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Inheritance And Variation Mcq
Quiz 17
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Q.1
on Q409) , has been mutated to , the type of gene mutation happened in DNA is:
0%
a) Transition only.
0%
b) Transversion only.
0%
c) Transition and transversion.
0%
d) Frame shift mutation.
Explanation
Answer:(d)
Q.2
on Q410) Which one of the following symbols and its representation, used in human pedigree analysis is correct?
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
Answer:(b)
Q.3
on Q411) The mapping of genes on chromosome was attempted by
0%
a) T. H. Morgan
0%
b) Walter Sutton
0%
c) Alfred Sturtevant
0%
d) Henking
Explanation
Answer:(c)
Q.4
on Q412) In the results, F2 generation of Mendelian dihybrid cross in pea plant. The ratio of green and wrinkled seeded pea plant is
0%
a) 1/16
0%
b) 3/16
0%
c) 4/16
0%
d) 7/16
Explanation
Yellow is dominant over green. Round is dominant over wrinkled. F2 generation will be, (yellow : green) × (round : wrinkled) = yellow, round : yellow, wrinkled : green, round : green, wrinkled (3 : 1) × (3 : 1) = 9 : 3 : 3 : 1 Green,wrinkled seeded plant = 1/16 Answer:(a)
Q.5
on Q413) If F2 dwarf plants were self-pollinated then the genotype of F3 & F4 plants is
0%
a) TT and Tt
0%
b) tt and tt
0%
c) tt and Tt
0%
d) Tt and Tt
Explanation
If homozygous recessive individual is self-pollinated, the offsprings will also be recessive. F3 and F4 plants genotype will be tt. Answer:(b)
Q.6
on Q414) How many laws of Mendel are based on monohybrid cross?
0%
a) 1
0%
b) 2
0%
c) 3
0%
d) 4
Explanation
There are two laws formulated on monohybrid cross. (1) Law of dominance and (2) Law of segregation. Law of independent assortment is devised from dihybrid cross. Answer:(b)
Q.7
on Q415) Who is known as father of physiological genetics or father of biochemical genetics?
0%
a) Slatyer
0%
b) Charles Elton
0%
c) Taylor
0%
d) Archibald Garrod
Explanation
Answer:(d)
Q.8
on Q416) Consider the cross AaBb x AaBb. If the alleles for both genes exhibit complete dominance, what genotypic ratio is expected in the resulting offspring?
0%
a) 1:1:1:1
0%
b) 9:3:3:1
0%
c) 3:6:3:1:2:1
0%
d) 1:2:1:2:4:2:1:2:1
Explanation
No. of traits (n)
Experiment
Types of gametes (2
n
)
No. of offsprings (gametes)
2
No. of phenotype (2
n
)
No. of genotype (3
n
)
Phenotypic ratio
Genotypic ratio
2
Dihybrid cross
4
16
4
9
9:3:3:1
2:4:2:1:2:1:1:2:1
Answer:(d)
Q.9
on Q417) The genotypes of a husband and wife are IA IB and IAi. Among the blood types of their children, how many different genotypes and phenotypes are possible?
0%
a) 3 genotypes; 3 phenotypes
0%
b) 3 genotypes; 4 phenotypes
0%
c) 4 genotypes; 3 phenotypes
0%
d) 4 genotypes; 4 phenotypes
Explanation
I
A
I
O
I
B
I
A
I
B
(AB blood group)
I
B
I
O
(Bblood group)
I
B
I
A
I
B
(AB blood group)
I
B
I
O
(B blood group)
3 phenotypes and 4 genotypes Answer:(c)
Q.10
on Q418) Self fertilising trihybrid plants from
0%
a) Eight different gametes and 64 different zygotes
0%
b) Four different gametes and sixteen different zygotes
0%
c) Eight different gametes and sixteen different zygotes
0%
d) Eight different gametes and thirty two different zygotes
Explanation
No. of traits (n)
Experiment
Types of gametes (2
n
)
No. of offsprings (gametes)
2
No. of phenotype (2
n
)
No. of genotype (3
n
)
Phenotypic ratio
Genotypic ratio
3
Trihybrid cross
8
64
8
27
(3:1)
3
(1:2:1)
3
Answer:(a)
Q.11
on Q419) If a dwarf pea plant which is treated with growth regulators becomes a tall plant, which on selfing gives:
0%
a) Only dwarf plants.
