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

 QUESTIONS 

1.Name one autosomal dominant and one autosomal recessive

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

3.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.

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.

Important Questions for NCERT Class 12 Biology Principles of Inheritance and Variation

Question. The law of dominance is used to explain the expession of only one of the parental characters in a monohybrid cross in ..................... and the expression of both in .......
(a) F1 and F2
(b) F2 and F3
(c) F1 and F3
(d) F2 and F1

Answer : A

Question. A diploid organism is heterozygous for 4 loci, how many types of gametes can be produced?
(a) 8
(b) 16
(c) 2
(d) 32

Answer : B

Question. When a cross is made between tall plant with yellow seed (TtYy) and tall plant with green seed (Ttyy), what proportion of phenotype in the offspring could be expected to be tall and green.
(a) 25%
(b) 12.5%
(c) 37.5%
(d) 50%

Answer : C

Question. The fact that the alleles donot show any blending and that both the characters are recovered as such in F2 generation, become the basis of
(a) Law of Dominance
(b) Law of paired factors
(c) Law of segregation
(d) Law of independent assortment

Answer : C 

Question. If there are four allelic forms for the gene controlling ABO blood group then what will be the number of possible genotypes
(a) 6
(b) 10
(c) 12
(d) 14

Answer : B

Question. Multiple alleles can be found during study of
(a) Gametes
(b) Individual
(c) Population
(d) All above

Answer : C

Question. In sickle cell anaemia which of the following genotype will show disease phenotype
(a) HbA HbA
(b) HbS HbS
(c) HbS HbA
(d) Both 1 and 2

Answer : B

Question. Which of the following is not concerned with sickle cell anaemia
(a) Sixth position of b-chain
(b) a chain of Hb
(c) Valine
(d) Haemoglobin

Answer : B

Question. Shape of seed depends on starch granules size, so inheritance of seed shape show ......... relationship while inheritance of starch grains show ...........
(a) Dominant recessive, codominance
(b) Incomplete dominance, codominance
(c) Dominant - recessive, incomplete dominance
(d) Codominance, incomplete dominance

Answer : C

Question. Inheritance of starch grains size shows
(a) Dominant recessive relationship
(b) Codominance
(c) Incomplete dominance
(d) Multiple allelism

Answer : C

Ques. Which one of the following conditions in humans is correctly matched with its chromosomal abnormality/linkage?
(a) Erythroblastosis fetalis - X-linked
(b) Down’s syndrome - 44 autosomes + XO
(c) Klinefelter’s syndrome - 44 autosomes + XXY
(d) Colour blindness - Y-linked 

Answer: C

Ques. Both sickle cell anaemia and Huntington’s chorea are
(a) virus-related diseases
(b) bacteria-related diseases
(c) congenital disorders
(d) pollutant-induced disorders. 

Answer: C

Ques. If a colour blind woman marries a normal visioned man, their sons will be
(a) all colour blind
(b) all normal visioned
(c) one-half colour blind and one-half normal
(d) three-fourths colour blind and one-fourth normal. 

Answer: A

Ques. Cri-du-chat syndrome in humans is caused by the
(a) trisomy of 21st chromosome
(b) fertilisation of an XX egg by a normal Y-bearing sperm
(c) loss of half of the short arm of chromosome 5
(d) loss of half of the long arm of chromosome 5.

Answer: C

Ques. Sickle cell anaemia has not been eliminated from the African population because
(a) it is controlled by dominant genes
(b) it is controlled by recessive genes
(c) it is not a fatal disease
(d) it provides immunity against malaria. 

Answer: D

Ques. 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?
(a) Sex-linked dominant
(b) Sex-linked recessive
(c) Sex-limited recessive
(d) Autosomal dominant 

Answer: B

Ques. A woman with 47 chromosomes due to three copies of chromosome 21 is characterised by
(a) superfemaleness
(b) triploidy
(c) Turner’s syndrome
(d) Down’s syndrome.

Answer: D

Ques. Haemophilia is more commonly seen in human males than in human females because
(a) a greater proportion of girls die in infancy
(b) this disease is due to a Y-linked recessive mutation
(c) this disease is due to an X-linked recessive mutation
(d) this disease is due to an X-linked dominant mutation. 

