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Inheritance And Variation Mcq
Quiz 16
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Q.1
on Q356) Which statement is incorrect about linkage ?
0%
a) It helps in maintaining the valuable traits of new varieties.
0%
b) It helps in forming new recombinants.
0%
c) Knowledge of linkage helps the breeder to combine all desirable traits in a single variety.
0%
d) It helps in locating genes on chromosome.
Explanation
Answer:(b)
Q.2
on Q357) Which among the following procedure during artificial cross pollination/hybridisation fixes the paternity of the cross?
0%
a) Bagging.
0%
b) Emasculation.
0%
c) Selection of pollen grains.
0%
d) Dusting the stigma with pollen grain.
Explanation
The stigma is first protected from foreign pollens by covering it with bag. This technique is called bagging. The pollen, at the dehiscent stage, is brought from the plant to be used as male parent, and is dusted on stigma of the emasculated flower. Dusting ensures that desired paternal trait is crossed. Answer:(a)
Q.3
on Q358) Which one of the following conditions of the Zygotic cell would lead to the birth of a normal human female child?
0%
a) Two X Chromosomes
0%
b) Only One Y Chromosomes
0%
c) Only One X Chromosomes
0%
d) One X chromosome and One Y chromosome
Explanation
Answer:(a)
Q.4
on Q359) Inheritance of skin colour in humans is an example of AIPMT - 2007
0%
a) Point Mutation
0%
b) Polygenic inheritance
0%
c) Codominance
0%
d) Chromosomal aberrations
Explanation
Skin colour, human height and eye colour show polygenic inheritance Answer:(b)
Q.5
on Q360) Which one of the following is an example of polygenic inheritance ?
0%
a) Skin colour in humans
0%
b) Flower colour in Miralibilis jalapa
0%
c) Production of male honey bee
0%
d) Pod shape in garden pea
Explanation
Characters
Expected ratios
Example
Monohybrid cross
Phenotypic = 3 : 1
Genotypic = 1 :2 : 1
Pisum sativum
Incomplete dominance
1 : 2 : 1
Mirabilis jalapa
Skin colour, human height and eye colour show polygenic inheritance Answer:(a)
Q.6
on Q361) Which one of the following traits of garden pea studied by Mendel was a recessive feature?
0%
a) Axial flower position
0%
b) Green Seed colour
0%
c) Green pod colour
0%
d) Round Seed shape
Explanation
Traits
Dominant
Recessive
Plant height
Tall (1.2 – 2.0m)(T)
Dwarf (0.25 – 0.5m) (t)
Flower position
Axillary (A)
Terminal (a)
Pod colour
Green (G) or (Y)
Yellow (g) or (y)
Pod shape
Full or inflated (I) or (C)
Constricted (i) or (c)
Flower colour
Violet (V) or (W)
White (v) or (w)
Seed shape
Round (R) or (W)
Wrinkled (r) or (w)
Seed colour
Yellow (Y) or (G)
Green (y) or (g)
Answer:(b)
Q.7
on Q362) What does the following pedigree chart represent ?
0%
a) Pedigree of sex - influenced disorder
0%
b) Pedigree of sex - linked disorder
0%
c) Pedigree of polydactylous in man
0%
d) Pedigree of gene mutation
Explanation
Answer:(c)
Q.8
on Q363) Name the disorder caused by trisomy of sex chromosomes.
0%
a) Down’s syndrome
0%
b) Kline felter’s syndrome
0%
c) Turner’s syndrome
0%
d) Edward’s syndrome
Explanation
Answer:(b)
Q.9
on Q364) In this group of plants male is heterogametic and female is homogametic type
0%
a) Gymnosperms
0%
b) Bryophytes
0%
c) pteridophytes
0%
d) angiosperms
Explanation
Answer:(d)
Q.10
on Q365) Crossing over during meiosis occurs between
0%
a) sister chromatids
0%
b) Non sister chromatids
0%
c) Centromeres
0%
d) Non homologous chromosomes
Explanation
Answer:(b)
Q.11
on Q366) If there were only parental combinations in F2 of a dihybrid cross then Mendel might have discovered ?
0%
a) Independent assortment
0%
b) Atavism
0%
c) Linkage
0%
d) Repulsion
Explanation
In the F2 generation, ther are two types of progenies, one which resembles parental combination and another which differes from parental combination i.e., recombinations. Recombinations occurs because alleles of both the trait segregate and assorts independently and is random. If only parental combination were found in F2 that would suggest that genes are tightly linked and Mendel might have discovered linkage. Answer:(c)
Q.12
on Q367) A change in number of chromosomes is:
0%
a) Polyploidy.
0%
b) Euploidy.
0%
c) Diploidy.
0%
d) Chromosomal mutation.
