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

Very Short Answer Type Questions

Question. Name the event during cell divison cycle that results in the gain or loss of chromosome.
Answer: Mutation 

Question. Why do normal red blood cells become elongated sickle shaped structures in a person suffering from sickle cell anaemia?
Answer: Sickle-cell anaemia is caused by the formation of an abnormal haemoglobin called haemoglobin-S.The mutant haemoglobin undergoes polymerisation under low oxygen pressure and causes change in the RBCs from biconcave disc to elongated sickle-shaped structure.

Short Answer Type Questions

Question. Explain the laws that Mendel derived from his monohybrid crosses.
Answer: Law of Dominance, states that factors that control characters occur in pairs and in dissimilar pair one factor dominates (expresses itself) whereas the other recessive factor does not express in the presence of the dominant one.
Law of Segregation, states that during gamete formation, the factors or alleles of a pair segregate/ separate (from each other) and one gamete receives only one of the factors.  

Question. Write three basic facts that are highlighted in Mendel’s Law of Dominance.
Answer: (i) Characters are controlled by discrete units called factors.
(ii) Factors occur in pairs.
(iii) In a dissimilar pair of factors, one member of a pair dominates (dominant) the other (recessive). 
Detailed Answer:
(i) Law of dominance states that characters are controlled by genes.
(ii) Genes occur in pairs.
(iii) When two alternate forms of a trait or character (genes or alleles) are present in an organism, only one factor express (dominant) itself in F1 generation. While the other factor remains hidden (Recessive).

Question. (i) Why is human ABO blood group gene considered a good example of multiple alleles?
(ii) Work out a cross up to F1 generation only, between a mother with blood group A (homozygous) and the father with blood group B (homozygous).
Explain the pattern of inheritance exhibited.
Answer: (i) In human ABO blood group, there are 4 possible phenotypes for this character i.e. AB, A, B, O.
More than two alleles govern the human blood group. The four blood groups result from various combination of 3 different alleles.
(ii) The cross exhibits co-dominance. When the two alleles IA and IB are present together both the alleles express themselves equally forming the blood group AB.

This cross exhibit Mendelian pattern of inheritance which states that the two factors for a trait segregate at the time of gamete formation and again come together at the time of fertilization in the zygote and off springs.

Question. A cross was carried out between a pea plant heterozygous for round and yellow seeds with a pea plant having wrinkled and green seeds.
(i) Show the cross in Punnett square.
(ii) Write the phenotype of the progeny of this cross.
(iii) What is this cross known as ? State the purpose of conducting such a cross.
Answer: 

(iii) Test cross, to identify the genotype of unknown if it is homozygous dominant or heterozygous dominant.

Question. What is the inheritance pattern observed in the size of starch grains and seed shape of Pisum sativum ?
Work out the monohybrid cross showing the above traits. How does this pattern of inheritance deviate from that of Mendelian Law of Dominance ?
Answer: A single gene controls the size of the starch grain and the seed shape.
The trait of size of starch grain shows incomplete dominance. Hence in heterozygous condition the starch grain are of intermediate size.
(Deviation from Mendelian Law of Dominance) :
The trait of seed shape follows Law of Dominance and the hybrid will show only dominant trait. 
Detailed Answer :
The starch synthesis in pea plant is controlled by a single gene. This gene in pea plant shows some degree of pleiotropy as it controls the shape of the seed and in addition the size of starch grain too. This gene has two alleles B and b. The BB homozygotes produce large starch grains as compared to those produced by bb homozygotes. Mature homozygous BB seeds were round while bb seeds were wrinkled.
The heterozygotes Bb form round seeds but the starch grains were of intermediate size. Thus if the size of starch grain is considered the Bb seeds show the phenomenon of incomplete dominance, but if seed shape is considered then allele B and b show dominant-recessive relationship. Thus in this case the pattern of inheritance deviates from the Mendelian Law of Inheritance in that here the dominance of an allele is not absolute but depends upon the product and the particular phenotype that it forms.

Question. In a dihybrid cross, white eyed, yellow bodied female Drosophila was crossed with red eyed, brown bodied male Drosophila. The cross produced 1.3 percent recombinants and 98.7 progeny with parental type combinations in the F₂ generation.
Analyze the above observation and compare with the Mendelian dihybrid cross.
Answer: Morgan observed that the two genes did not segregate independently of each other and the F2 ratio deviated very significantly from the 9:3:3:1 ratio. 
He attributed this to physical association or linkage of two genes, coined the term linkage and the term recombination to describe the generation of nonparental gene combinations. 
Morgan and his group found that even when the genes are grouped on the same chromosome, some genes are very tightly linked (show very low recombination) while others were loosely linked (showed higher recombination). 
In the Mendelian dihybrid cross, the phenotypes round, yellow; wrinkled, yellow; round, green and wrinkled, green appeared in the ratio 9:3:3:1.

Question. A particular garden pea plant produces only violet flowers.
(a) It is homozygous dominant for the trait or heterozygous?
(b) How would you ensure its genotype?
Explain with the help of crosses.
Answer: (a) Homozygous dominant.
(b) By performing test cross, genotype of the given plant can be determined.