Answer: C

Ques. Which of the following is not a hereditary disease?
(a) Cystic fibrosis
(b) Thalassaemia
(c) Haemophilia
(d) Cretinism 

Answer: D

Ques. A woman with normal vision, but whose father was colour blind, marries a colour blind man. Suppose that the fourth child of this couple was a boy. This boy
(a) may be colour blind or may be of normal vision
(b) must be colour blind
(c) must have normal colour vision
(d) will be partially colour blind since he is heterozygous for the colour blind mutant allele.

Answer: A

Ques. A male human is heterozygous for autosomal genes A and B and is also hemizygous for haemophilic gene h. What proportion of his sperms will be abh?
(a) 1/8
(b) 1/32
(c) 1/16
(d) 1/4 

Answer: A

Ques. A normal woman, whose father was colour-blind is married to a normal man. The sons would be
(a) 75% colour-blind
(b) 50% colour-blind
(c) all normal
(d) all colour-blind.

Answer: B

Ques. The recessive genes located on X-chromosome humans are always
(a) lethal
(b) sub-lethal
(c) expressed in males
(d) expressed in females.

Answer: C

Ques. Pattern baldness, moustaches and beard in human males are examples of
(a) sex linked traits
(b) sex limited traits
(c) sex influenced traits
(d) sex determining traits. 

Answer: C

Ques. Which one of the following conditions though harmful in itself, is also potential saviour from a mosquito borne infectious disease ?
(a) Thalassaemia
(b) Sickle cell anaemia
(c) Pernicious anaemia
(d) Leukaemia 

Answer: B

Ques. Down’s syndrome is caused by an extra copy of chromosome number 21. What percentage of offspring produced by an affected mother and a normal father would be affected by this disorder?
(a) 100 %
(b) 75 %
(c) 50 %
(d) 25 % 

Answer: C

160. Christmas disease is another name for
(a) haemophilia B
(b) hepatitis B
(c) Down’s syndrome
(d) sleeping sickness.

Answer: A

Ques. A diseased man marries a normal woman. They get three daughters and five sons. All the daughters were diseased and sons were normal. The gene of this disease is
(a) sex linked dominant
(b) sex linked recessive
(c) sex limited character
(d) autosomal dominant.

Answer: A

Ques. Which of the following is a correct match?
(a) Down’s syndrome - 21st chromosome
(b) Sickle cell anaemia - X-chromosome
(c) Haemophilia - Y-chromosome
(d) Parkinson’s disease - X and Y chromosome

Answer: A

Ques. Sickle cell anaemia induce to
(a) change of amino acid in a-chain of haemoglobin
(b) change of amino acid in b-chain of haemoglobin
(c) change of amino acid in both a and b chains of haemoglobin
(d) change of amino acid either a or b chains of haemoglobin. 

Answer: B

Ques. Mongolian Idiocy due to trisomy in 21st chromsome is called
(a) Down’s syndrome
(b) Turner’s syndrome
(c) Klinefelter’s syndrome
(d) Triple X syndrome. 

Answer: A

Ques. In Drosophila, the XXY condition leads to femaleness whereas in human beings the same condition leads to Klinefelter’s syndrome in male. It proves
(a) in human beings Y chromosome is active in sex determination
(b) Y chromosome is active in sex determination in both human beings and Drosophila
(c) in Drosophila Y chromosome decides femaleness
(d) Y chromosome of man have genes for syndrome.

Answer: A

Very Short Answer Type Questions

Question. A geneticist interested in studying variations and patterns of inheritance in living beings prefers to choose organisms for experiments with shorter life cycle. Provide a reason. 
Answer: Many generations can be obtained/variations can be exhibited/selected faster. 
Detailed Answer:
A geneticist choose organism with a short life cycle because the organism will produce the offspring quickly in a short period of time. These offsprings can be studied by the geneticist and can also be further mated to produce more generation of offsprings.

Question. If two genes are located far apart from each other on a chromosome, what will be its effect on the frequency of recombination ?
Answer: Frequency of recombination will be higher.
Detailed Answer :
Frequency of recombinations will be higher because of greater chances of crossing over and also because of the lesser strength of linkage as the strength of the linkage is inversely proportional to the distance between the genes located on the chromosome.

Question. Write the percentage of the pea plants that would be heterozygous tall in F₂ generation when tall heterozygous F₁ pea plants are selfed.
Answer: 50% heterozygous tall pea plants would be obtained in F₂ generation when tall heterozygous pea plants are selfed. 