0%
b) Only tall plants.
0%
c) Tall and dwarf in 3: 1 ratio.
0%
d) Tall and dwarf in 1: 1 ratio.
Explanation
On treating with growth regulator, dwarf pea plant becomes tall. This trait is acquired, there will be no change in genotype. The genotype of dwarf plant will be tt. When dwarf plant is self-pollinated, the progeny will be only dwarf plant. Answer:(a)
Q.12
on Q420) By which gene sex is determind in Spinach ?
0%
a) Single gene "m" located in the X- chromosome
0%
b) Single gene "m" located in the Y- chromosome
0%
c) Single gene "t" located in the X- chromosome
0%
d) Single gene "t" located in the Y- chromosome
Explanation
Answer:(a)
Q.13
on Q421) Which technique is used by Mendel for hybridization ?
0%
a) Emasculation
0%
b) Bagging
0%
c) Protoplast fusion
0%
d) Both A & B
Explanation
Mendel crossed two pea plants with alternate characters by artificial pollination. Artificial hybridization is the process in which only desired pollen grains are used for pollination and fertilization. This process was done using emasulation and bagging technique. Because garden pea is self-fertilizing, the anthers need to be removed before maturity. The process of removal of anther is called emasculation. The stigma is protected against any pollen by covering it with a bag. This technique is called bagging. Answer:(d)
Q.14
on Q422) Types of phenotypes of F2 generation of dihybrid cross ?
0%
a) 4
0%
b) 16
0%
c) 8
0%
d) 9
Explanation
No. of traits (n)
Experiment
Types of gametes (2
n
)
No. of offsprings (gametes)
2
No. of phenotype (2
n
)
No. of genotype (3
n
)
Phenotypic ratio
Genotypic ratio
2
Dihybrid cross
4
16
4
9
9:3:3:1
2:4:2:1:2:1:1:2:1
Answer:(a)
Q.15
on Q423) Genetic balance theory for sex determination in Drosophila was proposed by
0%
a) Prof. R. P. Roy
0%
b) H. E. Warmbe
0%
c) C.B. Bridges
0%
d) Mc. Chang
Explanation
Answer:(c)
Q.16
on Q424) A cross between hybrid and a parent is known as
0%
a) Test cross
0%
b) Back cross
0%
c) Monohybrid cross
0%
d) Reciprocal cross
Explanation
Backcross is a cross of F1 hybrid with either of two parents. Crossing of F1 individual having dominant phenotype with its homozygous recessive parents is called test cross. Reciprocal cross: A set of two reciprocal crosses means that the same two parents are used in two experiments in such a way that if in one experiment, ‘A’ is used as female parent and ‘B’ is used as male parents, in other experiments ‘A’ will be used as male parents and ‘B’ will be used as female parents. Thus the reciprocal crosses involve two crosses concerning the same characteristic, but with reverse sexes. Answer:(b)
Q.17
on Q425) What is the basis of pleiotropy?
0%
a) A spontaneous mutation during the replication of DNA.
0%
b) Interrelationship between various metabolic pathways in the body.
0%
c) Chromosomal aberration as chromosomes are the vehicles of genes.
0%
d) the behaviours of chromosomes during meiosis or gamete formation.
Explanation
Pleiotropy is the ability of gene to have multiple phenotypic effect because it influences a number of characters simultaneously. Thus, pleiotropy is generally caused by a single molecular function involved in multiple biological processes. Answer:(b)
Q.18
on Q426) Assertion (A) :- Crossing over leads to genetic variation. Reason (R) :- Crossing over causes recombination of characters.
