Extra Questions for Class 12 Biology Chapter 5 Principles of Inheritance and Variation
Q.1. A garden pea plant produced axial white flowers. Another of the same species produced terminal violet flowers. Identify the dominant traits.
Ans. Axial, violet flower.
Q.2. When a tall pea plant was self-pollinated, one-fourth of the progeny were dwarf. Give the genotype of the parent and dwarf progenies.
Ans. Genotype of parent is Tt and the genotype of dwarf progenies is tt.
Q.3. State a difference between a gene and an allele.
Ans. Gene contains information that is required to express a particular trait whereas alleles are alternating forms of a gene and are the code for a pair of contrasting traits for e.g., for plant height has two alleles – for tallness and dwarfness.
Q.4. Name the respective pattern of inheritance where F1 phenotype
(a) does not resemble either of the two parents and is in between the two.
(b) resembles only one of the two parents.
Ans. (a) Incomplete dominance (b) Dominance
Q.5. A cross was carried out between two pea plants showing the contrasting traits of height of the plant. The result of the cross showed 50% of parental characters. Name the type of cross.
Ans. Test cross
Q.6. 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.
Ans. Inflated green pod is the dominant trait.
Q.7. A garden pea plant produced round green seeds. Another of the same species produced wrinkled yellow seeds. Identify the dominant traits.
Ans. Round, yellow seed are the dominant traits.
Q.8. Mention any two contrasting traits with respect to seeds in pea plant that were studied by Mendel.
Ans. Round/Wrinkled, Yellow/Green
Q.9. Write the possible genotypes, Mendel got when he crossed F1 tall pea plants with a dwarf pea plant.
Ans. Possible genotypes: Tt and tt.
Q.10. How many kinds of phenotypes would you expect in F2 generation in a monohybrid cross?
Ans. Two (e.g., Tall and dwarf).
Q.11. Discuss is the genetic basis of wrinkled phenotype of pea seeds.
Ans. Wrinkled seed shape is a recessive trait. It expresses only under homozygous condition of alleles.
Q.12. Mention the type of allele that expresses itself only in homozygous state in an organism.
Ans. Recessive allele.
Q.13. Name the stage of cell division where segregation of an independent pair of chromosomes occurs.
Ans. Anaphase-I of Meiosis-I.
Q.14. Name the type of cross that would help to find the genotype of a pea plant bearing violet flowers.
Ans. Test cross.
Q.15. Why, in a test cross, did Mendel cross a tall pea plant with a dwarf pea plant only?
Ans. To determine the genotype of the tall plant, whether it is homozygous dominant or heterozygous, as dwarfness is a recessive trait which is expressed only in homozygous condition and he was sure of genotype of dwarf plant.
Q.16. If the frequency of a parental form is higher than 25% in a dihybrid test cross, what does that indicate about the two genes involved?
Ans. It shows that the two genes are linked.
Q.17. For the expression of traits, genes provide only the potentiality and the environment provides the opportunity. Comment on the veracity of the statement.
Ans. Phenotype = Genotype + Environment
(Trait) (Potentiality) (Opportunity)
Q.18. 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.
Ans. This is because many generations can be obtained (in a short time) and selection of character becomes faster.
Q.19. How many type of gametes are produced by the individual with genotype AABBCCDD and AaBbCcDd?
Ans. One type of gamete by individual (AABBCCDD) ABCD and sixteen (= 24 = 16) type of gametes by individual AaBbCcDd.
Q.20. Mention the combination(s) of sex chromosomes in a male and a female bird.
Ans. Male bird – ZZ, Female bird – ZW
Q.21. Give an example of a chromosomal disorder caused due to non-disjunction of autosomes.
Ans. Down’s Syndrome.
Q.22. Write the percentage of F2 homozygous and heterozygous populations in a typical monohybrid cross.
