Q 1 – Mendel conducted his famous breeding experiments by working on
(a) Drosophila
(b) Pisum sativum
(c) Escherichia coli
(d) all of these.
Q 2 – The main reason for Mendel’s success in discovering the principles of inheritance was
(a) he considered each character separately
(b) he was lucky not to encounter with linkage problem
(c) the plant was pure breeding
(d) all of these
Q 3 – Why were pea plants more suitable than dogs for Mendel’s experiments ?
(a) There were no pedigree records of dogs.
(b) Pea plants can be self-fertilised.
(c) All pea plants have only two chromosomes.
(d) Dogs have many genetic traits
Q 4 – An allele is said to be dominant if
(a) it is expressed in both homozygous and heterozygous conditions
(b) it is expressed only in second generation
(c) it is expressed only in heterozygous condition
(d) it is expressed only in homozygous condition.
Q 5 – In a dihybrid cross four phenotypes form in the ratio of 9 : 3 : 3 : 1, because of
(a) dominance of one phenotype in each pair of contrasting traits
(b) independent assortment of the genes of contrasting traits
(c) crossing over of genes
(d) mixed effect of dominance and independent assortment.
Q 6 – Which of the following represents the characteristic of a pleiotropic gene?
(a) Controls sexual characters.
(b) Present only in prokaryotes.
(c) Controls one character in association with the other.
(d) Control more than one character.
Q 7 – A segment of DNA providing information for a protein is called
(a) nucleus
(b) chromosomes
(c) trait
(d) gene.
Q 8 – Recessive mutations are expressed
(a) always since it is a mutation
(b) in heterozygous condition
(c) neither in homozygous nor in heterozygous condition
(d) in homozygous condition.
Q 9 – The reason why some mutations, which are harmful, do not get eliminated from gene pool is that
(a) they are recessive and carried by heterozygous individuals
(b) they are dominant and show up more frequently
(c) genetic drift occur because of a small population
(d) they have future survival value.
Q 10 – In Drosophila, red eye character is dominant over white eye character. When a homozygous red-eyed individual is crossed with a homozygous white-eyed individual, and individuals of F1 generation are intercrossed, 12 individuals
are produced. White-eyed individuals of these will be
(a) three
(b) six
(c) nine
(d) twelve.
Q 11 – A true breeding tall and smooth-seeded pea plant was crossed with a true breeding dwarf and wrinkled-seeded plant. All the F1 plants were tall and demonstrate
(a) principle of assortment of characters
(b) that recombination of characters appears in F2 generation
(c) that P tall plants were heterozygous
(d) that tallness was dominant over dwarfness
Q 12 – Two variants found in human population are shown in the given figure. Identify them and select the correct option.
(i) This is an inherited character.
(ii) This is acquired during lifetime.
(iii) This is a non-genetic trait.
(a) (ii) and (iii)
(b) Only (i)
(c) (i) and (iii)
(d) (i) and (iii)
Q 13 – Identify P, Q, R and S in the case of normal human from given flow chart and select the correct option.
P Q R S
(a) XX XY XY XX
(b) XY XX XX XY
(c) XXX XY XYY XXX
(d) XO XX XX XO
Q 14 – Refer to the given figure and select the correct statement(s) regarding it.
(i) It is a monohybrid cross.
(ii) Red flower colour trait is dominant over white flower colour trait.
(iii) Both the traits red flower colour and white flower colour were inherited in F1 plants, but only red flower trait was expressed.
(a) (ii) and (iii)
(b) Only (i)
(c) Only (ii)
(d) (i) (ii) and (iii)
Q 15 – Mendel performed a cross between two garden pea plants; one with round and green seed and the other with yellow and wrinkled seed as shown below.
Select the correct match(es) regarding it.
(a) (i) and (iii)
(b) (ii) and (iv)
(c) (ii) and (iii)
(d) (i) and (iv)
Q 16 – The information source for making proteins in the cell is the
(a) chromosome
(b) DNA
(c) enzyme
(d) nucleus.
Q 17 – A plant bearing purple flowers (RR) was cross pollinated with a plant bearing white flowers (rr). What would be the ratio of the plants bearing white flowers and purple flowers respectively in F2 generation when the F1 progeny were self pollinated?
(a) 1 : 3
(b) 3 : 1
(c) 1 : 1
(d) 2 : 1
Q 18 – What will be the percentage of purple stemmed plants in the F2 generation, when the F1 generation resulted due to cross breeding of green stemmed (GG) tomato plants with purple stemmed (gg) tomato plants, are self pollinated?
(a) 10%
(c) 75%
(b) 25%
(d) 50%
Q 19 – In human beings, the statistical probability of having a male child is
(a) 25%
(b) 50%
(c) 75%
(d) 60%.
Q 20 – Segregation of alleles takes place during
(a) meiosis
(b) cleavage
(c) fertilisation
(d) crossing over.
Q 21 – The genotypic ratio in F2 generation ofmonohybrid cross will be
(a) 1 : 2 : 1
(b) 3 : 1
(c) 1 : 1
(d) 1 : 2.
Q 22 – Mendel studied seven contrasting characters for his breeding experiment with Pisum sativum. Which of the following characters did he not use?
(a) Pod colour
(b) Pod shape
(c) Leaf shape
(d) Plant height
Q 23 – Mutation is a
(a) change that causes evolution when inherited
(b) change which affects the parents only but never inherited
(c) change which affects the offspring of F2 generation only
(d) factor responsible for plant growth.
Q 24 – Allele that cannot express itself in presence of another is
(a) codominant
(b) dominant
(c) recessive
(d) complementary.