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.19
on Q427) A woman with 47 chromosomes due to 3 copies of chromosome 21 is characterized by
0%
a) Super Femaleness
0%
b) Triploidy
0%
c) Turner's Syndrome
0%
d) Down's Syndrome
Explanation
Answer:(d)
Q.20
on Q428) Represented below is the inheritance pattern of a certain type of trait in humans. Which one of the following conditions could be an example of this pattern?
0%
a) Phenyl ketonuria
0%
b) Sickle Cellanaemia
0%
c) Haemophilia
0%
d) Thalassemia
Explanation
Answer:(c)
Q.21
on Q429) Name the abnormality seen in chromosome number – 2
0%
a) Interstitial Deletion
0%
b) Terminal Deletion
0%
c) Tandem Duplication
0%
d) Reverse Duplication
Explanation
Answer:(a)
Q.22
on Q430) Observe the following genotype of parents who are about to mate: RrYyTtGg × RRYyTtGg; if you are asked to draw a punnet square to represent their offsprings, how many boxes will you put in the square?
0%
a) 64
0%
b) 32
0%
c) 128
0%
d) 256
Explanation
Type of gametes = (2)n where n is no. of heterozygous allele. In RrYyTtGg case n = 4, types of gametes will be 16. In case of RRYyTtGg, n = 3, types of gametes will be 8. No. of offspring = no. of gametes from parent1 × no. of gametes from parent2 = 16 × 8 = 128 Answer:(c)
Q.23
on Q431) A homozygous grey bodied, long winged Drosophila is crossed with homozygous black bodied vestigial wings. The F1 is test crossed and we get; Grey bodied long wings – 470, Grey bodied vestigial wings – 185, Black bodied vestigial wings – 450 and black bodied long wings –What is the distance between the genes for body colour and wings shape of Drosophila?
0%
a) 14cM.
0%
b) 72cM.
0%
c) 28cM.
0%
d) 36cM.
Explanation
Answer:(c)
Q.24
on Q432) Which one of the following condition in humans is correctly matched with its chromosomal abnormality/linkage?
0%
a) Erythroblastosis foetalis - X-linked.
0%
b) Down syndrome - 44 autosomes + XO.
0%
c) Klinefelter's syndrome - 44 autosomes + XXY.
0%
d) Colour blindness - Y-linked.
Explanation
Answer:(c)
Q.25
on Q433) Which of the following terms best describes when the phenotype of the heterozygote differs from the phenotypes of both homozygotes?
0%
a) Incomplete dominance
0%
b) Multiple alleles
0%
c) Co dominance
0%
d) Epistasis
Explanation
When there is no dominant-recessive relationship, hybrid expresses intermediate phenotype. Example red flower and white is homozygous in Dog flower and hybrid is heterozygous which is pink flower. Answer:(a)
Q.26
on Q434) Law of Independent Assortment is not applicable to
0%
a) Linked genes
0%
b) Genes on different chromosomes
0%
c) Distantly located gene
0%
d) both (2) and (3)
Explanation
The law of independent assortment is applicable to only those factors or genes which are present on different chromosomes. The law applies only to gene pairs on different pairs of homologous. Answer:(a)
Q.27
on Q435) Trait expressed in F1 generation is
0%
a) Parental
0%
b) Recessive
0%
c) Dominant
0%
d) Recombinant
Explanation
According to law of dominance, only trait is expressed in F1 generation but recessive trait reappears in F2 generation. Answer:(c)
Q.28
on Q436) Mendel was born in
0%
a) 19th century.
0%
b) 18th century.
0%
c) 17th century.
0%
d) 20th century.
Explanation
Gregor Johann Mendel was born in year 1822 and died in year 1884. Answer:(a)
Q.29
on Q437) James Watson and Francis Crick got Nobel prize in the year
0%
a) 1960
0%
b) 1962
0%
c) 1959
0%
d) 1953
Explanation
Answer:(b)
Q.30
on Q438) Which of the following occurs due to monosomy of sex chromosome?