Explanation
Answer:(d)
Q.13
on Q368) Which among the following character is an example of Incomplete dominance?
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a) Flower colour of sweet pea.
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b) Flower colour of garden pea.
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c) Flower colour of Snapdragon.
0%
d) Pod colour of Garden pea.
Explanation
The inheritance of flower colour in the dog flower (snapdragon or Antirrhinum sp.), Mirabilis jalapa (four o’clock flower) is a good example to understand incomplete dominance. Answer:(c)
Q.14
on Q369) Identify the wrong statement:
0%
a) Transition and transversion leads to Frame shift mutation.
0%
b) Reading frame of mRNA is changed during Frame shift mutation.
0%
c) Deletion, addition, inversion and translocation leads to Frame shift mutation.
0%
d) Reciprocal translocation is the reason for Philadelphia disease.
Explanation
Answer:(a)
Q.15
on Q370) Identify the wrong statement.
0%
a) Alleles IA and IB produce sugars.
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b) Both IA and IB are present together and they express because of co-dominance.
0%
c) Alleles bb and cc also produce sugar
0%
d) When IB and bb or i are present only IB is expressed.
Explanation
Blood type (phenotype)
Genotype
Antigen
Antibodies
A
I
A
I
A
or I
A
I
O
A
b
B
I
B
I
B
or I
B
I
0
B
a
AB
I
A
I
B
Both A and B
Neither a nor b
O
I
O
I
O
Neither A nor B
Both a and b
Inheritance of ABO blood group is an example of dominance, codominance and multiple allele. Co-dominance because In ABO both allele expresses themselves in the organism. Multiple alleles because blood group is determined by combination of three alleles IA, IB and IO or i. Dominance because blood group is determined by antigen types. And IA and IB is codominant over recessive i. Answer:(c)
Q.16
on Q371) An offspring of two homozygous parents different from one another by alleles at only one gene locus is known as ...
0%
a) monohybrid
0%
b) dihybrid
0%
c) trihybrid
0%
d) backcross
Explanation
A monohybrid cross is a cross between two organisms with different variations at one genetic locus of interest. Answer:(a)
Q.17
on Q372) Which of the following shows linkage group in coupling phase?
0%
0%
0%
0%
Explanation
Answer:(a)
Q.18
on Q373) How many different types of gametes can be formed by F1 progeny, resulting from the following cross? AA BB CC X aa bb cc
0%
a) 3
0%
b) 8
0%
c) 27
0%
d) 64
Explanation
Cross of AABBCC and aabbcc produces AaBbCc. Number of gametes found in AaBbCc = (2)n where n is number of heterozygous alleles. Here n=3 No. of gametes = 23 = 8 Answer:(b)
Q.19
on Q374) Number of different type of phenotype and genotype in dihybrid F1 is:
0%
a) Four and nine.
0%
b) Nine and four
0%
c) Two and three.
0%
d) One type each.
Explanation
In dihybrid cross, F1 generation is heterozygous expressing dominant trait with genotype as TtRr. Only one phenotype and one genotype is produced in all offsprings. Answer:(d)
Q.20
on Q375) If the character in an organism is controlled by polygenic principle, where the characters is controlled by three genes namely A, B and C. Which among the following group of genotypes can produce the same phenotype?
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a) AABBCC, Aabbcc, AaBbCc.
0%
b) AABBCc, AaBBCC, AAbBCC.
0%
c) AABBCC, aabbcc, AABbcc.
0%
d) AaBBCC, aabbcc, aabbCC.
Explanation
The character is controlled by three genes A, B and C. Same phenotype will be expressed when allelic pair is homozygous dominant or heterozygous. Answer:(b)
Q.21
on Q376) In mice, black fur is dominant to white, and long fur is dominant too short. What is the probability that a white mouse that is homozygous for long fur and a mouse heterozygous for both traits can have an offspring with short white fur?
0%
a) 0.
0%
b) 0.25
0%
c) 0.5
0%
d) 0.75
Explanation
Gametes
BL
Bl
bL
bl
bL
BbLL (black, long fur)
BbLl (black, long fur)
bbLL (white, long fur)
bbLl (white, long fur)
White mouse homozygous for long fur (bbLL) × mouse heterozygous for both triat (BbLl) Answer:(a)
Q.22
on Q377) Match the following:
a. Genetics
Single set of chromosomes
b. Gene
Heredity and variations.
c. Haploid
Units of inheritance
d. Allele
Factors which control contrasting expression of a character
0%
a) a=1; b=3; c=2; d=4
0%
b) a=2; b=3; c=1; d=4
0%
c) a=3; b=2; c=1; d=4
0%
d) a=4; b=1; c=2; d=3
Explanation
Genetics is the branch of biological sciences which deals with the transmission of characteristics from one generation to another (heredity) and also the action of heredity unit called gene as they bring about the characteristics which they control. It also study of differences in characteristics shown by individuals of species (variation). Gene is the inherited factor that determine the biological character of an organism. Allele is a pair of contrasting character. Haploid describes a cell containing single set of chromosomes. Answer:(b)
Q.23
on Q378) Identify the type of chromosomal aberration given below:
0%
a) Reciprocal translocation.
