CBSE Class 12 Biology Principles Of Inheritance And Variation Worksheet Set B

Principles of Inheritance and Variation MCQ Questions with Answers Class 12 Biology

Question: The monohybrid genotypic ratio 1 : 2 : 1 in F2 generation indicates
(a) segregation
(b) independent assortment
(c) dominance
(d) incomplete dominance

Answer-A

Question: The distance between the genes is measured by
(a) angstrom
(b) map unit
(c) Dobson unit
(d) millimetre

Answer : B

Question: In a dihybrid cross, F2 phenotypic ratio is 13 : 3. It is case of
(a) complementary genes
(b) epistatic genes
(c) multigenic inheritance
(d) incomplete dominance

Answer : B

Question: In sickle-cell anaemia, shape of RBCs under oxygen tension becomes
(a) biconcave disc like
(b) elongated and curved
(c) circular
(d) spherical

Answer : B

Question: Linkage reduces the frequency of
(a) hybrids.
(b) all parental types.
(c) homozygous recessive parents.
(d) heterozygous recessive parents.

Answer : A

Question: In XO type of sex determination
(a) females produce two different types of gametes.
(b) males produce two different types of gametes.
(c) females produce gametes with Y chromosomes.
(d) males produce single type of gametes.

Answer : B

Question: HbA and HbS alleles of normal and sickle celled RBC are
(a) dominant-recessive alleles.
(b) polygenic alleles.
(c) codominant alleles.
(d) multiple alleles.

Answer : C

Question: The ‘X’ body of Henking was observed in
(a) all sperms during spermatogenesis.
(b) all eggs during oogenesis.
(c) half of the sperms during spermatogenesis.
(d) half of the eggs during oogenesis.

Answer : C

Question: The test cross is used to determine the
(a) genotype of the plant
(b) phenotype of the plant
(c) both (a) and (b)
(d) None of the above

Answer : A

Question: Sickle cell anaemia is
(a) caused by substitution of valine by glutamic acid in the beta globin chain of haemoglobin.
(b) caused by a change in a single base pair of DNA.
(c ) characterized by elongated sickle like RBCs with a nucleus.
(d) an autosomal linked dominant trait.

Answer : B

Question: The contrasting pairs of factors in Mendelian crosses are called
(a) multiple alleles
(b) alleles
(c) alloloci
(d) paramorphs

Answer : B

Question: Mental retardation in man associated with sex chromosomal abnormality is usually due to
(a) increase in size of X-chromosome.
(b) increase in size of Y-chromosome.
(c) increase in number of Y-chromosome.
(d) increase in number of X-chromosome.

Answer : D

Question: Which of the following crosses will give tall and dwarf pea plants in same proportions?
(a) TT × tt
(b) Tt × tt
(c) TT × Tt
(d) tt × tt

Answer : B

Question: The F2 generation of a cross produced identical phenotypic and genotypic ratio. It is not an expected Mendelian result, and can be attributed to
(a) independent assortment
(b) linkage
(c) incomplete dominance
(d) none of the above

Answer : C

Question._________ is a type of trait whose phenotype is influenced by more than one gene
(a) Oncogenic Trait
(b) Monogenic trait
(c) Polygenic trait
(d) None of the above

Answer : C

Question. Two genes very close on a chromosome will show:
(a) No crossing over
(b) High crossing over
(c) Hardly an crossing over
(d) Only double crossing over

Answer : A

Question. A point mutation is:
(a) Thalassemia
(b) Sickel-cell anaemia
(c) Down‟s syndrome
(d) Nightblindness

Answer : B

Question. A man marries a woman and both do not show any apparent traits of inherited disease.
Five sons and two daughters
are born, and three of their sons suffer from a disease. However, none of the daughters is affected. The following mode of inheritance for the disease is
(a) Sex-linked recessive
(b) Sex-linked dominant
(c) Autosomal dominant
(d) None of the above

Answer : A

Question. A true hybrid condition is:
(a) tt Rr
(b) Tt rr
(c) tt rr
(d) Tt Rr

Answer : D

Question. If a F1 expresses a character, it is called _________
(a) Incomplete dominance
(b) Dominant
(c) Co-dominant
(d) Recessive

Answer : B

Question. Failure of segregation of chromatids during cell division Cycle result in the gain or loss of a cromosome (s) is called :
(a) Female heterogamety
(b) Male heterogamety
(c) Aneuploidy
(d) None of these

Answer : C

Question. Down‟s syndrome is a:
(a) Mendelian disorder
(b) Chromosomal disorder
(c) can be both
(d) None of these

Answer : B

Question. Colour blindness is an _________ linked recessive trait
(a) Z chromosome
(b) Y chromosome
(c) X chromosome
(d) None of the above

Answer : C

Question. Mendel‟s laws were rediscovered by:
(a) Correns
(b) TShermak
(c) De vries
(d) All of these.

