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Assignment for Class 12 Biology Chapter 5 Principles Of Inheritance And Variation
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Chapter 5 Principles Of Inheritance And Variation Class 12 Biology Assignment
POINTS TO REMEMBER
Allele : Various or slightly different forms of a gene, having same position on chromosomes.
Phenotype : The observable or external characteristics of an organism Genotype : The genetic constitution of an organism.
Monohybrid cross : A cross between two individuals of species, considering the inheritance of single pair of contrasting character e.g., a cross between pure tall (TT) and Dwarf (tt).
Dihybrid cross : A cross between two individuals of a species, considering the inheritance of two pairs of contrasting traits/characters e.g., a cross between Round and Yellow (RRYY) and wrinkled and green (rryy) pea seeds Incomplete dominance : When one of the two alleles of a gene is incompletely dominant over the other allele.
Co-dominance : When two alleles of a gene are equally dominant and express themselves even when they are together.
Multiple allelism : When a gene exists in more than two allelic forms e.g., gene for blood group exist in three allelic forms, IA, IB and i.
Aneuploidy : The phenomenon of gain or loss of one or more chromosome(s), that results due to failure of separation of homologous pair of chromosomes during meiosis. Trisomy : The condition in which a particular chromosome is present in three copies in a diploid cell/ nucleus.
Male heterogamety : When male produces two different types of gametes/ sperms e.g., In human beings X and Y.
Important Questions for NCERT Class 12 Biology Principles of Inheritance and Variation
Question. Mendel found that the F1 always resembled either one of the parents and that the trait of the other parent was not seen in them. This is due to :
(a) Segregation
(b) Dominance
(c) Partial dominance
(d) Unit factor
Answer : B
Question. Study of family history about inheritance of a particular trait in several generations of a family called
(a) Phylogeny
(b) Ontogeny
(c) Pedigree analysis
(d) Cladistics
Answer : C
Question. Symbols 5 used in pedigree analysis, represents
(a) Five offspring with unspecified sex
(b) Five diseased offspring
(c) Five unaffected offspring
(d) Five affected offsprings
Answer : C
Question. It was found that sometimes the F1 - had a phenotype that did not resemble either of the two parents and was in between the two. This is the case of :
(a) Dominance
(b) Incomplete dominance
(c) Codominance
(d) Pleiotropism
Answer : B
Question. Theoratically, the modified allele could be responsible for the production of :
(a) less efficient enzyme
(b) A non functional enzyme
(c) Non enzyme at all
(d) All the above
Answer : D
Question. How many true breeding pea plant varieties were selected by Mendel
(a) 7
(b) 14
(c) 21
(d) 28
Answer : B
Question. Regarding to pair of dominant and recessive trait which of the following combination is wrong
(a) Flower colour - Violet / white
(b) Flower position - Axial / terminal
(c) Pod shape - Inflated / constricted
(d) Seed colour - Green / yellow
Answer : D
Question. If F1 individual of genotype (Tt) go through sexual reproduction, then it's gamete (pollengrain) with genotype (T) have what chances to pollinate eggs of the genotype (T)
(a) 25 %
(b) 50 %
(c) 75 %
(d) 100 %
Answer : B
Question. Mendel proposed how many conclusions to consolidate his understanding of inheritance in monohybrid cross
(a) One
(b) Two
(c) Three
(d) None of the rules, he proposed laws / principles
Answer : C
Question. Which of the following is correct ?
(a) When genes are grouped on the same chromosome, some genes are very tightly linked and showed very low recombination
(b) When genes are loosely linked show very low recombination
(c) When genes are tightly linked show higher recombination
(d) When genes are loosely linked show no recombination
Answer : A
Question. In Morgan's experiment, what will be percentage of recombination in case of body colour and eye colour in Drosophila ?
(a) 37.2%
(b) 1.3%
(c) 98.7%
(d) 37.2%
Answer : B
Ques. How many true breeding pea plant varieties did Mendel select as pairs, which were similar except in one character with contrasting traits?
