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Inheritance And Variation Mcq
Quiz 11
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Q.1
on Q103) The two alleles of a gene pair are located on:
0%
a) Homologous sites on homologous chromosomes
0%
b) Heterologous sites on homologous chromosomes
0%
c) Homologous sites on heterologous chromosomes
0%
d) Heterologous sites on heterologous chromosomes
Explanation
Answer:(a)
Q.2
on Q104) Which of the following statements is not true of two genes that show 50% recombination frequency?
0%
a) the genes may be on different chromosomes
0%
b) the genes are tightly linked
0%
c) the genes show independent assortment
0%
d) If the genes are present on the same chromosome, they undergo more than one crossovers in every meiosis
Explanation
Answer:(b)
Q.3
on Q105) 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?
0%
a) Only round seeds with green cotyledons
0%
b) Only wrinkled with yellow cotyledons
0%
c) Only wrinkled seeds with green cotyledons
0%
d) Round seeds with yellow cotyledons and wrinkled seeds with yellow cotyledons.
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 Answer:(d)
Q.4
on Q106) Percentage of recombination between A and B is 9%, A and C is 17%, B and C is 26%, then the arrangement of genes is
0%
a) ABC
0%
b) ACB
0%
c) BCA
0%
d) BAC
Explanation
Answer:(d)
Q.5
on Q107) A colour-blind man marry with a daughter of colour-blind father, the generation will be
0%
a) there will be no daughter colour-blind
0%
b) all sons will be colour-blind
0%
c) all daughter will be colour-blind
0%
d) half sons will be colour-blind
Explanation
Gametes
X
c
Y
X
c
X
c
X
c
(colour blind daughter)
XcY (colour blind son)
X
X
c
X (carrier daughter)
XY (normal son)
It is a sex-linked recessive disorder due to defect in either red or green cone of eye resulting in failure to discriminate between red and green colour. This defect is due to mutation in certain genes present in the X chromosome. A color blind man has genotype XcY. Woman will be carrier of defective genes since her father is color blind. Her genotype will be XcX. XcY × XcX 50% of progeny will be colour blind and 50% will be normal. Answer:(d)
Q.6
on Q108) A cross between tall and dwarf plant was performed and 120 offspring were produced in which 90 plants were tall and 30 were dwarf. Find out the genotype of their parents.
0%
a) Tt and TT
0%
b) Tt and Tt
0%
c) TT and tt
0%
d) tt and tt
Explanation
Ratio of tall : dwarf = 3 : 1. This ratio is seen in F2 generation of monohybrid cross. Genotype of F1 generation is Tt and Tt. Answer:(b)
Q.7
on Q109) A cross was performed between pea plants P1 (violet flower) and P2 (white flower). If obtained progeny showed ratio of 1 (violet): 1 (white), then determine the genotype of P1.
0%
a) Homozygous recessive
0%
b) Heterozygous
0%
c) Homozygous dominant
0%
d) Can’t be determined
Explanation
Gametes
W
w
w
Ww (violet flower)
ww (white flower)
This ratio is obtained in test cross. One parent is heterozygous expressing dominant trait and another parent is recessive. Ww (violet flower) × ww (white flower) Answer:(b)
Q.8
on Q110) Chromosomal theory of inheritance is a combination of
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a) Knowledge of chromosomal segregation with Mendelian principle
0%
b) Knowledge of Dihybrid cross with motion of chromosome
0%
c) Knowledge of Law of dominance with chromosomal appearance
0%
d) Knowledge of classical genetics with modern molecular biology
Explanation
Answer:(a)
Q.9
on Q111) The concept of X body was given by
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a) T. H. Morgan
0%
b) Walter Sutton
0%
c) Alfred Sturtevant
0%
d) Henking
Explanation
Henking (1891) could trace a specific nuclear structure all through spermatogenesis in a few insects, and it was also observed by him that 50 per cent of the sperm received this structure after spermatogenesis, whereas the other 50 per cent sperm did not receive it. Henking gave a name to this structure as the X body but he could not explain its significance. Further investigations by other scientists led to the conclusion that the ‘X body’ of Henking was in fact a chromosome and that is why it was given the name X-chromosome. Answer:(d)
Q.10
on Q112) Study the pedigree chart given below to identify the disorder
0%
a) Sickle cell anaemia
0%
b) Haemophilia
0%
c) Myotonic dystrophy
0%
d) Phenylketonuria
Explanation
Answer:(c)
Q.11
on Q113) In humans, Albinism (c) is recessive to normal condition (C). If two heterozygous parents have a progeny. What is the probability that it will be albino?