Answer : D

Assertion Reason Type Questions:

A. Both A and R are true and R is the correct explanation of A
B. Both A and R are true and R is not the correct explanation of A
C. A is true but R is false
D. A is False but R is true

Question. Assertion: In dog flowers F1 plants produce pink flowers.
Reason: It is due to codominance of flower colour alleles with both genes expressing themselves equally.

Answer : C

Question. Assertion: In monohybrid cross, at F2 stage, both parental traits are expressed in 3 : 1 proportion.
Reason: At F2 stage, the contrasting parental traits show blending.

Answer : C

Question. Assertion: Sickle cell anaemia occurs due to a point mutation.
Reason: The mRNA produced from HbS gene has GUG instead of GAG.

Answer : A

Question. Assertion: The possibility a human male becoming haemophilic is extremely rare.
Reason: Mother of such a male should be normal and the fathers should be haemophilic.

Answer : D

Question. Assertion: Gametes receives only one allele of a gene.
Reason: During gamete formation, mitosis takes place leads to formation of haploid cells.

Answer : C

Question. Assertion: Mendel used true-breeding pea lines for artificial pollination experiments for his genetic studies.
Reason: For several generations, a true-breeding line shows the stable trait inheritance and expression.

Answer : A

Question. Assertion: Cross of F1 individual with recessive homozygous parent is test cross.
Reason: No recessive individual is obtained in the monohybrid test cross progeny.

Answer : C

Question. Assertion: A test cross is used to determine the phenotype of an organism.
Reason: F2 generation of a monohybrid test cross produces one or two phenotypes depending upon the genotype of the unknown organism.

Answer : D

Question. Assertion: A good example of multiple alleles is ABO blood group system.
Reason: When I^A and I^B are present together in ABO blood group system, they both express their own types.

Answer : B

Question. Assertion: XO type of sex determination is found in large number of insects.
Reason: 50 % of sperms contain X chromosome and the other 50% contain “O” chromosome.

Answer : C

Very Short Answer Type Questions:

Question. How will you find out whether a given plant is homozygous or heterozygous?
Answer : To test whether a plant is homozygous or heterozygous, test cross is performed in which individual is crossed with homozygous recessive for the trait. If plant is heterozygous, progeny of test cross consists of tall and dwarf plants in the ratio l:l If plant is homozygous, progeny of test cross will have all tall plants

Question. Identify the sex of organism as male or female in which the sex chromosome are found as
(i) ZW in bird (ii) XY in Drosophila (iii) ZZ in birds. (iv) XO in grasshopper.
Answer : (i) Female; (ii) Male; (iii) Female (iv) Male

Question. Why do sons of haemophilic father never suffer from this trait?
Answer : Since haemophilic is a sex – linked character, it shows criss – cross inheritance i-e from father to his daughter therefore son of haemopilic father is never haemophilic.

Question.The map distance in certain organism between genes A & B is 4 units, between B & C is units, &between C &D is 8 units which one of these gene pairs will show more recombination frequency? Give reason.
Answer : C& D will show maximum gene recombination because genes which are more closely linked, frequency of recombination is least & vice versa.

Question. The human male never passes on the gene for haemophilia to his son. Why is it so?
Answer : The gene for haemophilia is present on X chromosome. A male has only one X chromosome which he receives from his mother and Y chromosome from father. The human male passes the X chromosome to his daughters but not to the male progeny (sons).

Question. How is the child affected if it has grown from the zygote formed by an XX-egg fertilized by Y-carrying sperm? What do you call this abnormality?
Answer : If a child has grown from the zygote formed by XX-egg fertilized by Y-sperm, the child will suffer from klinefiter syndrome & will have XXY genotype. It is characterized by prominent feminine characters e.g. tall stature with feminised physique, Breast development pubic hair pattern, poor beard growth &sterility.

Case Based Questions:

1. Nondisjunction is the failure of homologous chromosomes to disjoin correctly during meiosis. It leads to the formation of a new cell with an abnormal amount of genetic material. A number of clinical conditions are the result of this type of chromosomal mutation. This results in the production of gametes containing a greater or lesser chromosomal amount than normal ones. Consequently, the individual may develop a trisomal or monosomal syndrome. Non disjunction can occur in both Meiosis I and Meiosis II of the cellular division. Non disjunction can occur in both Meiosis I and Meiosis II of the cellular division. It is also the main cause of many genetic disorders; however, its origin and process remain vague. Although it results in the majority of cases from errors in the maternal meiosis II, both paternal and maternal meiosis I do influence it. The maternal age, is considered a risk factor of trisomy‟s, as well as recombination alterations and many others that can affect the chromosomal segregation.

