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

Very Short Answer Type Questions

Question. A haemophilic son was born to normal parents.
Give the genotypes of the parents.
Answer: Genotypes of parents : XXh and XY

Question. State the chromosomal defect in individuals with Turner’s syndrome. 
Answer: Turner’s syndrome is due to monosomy.
It occurs due to union of an allosome free egg (22 + 0) and a normal X sperm or a normal egg and an allosome free sperm (22 + 0). The individual has 2n = 45 chromosomes (44 + X0) instead of 46.

Question. The son of a haemophilic man may not get this genetic disorder. Mention the reason.
Answer: The gene for haemophilia is located on the X-chromosome. A male receives the X-chromosome from his mother so haemophilic father does not pass X-chromosome or haemophilia to his son.

Question. Write the chromosomal defect in individuals affected with Klinefelter’s syndrome.
Answer: Klinefelter’s syndrome is caused by union of an abnormal XX egg and a normal Y sperm or normal X and abnormal XY sperm. The individual has 47 (44 + XXY) chromosomes.

Question. Give an example of a human disorder that is caused due to a single gene mutation.
Answer: Sickle cell anaemia is due to inheritance of a defective allele coding for b-globin. It results in the transformation of HbA into HbS in which glutamic acid is replaced by valine at sixth position in each of two b-chains of haemoglobin. The substitution of amino acid in the globin protein results due to the single base substitution at the sixth codon of the beta globin gene from GAG to GUG.

Question. Name one autosomal dominant and one autosomal recessive Mendelian disorder in human. 
Answer: Huntington’s disease is an autosomal dominant and sickle-cell anaemia is an autosomal recessive Mendelian disorder.

Question. Write the genotype of (a) an individual who is carrier of sickle cell anaemia gene but apparently unaffected, and (b) an individual affected with the disease. 
Answer: (a) Hb^AHb^S
(b) Hb^SHb^S 

Short Answer Type Questions

Question. How are dominance, co-dominance and incomplete dominance patterns of inheritance different from each other ?
Answer: Dominance : One allele expresses itself in the hybrid heterozygous condition, other is suppressed.
Co – dominance : Both the alleles of a gene express in a heterozygous hybrid containing two dominant alleles.
Incomplete dominance : Neither of the two alleles of a gene is completely dominant over the other in heterozygous, the hybrid is Intermediate.

Question. During his studies on genes in Drosophila that were sex-linked T.H.Morgan found F2 – population phenotypic ratios deviated from expected 9 : 3 : 3 :
1. Explain the conclusion he arrived at.
Answer: (i) Linkage, genes on the same chromosome were either closely associated or far apart.
(ii) Higher percentage of parental combination and fewer percentage of recombinants are observed when two genes are located very close / tightly linked on the same chromosome.
(iii) Higher percentage of recombinants and fewer percentage of parental combinations are observed when two genes are located far apart / loosely / linked on the same chromosome.

Question. Given below is a table showing the genotypes and the phenotypes of blood groups in the human population :

(a) Identify the genotypes (W) and (X) and the phenotypes (Y) and (Z).
(b) How is co-dominance different from incomplete dominance and dominance?
(c) Name the pattern of inheritance exhibited by the phenotypes (Y) and (Z) in the table.
Answer: (a) Genotypes : W – I^A I^° or I^AI^A and X – I^°I^°
Phenotypes : Y – B and Z – AB
(b) In dominance, F₁ is similar to the dominant parent, phenotypic ratio is different from genotypic ratio. In incomplete dominance, F₁ is different from either of the two parents. Phenotypic and genotypic ratios are the same. In codominance, the effect of both the alleles is equally conspicuous. Both the
alleles produce their effect independently.
(c) ‘Y’ exhibits dominance while ‘Z’ exhibits codominance pattern of inheritance.

Question. The cell division involved in gamete formation is not of the same type in different organisms. Justify.
Answer: The parent organism may be haploid or diploid but the gametes produced by them are always haploid. This is because of the different cell division that take place during gamete formation. The diploid parents undergo meiosis to produce haploid gametes whereas the haploid parents undergo mitosis to produce haploid gametes. 

