) Sex-limited disease
Sol. Answer (3) XDY × XY
5 Normal son (XY), 3(XXD) carrier daughter
89. A person whose father is colour blind marries a lady whose mother is daughter of a colour blind man. Their children will be
(1) All sons colour blind
(2) Some sons normal and some daughters colour blind (3) All sons and daughters colour blind
(4) All daughter normal Sol. Answer (4)
XY × XX / X XC All normal daughter
90. In which of the following disease, the individual has one less X-chromosome?
(1) Turner’s syndrome (2) Klinefelter’s syndrome
(3) Bleeder’s disease (4) Down’s syndrome
Sol. Answer (1)
AA – XO (2n – 1) Monosomic
91. H.J. Muller had received Nobel Prize for
(1) His studies on Drosophila for genetic study (2) Proving that the DNA is a genetic material (3) Discovering the linkage of genes (4) Discovering the induced mutations by X-rays Sol. Answer (4)
92. The polygenic genes show
(1) Different karyotypes (2) Different genotypes (3) Different phenotypes (4) None of these Sol. Answer (3)
Skin shade of human being.
93. Foetal sex can be determined by examining cells from the amniotic fluid by looking for
(1) Chiasmata (2) Kinetochore (3) Barr bodies (4) Autosomes
Sol. Answer (3)
+ contain 1 barr body
94. A fruit fly is hemizygous for sex-linked genes, mated with normal female fruit fly, the males specific chromosome will enter egg cell in the proportion
(1) 3 : 1 (2) 7 : 1 (3) 1 : 1 (4) 2 : 1
Sol. Answer (3) XAY × XX
XAX × XY
1 : 1
95. Genetic identity of a human male is determined by
(1) Sex-chromosome (2) Cell organelles (3) Autosome (4) Nucleolous
Sol. Answer (1) Y-Chromosome
96. Different forms of a gene located at the same locus of chromosomes are called
(1) Multiple alleles (2) Polygenes (3) Oncogenes (4) None of these
Sol. Answer (1)
They are mutated genes occupy same gene locus on homologous chromosome.
97. After crossing two plants, the progenies are found to be male sterile. This phenomenon is found to be maternally inherited and is due to some genes which reside in
(1) Mitochondria (2) Cytoplasm (3) Nucleus (4) Chloroplast
Sol. Answer (1)
98. Albinism is known to be due to an autosomal recessive mutation. The first child of a couple with normal skin pigmentation was an albino. What is the probability that their second child will also be an albino?
(1) 50% (2) 75% (3) 100% (4) 25% Sol. Answer (4) AA : AA : A Aa a a 1 2 1 25% AAa × AAa
99. How many different types of genetically different gametes will be produced by a heterozygous plant having the genotype AABbCc?
(1) Six (2) Nine (3) Two (4) Four
Sol. Answer (4) AA Bb Cc 3n 32 9
100. When a single gene influences more than one traits it is called
(1) Pseudodominance (2) Pleiotropic (3) Epistasis (4) None of these
Sol. Answer (2)
101. Mental retardation in man, associated with sex chromosomal abnormality is usually due to (1) Moderate increase in Y complement (2) Large increase in Y complement
(3) Reduction in X complement (4) Increase in X complement
Sol. Answer (4)
Increase in X complement
102. Loss of a X-chromosome in a particular cell, during its development, results into
(1) Gynandromorphs (2) Meta female (3) Triploid individual (4) Myotonic dystrophy Sol. Answer (1)
Which contain both and + character.
103. If Mendel had studied the seven traits using a plant with 12 chromosomes instead of 14, in what way would his interpretation have been different?
(1) He would not have discovered the law of independent assortment (2) He would have discovered sex linkage
(3) He could have mapped the chromosome
(4) He would have discovered blending or incomplete dominance Sol. Answer (1)
104. A woman with two genes for haemophilia and one gene for colour blindness on one of the ‘X’ chromosomes marries a normal man. How will the progeny be?