(a) 4
(b) 2
(c) 14
(d) 8
Answer: C
Ques. Among the following characters, which one was not considered by Mendel in his experiments on pea?
(a) Trichomes-Glandular or non-glandular
(b) Seed-Green or yellow
(c) Pod-Inflated or constricted
(d) Stem-Tall or dwarf
Answer: A
Ques. Which one from those given below is the period for Mendel’s hybridisation experiments?
(a) 1840-1850
(b) 1857-1869
(c) 1870-1877
(d) 1856-1863
Answer: D
Ques. In his classic experiments on pea plants, Mendel did not use
(a) seed shape
(b) flower position
(c) seed colour
(d) pod length.
Answer: D
Ques. How many pairs of contrasting characters in pea plants were studied by Mendel in his experiments?
(a) Eight
(b) Seven
(c) Five
(d) Six
Answer: B
Ques. Which one of the following traits of garden pea studied by Mendel was a recessive feature?
(a) Axial flower position
(b) Green seed colour
(c) Green pod colour
(d) Round seed shape
Answer: B
Ques. The genes controlling the seven pea characters studied by Mendel are now known to be located on how many different chromosomes?
(a) Seven
(b) Six
(c) Five
(d) Four
Answer: D
Ques. According to Mendelism, which character shows dominance?
(a) Terminal position of flower
(b) Green colour in seed coat
(c) Wrinkled seeds
(d) Green pod colour
Answer: D
Ques. First geneticist/father of genetics was
(a) De Vries
(b) Mendel
(c) Darwin
(d) Morgan.
Answer: B
Ques. Which contribute to the success of Mendel?
(a) Qualitative analysis of data
(b) Observation of distinct inherited traits
(c) His knowledge of biology
(d) Consideration of one character at one time
Answer: D
Ques. Identify the wrong statement with reference to the gene ‘I’ that controls ABO blood groups.
(a) The gene (I) has three alleles.
(b) A person will have only two of the three alleles.
(c) When IA and IB are present together, they express same type of sugar.
(d) Allele i does not produce any sugar.
Answer: C
Ques. In Antirrhinum (Snapdragon), a red flower was crossed with a white flower and in F1 generation all pink flowers were obtained. When pink flowers were selfed, the F2 generation showed white, red and pink flowers. Choose the incorrect statements from the following.
(a) Law of segregation does not apply in this experiment.
(b) This experiment does not follow the Principle of Dominance.
(c) Pink colour in F1 is due to incomplete dominance.
(d) Ratio of F2 is 1/4 (red) : 2/4 (pink) : 1/4 (white).
Answer: A
Ques. The genotypes of a husband and wife are IAIB and IAi. Among the blood types of their children, how many different genotypes and phenotypes are possible?
(a) 3 genotypes; 4 phenotypes
(b) 4 genotypes; 3 phenotypes
(c) 4 genotypes; 4 phenotypes
(d) 3 genotypes; 3 phenotypes
Answer: B
Ques. A tall true breeding garden pea plant is crossed with a dwarf true breeding garden pea plant. When the F1 plants were selfed the resulting genotypes were in the ratio of
(a) 3 : 1 : : Tall : Dwarf
(b) 3 : 1 : : Dwarf : Tall
(c) 1 : 2 : 1 : : Tall homozygous : Tall heterozygous : Dwarf
(d) 1 : 2 : 1 : : Tall heterozygous : Tall homozygous : Dwarf.
Answer: C
Ques. A gene showing co-dominance has
(a) alleles that are recessive to each other
(b) both alleles independently expressed in the heterozygote
(c) one allele dominant on the other
(d) alleles tightly linked on the same chromosome.
Answer: B
Ques. Alleles are
(a) different molecular forms of a gene
(b) heterozygotes
(c) different phenotype
(d) true breeding homozygotes.
Answer: A
Ques. Multiple alleles are present
(a) at the same locus of the chromosome
(b) on non-sister chromatids
(c) on different chromosomes
(d) at different loci on the same chromosome.