Ans. The ratio of a typical monohybrid cross is 1 : 2 : 1 where 50% are homozygous and 50% are
heterozygous populations. (25% homozygous dominant, 25% homozygous recessive)
Q.23. Write the types of sex determination mechanisms the following crosses show. Give an example of each type.
(i) Female XX with Male XO
(ii) Female ZW with Male ZZ
Ans. (i) Male heterogamety, Grasshopper
(ii) Female heterogamety, Birds
Q.24. A male honeybee has 16 chromosomes whereas its female has 32 chromosomes. Give one reason.
Ans. Male honeybee develops from unfertilised female gamete (Parthenogenesis) and thus has 16 chromosomes whereas female develops by fertilisation and thus has 32 chromosomes.
Q.25. Name a human genetic disorder due to the following:
(i) An additional X-chromosome in a male
(ii) Deletion of one X-chromosome in a female
Ans. (i) Klinefelter’s Syndrome (ii) Turner’s Syndrome
Q.26. How many chromosomes do drones of honeybee possess? Name the type of cell division involved in the production of sperms by them.
Ans. Drones possess 16 chromosomes. Mitosis is involved in the production of sperms.
Q.27. State what does aneuploidy lead to.
Ans. Aneuploidy leads to individuals with abnormal number of chromosomes. Some disorder due to aneuploidy are Down’s Syndrome, Turner’s Syndrome, Klinefelter’s Syndrome.
Q.28. Give an example of a human disorder that is caused due to a single gene mutation.
Ans. Sickle-cell anaemia.
Q.29. A haemophilic man marries a normal homozygous woman. What is the probability that their daughter will be haemophilic?
Ans. 0% because only one X chromosome will carry the haemophilia gene. So, she will be a carrier.
Q.30. Observe the pedigree chart and answer the following questions:
(a) Identify whether the trait is sex-linked or autosomal.
(b) Give an example of a disease in human beings which shows such a pattern of inheritance.
Ans. (a) The trait is sex-linked.
(b) Haemophilia, Colour blindness (Any one)
Q.31. A haemophilic son was born to normal parents. Give the genotypes of the parents and son.
Ans. Father : 44 + XY
Mother : 44 + XXh
Son : 44 + XhY.
(Xh= X chromosome with gene for haemophilia)
Q.32. State the chromosomal defect in individuals with Turner’s syndrome.
Ans. Monosomy of sex chromosome in females (XO condition).
Q.33. The egg of an animal contains 10 chromosomes, of which one is X-chromosome. How many autosomes would there be in the karyotype of this animal?
Ans. There will be 9 pairs of autosomes in the karyotype of this animal.
Q.34. A human being suffering from Down’s syndrome shows trisomy of 21st chromosome. Mention the cause of this chromosomal abnormality.
Ans. Due to non-disjunction i.e., 21st pair of chromosomes fail to separate during gametogenesis.
Therefore, the gamete possesses 24 chromosomes instead of 23. When such a gamete fuses with
another gamete, the zygote will have three copies of chromosome 21 causing trisomy.
Q.35. Name the event, during cell division cycle that results in the gain or loss of chromosome.
Ans. Failure of segregation of chromosomes.
Q.36. Why do normal red blood cells become elongated sickle shaped structures in a person suffering from sickle cell anaemia?
Ans. Due to point mutation, glutamic acid (Glu) is replaced by valine (Val) at the sixth position of β-globin chain of haemoglobin molecule. Under oxygen stress erythrocytes lose their circular shape and become sickle-shaped.
Q.37. Name one autosomal dominant and one autosomal recessive Mendelian disorder in humans.
Ans. Huntington’s disease is an autosomal dominant disorder and sickle-cell anaemia is an autosomal recessive disorder.
Q.38. Why is it that the father never passes on the gene for haemophilia to his sons? Explain.
Ans. Haemophilia is a sex-linked recessive disease and the defective gene is present on X chromosome only and not on Y chromosome. Father never passes X chromosome to the son as father only contributes Y chromosome to the son.
