HOTs Principles of Inheritance and Variation Class 12 Biology

HOTs for Class 12

Please refer to Principles of Inheritance and Variation HOTs Class 12 Biology provided below with Principles of Inheritance and Variation. All HOTs for Class 12 Biology with answers provided below have been designed as per the latest syllabus and examination petter issued by CBSE, NCERT, KVS. Students of Standard 12 Biology should learn the solved HOTS for Class 12 Biology provided below to gain better marks in examinations.

Principles of Inheritance and Variation Class 12 Biology HOTs

Question. Phenylketonuria is caused due to
(a) pleiotropy
(b) multiple alleles
(c) codominance
(d) incomplete dominance

Answer

A

Question. Phenylketonuria in human
(a) manifests through phenotypic expressions
(b) is characterised by mental retardation
(c) leads to hair reduction and skin pigmentation
(d) All of the above

Answer

D

Question. Which one of the following pairs is wrongly matched? 
(a) XO type of sex-determination – Grasshopper
(b) ABO blood grouping – Codominance
(c) Starch synthesis in pea – Multiple allele
(d) TH Morgan – Linkage

Answer

C

Question. Sex-determination is controlled by …A… and the remaining chromosomes which are not involved in sex-determination are …B… .
                A                     B
(a) Allosomes              Autosomes
(b) Allosomes              Sex-chromosomes
(c) Sex-chromosomes  Allosomes
(d) Autosomes            Sex-chromosomes

Answer

A

Question. Identify the type of inheritance in the given diagram. (Image 116)
(a) Dominant X-linked
(b) Recessive X-linked
(c) Dominant Y-linked
(d) Cytoplasmic or Mitochondrial inheritance

Answer

D

Question. Following pedigree chart shows (Image 117)
(a) character is carried by Y-chromosome
(b) character is sex-linked recessive
(c) character is sex-linked dominant
(d) character is recessive autosomal

Answer

A

Question. Choose the incorrect pair with respect to sex determination in different organisms.
(a) Grasshopper = XO type
(b) Birds = ZZ-ZW type
(c) Drosophila = XX-XO type
(d) Human = XX-XY type

Answer

C

Question. In XX and XY type of sex-determination,
(a) males are heterogametic
(b) females are isogametic
(c) Both (a) and (b)
(d) None of the option is correct

Answer

A

Question. Male heterogamety is seen in
(a) Humans
(b) Grasshopper
(c) Drosophila
(d) All of these

Answer

D

Question. Choose the incorrect pair amongst the following.
(a) Male bird – Homogametic
(b) Female bird – Heterogametic
(c) Male Drosophila – Heterogametic
(d) None of the above

Answer

D

Question. The chromosomal denotation for heterogameticfem ale and homogametic males are
(a) ZW and ZZ
(b) ZO–ZZ
(c) XX–XO
(d) Both (a) and (b)

Answer

D

Question. A human male contains the karyotype of …A… and ahuman fe male has …B… chromosomes.
          A               B
(a) 44 + XX     44 + XY
(b) 44 + XY     44 + XX
(c) 44 + XO     44 + XX
(d) 44 + XX     44 + XO

Answer

B

Question. The number of chromosomes in females and males honeybees are
(a) 32
(b) 16
(c) 32 and 16, respectively
(d) 16 and 32, respectively

Answer

C

Question. Both chromosome and gene (Mendelian factors) whether dominant or recessive are transmitted from generation to generation in
(a) changed form
(b) unaltered form
(c) altered form
(d) disintegrated form

Answer

B

Question. The figure depicts. (Image 73) 
(a) Linkage
(b) Independent assortment
(c) Law of dominance
(d) Equational division

Answer

B

Question. The unfertilised eggs in honeybees develop into
(a) males
(b) queen
(c) worker
(d) Both (a) and (c)

Answer

A

Question. In honeybees, male and female gametes are produced through
(a) mitosis
(b) mitosis and meiosis, respectively
(c) meiosis
(d) meiosis and mitosis, respectively

Answer

B

Question. Chromosomal abberation is commonly found in the
(a) cancer cells
(b) normal cells
(c) healthy cells
(d) autosomal cells

Answer

A

Question. Point mutation arises due to the change in
(a) single base DNA
(b) single base pair of DNA
(c) segment of DNA
(d) double base pair of DNA

Answer

B

Question. Polygenic traits are controlled by
(a) one gene
(b) two genes
(c) three or more genes
(d) mutant genes

Answer

C

Question. In human skin colour inheritance, the genotype with three dominant and three recessive alleles will produce
(a) darkest skin colour
(b) lightest skin colour
(c) intermediate skin colour
(d) patches of black and white

Answer

C

Question. A pleiotropic gene
(a) is not found in humans
(b) is a single gene which exhibit multiple phenotypic expressions
(c) show effect on metabolic pathways, so as to produce various phenotypes
(d) Both (b) and (c)

Answer

D

Question. The polygenic traits
(a) are influenced by environment
(b) phenotype reflect the contribution of each allele
(c) effect of each allele is additive
(d) All of the above

Answer

D

Question. If there are four different types of nitrogenous bases (A, T, G and C) then how many different types of transitions and transversion are possible?
(a) Transition = 8, Transversion = 4
(b) Transition = 4, Transversion = 4
(c) Transition = 8, Transversion = 4
(d) Transition = 4, Transversion = 8

Answer

D

Question. Choose the incorrect pairing among the following.
(a) Sutton and Boveri – Chromosome theory
(b) Walter and Boveri – Behaviour of chromosome during cell divisions
(c) TH Morgan – Mutation
(d) Henking – Barr bodies

Answer

C

Question. Linked genes that were observed by Morgan were present on
(a) X-chromosome
(b) different chromosome
(c) heterologous chromosome
(d) paired chromosome