0%
a) 3/4
0%
b) 1/2
0%
c) 1/4
0%
d) 1
Explanation
When heterozygous parents have progeny, 25% progeny will express recessive trait and 75% of progeny will express dominant trait. Since albinism is recessive, probability of progeny will be albino is ¼. Answer:(c)
Q.12
on Q114) In a person affected with Phenylketonuria, phenylalanine is converted into:
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a) Phenyl pyruvic acid.
0%
b) Pyruvic acid.
0%
c) Tyrosine.
0%
d) α – keto pyruvic acid.
Explanation
Answer:(a)
Q.13
on Q177) Holandric genes are located
0%
a) both in X and Y-chromosome
0%
b) only in Y-chromosome
0%
c) only in X-chromosome
0%
d) only in autosomes
Explanation
The genes that are carried on the Y chromosome are called holandric genes. Holandric genes can only be passed by males onto their sons; they code for 'maleness' but sometimes cause rare conditions like hypertrichosis pinnae and color blindness. Answer:(b)
Q.14
on Q115) A polygenic trait is controlled by 3 genes A, B and C. In a cross AaBbCc × AaBbCc, the phenotypic ratio of the offspring’s was observed as 1 : 6 : x : 20 : x : 6 :What is the possible value of x?
0%
a) 3.
0%
b) 6.
0%
c) 15.
0%
d) 25.
Explanation
Polygene results in quantitative inheritance which is characterised by occurrence of intermediate forms between the parental type. In case of crossing between AABBCC (dark colour) and aabbcc (light colour) in F2 generation seven phenotypes will obtain with ratio of 1:6:15:20:15:6:1. Answer:(c)
Q.15
on Q116) The idea of mutation was brought forth by:
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a) Gregor Mendel, who worked on Pisum sativum.
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b) Hardy Weinberg, who worked on allele frequencies in a population.
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c) Charles Darwin, who observed a wide variety of organisms during sea voyage.
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d) Hugo de Vries, who worked on Oenothera lamarkiana.
Explanation
Answer:(d)
Q.16
on Q117) In a family, father has a blood group 'A' and mother has a blood group 'B', Children show 50 % probability for a blood group "AB" indicate that -
0%
a) Father is heterozygous
0%
b) Mother is heterozygous
0%
c) Either of parent is heterozygous
0%
d) Mother is homozygous
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)
Blood group AB has genotype IAIB. Answer:(c)
Q.17
on Q118) Which law of Mendel can be explained on chromosomal basis of inheritance ?
0%
a) Law of dominance
0%
b) Law of segregation
0%
c) Law of independent assortment
0%
d) All the above
Explanation
Answer:(c)
Q.18
on Q119) Assertion (A) :- strong linkage allow the variations to come into population Reason (R) :- Linkage does not maintain parental gene combination
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:(d)
Q.19
on Q120) Cri - du -chat Syndrome in humans is caused by the AIPMT - 2006
0%
a) trisomy of 21st chromosome
0%
b) Fertilization of an XX egg by a normal Y - bearing sperm
0%
c) loss of half of the short arm of chromosome 5
0%
d) loss of half of the long arm of chromosome 5
Explanation
Answer:(c)
Q.20
on Q121) Study the pedigree chart of a certain family given below and select the correct conclusion which can be drawn for the character
0%
a) The female parent is heterozygous
0%
b) The parent could not have had a normal daughter for this character
0%
c) The trait under study could not be colourblindness
0%
d) The male parent is homozygous dominant
Explanation
Answer:(a)
Q.21
on Q122) The possibilities of hereditary and evolutionary changes are greatest in species that reproduce by
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a) fission
0%
b) budding
0%
c) asexual means
0%
d) sexual means
Explanation
Sexual reproduction was an early evolutionary innovation after the appearance of eukaryotic cells. During sexual reproduction, the genetic material of two individuals is combined to produce genetically-diverse offspring that differ from their parents. The fact that most eukaryotes reproduce sexually is evidence of its evolutionary success. In many animals, it is actually the only mode of reproduction. The genetic diversity of sexually-produced offspring is thought to give species a better chance of surviving in an unpredictable or changing environment. Answer:(d)
Q.22
on Q123) A dihybrid test cross produces the following progeny - AaBb = 240; Aabb =754; aaBb = 746; aabb =What is the distance between the two genes on the chromosome?