0%
b) Simple translocation.
0%
c) Pericentric inversion.
0%
d) Paracentric inversion.
Explanation
Answer:(c)
Q.24
on Q379) In human being sex chromosomal complement is
0%
a) XX - XY
0%
b) XX - XO
0%
c) ZO - ZZ
0%
d) ZW – ZZ
Explanation
Answer:(a)
Q.25
on Q380) Which condition among the following is lethal ?
0%
a) (2n-2)
0%
b) (2n +2)
0%
c) (2n +1)
0%
d) (2n- 1)
Explanation
Nullisomic is a genetic condition involving the lack of both the normal chromosomal pairs for a species (2n-2). Humans with this condition will not survive. Answer:(a)
Q.26
on Q381) The number of characters investigated by Mendel was
0%
a) four
0%
b) seven
0%
c) one
0%
d) six
Explanation
Mendel conducted experiments on garden pea. He studied inheritance of seven different pairs of contrasting characters in this plant but considered only one pair at a time. Answer:(b)
Q.27
on Q382) Assertion: When the two genes under consideration are present on different pair of chromosomes , segregation of one gene is independent of another gene. Reason: Law of independent assortment is not applicable to linked genes.
0%
a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
0%
b) Both Assertion and Reason are true but Reason is NOT the correct explanation of Assertion
0%
c) Assertion is true but Reason is false.
0%
d) Both Assertion and Reason are false.
Explanation
When a cross is made between two individuals different from each other in two or more characters, then the inheritance of one character is independent of the inheritance of another character. It is called Law of independent assortment. This is correct if the genes are present on two different chromosomes. But if the genes are present on the same chromosomes, they may or may not show independent assortment, i.e. Linkage. Answer:(b)
Q.28
on Q383) What is called Free martin ?
0%
a) Sterile male born along with a fertile female
0%
b) Sterile intersex born along with a fertile female
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c) Sterile super male born along with a intersex
0%
d) Sterile female born along with a fertile male
Explanation
Answer:(d)
Q.29
on Q384) Assertion (A) :- Child is known as Thalassaemia major Reason (R) :- Effective gene from both the parents (Thalassaemia minor) passed to the child
0%
a) If both (A) and (R) are true and (R) is the correct explanation of (A)
0%
b) If both (A) and (R) are true but (R) is not the correct explanation of (A)
0%
c) If (A) is true but (R) is false
0%
d) If both (A) and (R) are false.
Explanation
Answer:(a)
Q.30
on Q385) Which of the following is non - heritable?
0%
a) Point mutation
0%
b) Chromosomal mutation
0%
c) Gene mutation
0%
d) Somatic mutation
Explanation
Answer:(a)
Q.31
on Q386) 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?
0%
a) 1/8
0%
b) 1/32
0%
c) 1/16
0%
d) 1/4
Explanation
Answer:(a)
Q.32
on Q387) which statement is incorrect regarding gene mutation
0%
a) Mutations may be gradual or they may not be induced.
0%
b) Mutation is an evolutionary agent
0%
c) Any gene can undergo mutation
0%
d) Mutated gene is harmful to the individual
Explanation
Answer:(a)
Q.33
on Q388) F2 generation is produced as a result of
0%
a) Crossing F1 individuals with dominant parent
0%
b) Crossing F1 individuals with recessive parent
0%
c) Crossing one of the parental individual with dominant individual.
0%
d) Crossing F1 individuals amongst them selves.
Explanation
Selfing of F1 generation plants produces filial second or F2 generation. Answer:(d)
Q.34
on Q389) Genes do not occur in pairs in
0%
a) Zygote
0%
b) Somatic cell
0%
c) Endosperm cell
0%
d) Gametes
Explanation
Genes always occur in pair in individuals. During gamete formation, these pairs segregate and only allele is present in each gamete. Answer:(d)
Q.35
on Q390) An organism is able to live on a culture medium containing nutrient A, by the enzyme catalysed reaction: A mutant failed to survive in this medium but grew when nutrient B was added to it. Which gene of this mutant was defective?
0%
a) Only X.
0%
b) Only Y.
0%
c) X and Y both.
0%
d) Neither X or Y.