(1) 50% haemophilic colour-blind sons and 50% haemophilic sons (2) 50% haemophilic daughters and 50% colour blind daughters (3) All sons and daughters haemophilic and colour-blind
(4) Haemophilic and colour-blind daughters Sol. Answer (1)
XhXC
105. In human beings, multiple genes are involved in the inheritance of
(1) Sickle cell anaemia (2) Skin colour
(3) Colour blindness (4) Phenylketonuria
Sol. Answer (2)
Polygenic inheritance
106. Haemophilic man marries a normal woman. Their offsprings will be
(1) All haemophilic (2) All boys haemophilic (3) All girls haemophilic (4) All normal Sol. Answer (4)
XhY × XX
All normal
107. A marriage between normal visioned man and colour blind woman will produce offspring
(1) Colour blind sons and 50% carrier daughters (2) 50% colourblind sons and 50% carrier daughters (3) Normal males and carrier daughters (4) Colour blind sons and carrier daughters
Sol. Answer (4) XY × XCXC
108. In hybridization, Tt × tt gives rise to the progeny in the ratio (1) 2 : 1 (2) 1 : 2 : 1 (3) 1 : 1 (4) 1 : 2 Sol. Answer (3) Tt × tt Tt : tt 1 : 1
109. According to Mendelism, which character shows dominance?
(1) Terminal position of flower (2) Green colour in seed coat
(3) Wrinkled seeds (4) Green pod colour
Sol. Answer (4)
110. Due to the cross between TTRr × ttrr the resultant progenies show what percent of tall, red flowered plants?
(1) 50% (2) 75% (3) 25% (4) 100% Sol. Answer (1) TTRrr × ttrr TR Tr × br TtRrTR ttrrTr tr 1 1
111. In Drosophila, the XXY condition leads to femaleness whereas in human beings the same condition leads to Klienfelter’s syndrome in male. It proves
(1) In human beings, Y chromosome is active in sex determination
(2) Y chromosome is active in sex determination in both human beings and Drosophila (3) In Drosophila, Y chromosome decides femaleness
(4) Y chromosome of man has genes for syndrome Sol. Answer (1)
112. Independent assortment of genes does not take place when (1) Genes are located on homologous chromosomes (2) Genes are linked and located on same chromosome (3) Genes are located on non-homogenous chromosomes (4) All of these
Sol. Answer (2)
113. Mendel obtained wrinkled seeds in pea due to deposition of sugars instead of starch. It was due to which enzyme?
(1) Amylase (2) Invertase
(3) Diastase (4) Absence of starch branching enzyme
Sol. Answer (4)
114. Ratio of complementary genes is
(1) 9 : 3 : 4 (2) 12 : 3 : 1 (3) 9 : 3 : 3 : 4 (4) 9 : 7
Sol. Answer (4) 9 : 7
115. When both parental alleles are expressed together, it is called
(1) Co-dominance (2) Dominance (3) Incomplete dominance (4) Pseudodominance Sol. Answer (1)
116. A and B genes are linked. What shall be genotype of progeny in a cross between AB/ab and ab/ab?
(1) AAbb and aabb (2) AaBb and aabb (3) AABB and aabb (4) aaBB × Aabb
Sol. Answer (2) A B ab × ab ab A B ab 1 a b ab A B ab a b : 1
117. Probability of four sons to a couple is
(1) 1/4 (2) 1/8 (3) 1/16 (4) 1/32
Sol. Answer (3)
1 1 1 1 1
2 2 2 2 ⇒16
118. If recombination frequency between AB genes is 20% and BC gene is 40% and interference is 30% in the case of double cross over then what will be coincidance under this condition?
(1) 2.4 (2) 8 (3) 5.6 (4) 0.7
Sol. Answer (4) C + I = 1 0.3 + 0.7 = 1
119. Male XX and female XY sometime occur due to
(1) Deletion (2) Transfer of segments in X and Y chromosome
(3) Aneuploidy (4) Hormonal imbalance
Sol. Answer (2)
120. Number of Barr body in XXXX female is
(1) 1 (2) 2 (3) 3 (4) 4
Sol. Answer (3) X X X X
Number of X-chromosome – 1 4 – 1 3
121. Extranuclear inheritance occurs in
(1) Killer Paramecium (2) Killer Amoeba (3) Euglena (4) Hydra
122. Which of the following is correct match?
(1) Down’s syndrome - 21st chromosome (2) Sickle cell anaemia – X-chromosome
(3) Haemophilia – Y-chromosome (4) Parkinson’s disease – X & Y chromosome Sol. Answer (1)