0%
a) 12.5 map units
0%
b) 1 map unit
0%
c) 20 map units
0%
d) 25 map units
Explanation
Answer:(d)
Q.23
on Q124) A woman receives her X chromosomes from:
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a) Her mother only
0%
b) Her father only
0%
c) Both her mother and father
0%
d) Mitochondria of mother only
Explanation
In human, female has pair of X chromosomes. One is received from mother and another from father. Answer:(c)
Q.24
on Q125) A human female with turner's syndrome
0%
a) has 45 chromosomes with XO
0%
b) has one additional X-chromosome
0%
c) exhibits male characters
0%
d) is able to produce children with normal husband
Explanation
Answer:(a)
Q.25
on Q126) Test cross involves:
0%
a) crossing between two genotypes with recessive trait
0%
b) crossing between two F1 hybrids
0%
c) crossing the F1 hybrid with a double recessive genotype
0%
d) crossing between two genotypes with dominant trait
Explanation
Gametes
T
t
t
Tt (tall)
tt (dwarf)
t
Tt (tall)
tt (dwarf)
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. Tt × tt Answer:(c)
Q.26
on Q127) A point mutation comprising the substitution of a purine by pyramidine is called
0%
a) transition
0%
b) translocation
0%
c) deletion
0%
d) transversion
Explanation
Answer:(d)
Q.27
on Q128) If both parents are carriers for Thalassemia, which is an autosomal recessive disorder, what are the chances of pregnancy resulting in an affected child?
0%
a)25%
0%
b) 100%
0%
c) No chance
0%
d) 50%
Explanation
Answer:(a)
Q.28
on Q129) A woman is married for the second time. Her first husband was ABO blood type A, and her child by that marriage was type O. Her new husband is type B and their child is type AB. What is the woman's ABO genotype and blood type?
0%
a) IAIO; Blood type A
0%
b) IAIB; Blood type AB
0%
c) IBIO; Blood type B
0%
d) IOIO; Blood type O
Explanation
First child has blood group O with genotype IOIO. This means that one of IO is present from mother. Her second child has blood group AB with genotype IAIB and her husband has blood group B. That implies that allele IA is obtained from mother’s side. Combining the above information, we can infer that woman has blood group A with genotype IAIO. Answer:(a)
Q.29
on Q130) Which of the following was not explained by the Mendel’s law of dominance?
0%
a) Characters are controlled by discrete units called factors.
0%
b) Factors occur in pairs.
0%
c) Every factor has alternate forms known as alleles
0%
d) In a dissimilar pair of factors one member of the pair dominates the other.
Explanation
Law of Dominance: (i) Characters are controlled by discrete units called factors. (ii) Factors occur in pairs. (iii) In a dissimilar pair of factors one member of the pair dominates (dominant) the other (recessive). The law of dominance is used to explain the expression of only one of the parental characters in a monohybrid cross in the F1 and the expression of both in the F2 . It also explains the proportion of 3:1 obtained at the F2 Answer:(c)
Q.30
on Q131) Assertion: Aneuploidy is due to the non – disjunction of homologous chromosomes. Reason: Disjunction of homologous chromosomes occurs during Anaphase II.
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
Answer:(c)
Q.31
on Q132) If a man consists of an extra copy of chromosome number 18, then the condition is known as
0%
a) Monosomy
0%
b) Trisomy
0%
c) Nullisomy
0%
d) Polyploidy
Explanation
Answer:(b)
Q.32
on Q133) Linkage was described by Morgan in Drosophila. Which of the following combinations are correct? (i) He hybridised yellow bodied, white-eyed females to brown-bodied, red eyed males and then intercrossed their F1 (ii) The ratio of F2 generation is same as that of 9:3:3:1 in this cross. (iii) The parental combinations are observed more than recombinants. (iv) According to Morgan Linkage was observed due to chemical association of genes. (v) All the genes present on same chromosomes have same efficiency to link.