0%
a) Down's syndrome
0%
b) Turner's syndrome
0%
c) Haemophilia
0%
d) Sickle cell anaemia
Explanation
Answer:(b)
Q.31
on Q439) Base substitutions from base analogues I are called
0%
a) transitions
0%
b) transversions
0%
c) intragenic complementation
0%
d) intergenic complementation
Explanation
Answer:(a)
Q.32
on Q440) A woman with normal vision, but whose father was colourblind, marries a colourblind man. Suppose that the fourth child of this couple was a boy. This boy
0%
a) May be colourblind or may be of normal vision
0%
b) Must be colourblind
0%
c) Must have normal colour vision
0%
d) Will be partially colourblind since he is heterozygous for the colourblind mutant allele.
Explanation
Answer:(a)
Q.33
on Q441) F2 generation in a Mendelian cross showed that both genotypic and phenotypic ratios are same as 1 : 2 : 1 . It represents in case of AIPMT - 2012
0%
a) Co - Dominance
0%
b) Dihybrid Cross
0%
c) Monohybrid cross with complete dominance
0%
d) Monohybrid cross with incomplete dominance
Explanation
Phenotypic and genotypic ratio is same. This occurs when there is no dominant-recessive relationship and heterozygous or hybrid expresses intermediate. This phenomenon is called incomplete dominance. Answer:(d)
Q.34
on Q442) The value of X/A is given in column A and sex of the fly is written in column B. Match the following :-
Column I
Column II
1
(a) Normal Male
1.5
(b) super male Sterile
0.67
(c) inter sex
0.33
(d) super Female sterile
0.5
(e) Normal Female
0%
a) (1 - a) (2 - d) (3 - c) (4 - e) (5 - b)
0%
b) (1 - e) (2 - d) (3 - c) (4 - b) (5 - a)
0%
c) (1 - b) (2 - a) (3 - c) (4 - d) (5 - e)
0%
d) (1 - e) (2 - c) (3 - d) (4 - a) (5 - b).
Explanation
Answer:(b)
Q.35
on Q443) which cross yielded a ratio of 7:1 :1:7 ?
0%
a) Test cross (Dihybrid) - Bateson and punnet
0%
b) Test cross (Monohybrid) - Bateson and punnet
0%
c) Test cross (Dihybrid) - Carrel correns
0%
d) Test cross ( Dihybrid) - Mendel & Morgan
Explanation
Normally, we should expect a 1 : 1: 1:1 ratio in a testcross if independent assortment takes place. But actually, we get a 7: 1: 1:7 ratio, this indicates that dominant alleles are having a tendency to remain together. Similarly was the case with recessive alleles also. Therefore this deviation in the test cross ratio was, explained by Bateson as the gametic coupling. Answer:(a)
Q.36
on Q444) “Gametes are never hybrid”. This is a statement of:
0%
a) Law of dominance.
0%
b) Law of independent assortment.
0%
c) Law of segregation.
0%
d) Law of random fertilization.
Explanation
The gametes which are formed are always pure for a particular character. A gamete may carry either dominant or recessive factor but not both. This is why it is called ‘principle of segregation’ or ‘law of purity of gametes’. Answer:(c)
Q.37
on Q445) Percentage of recombination between A and B is 9% and C is 17%, B and C is 26%, then what is the arrangement of genes?
0%
a) ABC.
0%
b) ACB.
0%
c) BCA.
0%
d) BAC.
Explanation
Answer:(d)
Q.38
on Q446) Excessive growth of hair on the pinna is a feature found only in males because:
0%
a) The gene responsible for the character is recessive in females and dominant only in males.
0%
b) The character is induced in males as males produce progesterone.
0%
c) The female sex hormone oestrogen suppresses the character in females.
0%
d) The gene responsible for the character is present on the Y chromosome only.
Explanation
Answer:(d)
Q.39
on Q447) ABO blood groups is determined by three different alleles. How many genotypes and phenotypes are possible?
0%
a) Genotype – 3, phenotype – 1.
0%
b) Genotype – 6, phenotype – 4.
0%
c) Genotype – 4, phenotype – 6.