Question.

This is the pedigree of a family tracing the movement of the gene for haemophilia. Explain the pattern of inheritance of the disease in the family. 
Answer: Haemophilia is a sex-linked, recessively inherited disorder whose gene is present on the X chromosome. The pedigree shows criss-cross inheritance. Here, a parent passes the traits to the grand child of the same sex through offspring of the opposite sex, i.e., in the given case, father passes the traits to grandson through his daughter.

Question. A relevant portion of b – chain of haemoglobin of a normal human is given below :

The codon for sixth amino acid is GAG. The sixth codon GAG mutates to GAA as a result of mutation ‘A’ and into GUG as a result of mutation ‘B’. Haemoglobin structure did not change as a result of mutation ‘A’ whereas haemoglobin structure changed because of mutation ‘B’ leading to sickle shaped RBCs.
Explain giving reasons how could mutation ‘B’ change the haemoglobin structure and not mutation ‘A’. 
Answer: Mutation A leads to change in the codon GAG to GAA. This does not lead to the change in haemoglobin structure because GAG and GAA both codes for amino acid glutamic acid, i.e, the mutation A does not change the amino acid.
In mutation B, the codon GAG is changed to GUG,where GUG codes for valine, while the original codon GAG codes for glutamic acid; hence, there is a change in the haemoglobin structure and it leads to sickle-cell anaemia.

Question. A cross between a normal couple resulted in a son who was haemophilic and a normal daughter. In course of time, when the daughter was married to a normal man, to their surprise, the grandson was also haemophilic.
(a) Represent this cross in the form of a pedigree chart. Give the genotypes of the daughter and her husband.
(b) Write the conclusion you draw of the inheritance pattern of this disease.
Answer: 

(b) It a sex-linked inheritance showing criss-cross pattern where a parent passes the traits to the grand child. Here , the female is the carrier of the disease.

Question. Why is pedigree analysis done in the study of human genetics? State the conclusions that can be drawn from it. 
Answer: Pedigree analysis is study of pedigree for the transmission of particular trait. It is done to study human genetics because control crosses are not possible in human being as the generation time is more in humans. Pedigree analysis is useful in following ways:
(i) It is useful in finding the possibility of absence or presence of that trait in homozygous or heterozygous state in a particular individual and his family members.
(ii) It is useful in detecting genetic defects like haemophilia, colourblindness, alkaptonuria, phenylketonuria, thalassemia, sickle cell anaemia (recessive traits), brachydactyly and syndactyly (dominant traits).
(iii) It helps to detect sex-linked characters and other linkages.

Question. Given below is the representation of amino acid composition of the relevant translated portion of b-chain of haemoglobin, related to the shape of human red blood cell

(a) Is this representation indicating a normal human or a sufferer from certain related genetic disease? Give reason in support of your answer.
(b) What difference would be noticed in the phenotype of the normal and the sufferer related to this gene?
(c) Who are likely to suffer more from the defect related to the gene represented the males, the females or both males and females equally? And why? 
Answer: (a) The representation indicates a normal human. HbA is a normal peptide with glutamic acid at the sixth position of beta globin chain.
(b) The normal individual has biconcave, disclike RBCs whereas the suferer of the disease has elongated sickle-shaped RBCs.
(c) Both males and females suffer equally because sickle-cell anaemia is an autosomal disease and not a sex-linked one. The sickle-shaped RBC will cause equal oxygen deficiency in both males and females.

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 where the individual has 22 pairs of autosomes and XO sex chromosomes i.e., 45 chromosomes. So, the karyotype will be 44 +XO.
Symptoms : (i) Sterile females
(ii) Rudimentary ovaries
(iii) Lack of secondary sexual characters
(iv) Webbed neck and broad chest
(v) Underdeveloped breasts

Question. Recently a girl baby has been reported to suffer from haemophilia. How is it possible? Explain with the help of a cross.
Answer: Haemophilia is an X-linked recessive disorder.It is possible to have a haemophilic girl when a carrier woman marries a haemophilic man.

Question. Name a disorder, give the karyotype and write the symptoms a human suffer from as a result of an additional X-chromosome.
Answer: Klinefelter’s syndrome is a disorder caused due to an additional X – chromosome. The individual has 22 pairs of autosomes and XXY sex chromosomes i.e., the karyotype is 44 + XXY.
Symptoms: (i) The individual is a male.
(ii) The male shows development of feminine characters like development of breasts.
(iii) Body hair is sparse.
(iv) The individual is sterile.

Question. Study the given pedigree chart and answer the question that follow.