Question. Write technical term used in human ABO blood groups for IA, IB and i 
Answer : IA, IB and i are the alleles of the gene I.

Question. Name the respective pattern of inheritance where F₁ phenotype.
(i) does not resemble either of the two parents and is in between the two.
(ii) resemble only one of the parents.
Answer : (i) Incomplete dominance
(ii) Dominance. 

Question. AaBb was crossed with aabb. What would be the phenotypic ratio of the progeny ? Mention the term used to denote this kind of cross ?
Answer : Phenotypic ratio would be 1:1:1:1. It is test cross.

Question. Name the type of cross that would help to find the genotype of a plant bearing violet flowers.
Answer : Crossing of an F₁ hybrid with it’s recessive plant is called test cross. It is used to find out the unknown genotype.

Question. How many types of phenotypes would you expect in F₂ generation in a monohybrid cross ? 
Answer : Two types in the ratio of 3 : 1. 1

Question. Mention two contrasting flower-related traits studied by Mendel in his pea plant experiments.
Answer : Contrasting flower-related traits :
(i) Flower colour : violet/white or red/white.
(ii) Flower position : Axial/Terminal. 

Question. Give an example of a polygenic trait in humans.
Answer : Skin colour/height in humans (any other suitable example)

Question. What are true breeding lines’ that are used to study inheritance pattern of traits in plants ?
Answer : Self pollination continuous for several generations / homozygous. 
Detailed Answer :
True breeding lines for a trait are those plants that have been generated through repeated continuous self-pollination and have become homozygous for a particular trait. They show stability in the inheritance of trait for several future generation. 

Question. A garden pea plant (A) produced inflated yellow pod, and another plant (B) of the same species produced constricted green pods. Identify the dominant traits. 
Answer : Inflated, green pods. 

Short Answer Type Questions

Question. Write the scientific name of the fruit-fly. Why did Morgan prefer to work with fruit-flies for his experiments ? State any three reasons.
Answer : Drosophila melanogaster 
Grown in simple synthetic medium, complete the life cycle in two weeks / short life cycle, single mating produce more progeny, dimorphism, many heritable variations / easy to handle.
Detailed Answer :
Morgan preferred to work with fruit flies because of the following reasons:
(i) It can be grown in simple synthetic medium.
(ii) It can complete its life cycle within two weeks i.e., it has short life cycle.
(iii) Single mating produce more progenies.
(iv) It shows dimorphism and many heritable variations.
(v) It is easy to handle. 

Question. State the Mendelian principle which can be derived from a dihybrid cross and not from monohybrid cross.
Answer : From the dihybrid cross, law of independent assortment can be derived which states that when two pairs of traits are combined in a hybrid, segregation of one pair of characters is independent of the other pair of characters.

Question. With the help of one example, explain the phenomena of co-dominance and multiple allelism in human population.
Answer : ABO blood group in human being is an example of multiple allelism.
Three alleles for the gene I i.e. I^A, I^B, i.
When I^A and I^B are present together the blood group is AB.
Both A and B are expressed and is called co-dominance.
Detailed Answer :
ABO blood group in human population is an example of multiple allelism and co-dominance.
There are three alleles of gene I, i.e., IA, IB and i. thus exhibiting multiple allelism. Out of these three alleles any two alleles may occur in human being, one alleles is contributed from each parent. The alleles I^A and I^B are dominant over allele i and produce A and B phenotypes respectively. But however when alleles I^A and I^B are present together both of them express equally and independently producing AB phenotype. This is a phenomenon called co-dominance.

Question. ABO blood groups is a good example of codominance. Justify.
Answer : (i) ABO blood group in humans is contributed by gene ‘I’ that has 3 alleles ‘I^A‘ ‘I^B‘ and ‘i.’
(ii) Because human beings are diploid and each person has two of the three alleles.
(iii) I^A and I^B produce two different types of sugar while allele i does not produce sugar on the plasma membrane of RBC.
(iv) When I^A and I^B present they produce their own type of sugar-this is called co-dominance.

Question. Differentiate between multiple allelism and pleiotropy with the help of an example each.
Answer : Multiple alleles : More than two alternate forms of a gene present on the same locus of a homologous pair of chromosomes in a population are called multiple alleles. They control the single trait. For example, ABO blood group in humans.
Pleiotropic genes : The gene having a multiple phenotypic effect because of its ability to control the expression of a number of characters is called pleiotropic gene. For example, skin pigmentation and phenylketonuria. 