Explanation
Answer:(a)
Q.36
on Q391) Alpha Thalassemia is controlled by ------- on chromosome number----
0%
a) Linked gene, 11
0%
b) Linked gene, 16
0%
c) Unlinked gene, 11
0%
d) Unlinked gene, 16
Explanation
Answer:(b)
Q.37
on Q392) Eye colour in humans is controlled by a single pair of gene where B i.e. brown is dominant to b i.e. blue. What will be the genotype of a brown eyed man who marries a blue eyed lady and has blue eyed child?
0%
a) BB
0%
b) Bb
0%
c) BB and Bb
0%
d) bb
Explanation
Answer:(b)
Q.38
on Q393) The gene for eye colour in a certain organism has seven alleles. How many genotypic combinations are possible?
0%
a) 2
0%
b) 14
0%
c) 28
0%
d) 56
Explanation
The number of genotypes can be found by the formula N×[(N+1)/2] where N is the number of alleles Thus the answer is 7×8/2=28 Answer:(c)
Q.39
on Q394) Down’s syndrome was first described in the year
0%
a) 1962
0%
b) 1866
0%
c) 1856
0%
d) 1863
Explanation
Answer:(b)
Q.40
on Q395) Pure line breed refers to
0%
a) heterozygosity only
0%
b) heterogyzosity and linkage
0%
c) homozygosity only
0%
d) homozygosity and self assortment.
Explanation
Pure line or true breed is a strain of individuals homozygous for all the genes considered. Answer:(c)
Q.41
on Q396) Sickle cell anaemia is
0%
a) an autosomal linked dominant trait
0%
b) caused by substitution of valine by glutamic acid in the β-globin chain of haemoglobin.
0%
c) caused by a change in base pair of DNA
0%
d) characterized by elongated sickle like RBCs with a nucleus.
Explanation
Answer:(c)
Q.42
on Q397) In humans, the dominance relationship between the A and B alleles of ABO blood group gene is an example of
0%
a) complete dominance.
0%
b) incomplete dominance.
0%
c) co dominance.
0%
d) epistasis
Explanation
There is no dominance – recessive relationship between A and B alleles. Both express themselves and this phenomenon is called codominance. Answer:(c)
Q.43
on Q398) Pedigree analysis is resorted to for genetic analysis in humans rather than conventional genetic methods because: I. Choice mating is not possible II. Number of progeny is limited Of the two statements:
0%
a) Only I is correct
0%
b) Only II is correct
0%
c) Both I and II are correct
0%
d) Both I and II are incorrect
Explanation
Answer:(c)
Q.44
on Q399) A normal - Visioned man whose father was colourblind, marries a woman whose father was also colourblind. They have their first child as a daughter. What are the chances that this child would be colourblind?
0%
a) 100%
0%
b) 0%
0%
c) 25%
0%
d) 50%
Explanation
Answer:(b)
Q.45
on Q400) The genotype of a plant showing the dominatnt phenotype can be determined by
0%
a) test cross
0%
b) dihybrid cross
0%
c) pedigree analysis
0%
d) Back Cross
Explanation
Crossing of F1 individual having dominant phenotype with its homozygous recessive parents is called test cross. The test cross is used to determine whether the individuals exhibiting dominant characters are homozygous or heterozygous. Answer:(a)
Q.46
on Q401) If a colourblind woman marries a normal visioned man, their sons will be
0%
a) all colourblind
0%
b) all normal visioned
0%
c) one - half colourblind and one - half normal
0%
d) three - fourths colourblind and one - fourth normal
Explanation
Answer:(a)
Q.47
on Q402) Which of the following is not a hereditary disease?
0%
a) Cystic Fibrosis
0%
b) Thalassaemia
0%
c) Haemophilia
0%
d) Cretinism
Explanation
Answer:(d)
Q.48
on Q403) The genes Controlling the seven characters of a pea plant studied by Mendel are now known to be located on how many different chromosomes?
0%
a) Seven
0%
b) Six
0%
c) Five
0%
d) Four
Explanation
Mendel was lucky in choosing those characters which showed compete independent assortment despite the seven characters chosen by him being present on four chromosomes – 1, 4, 5 and 7. Answer:(d)
Q.49
on Q404) Incomplete linkage in Drosophila produces off springs with parental and non-parental combinations.State the percentage of non parental combination of Drosophila.
0%
a) 83%
0%
b) 17 %
0%
c) 15 %
0%
d) 85 %
Explanation
Answer:(b)
Q.50
on Q408) If a cross is made between two individuals, each having genotype Bb, two offspring’s are obtained. Out of these first has dominant trait. What is the probability that the second offspring will exhibit recessive trait?
0%
a) 1/4
0%
b) 100
0%
c) 0
0%
d) 3/4
Explanation
Cross between heterozygous Bb individuals will produce 1 homozygous dominant, 2 heterzygous, 1 homozygous recessive. Answer:(a)
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