Question. A human being suffering from Down’s syndrome shows trisomy of 21st chromosome. Mention the cause of this chromosomal abnormality.
Answer: Down’s syndrome is an autosomal aneuploidy,caused by presence of an extra chromosome number 21. Both the chromosomes of the 21st pair pass into a single egg due to non-disjunction during oogenesis.Thus the egg possess 24 chromosomes instead of 23 and offspring has 47 chromosomes (45 + XY in male, 45 + XX in female) instead of 46.
Down’s syndrome is also called 21- trisomy.

Question. State and explain with the help of a cross, the law of segregation as proposed by Mendel.
Answer: Law of segregation states that, “when a pair of contrasting factor or gene are brought together in a hybrid; these factors do not blend or mix up but simply associate themselves and remain together and, separate at the time of gamete formation”. The above law is also known as “law of purity of gametes” because each gamete is pure in itself i.e., having either G (i.e., gene for yellow seed coat) or g (i.e., gene for green seed coat) as illustrated below.

Question. Human blood group is a good example of multiple allelism and co-dominance. Justify.
Answer: Multiple allelism : Generally in an individual/ population only two alleles of a trait govern the character but in case of ABO blood group, three alleles IA, IB and i are found to govern blood group in human population.
Co-dominance : Allele IA and IB when present in an individual, both being dominant express their own types of sugars/antigen (no marks for the second step if two alleles are not given correctly).
Detailed Answer :
In human blood group, there are four possible phenotypes A, B, AB & O. These blood groupings are controlled by gene I. There are three instead of normally two alleles of this gene namely IA, IB & i which control these four blood groups. Hence it is an example of multiple allelism.
Out of these three alleles IA and IB are dominant over i. Each person in a population possesses any two of the three alleles, one from each of the two parents. When allele IA and IB both are present together in an individual, both being dominant express equally and independently and hence are co-dominant.
Thus, human blood is a good example of both multiple allelism as well as co-dominance.

Question. (i) What is polygenic inheritance ? Explain with the help of suitable example.
(ii) How are pleiotropy and Mendelian pattern of inheritance different from polygenic pattern of inheritance? 
Answer: (i) Inheritance in which traits are controlled by three or more genes, e.g., human skin colour / height, the inheritance depends upon the additive / cumulative effect of alleles, more the number of dominant alleles the expression of the trait will be more distinct / prominent, more the number of recessive alleles the trait will be diluted, if member of dominant and recessive alleles are equal the effect is intermediate.
Same explanation with the help of any suitable example.
Single gene controls multiple phenotypic expression (Pleiotropy), one gene controls one phenotypic expression (Mendelian).

Question. (i) How are polygenic inheritance and multiple allelism different ? Explain with the help of an example each.
(ii) List the criteria a chemical molecule must fullfil to be able to act a genetic material.
Answer: 

(ii) (a) It should be able to generate its replica / replication.
(b) It should be chemically and structurally stable.
(c) It should provide the scope for slow changes / mutation that are required for evolution.
(d) It should be able to express itself in the form of a Mendelian characters.

Question. Identify ‘a’, ‘b’, ‘c’, ‘d’, ‘e’ and ‘f ’ in the table given below.

Answer: (a) (i) Partially opened mouth with furrowed tongue.
(ii) Palm is broad with palm crease
(b) Both
(c) Klinefelter’s
(d) Male
(e) (i) Sterile female with poorly developed ovaries and under developed breasts.
(ii) Webbed neck and broad chest
(f) Female 

Question. A couple with normal vision bear a colourblind child. Work out a cross to show how it is possible and mention the sex of the affected child. 
Answer: Colourblindness is a X-linked recessive disorder which shows transmission from carrier female to male progeny and hence, usually males are affected and females remain carriers.
If the given couple is normal with a colourblind child, their genotypes will be

Hence, a colourblind male child is born to the given couple.