123. How many genome types are present in a typical green plants cell?
(1) More than five (2) More than ten (3) Two (4) Three
Sol. Answer (3) 2n
124. Which of the following is an example of sex linked disease?
(1) AIDS (2) Colour blindness (3) Syphilis (4) Gonorrhoea
Sol. Answer (2) AIDS – Viral
Syphilis, Gonorrhoea Spirochetes
125. Which of the following is an example of pleiotropy?
(1) Haemophilia (2) Thalassemia (3) Sickle cell anaemia (4) Colour blindness Sol. Answer (3)
One gene control multiple feature. 126. A gene is said to be dominant if
(1) It expresses its effect only in homozygous state (2) It expresses its effect only in heterozygous condition
(3) It expresses its effect both in homozygous and heterozygous condition (4) It never expresses it’s effect in any condition
Sol. Answer (3) Tt Tall
127. On selfing a plant of F1-generation with genotype ‘’AABbCC’’, the genotypic ratio in F2-generation will be
(1) 1 : 2 : 1 (2) 1 : 1
(3) 9 : 3 : 3 : 1 (4) 27 : 9 : 9 : 9 : 3 : 3 : 3 : 1
Sol. Answer (1) AABbCC
ABC
2 type gamete AbC 1 : 2 : 1
AABBCC AABbCC AABbCC AAbbCC ABC ABC ABC AbC
AABBCC : AABbCC : AAbbCC
128. A diseased mans marries a normal woman. They get three daughters and five sons. All the daughters were diseased and sons were normal. The gene of this disease is
(1) Sex linked dominant (2) Sex linked recessive (3) Sex limited character (4) Autosomal dominant Sol. Answer (1)
XDY × XX
(3. ) X X : XY (5 Sons)D +
+
129. Down’s syndrome is caused by an extra copy of chromosome number 21. What percentage of offspring is produced by an affected mother and a normal father?
(1) 100% (2) 75% (3) 50% (4) 25%
Sol. Answer (3)
130. Which one of the following discoveries resulted in a Nobel Prize?
(1) X-rays induce sex-linked recessive lethal mutations (2) Cytoplasmic inheritance (3) Recombination of linked genes (4) Genetic engineering Sol. Answer (1)
131. Two crosses between the same pair of genotypes or phenotypes in which the sources of the gametes are reversed in one cross, is known as
(1) Test cross (2) Reciprocal cross (3) Dihybrid cross (4) Reverse cross Sol. Answer (2)
+
132. The genes controlling the seven pea characters studied by Mendel are now known to be located on how many different chromosomes?
(1) Seven (2) Six (3) Five (4) Four
Sol. Answer (4) 1, 4, 5, 7
133. Which one of the following traits of garden pea studied by Mendel was a recessive feature ?
(1) Axial flower position (2) Green seed colour (3) Green pod colour (4) Round seed shape Sol. Answer (2)
134. Moustaches, beard and horseness in voice in human males are examples of
(1) Sex-linked traits (2) Sex limited traits
(3) Sex differentiating traits (4) Sex determining traits Sol. Answer (2)
135. In Drosophila, the sex is determined by
(1) The ratio of number of X chromosomes to the sets of autosomes (2) X and Y chromosomes
(3) The ratio of X-chromosomes to the pairs of autosomes (4) Whether the egg is fertilized or develops parthenogenetically Sol. Answer (1)
Number of X-chromosome Set of autosomes
136. One of the parents of a cross has a mutation in its mitochondria. In that cross, that parent is taken as a male. During segregation of F2 progenies that mutation is found in
(1) One-third of the progenies (2) None of the progenies
(3) All the progenies (4) Fifty percent of the progenies
Sol. Answer (2)
Mitochondrial inheritance is cytoplasmic inheritance
137. Lack of independent assortment of two genes A and B in fruit fly Drosophila is due to
(1) Repulsion (2) Recombination (3) Linkage (4) Crossing over
Sol. Answer (3)
Two genes are present on same chromosome.
138. What kind of evidence suggested that man is more closely related with chimpanzee than with other hominoid apes?
(1) Evidence from DNA from sex chromosomes only (2) Comparison of chromosomes morphology only
(3) Evidence from fossil remains, and the fossil mitochondrial DNA alone (4) Evidence from DNA extracted from sex chromosomes, autosomes Sol. Answer (4)
139. The recessive genes located on X-chromosome of humans are always
(1) Lethal (2) Sub-lethal
(3) Expressed in males (4) Expressed in females
Sol. Answer (3)
Hemizygous condition in XCY
140. 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?
(1) 1/8 (2) 1/32 (3) 1/16 (4) 1/4
Sol. Answer (1) Aa Bb XhY
1 1 1 1
2 2 2 ⇒8
141. A self-fertilizing trihybrid plant forms
(1) 8 different gametes and 64 different zygotes (2) 4 different gametes and 16 different zygotes (3) 8 different gametes and 16 different zygotes (4) 8 different gametes and 32 different zygotes Sol. Answer (1)
142. There are three genes a, b, c. Percentage of crossing over between a and b is 20%, b and c is 28% and a and c is 8%. What is the sequence of genes on chromosome?