0%
a) (ii), (iv), (v)
0%
b) (i), (iii)
0%
c) (i), (iii), (v)
0%
d) (i) and (iv)
Explanation
Answer:(b)
Q.33
on Q134) In which of the following male heterogametey is not observed?
0%
a) Insects
0%
b) Birds
0%
c) Humans
0%
d) (1) and (2)
Explanation
Answer:(b)
Q.34
on Q135) Which of the following disorder can’t be transmitted from a single diseased parent?
0%
a) Sickle cell anaemia
0%
b) Haemophilia
0%
c) Cystic fibrosis
0%
d) Myotonic dystrophy
Explanation
Answer:(a)
Q.35
on Q136) In case of Pea plant, grey seed colour (G) is dominant to white (g). If the progenies are formed in the following phenotypic ratio, find the genotypes of parents. Grey was crossed with white, 49 grey and 54 white were formed. The genotype of the grey parent will be
0%
a) Gg
0%
b) GG
0%
c) gg
0%
d) GGG
Explanation
The ratio of grey : white is almost 1 : 1. This ratio is seen in test cross where one parent is heterozygous expressing dominant trait and another is recessive. Genotype of grey parent will be Gg and white parent will be gg. Answer:(a)
Q.36
on Q137) What phenotypic ratio will be there when selfing of a trihybrid is performed in which two gene pairs are completely dominant and one gene pair is co dominant?
0%
a) 3:9:1:2:1:3
0%
b) 27:9:3:1:3:3:1
0%
c) 9:3:3:1:18:6:6:2:9:3:3:1
0%
d) 18:6:6:2:1:2:1:2:1
Explanation
Answer:(c)
Q.37
on Q138) Mendel formulated the law of segregation on the basis of:
0%
a) Monohybrid cross.
0%
b) Test cross.
0%
c) Back cross.
0%
d) Dihybrid cross.
Explanation
Law of segregation and law of dominance is based on monohybrid cross Answer:(a)
Q.38
on Q199) Diandric inheritance means:
0%
a) Father’s X linked character passes to the daughter.
0%
b) Mother’s X linked character passes to the granddaughter through son.
0%
c) Father’s X linked character passes to the grandson through daughter.
0%
d) Father’s Y linked character passes to the grandson through son.
Explanation
Diandric :- Inheritance in which characters are inherited from mother to the son and from son to grand daughter Answer:(b)
Q.39
on Q139) Which one of the following can not be explained on the basis of Mendel's law of dominance?
0%
a) The discrete unit controlling a particular character is called a factor
0%
b) Out of one pair of factors one is dominant, and the other recessive.
0%
c) Alleles do not show any blending and both the characters recover as such in F2 generation.
0%
d) Factors occur in pairs
Explanation
Law of Dominance: (i) Characters are controlled by discrete units called factors. (ii) Factors occur in pairs. (iii) In a dissimilar pair of factors one member of the pair dominates (dominant) the other (recessive). The law of dominance is used to explain the expression of only one of the parental characters in a monohybrid cross in the F1 and the expression of both in the F2 . It also explains the proportion of 3:1 obtained at the F2 Answer:(c)
Q.40
on Q140) In a cross between a pure tall plant with green pod and a pure short plant with yellow pod. How many short plants are produced in F2 generation out of 16?
0%
a) 1
0%
b) 3
0%
c) 4
0%
d) 9
Explanation
Gametes
TG
Tg
tG
tg
TG
TTGG (tall,green)
TTGg (tall,green)
TtGG (tall,green)
TtGg (tall,green)
Tg
TTGg (tall,green)
TTgg (tall,yellow)
TtGg (tall,green)
Ttgg (tall,yellow)
tG
TtGG (tall,green)
TtGg (tall,green)
ttGG (dwarf,green)
ttGg (dwarf,green)
tg
TtGg (tall,green)
Ttgg (tall,yellow)
ttGg (dwarf,green)
ttgg (dwarf,yellow)
TTGG (tall plant with green pod) crossed with ttgg (dwarf plant with yellow) produces F1 generation of tall plant with green pod with genotype TtGg. When selfed F2 generation is produced. TtGg × TtGg Dwarf plants = 4/16 Answer:(c)
Q.41
on Q141) The most popularly known blood grouping is the ABO grouping. It is named ABO and not ABC, because "O" in it refers to having ... ..