Question. By interpreting the graph of Down syndrome frequency and mothers age, select the best conclusion (s) from the following options. 

i. Aneuploidy is not influenced by mother‟s age.
ii. Delivery before 30 years of age can decrease the incidence of aneuploidy in most cases
iii. The chance of aneuploidy increases up to 22 years of age.
iv. There is a dramatic increase in aneuploidy if maternal age exceeds 30
(a) I only
(b) both I and iii
(c) both ii and iv
(d) I, iii and iv.

Answer : C

Question. The type of genetic disorders mainly caused by chromosomal non-disjunction is
(a) Chromosomal disorders
(b) Mendelian disorders
(c) Incomplete dominance
(d) All the above

Answer : A

Question. Assertion: All types of genetic disorders are caused by chromosomal nondisjunction
Reason: Chromosomal nondisjunction always affects female individuals.
(a) Both assertion and reason are correct and reason is the correct explanation of assertion
(b) Both assertion and reason are correct but reason is not the correct explanation of assertion
(c) Assertion is correct but reason is incorrect
(d) Both Assertion and reason are incorrect.

Answer : D

Question. Which of the following conclusions can be true regarding aneuploidy?
i. It is the presence of an extra chromosome in a diploid cell.
ii. An aneuploid cell differs from other cells only in size.
iii. It can be less number of chromosomes in a diploid cell.
iv. Aneuploidy always affects female individuals.
(a) I only
(b) both I and iii
(c) both ii and iii
(d) I, iii and iv.

Answer : B

Question. Considering the different phases of meiosis, select the correct statements from the following.
i. Errors in meiosis I is the only cause of aneuploidy
ii. Aneuploidy always affects sex chromosomes.
iii. Most of the aneuploidy results from errors in cell division involved in egg formation.
iv. Nondisjunction in meiosis I can lead to more abnormal cells than disjunction in meiosis II.
(a) I only
(b) both I and iii
(c) both iii and iv
(d) I, iii and iv.

Answer : C

2.The word parthenogenesis originates from the Greek language meaning virgin birth. In honeybees, the drones are entirely derived from the queen, their mother. Parthenogenesis is a form of reproduction where the unfertilized egg will develop into a DRONE BEE while fertilized eggs will hatch into the WORKER BEES. This type of reproduction occurs in various species in nature.
Fertilized eggs will develop into QUEEN BEES or WORKER BEES. It will depend on the size and type of the cells and on the composition of food fed by WORKER BEES to BEE LARVAE. The DRONE BEES are always born from unfertilized eggs. Drones produce sperm cells that contain their entire genome, so the sperm are all genetically identical except for mutations. The male bees‟ genetic makeup is therefore entirely derived from the mother, while the genetic makeup of the female worker bees is half derived from the mother, and half from the father. The QUEEN BEES mate with numerous DRONE BEES high in the air. This polyandry and the phenomenon of parthenogenesis in honey bees create a super-organism in the beehive populations. The WORKER BEES who share the same father and mother are called SUPER SISTERS because they are more closely related to each other than their sisters who have different fathers.

Question. In a honey bee colony, the queen is different from workers in
(a) Chromosome number
(b) The way of production in sexual or asexual method
(c) The type of gametes involved in production
(d) The type of food given in larval stage.

Answer : D

Question. Some of the members in a honey bee colony have no father but have grandfather, they are
(a) Workers
(b) Drones
(c) Queen
(d) Both drones and workers.

Answer : B

Question. What can be the advantage of parthenogenesis for species survival?
(a) It leads to much variation in offspring
(b) It helps isolated female individuals to reproduce
(c) It helps to evolve clones
(d) It conserves same chromosome number in generations.

Answer : B

i. Identify the members of a bee colony which possess same chromosome number.
(a) Drone and Queen
(b) Drone and Worker
(c) Queen and Worker
(d) Both a and c.

Answer : C

Question. The cell division involved in the formation of egg and sperm in honey bees respectively
(a) Mitosis and meiosis
(b) Mitosis only
(c) Meiosis and mitosis
(d) Meiosis only.

Answer : C

3. Study the given figure and answer the questions that follow. 

Question. What is the cause of this disease?
Answer : Substitution of Glutamic acid by Valine at the sixth position of the beta globin chain.