Answer: A
Ques. A man with blood group ‘A’ marries a woman with blood group ‘B’. What are all the possible blood groups of their offspring?
(a) A, B, AB and O
(b) O only
(c) A and B only
(d) A, B and AB only
Answer: A
Ques. If two persons with ‘AB’ blood group marry and have sufficiently large number of children, these children could be classified as ‘A’ blood group: ‘AB’ blood group : ‘B’ blood group in 1 : 2 : 1 ratio. Modern technique of protein lectrophoresis reveals presence of both ‘A’ and ‘B’ type proteins in ‘AB’ blood group individuals. This in an example of
(a) partial dominance
(b) complete dominance
(c) codominance
(d) incomplete dominance.
Answer: C
Ques. Which idea is depicted by a cross in which the F1 generation resembles both the parents?
(a) Inheritance of one gene
(b) Co-dominance
(c) Incomplete dominance
(d) Complete dominance
Answer: B
Ques. F2 generation in a Mendelian cross showed that both genotypic and phenotypic ratios are same as 1 : 2 : 1. It represents a case of
(a) co-dominance
(b) dihybrid cross
(c) monohybrid cross with complete dominance
(d) monohybrid cross with incomplete dominance.
Answer: D
Ques. A test cross is carried out to
(a) determine the genotype of a plant at F2
(b) predict whether two traits are linked
(c) assess the number of alleles of a gene
(d) determine whether two species or varieties will breed successfully.
Answer: A
Ques. Test cross in plants or in Drosophila involves crossing
(a) between two genotypes with recessive trait
(b) between two F1 hybrids
(c) the F1 hybrid with a double recessive genotype
(d) between two genotypes with dominant trait.
Answer: C
Ques. ABO blood groups in humans are controlled by the gene I. It has three alleles - IA, IB and i. Since there are three different alleles, six different genotypes are possible. How many phenotypes can occur?
(a) Three
(b) One
(c) Four
(d) Two
Answer: C
Very Short Answer Type Questions
Question. What is the cross between the progeny of F1 and the homozygous recessive parent called? How is it useful?
Answer : When a progeny of F1 is crossed with the homozygous recessive parent, it is called test cross.
Test cross between pure dominant (A) and hybrid dominant (B) individuals with recessive parent is shown below Such a cross is useful to determine the genotype of an unknown trait, i.e., whether it is heterozygous or homozygous dominant for the trait.
Question. A person has to perform crosses for the purpose of studying inheritance of a few traits/characters. What should be the criteria for selecting the organisms?
Answer : The criteria for selecting the organism to study inheritance are
(i) Easily visible and different traits
(ii) Short life span
(iii) Simple pollination procedure
(iv) Organisms must be true breeds
(v) Mating of gametes has to be random
(vi) Can be easily manipulated
Question. The pedigree chart given below shows a particular trait which is absent in parents but present in the next generation irrespective of sexes. Draw your conclusion on the basis of the pedigree.
Answer : The pedigree chart shows that the trait is autosome linked and recessive in nature. But, the parents are carriers (i.e., heterozygous) hence, among the offsprings only few show the trait irrespective of sex. The other offsprings are either normal or carrier.
Question. In order to obtain the F1-generation Mendel pollinated a pure-breeding tall plant with a pure-breeding dwarf plant. But for getting the F2-generation, he simply self-pollinated the tall F1 plants. Why?
Answer : Characters segregate during gamete formation. Pure-breeding parents give rise to F1 with heterozygous conditions. Only self-pollination of heterozygotes can result in all possible recombinations of characters in progeny as mating is random.
Question. Genes contain the information that is required to express a particular trait.’ Explain.
Answer : Genes contain the information required to express a particular trait can be explained by the following experiment.
G Beadle and E Tatum set an experiment to prove that one gene possess a particular trait and is responsible for the production of one enzyme or protein. They performed their experiment on Neurospora crassa which were nutritionally mutant. It was proved that a single protein contains several polypeptide and each polypeptide is controlled by separate gene. Thus, each gene expresses a particular trait. This theory was called one-gane-one enzyme or one gene-one polypeptide hypothesis.