0%
a) Overdominance of this type on the genes for A and B types
0%
b) One antibody only - either anti - A or anti - B on the RBCs
0%
c) no antigens A and B on RBCs
0%
d) other antigens besides A and B on RBCs
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
Answer:(c)
Q.42
on Q142) 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 affected. Which of the following mode of inheritance do you suggest for this disease?
0%
a) Sex - linked dominant
0%
b) Sex - linked recessive
0%
c) Sex - limited recessive
0%
d) Autosomal dominant
Explanation
Parents are not showing inherited disease. This means the disease is not autosomal. In the progeny only sons show this disease, this means it is sex linked recessive. Daughters are not diseased All daughters will receive at least one normal copy of X chromosome from father. Hence, none of them shows signs of inherited disease. However, three out of five sons received defective X chromosome from mother and suffered from disease. Other two sons received normal X chromosome from mother and showed no signs of disease. Answer:(b)
Q.43
on Q143) Assertion (A) :- Human beings are not suitable breeding experiments to Study the inheritance of human traits. Reason (R) :- In Human beings controlled hybridization is not possible
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.44
on Q144) Which kind of structural abnormality is seen in the above chromosome?
0%
a) Terminal deletion
0%
b) Duplication (Tandem)
0%
c) Duplication (Reverse)
0%
d) Inversion
Explanation
Answer:(b)
Q.45
on Q145) Match the following
Column I
Column II
Johansen
a. Coined the term gene
Mendel
b. Crossing over in drosophila
T. H. Morgan
c. Linkage in lathyrus odoratus
Bateson & punnett
d. Law of segregation
0%
a) (1 - d) (2 - c) (3 - b) (4 - a)
0%
b) (1 - a) (2 - d) (3 - b) (4 - c)
0%
c) (1 - c) (2 - d) (3 - b) (4 - a)
0%
d) (1 - b) (2 - d) (3 - a) (4 - c)
Explanation
Answer:(b)
Q.46
on Q146) Which factor in nature causes discontinuous Variation in a population ?
0%
a) Recombination
0%
b) Shufflng of parental genes
0%
c) Mutation
0%
d) Geneflow
Explanation
Answer:(c)
Q.47
on Q147) Checkerboard method of calculations was developed by
0%
a) Mendel
0%
b) Bateson
0%
c) punnett
0%
d) Morgan
Explanation
Punnett square is a checker board which is devised by R.C. Punnett and used to show the result of a cross between two organism. Answer:(c)
Q.48
on Q148) Blood grouping in humans is controlled by
0%
a) 4 alleles in which IA is dominant
0%
b) 3 alleles in which IA and IB are dominant
0%
c) 2 alleles in which none is dominant
0%
d) 3 alleles in which IA is recessive
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
4 phenotype and 6 genotype. 2 alleles IA and IB are codominant and both are dominant over IO or i. Answer:(b)
Q.49
on Q149) In Mirabilis plant the appearance of the pink hybrid (Rr) between cross of a red (RR) and white (rr) flower parent indicates
0%
a) Segregation
0%
b) Dominance
0%
c) Incomplete dominance
0%
d) Heterosis
Explanation
In Mirabilis jalapa, there is no dominant-recessive relationship and intermediate hybrid trait is expressed in F1 generation. Red is not completely dominant over white and thus intermediate hybrid, pink flower is produced. This phenomenon is called incomplete dominance. Answer:(c)
Q.50
on Q150) Genetic recombinations occur through
0%
a) Mitosis & fertilization
0%
b) Mitosis & Meiosis
0%
c) Meiosis & fertilization
0%
d) None
Explanation
Answer:(c)
0 h : 0 m : 1 s
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8
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12
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