CBSE Class 12 Biology Principles of Inheritance and Variation Assignment Set B

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)