0%
d) Genotype – 9, phenotype – 7.
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
In ABO blood grouping, there are 3 alleles IA, IB and IO or i. Combination of 2 out of these 3 alleles gives 6 genotype and 4 phenotype. Answer:(b)
Q.40
on Q448) Conditions of a karyotype 2n ± 1 and 2n ± 2 are called:
0%
a) Aneuploidy.
0%
b) Polyploidy.
0%
c) Allopolyploidy.
0%
d) Monosomy.
Explanation
Answer:(a)
Q.41
on Q449) If a character is controlled by 5 alleles, what will be the number of genotype in this case?
0%
a) 12
0%
b) 18
0%
c) 30
0%
d) 15
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 5×5/2=15 Answer:(d)
Q.42
on Q450) Mendelian disorders are caused by
0%
a) Mutation in the single chromosome
0%
b) Mutation in the single gene
0%
c) Polyploidy
0%
d) Aneuploidy
Explanation
Answer:(b)
Q.43
on Q451) Mutation causes
0%
a) alterations at protein level
0%
b) changes in genotype only
0%
c) changes in phenotype
0%
d) both (2) and (3)
Explanation
Answer:(d)
Q.44
on Q452) Who was the first to use statistical analysis and mathematical logics for solving biological problems?
0%
a) Francis crick
0%
b) Gregor Mendel
0%
c) Hugo de Vries
0%
d) W. Bateson
Explanation
During Mendel’s investigations into inheritance patterns it was for the first time that statistical analysis and mathematical logic were applied to problems in biology Answer:(b)
Q.45
on Q453) Two genes R and Y are located very close on the chromosomal linkage map of maize plant. When RRYY and rryy genotypes are hybridised, then F2 segregation will show:
0%
a) higher number of the recombinant types
0%
b) segregation in the expected 9: 3: 3: 1 ratio
0%
c) segregation in 3: 1
0%
d) higher number of the parental types
Explanation
Answer:(d)
Q.46
on Q454) Mutations can be induced with
0%
a) IAA
0%
b) ethylene
0%
c) gamma radiations
0%
d) infra red radiations
Explanation
Answer:(c)
Q.47
on Q455) Among the following characters, which one was not considered by Mendel in his experiments on pea?
0%
a) Stem - Tall or Dwarf
0%
b) Trichomes - Glandular or Non-glandular
0%
c) Seed - Green or Yellow
0%
d) Pod - Inflated or Constricted
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.48
on Q456) When Morgan hybridised yellow-bodied, white-eyed females to brown-bodied, red-eyed males and intercrossed their F1 progeny, the F2 ratio deviated very significantly from the 9:3:3:1 ratio. this can be attributed to the fact that:
0%
a) The genes are located on X and Y chromosomes
0%
b) Fruit fly has abnormal chromosomes
0%
c) The genes are located on the X chromosome
0%
d) the genes exhibit incomplete dominance
Explanation
The pollination taking place when the flower is in the bud condition is called as the bud pollination. It usually occurs in the flowers in which the stigma and the pollen grains mature or ripens even before opening of the bud. Example Pea, Grass etc. Pisum sativum (garden pea) has perfect bisexual flowers containing male and female parts. The flowers are predominately self pollinating. Answer:(c)
Q.49
on Q457) Genes which code for a pair of contrasting traits or slightly different forms of the same gene are known as
0%
a) Alleles
0%
b) Loci
0%
c) Cistrons
0%
d) Introns
Explanation
A pair of contrasting character is called allele. The portion or region on chromosomes representing a single gene is called gene locus. Loci is plural of locus. Cistron is a section of a DNA or RNA molecule that codes for a specific polypeptide in protein synthesis. An intron is a portion of a gene that does not code for amino acids. Answer:(a)
Q.50
on Q458) Mutations which are not dominant are not lost by a gene pool this is known as
0%
a) Survival of the dominant
0%
b) Survival of the recessive
0%
c) Hardly-Weinberg law
0%
d) None of these
Explanation
Answer:(c)
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