Question. In Snapdragon, a cross between true breeding red flower (RR) plants and true breeding white flower (rr) plants showed a progeny of plants with all pink flowers.
(i) The appearance of pink flowers is not known as blending. Why ?
(ii) What is the phenomenon known as ?
Answer : (i) R (dominant allele red colour) is not completely dominant over r (recessive allele white colour) / r maintains its originality and reappear in F₂ generation. 
(ii) Incomplete dominance. 
Detailed Answer :
(i) The appearance of pink flowers in snapdragon in F₁ generation is not due to blending of genes Rr which are although together in hybrid, because the parental characters i.e. red and white flowers appear again in F₂ generation without any change.
(ii) Neither of the two alleles are completely dominant over each other therefore the phenomenon is called as incomplete dominance.

Long Answer Type Questions

Question. (i) How does a chromosomal disorder differ from a Mendelian disorder ?
(ii) Name any two chromosomal aberrations associated disorders.
(iii) List the characteristics of the disorders mentioned above that help in their diagnosis.
Answer : (i) Mendelian disorders are mainly determined by alteration or mutation in the single gene, chromosomal disorders are due to absence / excess / abnormal arrangement of one or more chromosomes. 
(ii) Turner’s syndrome, Klinefelter’s syndrome, Down’s syndrome. (Name any two) 
(iii) Turner’s syndrome : Such a disorder is caused due to the absence of one of the X chromosomes i.e. 44 with XO, Such females are sterile as ovaries are rudimentary besides
other features including lack of other secondary sexual characters.
Klinefelter’s syndrome : This genetic disorder is also caused due to the presence of an additional copy of X-chromosome resulting into a karyotype of 47, XXY.
Such an individual has overall masculine development, however, the feminine development (development of breast i.e. Gynaecomastia) is also expressed. Such individuals are sterile.
Down’s syndrome : The cause of this genetic disorder is the presence of an additional copy of the chromosome number 21 (trisomy of 21)
This disorder was first described by Langdon Down (1866).
The affected individual is short statured with small round head, furrowed tongue and partially open mouth. Palm is broad with characteristic palm crease. Physical, psychomotor and mental development is retarded. 

Question. (i) State the cause and symptoms of colour blindness in humans.
(ii) Statistical data has shown that 8% of the human males are colour-blind whereas only 0.4% of females are colour-blind. Explain giving reasons how is it so. 
Answer : (i) Cause-sex-linked recessive disorder.
Symptoms-failure to discriminate between red and green colour.
(ii) Since males have only one X chromosome gene for colour blindness, if present in any one parent will always be expressed, whereas in female it will be expressed only if it is present on both the X chromosomes, when both parents are carrying gene for colour blindness. 
Detailed Answer :
(i) The colour blindness is due to recessive sex linked disorder. In this disorder of vision the patient is unable to distinguish between red and green colour. The gene for colour blindness
is located on the X chromosome.
(ii) The colour blindness is found in about 8% of the males and only 0.4% of the females. The greater prevalence of the colour blindness in males is due to the presence of only one X chromosome and the hemizygous (X & Y) expression of the allele for colour blindness i.e. if gene for colour blindness is present on X-chromosome of male then it will always express while in case of females the incidence of disease of colour blindness is possible only in homozygous condition (X^CX^C) i.e. if both the X-chromosomes carry the allele (X^CX^C) for colour blindness. Occurrence of allele for colour blindness on one of the X chromosome makes the female a carrier X^CX.

Question. (i) Why is haemophilia generally observed in human males ? Explain the condition under which a human female can be haemophilic.
(ii) A pregnant human female was advised to undergo M.T.P. It was diagnosed by her doctor that the foetus she is carrying has developed from a zygote formed by an XX-egg fertilized by Y-carrying sperm. Why was she advised to undergo M.T.P. ?
Answer : (i) Haemophilia is caused due to the recessive gene on X chromosome. 
Y has no allele for this / if a male is XhY then he is haemophilic / if male inherits Xh from the mother he is haemophilic (with the genotype X^hY). 
If female inherits X^hX^h, one from the carrier mother and one from her haemophilic father then she can be haemophilic. 
(ii) Embryo has (trisomy of sex chromosome) XXY / Klinefelter’s syndrome. 
Advised MTP since child will have the following problems :
Male with feminine traits / like gynecomastia / under developed testes / sterile.