But after the discovery of cistron (the functional unit of gene), the theory was named as one-cistron-one polypeptide hypothesis.
Question. Do you think Mendel’s Laws of inheritance would have been different in the characters that he chose were located on the same chromosome.
Answer : If the characters are present on the same chromosome they would not assort independently as they are linked on the same chromosome. Percentage of linkage depends on the distance between the genes. With linkage no conclusive laws can be drown.
Question. For the expression of traits genes provide only the potentiality and the environment provides the opportunity. Comment on the veracity of the statement.
Answer : Obviously, genes are not the only factors that determine phenotype. Environment also plays an important role in the expression of traits. Genes are actually quite active
throughout our lives, switching their expression on and off in response to the environment.
Besides the effect of internal factors like hormones and metabolism on gene expression, external factors like temperature, light, nutrition, etc., also affect the gene expression and
ultimately exhibiting phenotypic changes. So, we can say that genes provide only the potentiality and the environment provides the opportunity for the expression of traits.
Question. A, B, D are three independently assorting genes with their recessive alleles a, b, d, respectively. A cross was made between individuals of Aa bb DD genotype with aa bb dd. Find out the type of genotypes of the offspring produced.
Answer : The given cross Aa bb DD X aa bb dd, is a trihybrid cross, Accordingly the type of offspring produced would be, (Image 91)
Short Answer Type Questions
Question. In a Mendelian monohybrid cross, the F2-generation shows identical genotypic and phenotypic ratios. What does it tell us about the nature of alleles involved? Justify your answer.
Answer : In case of incomplete dominance, a monohybrid cross shows the result as follows Here, the phenotypic and genotypic both ratios are the same. So, we can conclude that when genotypic and phenotypic ratios are the same, alleles show incomplete dominance. i.e., none of the two alleles shows dominance thus producing hybrid intermediate from the expression of two homozygous alleles.
Question. Name a disorder, give the karyotype and write the symptoms, a human suffers from as a result of monosomy of the sex chromosome.
Answer : Turner’s syndrome is a disorder caused by the absence of one of the X-chromosomes. Its karyotype will be 45 + XO. Symptoms are:
(i) Sterile females
(ii) Rudimentary ovaries
(iii) Lack of secondary sexual characters.
Question. Name a disorder, give the karyotype and write the symptoms where a human male suffers as a result of an additional X-chromosome.
Answer : Klinefelter’s syndrome. The karyotype is 44 + XXY. Symptoms are:
(i) Sex of the individual is masculine but possesses feminine characters.
(ii) Gynaecomastia, i.e., development of breasts.
(iii) Poor beard growth and often sterile.
(iv) Feminine pitched voice.
Question. Name the phenomenon that leads to situations like ‘XO’ abnormality in humAnswer : How do humans with ‘XO’ abnormality suffer? Explain.
Answer : Absence of one X chromosome due to non segregation of chromatids during cell division leads
to XO abnormality. These are sterile female with rudimentary ovaries. They have shield-shaped
thorax, webbed neck, poor development of breasts, short stature, small uterus and puffy fingers.
Question. Which chromosome carries the mutated gene causing β-thalassemia? What are the problems caused by the mutation?
Answer : Chromosome number 11 carries the mutant gene causing β-thalassemia. It causes formation of abnormal haemoglobin molecules, resulting into anaemia.
Question. Give the chromosomal constitution and the resulting sex in each of the following syndromes:
(i) Turner’s syndrome
(ii) Klinefelter’s syndrome
Answer : (i) XO, female
(ii) XXY, male with female characters
Question. If a father and son are both defective in red-green colour vision, is it likely that the son inherited the trait from his father? Comment.
Answer : Gene for colourblindness is X-chromosome linked, and sons receive their sole X-chromosome from their mother, not from their father. Male to male inheritances is not possible for X-linked traits in humans
In the given case the mother of the son must be a carrier (heterozygous) for colour blindness gene, thus transmitting the gene to her son.