Question. Why is haemophilia rare in human females?
Mention a clinical symptom for the disease.
Answer: Haemophilia is genetically due to the presence of a recessive sex linked gene ‘h’, carried by X chromosome. It is generally observed in males as a single gene for the defect is able to express itself as the Y chromosome is devoid of any corresponding allele (XhY). Women will suffer from this disorder when a carrier woman (XXh) marries with haemophilic man (XhY). 50% girl babies will be carriers (XXh) while the remaining 50% will be haemophilic (XhXh). The patient having haemophilia will continue to bleed even from a minor cut since he/she does not possess the natural phenomenon of blood clotting due to the absence of antihaemophilic globulin or factor responsible for clotting.

Question. Which chromosomes carry the mutant genes causing thalassemia in humans? What are the problems caused by these mutant genes?
Answer: α alassemia is an autosomal, recessively inherited disorder. The defect can occur due to mutation or deletion of the genes controlling the formation of globin chains (commonly a and b) of haemoglobin.
a thalassemia is caused by the defective formation of a-globin which is controlled by two genes HBA1 and HBA2 present on chromosome 16. The mutant gene cause anaemia, jaundice, hepatoseplenomegaly and bone changes. All the defective alleles kill the foetus resulting in still birth or death soon after delivery. b thalassemia is caused due to decreased synthesis of b globin. The defect is due to alleles of HBB gene present on chromosome 11. It results in severe haemolytic anaemia, hepatosplenomegaly, cardiac enlargement and skeletal deformities.

Question. (a) Name the kind of diseases/disorders that are likely to occur in humans if
(i) mutation in the gene that codes for an enzyme phenylalanine hydrolase occurs
(ii) there is an extra copy of chromosome 21
(iii) the karyotype is XXY.
(b) Mention any one symptom of the diseases/ disorders named above.
Answer: (a) (i) Phenylketonuria
(ii) Down’s syndrome
(iii) Klinefelter’s syndrome
(b) Symptoms:
(i) Phenylketonuria-mental retardation
(ii) Down’s syndrome-partially open mouth with furrowed tongue
(iii) Klinefelter’s syndrome-development of feminine characters like development of breasts in male.

Question. Why is the possibility of a human female suffering from haemophilia rare ? Explain.
Answer: Haemophilia is genetically due to the presence of a recessive sex linked gene ‘h’, carried by X chromosome. It is generally observed in males as a single gene for the defect is able to express itself as the Y chromosome is devoid of any corresponding allele (XhY). Women will suffer from this disorder when a carrier woman (XXh) marries with haemophilic man (XhY). 50% girl babies will be carriers (XXh) while the remaining 50% will be haemophilic (XhXh).

Long Answer Type Questions

Question. (i) Work out a dihybrid cross upto F₂ generation between pea plants bearing violet coloured axial flowers and white coloured terminal flowers. Give their phenotypic ratio.
(ii) State the Mendel’s law of inheritance that was derived from such a cross.
Answer: (i)

Phenotypes – violet axial : white axial : violet terminal : white terminal Phenotype ratio – 9 : 3 : 3 : 1 
(ii) Law of Independent Assortment : When two pairs are combined in a hybrid segregation of
one pair of characters is independent of the other pair of characters. 

Question. (i) Work out a dihybrid cross upto F₂ generation between homozygous tall pea plant bearing violet flowers and dwarf pea plants bearing white flowers.
(ii) Name the law that Mendel deduced from such a dihybrid cross.
Answer: (i) A dihybrid cross between a homozygous :
(a) Tall pea plant bearing violet flowers (Dominant). Genotype TTVV.
(b) Dwarf pea plant bearing white flowers (recessive). Genotype ttvv.

(ii) Law of Independent Assortment. 

Question. (i) Dihybrid cross between two garden pea plant one homozygous tall with round seeds and the other dwarf with wrinkled seeds was carried.
(a) Write the genotype and phenotypes of the F₁ progeny obtained from the cross.
(b) Give the different types of gametes of the F₁ progeny.
(c) Write the phenotypes and its ratios of the F₂ generation obtained in this cross along with the explanation provided by Mendel.
(ii) How were the observations of F₂ progeny of dihybrid crosses in Drosophila by Morgan different from that of Mendel carried out in pea plants ? Explain giving reasons.
Answer: 

Explanation : The Law of Independent Assortment states that when two pairs of traits are combined in a hybrid, segregation of one pair of character is independent of the segregation of other pair of characters.
(ii) Morgan observed the result of linkage of genes on a chromosome but Mendel did not observe phenomenon of linkage in pea plants /F₂ ratio of Morgan deviated significantly from 9 : 3 : 3 : 1 ratio (Mendelian ratio). 