Q 25 – XX-XO type of sex determination and XX-XY type of sex determination are the examples of
(a) male heterogamety
(b) female heterogamety
(c) male homogamety
(d) both (b) and (c).
Q 26 – Select the incorrect statement.
(a) In male grasshoppers, 50% of sperms have no sex chromosome.
(b) Female fruitfly is heterogametic.
(c) Human male produces two types of sperms 50% having X chromosome and 50% having Y chromosomes.
(d) In turtle, sex determination is regulated by environmental factors.
Q 27 – Some of the dominant traits studied by Mendel were
(a) round seed shape, green seed colour and axial flower position
(b) terminal flower position, green pod colour and inflated pod shape
(c) violet flower colour, green pod colour and round seed shape
(d) wrinkled seed shape, yellow pod colour and axial flower position.
Q 28 – In plant, tall phenotype is dominant over dwarf phenotype, and the alleles are designated as T and t, respectively. Upon crossing one tall and one dwarf plant, total 250 plants were obtained, out of which 124 displayed tall phenotype and rest were dwarf. Thus, the genotype of the parent plants were
(a) TT × TT
(b) TT × tt
(c) Tt × Tt
(d) Tt × tt
Q 29 – The percentage of yr gamete produced by YyRr parent will be
(a) 25%
(b) 50%
(c) 75%
(d) 12.5%.
Q 30 – If a genotype consists of different types of alleles, it is called
(a) homozygous
(b) heterozygous
(c) monoallelic
(d) uniallelic.
Read the following and answer the following questions from 31 to 35. Refer to the schematic representation of the albinism that is an inherited condition caused by recessive allele (a). ‘A’ is the dominant allele for the normal condition. The inheritance of certain genetic traits for two or more generations is represented in a pedigree or family tree.
Study the given pedigree chart and answer the following questions.
Q 31 – Which of the following could be the genotypes of X and Y?
X Y
(a) AA AA
(b) AA Aa
(c) Aa Aa
(d) aa aa
Q 32 – Which of the following could be the genotype of generation – 1 male and female?
Male Female
(a) AA aa
(b) aa AA
(c) Aa aa
(d) AA AA
Q 33 – If X married an albino female, then what is the probability that their children would be is the probability that their children would be albino ?
(a) 0
(b) (b) 0.125
(c) 0.25
(d) 0.5
Q 34 – If Y married a normal homozygous male, then what is the probability that their children would be albino?
(a) (a) 0
(b) 0.125
(c) 0.25
(d) 0.5
Q 35 – Which of the following could be the genotype of offsprings produced by cross of X and Y?
(a) AA, Aa, aa
(b) aa, aa
(c) Aa, Aa
(d) AA, AA
Case II : Read the following and answer the following questions from 36 to 40. Refer to the given table regarding results of F2 generation of Mendelian cross.
Q 36 – Which of the following would be the phenotype of F1 generation regarding given data of F2 generation?
(a) Plants with round and yellow coloured seeds.
(b) Plants with round and green coloured seeds.
(c) Plants with wrinkled and yellow coloured seeds.
(d) Plants with wrinkled and green coloured seeds.
Q 37 – Which of the following would be the geno type of parental generation regarding given result of F2 generation?
(a) YYRR and yyrr
(b) YYRR and YYRR
(c) YYRR and YyRr
(d) YyRr and YyRr
Q 38 – If plant with wrinkled and green coloured seeds (S) is crossed with plant having wrinkled and yellow coloured seeds (R), what will be the probable phenotype of offsprings?
(a) All plants with wrinkled and yellow coloured seeds.
(b) 50% plants with wrinkled and yellow coloured seeds and 50% plants with wrinkled and green coloured seeds.
(c) All plants with wrinkled and green coloured seeds.
(d) Both (a) and (b)
Q 39 – Which of the following will result when plant YyRr is self-pollinated?
(a) 9 : 3 : 3 : 1 ratio of phenotypes only
(b) 9 : 3 : 3 : 1 ratio of genotypes only
(c) 1 : 1 : 1 : 1 ratio of phenotypes only
(d) 1 : 1 : 1 : 1 ratio of phenotypes and genotypes
Q 40 – The percentage of yR gamete produced by YyRR parent will be
(a) 25%
(b) 50%
(c) 75%
(d) 12.5%
Q 41 – Assertion : Mendel successfully postulated laws of heredity.
Reason : He recorded and analysed results of breeding experiments quantitatively.
Q 42 – Assertion : The principle of segregation given by Mendel is the principle of purity of gametes.
Reason : Gametes are pure for a character and do not mix up.
Q 43 – Assertion : Test cross is a back cross.
Reason : In test cross, individual is crossed with recessive parent.
Q 44 – Assertion : Pure lines are called true breeds.
Reason : True breeds are used for cross breeding.
Q 45 – Assertion : In a monohybrid cross, offspring of F1 generation express dominant character.
Reason : Dominance occurs only in heterozygous state.
Q 46 – Assertion : The traits that are obtained from parents are inherited traits.
Reason : These traits were developed in the parents during their lifetime.
Q 47 – Assertion : If blood group of both mother and father is ‘O’ then the blood group of children will also be O.
Reason : Blood group in humans is determined by many alleles of a gene viz. IA, IB, IO
Q 48 – Assertion : In grasshoppers, females are heterogametic and males are homogametic.
Reason : In grasshoppers, male has only one sex chromosome (XO) whereas the female has sex chromosomes (XX).
Q 49 – Assertion : If mother is homozygous for black hair and father has red hair then their child can inherit black hair.
Reason : Gene for black hair is recessive to gene for red hair in humans.
Q 50 – Assertion : A child which has inherited X chromosome from father will develop into a girl child.
Reason : Girl child inherits X chromosome from father and Y chromosome from mother.