(1) b, a, c (2) a, b, c (3) a, c, b (4) None of these
Sol. Answer (1)
b 20 a 8 c
28% bac
143. The linkage map of X-chromosome of fruit fly has 66 units, with yellow body gene (y) at one end and bobbed hair (b) gene at the other end. The recombination frequency between these two genes (y and b) should be
(1) 66% (2) > 50% (3) 50% (4) 100%
Sol. Answer (3)
y b
Recombination frequency will be less than 50%.
144. In a plant, red fruit (R) is dominant over yellow fruit (r) and tallness (T) is dominant over shortness (t). If a plant with RRTt genotype is crossed with a plant that is rrtt,
(1) 25% will be tall with red fruit (2) 50% will be tall with red fruit
(3) 75% will be tall with red fruit (4) All the offsprings will be tall with red fruit Sol. Answer (2) RRTt × rrtt RT Rt rt RT RrTtRrttRt rt + 50 50
145. A normal woman, whose father was colour blind is married to a normal man. The sons would be (1) 75% colour blind (2) 50% colour blind (3) All normal (4) All colour blind Sol. Answer (2)
XXC × XY
X Y, XYC Colour blind
146. De Vries gave his mutation theory on organic evolution while working on
(1) Pisum sativum (2) Drosophila melanogaster
(3) Oenothera lamarckiana (4) Althea rosea
Sol. Answer (3)
147. Triticale, the first man-made cereal crop, has been obtained by crossing wheat with
(1) Barley (2) Rye (3) Pearl millet (4) Sugarcane
Sol. Answer (2)
Rye Scalea cereals
148. Normally DNA molecule has A-T, G-C pairing. However, these bases can exist in alternative valency status owing to rearrangements called
(1) Frame-shift mutation (2) Tautomerisational mutation
(3) Analog substitution (4) Point mutation
Sol. Answer (2)
A T Tautomerisational mutation G C Tautomerisational mutation A T Amino Imino group
149. The most striking example of point mutation is found in a disease called
(1) Down’s syndrome (2) Sickle cell anaemia (3) Edward syndrome (4) Night blindness Sol. Answer (2)
150. Identify the one, which causes gene mutation
(1) Granoson (2) Colchicine (3) Crossing over (4) X-rays
Sol. Answer (4) Ionised radiation
151. The mutations are mainly responsible for
(1) Increasing the population rate (2) Maintaining genetic continuity
(3) Constancy in organisms (4) Variation in organisms
Sol. Answer (4)
152. Which of the following is the main category of mutation?
(1) Somatic mutation (2) Genetic mutation (3) Heterosis (4) None of these Sol. Answer (2)
153. Change in sequence of nucleotide in DNA is called
(1) Mutagen (2) Mutation (3) Recombination (4) Translation
Sol. Answer (2)
154. When a cluster of genes show linkage behaviour they
(1) Do not show a chromosome map (2) Show recombination during meiosis (3) Do not show independent assortment (4) Induce cell division
Sol. Answer (3)
Linked genes will not show independent assortment. 155. Genetic map is one that
(1) Establishes sites of the genes on a chromosome (2) Establishes the various stages in gene evolution (3) Shows the stages during the cell division
(4) Shows the distribution of various species in a region Sol. Answer (1)
Linear arrangement of gene on chromosome.
156. In mutational event, when adenine is replaced by guanine, it is a case of
(1) Frame shift mutation (2) Transcription (3) Transition (4) Transversion Sol. Answer (3)
Replaced by
AG Transition
157. The most likely reason for the development of resistance against pesticides in insects damaging a crop is
(1) Random mutations (2) Genetic recombination
(3) Directed mutations (4) Acquired heritable changes
Sol. Answer (1)
158. When two genetic loci produce identical phenotypes in cis and trans position, they are considered to be
(1) Multiple alleles (2) The parts of same gene
(3) Pseudoalleles (4) Different genes
Sol. Answer (3)
When two adjacent genes occupy different gene locus but performer regulate same function. 159. What base is responsible for hot spots for spontaneous point mutations?
(1) 5-bromouracil (2) 5-methylcytosine (3) Guanine (4) Adenine
Sol. Answer (2)
160. Nucleus of a donor embryonal cell/somatic cell is transferred to an enucleated egg cell. Then after the formation of organism, what shall be true?