Question. Discuss why Drosophila has been used extensively for genetical studies?
Answer : Morgan worked with the tiny fruit flies, Drosophila melanogaster, which were found to be suitable for genetical studies due to the following characteristics
(i) They could be grown on simple synthetic medium in the laboratory.
(ii) They complete their life-cycle in about two weeks.
(iii) A single mating could produce a large number of progeny flies.
(iv) A clear differentiation of the sexes– the male and female flies are easily distinguishable.
(v) It has many types of variations (hereditary) that can be seen with low power microscopes.
Question. Can a child have blood group ‘O’ if his parents have blood group ‘A’ and ‘B’ Explain.
Answer : A child have blood group O in the following two cases Case I When father is IA i and mother is IB i. (Image 93)
The offsprings will have the above possible blood groups, i.e., AB, A, B and O. Thus, a child can have blood group ‘O’ if parents have heterozygous alleles for group ‘A’ and ‘B’.
Question. What is Down's syndrome? Give its symptoms and cause. Why is it that the chances of having a child with Down's syndrome increases if the age of the mother exceeds forty years?
Answer : Down's syndrome is a human genetic disorder caused due to trisomy of chromosome number 21. Such individuals are aneuploid and have 41 chromosomes, i.e., (2n+1)
Symptoms of down’s syndrome are
(i) Mental retardation
(ii) Growth abnormalities
(iii) Constantly open mouth
(iv) Dwarfness, etc., gonads and genitalia under developed The reason for the disorder is the non-disjunction (failure to separate) of homologous
chromosome (a pair 21 during meiotic division. The chances of having a child with Down's syndrome increases with the age of the mother (+40) because age adversely affects meiotic chromosome behaviour. Meiosis in the egg cells is not completed, until after fertilisation. During this long gap (till meiosis is not completed) egg cells are arrested in prophase I and chromosomes are unpaired. The greater the time they remain upaired greater the chance for unpairing and chromosome non-disjunction.
Question. What are the characteristic features of a true-breeding line?
Answer : True breeding is a stable trait inheritance and expression for several generations as a result of continuous self-pollination.
Characteristic features of a true-breeding line
(i) They are used as parents in artificial hybridisation as they provide gametes with all similar traits.
(ii) Homozygous recessive plants are used in test cross to determine the genotype.
Question. In peas,tallness is dominant over dwarfness, and red colour of flowers is dominant over the white colour. When a tall plant bearing red flowers was pollinated with a dwarf plant bearing white flowers, the different phenotypic groups were obtained in the progeny in numbers mentioned against them
Tall, Red = 138
Tall, White = 132
Dwarf, Red = 136
Dwarf, White = 128
Mention the genotypes of the two parents and of the four offspring types.
Answer : The result shows that the four types of offspring are in a ratio of 1:1:1:1. Such a result is observed in a test cross progeny of a dihybrid cross.
The cross can be represented as
Parents Tall and red (TtRr) × Dwarf and white (ttrr)
QUESTIONS
1. Give any two reasons for the selection of pea plants by Mendel for his experiments.
2. Name any one plant that shows the phenomenon of incomplete dominance during the inheritance of its flower colour.
3. Name the base change and the amino acid change, responsible for sickle cell anaemia.
4. Name the disorder with the following chromosome complement.
(i) 22 pairs of autosomes + X X Y
(ii) 22 pairs of autosomes + 21st chromosome + XY.
5. A haemophilic man marries a normal homozygous woman. What is the probability that their daughter will be haemophilic?
6. A test is performed to know whether the given plant is homozygous dominant or heterozygous. Name the test and phenotypic ratio of this test for a monohybrid cross.
7. 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.
8. Mention two differences between Turner's syndrome and Klinefelter's syndome.
9. The human male never passes on the gene for haemophilia to his son. Why is it so?
10. Mention four reasons why Drosophila was chosen by Morgan for his experiments in genetics.
11. Differentiate between point mutation and frameshift mutations.
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