Question. What are the possible phenotypes of children born to a couple of Carrier woman and hemophilic man.

Answer : Normal, Carrier and Sickle celled children.

Question. Name disorder that is shown in the figure.
Answer : Sickle cell anemia.

Question. How is this disorder inherited?
Answer : Inherited as Autosome linked recessive trait.

4. In 1911, while studying the chromosome theory of heredity, biologist Thomas Hunt Morgan had a major breakthrough. Morgan occasionally noticed that "linked" traits would separate. Meanwhile, other traits on the same chromosome showed little detectable linkage. Morgan considered the evidence and proposed that a process of crossing over, or recombination, might explain his results.
Specifically, he proposed that the two paired chromosomes could "cross over" to exchange information. When proposing the idea of crossing over, Morgan also hypothesized that the
frequency of recombination was related to the distance between the genes on a chromosome, and that the interchange of genetic information broke the linkage between genes. Morgan imagined that genes on chromosomes were similar to pearls on a string; in other words, they were physical objects. The closer two genes were to one another on a chromosome, the greater their chance of being inherited together. In contrast, genes located farther away from one another on the same chromosome were more likely to be separated during recombination. Therefore, Morgan correctly proposed that the strength of linkage between two genes depends upon the distance between the genes on the chromosome. This proposition became the basis for construction of the earliest maps of the human genome.

Question. How does the concept of linkage help in gene mapping?
(a) Distance between the linked genes can be measured by frequency of recombination.
(b) Linked genes express the characters easily than non-linked genes
(c) Linked genes are larger in size
(d) Position of linked genes can be observed in ultra-microscopy.

Answer : A

Question. The linked genes behave differently in the way that
(a) Linked genes assort easily and enter different gametes
(b) Linked genes don‟t segregate easily and inherit as a unit
(c) Linked genes lead to high cross over and recombination
(d) Linked genes segregate only at the time of cell division

Answer : B

Question. In Morgan‟s experimental crosses, the white eyed flies were all males. The clue that it gives 

(a) White eye colour is X-linked dominant trait
(b) White eye colour is Y linked dominant trait
(c) White eye colour is X-linked recessive trait
(d) White eye color is Y-linked dominant trait.

Answer : C

Question. The traits which are found tightly linked in Morgan‟s experiments were
(a) Body colour and wing size
(b) Eye colour and wing size
(c) Body colour and eye colour
(d) Body size and wing size.

Answer : C

Question. Which of the following condition favours linkage?
(a) Genes on the same chromosome and are closely placed.
(b) Genes on different chromosomes and distantly placed.
(c) Genes on the same chromosome but distantly placed
(d) Genes present on different autosomes.

Answer : A

Unit VII
Chapter 5
Principles of Inheritance and Variation
Pleiotropy

Pleiotropism is defined as a phenomenon when single gene may produce more than one effect (the multiple effect of a gene) or control several phenotypes depending on its position.
The basis of Pleiotropy is the interrelationship between the metabolic pathways that may contribute towards different phenotypes. In phenylketonuria, mutation of a gene that codes for the enzyme phenyl alanine hydroxylase.

This results in a phenotypic expression characterized by mental retardation and a reduction in hair and skin pigmentation.
In drosophila white eye mutation leads to depigmentation in many other parts of the body, giving a pleitropic effect.
In transgenic organisms, the introduced gene can produce different effects depending on where the gene has introgressed.

More Question

1.Name one autosomal dominant and one autosomal recessive Mendelian disorder in humans.

2.Write the genotype of i) an individual who is carrier of sickle cell anaemia gene but apparently unaffected, and ii) an individual affected with the disease.

3.A human being suffering from Down’s syndrome shows trisomy of 21st chromosome. Mention the cause of this chromosomal abnormality.

4.Write the percentage of F2 homozygous and heterozygous populations in a typical monohybrid cross.

5.A man with blood group A married a woman with B group. They have a son with AB blood group and a daughter with blood group O. Workout the cross and show the possibility of such inheritance.

6.The male fruit fly and female fowl are heterogametic while the female fruit fly and male fowl are homogametic. Why are they called so?

7.A plant of Antirrhinum majus with red flowers was crossed with another plant of the same species with white flowers. The plants of the F1 generation bore pink flowers. Explain the pattern of inheritance with the help of a cross.

8.A woman with blood group ‘O’ married a man with AB group. Show the possible blood groups of the progeny. List the alleles involved in this inheritance.

9.How does a test cross help in identifying the genotype of the organism? Explain.

10.When a tall pea plant was selfed, it produced one fourth of its progeny as dwarf. Explain with the help of a cross.

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