(1) Organism will have extranuclear genes of the donor cell (2) Organism will have extranuclear genes of recipient cell
(3) Organism will have extranuclear genes of both donor and recipient cell (4) Organism will have nuclear genes of recipient cell
Sol. Answer (2)
Organism contain Extra chromosome DNA of recepient cell
161. Genes for cytoplasmic male sterility in plants are generally located in
(1) Chloroplast genome (2) Mitochondrial genome (3) Nuclear genome (4) Cytosol Sol. Answer (2)
162. Extranuclear inheritance is the consequence of presence of genes in
(1) Mitochondria and chloroplasts (2) Endoplasmic reticulum and mitochondria
(3) Ribosomes and chloroplast (4) Lysosomes and ribosomes
Sol. Answer (1)
SECTION - D
Assertion-Reason Type Questions 1. A : Turner's syndrome generally does not occur in males.
R : Foetus with 44 + YO complement generally dies. Sol. Answer (1)
Because absence of X chromosome
2. A : Sickel cell anaemia occurs due to the point mutation. R : mRNA produced from Hb(s) gene has GAG instead of GUG. Sol. Answer (3)
GUG Valine GAG Glutamine
3. A : Holandric traits are passed from one generation to the next generation. R : These traits appear more frequently in one sex than in other.
Sol. Answer (3)
Hoalandric genes are Y linked.
4. A : Dominance is not an autonomous feature of a gene.
R : It depends as much on the gene product and the production of a particular phenotype from this product. Sol. Answer (1)
5. A : The posssibility of a female becoming a haemophilic is extremely rare. R : Mother of such a female has to be carrier and father should be haemophilic. Sol. Answer (1)
It is X linked recessive
6. A : Polyploids with odd number of chromosomes are propagated vegetatively. R : Seed formation is absent due to meiotic abnormality.
Sol. Answer (1)
7. A : Pseudoalleles are actually closely linked genes.
R : These can be identified easily as both affect different characters, providing separate phenotypes. Sol. Answer (3)
These genes occupy different location on a chromosome but they regulate same feature. 8. A : The heterozygotic female for haemophilia may transmit the disease to sons.
R : Such traits show criss-cross inheritance. Sol. Answer (2)
9. A : Non-allosomic genic determination of sex is found in bacteria. R : Sex is dependent on some environmental factors in prokaryotes. Sol. Answer (3)
Sex is regulated by F-plasmid or fertility genes.
10. A : Crossing over is exchange of genetic material between non-homologous chromosomes. R : It produces new linkages.
Sol. Answer (4)
Exchange of genetic material between homologous chromosome.
11. A : Mendel gave postulates like "principles of segregation and principles of independent assortment" after studying seven pairs of contrasting traits in garden pea.
R : He was lucky in selecting seven characters in pea that were located on seven different chromosomes. Sol. Answer (3)
12. A : Test cross is the tool for knowing linkage between genes. R : Monohybrid test cross gives two phenotypes and two genotypes. Sol. Answer (2)
13. A : Myotonic dystrophy is caused by recessive mutant pleiotropic gene. R : Gene mutation leads to more synthesis of fibrillin proteins.
Sol. Answer (4)
14. A : In snapdragon, F1 plants do not have red or white flowers.
R : It is intermediate inheritance with neither of the two alleles of a gene being dominant over each other. Sol. Answer (1)
Because of incomplete dominance they allele cannot give its expression completely in figuration. 15. A : en block inheritance of all genes located on the same chromosome may occur in some organisms.
R : Dihybrid test cross will have only two phenotypes. Sol. Answer (1)
16. A : Morgan's cross III was conducted in Drosophila to locate genes on chromosome for white eye colour. R : The cross was done between red eyed hybrid female and white eyed male.
Sol. Answer (3)
It was a reciprocal cross so + white x red eye. 17. A : Antlers in male deer are sex influenced traits.
R : These are controlled by autosomal genes which are influenced by the sex of bearer. Sol. Answer (4)
Antler of deer is sex limited feature.
18. A : One drum stick per nucleus is present in the neutrophil of normal female. R : It is absent in the neutrophil of male.
Sol. Answer (2)
19. A : Blood group phenotype is controlled by presence or absence of antigens present on surface coating of RBC.
R : These antigens are of three types and found in the oligosaccharides rich head regions of a glycophorin. Sol. Answer (2)
20. A : XO type sex determination is found in large number of insects. R : Some of the sperms bear the X-chromosome whereas some do not. Sol. Answer (2)
XX – XO – Type is found in some insects.
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