Question. (i) A pea plant bearing axial flowers is crossed
with a pea plant bearing terminal flowers. The cross is carried out to find the genotype of pea plant bearing axial flowers. Work out the cross to show the conclusions you arrive at.
(ii) State the Mendel’s law of inheritance that is universally acceptable.
Answer: (i) If the plants is homozygous for the dominant trait

Conclusion : If all progeny show axial flowers (dominant) the plant is homozygous (AA), If 50% of progeny show Axial flower (Dominant) and 50% Terminal flower (Recessive) the plant is heterozygous. 
(ii) Law of Segregation : allelic pair segregate (separates) during gamete formation (do not loose their identity).

Question. Explain the genetic basis of blood grouping in human population.
Answer: There are four types of blood groups in human population namely A, B, AB and O. They are determined by presence or absence of two types of RBC surface antigens / sugar polymer A and B. Individuals with blood group A have antigen A, while those with group B have antigen B, AB have both the antigens and ‘O’ persons do not have any antigen. The type of antigens and their presence or absence is controlled by gene I which has three alleles I^A, I^B and i. IA produces antigen A, IB antigen B whereas allele i (i°) does not form any antigen and is recessive. IA and IB are dominant over i and show dominant-recessive relationship. When IA and IB both are present together in a person, both express themselves equally, independently and produce the surface antigen A and B and therefore show the phenomenon of co-dominance. Such genes are called as co-dominant because human beings are diploid individuals, each person therefore have any two of these three alleles of gene I. This results into six different genotypic combination and four phenotypic expressions as follows :
Table showing genetic basis of blood groupings

Blood group alleles thus show both co-dominance and dominance relationship. 

Question. In a cross between a true-breeding red-flowered and a true-breeding white-flowered snapdragon plant, the F1 plants produced pink flowers. Name and explain the type of inheritance.
Answer: 

It shows incomplete dominance. In a cross between true-breeding red flower (RR) and true-breeding white-flowered plants (rr) in F₁ generation we get pink (Rr) flowers. When F₁ was self-pollinated, we get 1 : 2 : 1 (Red : Pink : White) ratio of flowers.
Genotypic ratio was exactly as we would expect but phenotype ratio had changed from 3 : 1 (dominant : recessive) ratio to 1 : 2 : 1. This is because R was not completely dominant over r and it made it possible to distinguish Rr as pink from RR (red) and rr (white). 

Question. A teacher wants his/her students to find the genotype of pea plants bearing purple coloured flowers in their school garden. Name and explain the cross that will make it possible. 
Answer: Purple colour of flower is a dominant trait in pea plants. The genotype of such plants can be determined by test cross. Test cross determines that the dominant character is coming from homozygous dominant genotype or heterozygous genotype e.g. purple flower coming from PP or Pp. It can be done by crossing plants having purple coloured flowers with plants having white coloured flowers, which will always have homozygous recessive genotype.
If the progenies obtained, all have purple flowers, the genotype of purple flower would be PP as shown in the following cross 

Question. What is a test cross ? How can it decipher the heterozygosity of a plant ? 
Answer: Test cross : A cross to analyse whether genotypes of dominant individual is homozygous or heterozygous.
On crossing with a recessive parent, if 50% of progeny have dominant trait and 50% have recessive trait then the plant is said to be heterozygous. 
Detailed Answer :
Test cross is a cross between an organisms with unknown genotype and a recessive parent. It is used to determine whether the genotype of the individual with a dominant trait is homozygous or heterozygous.

If the progenies produced by a test cross show 50% dominant trait and 50% recessive trait, then the unknown individual is heterozygous for the trait. Thus on basis of this ratio 50% : 50% or 1 : 1 the heterozygosity of the plant can be deciphered. On the other hand, if the progeny produced shows only dominant trait, then the unknown individual is homozygous for a trait.