Answer

A

Question. Sickle-cell anaemia is a classical example of
(a) frame-shift mutation
(b) point mutation
(c) Both (a) and (b)
(d) None of the above

Answer

B

Question. Identify the correct choice for given symbols (A and B). (Image 114)
(a) A–Consanguineous mating; B–Mating
(b) A–Mating; B–Mating between relatives
(c) A–Mating; B–Consanguineous mating
(d) Both (b) and (c)

Answer

D

Question. Genes A, B and C are linked. Genes A and B are more close than A and C.
I. Amight be beforeBand C.
II. Bmight be betweenAand C.
III. Cmight be betweenAand B.
IV. More crosses cannot occur betweenAandCthan Aand B.
Find out the correct option for the given information.
(a) I and II
(b) II and III
(c) III and IV
(d) I, II and IV

Answer

A

Question. In human skin colour which is a polygenic trait, all dominant and all recessive alleles show
(a) Darkest and lightest skin colour, respectively
(b) Lightest and darkest skin colour, respectively
(c) Only darkest skin colour
(d) Only lightest skin colour

Answer

A

Question. Identify the symbols given below and the correct option with respect of A, B, C and D. (Image 113)
(a) A–Male, B–Female, C–Sex unspecified, D–Affected male
(b) A–Male, B–Female, C–Sterile, D–Carrier male
(c) A–Male, B–Female, C–Fertile, D–Affected female
(d) A–Female, B–Male, C–Sex unspecified, D–Carrier female

Answer

A

Question. In the following human pedigree, the filled symbols represent the affected individuals. Identify the type of given pedigree. (Image 118)
(a) X-linked dominant
(b) Autosomal recessive
(c) Autosomal dominant
(d) X-linked recessive

Answer

B

Question. In a family, father had a trait but mother did not. All their sons and daughter had this trait. The same trait was found in some grand daughters, though daughter were married to the normal persons. Choose the correct pedigree chart for this condition. (Image 119)

Answer

A

Question. Observe the pedigree chart given below. Find out the cause of trait, i.e. it is due to (Image 120)
(a) Incompletely dominant allele
(b) Dominant allele
(c) Either dominant or recessive allele
(d) Recessive allele

Answer

C

Question. The diagram given below show the inheritance of haemophilia in a family. What will be the genotype of the individual marked M? (Image 121)
(a) M− XY
(b) M − XX
(c) M X X − h h
(d) M X X − h

Answer

D

Question. Given below is a pedigree chart of a family with five children. It shows the inheritance of attached ear lobes as opposed to the free ones.Which of the following condition can be drawn?
(a) Parents are heterozygous
(b) Parents are homozygous recessive
(c) Parents are homozygous dominant
(d) All are incorrect

Answer

A

Question. Colour blindness in humans
(a) results in defect in either red or green cone of eyes
(b) is caused due to the mutation in gene found on X-chromosome
(c) affects males more frequently than females 
(d) All of the above

Answer

D

Question. A woman has an X-linked condition on one of her X-chromosomes. This chromosome can be inherited by NEET 2018
(a) Only grand children
(b) Only sons
(c) Only daughters
(d) Both (b) and (c)

Answer

D

Question. A normal woman whose father was colourblind, marries a normal man. What kinds of children can be expected and in what proportion ? 
(a) All daughters normal, 50% of sons colourblind
(b) All daughters normal, all sons colourblind
(c) 50% daughters colourblind, all sons normal
(d) All daughters colourblind, all sons normal

Answer

A

Question. Which of the following most appropriately describes haemophilia? 
(a) X-linked recessive gene disorder
(b) Chromosomal disorder
(c) Dominant gene disorder
(d) Recessive gene disorder

Answer

A

Question. In haemophilia, the affected protein is a part of a cascade of protein which is involved in the
(a) formation of RBCs
(b) formation of WBCs and platelets
(c) coagulation of blood
(d) anticoagulation

Answer

C

Question. Sickle-cell anaemia is an autosomal linked recessive trait that can be transmitted from parents to the offspring when both the partners are carriers for all the genes or heterozygous.
The disease is controlled by a single pair of allele, HbA and HbS . Identify X, Y and Z. (Image 130)
       X        Y          Z
(a) GTG   CAC     Val (GUG)
(b) CAC   CTC      Val (GUG)
(c) GTA   GAG      Val (GUG)
(d) GTC   GAC      Val (GUG)

Answer

A

Question. In sickle-cell anaemia,
(a) Both parents are heterozygous carriers, but are unaffected
(b) Single pair of allele controls the disease
(c) Only Hb Hb s s show diseased phenotype
(d) All of the above

Answer

D

Question. In individual suffering from phenylketonuria,
(a) enzyme phenylalanine hydroxylase is absent
(b) phenylalanine do not convert to tyrosine
(c) phenylpyruvic acid is formed
(d) All of the above

Answer

B

Question. b-thalassemia in humans is controlled by
(a) HBA2 gene on chromosome 16
(b) HBB gene on chromosome 11
(c) HBA1 gene on chromosome 15
(d) HBA1 and HBA2 gene on chromosome 8

Answer

B

Question. Failure of segregation of chromatid during cell division cycle results in the gain or loss of chromosome which as called
(a) aneuploidy
(b) hypopolyploidy
(c) hyperpolyploidy
(d) polyploidy

Answer

A

Question. A cell telophase stage is observed by a student in a plant brought from the field. He tells his teacher that this cell is not like other cells at telophase stage.
There is no formation of cell plate and thus the cell is containing more number of chromosomes as compared to other dividing cells. This would result in
(a) polyploidy
(b) somaclonal variation
(c) polyteny
(d) aneuploidy

Answer

A