(a) Is the trait recessive or dominant ?
(b) Is the trait sex-linked or autosomal ?
(c) Give the genotypes of the parents in generation I and of their third and fourth child in generation II.
Answer: (a) It is a recessive trait.
(b) It is an autosomal trait.
(c) Generation I → Aa and Aa
Generation II → Third child – aa
Fourth child – Aa

Question. A child suffering from thalassemia is born to a normal couple. But the mother is being blamed by the family for delivering a sick baby.
(a) What is thalassemia ?
(b) How would you counsel the family not to blame the mother for delivering a child suffering from this disease? Explain.
(c) List the values your counselling can propagate in the families. 
Answer: (a) Thalassemia is a disease characterised by reduced synthesis of either the a or b chains of haemoglobin and likewise designated as a or b thalassemia. It may lead to haemolytic anaemia.
(b) Thalassemia is an autosome linked recessive blood disease transmitted from parents to the offspring when both the partners are unaffected carrier for the gene i.e. have heterozygous condition for thalassemia gene.
A child born to a diseased mother and normal father will have only one gene for the disease and will be an
unaffected carrier. A thalassemic baby can be born only to a carrier couple. Thus, mother only cannot be blamed for the birth of a thalassemic baby, its father is equally responsible for the disease.
(c) The counselling can encourage families to understand that only mother is not responsible for any disease or defect found in the baby. A baby inherits its characters from both of its parents not just only from mother. It will help to upgrade condition of women in such orthodox families.

Question. Explain the pattern of inheritance of haemophilia in humans. Why is the possibility of a human female becoming haemophilic extremely rare? Explain. 
Answer: Haemophilia is a sex-linked recessive disorder. Haemophilia (= hemophilia) is genetically due to the presence of a recessive sex linked gene h, carried by X chromosome. A female becomes haemophilic only when both its X chromosomes carry the gene (XhXh). However, such females generally die before birth because the combination of these two recessive alleles is lethal. A female having only one allele for haemophilia (XXh) appears normal because the allele for normal blood clotting present on the other X chromosome is dominant. Such females are known as carriers. In case of males, a single gene for the defect is able to express itself as the Y chromosome is devoid of any corresponding allele (XhY).The possibility of human female becoming haemophilic is extremely rare because she has to be homozygous recessive for the trait, i.e., her father must be a haemophilic and mother must be atleast a carrier.

Question. (a) Why are colourblindness and thalassemia categorised as Mendelian disorders? Write the symptoms of these diseases seen in people suffering from them.
(b) About 8% of human male population suffers from colourblindness whereas only about 0.4% of human female population suers from this disease. Write an explanation to show how it is possible.
Answer: (a) Colourblindness and thalassemia are categorised as Mendelian disorders because of the following reasons:
(i) They are mainly due to alteration or mutation in a single gene.
(ii) These disorders are transmitted to the offspring in the same line as Mendelian principles of inheritance, i.e., by the parents who are carriers and are apparently normal.
(iii) The pattern of inheritance of these disorders can be traced in a family by pedigree analysis.
Symptoms of Colourblindness : The person fails to discriminate between red and green colour due to the defect in either red or/and green cone cells of retina.
Symptoms of Thalassemia : The person suffers from anaemia as the synthesis of either alpha globin chain(s) or beta globin chain (s) of haemoglobin is impaired.
(b) Colourblindness is a X-linked recessive disorder which shows transmission from carrier female to male progeny. In females colour blindness appears only when both the sex chromosomes carry the recessive gene (XcXc). The females have normal vision but function as carrier if a single recessive gene for colourblindness is present (XXc). However , in human males the defect appears in the presence of a single recessive gene (XcY) because Y-chromosome of male does not carry any gene for colour vision. As a result colour blindness is more common in males (8%) as compared to females (0.4%).

Question.

Study the pedigree chart given above, showing the inheritance pattern of blood groups in a family and answer the following questions.
(a) Give the possible genotypes of the individuals 1 and 2.
(b) Which antigen or antigens will be present on the plasma membranes of the RBCs of individuals 5 and 9?
(c) Give the genotypes of the individuals 3 and 4. 
Answer: (a) Individual 1 – I^OI^B
Individual 2 – I^OI^A
(b) Individual 5 will have both antigens A and B while individual 9 will neither have the two on the plasma membranes of the RBCs.
(c) Individual 3 – I^BI^O
Individual 4 – I^AI^O

Long Answer Type Questions

Question. Study the pedigree chart given below showing the inheritance pattern of a human trait and answer the question that follow:

(a) Give the genotype of the parents shown in generation I and of the son and daughter shown in generation II.
(b) Give the genotype of the daughters shown in generation III.
(c) Is the trait sex-linked or autosomal? Justify your answer.
Answer: (a) Genotype of the parents in generation I : Aa and Aa; Son (Generation II) – Aa; Daughter (Generation II) – aa
(b) Genotypes of the daughters shown in generation III – Aa and aa.
(c) It is an autosomal trait, because if the sex-linked trait has to appear in the daughter (generation III), the father must have it; but he does not show the trait and so it is not sex-linked.
Question. (a) State the cause and symptoms of Down’s syndrome. Name and explain the event responsible for causing this syndrome.
(b) Haemophilia and thalassemia are both examples of Mendelian disorder, but show difference in their inheritance pattern. Explain how. 
Answer. (a) 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.
Symptoms of Down’s syndrome : It is characterised by round face, broad forehead, permanently open mouth, protruding tongue, projecting lower lip, short neck, That hands and stubby (small) fingers, many ‘loops’ on finger tips, coarse and straight hair, furrowed tongue, broad palm with characteristic palmer crease, which runs all the way across the palm and monogolian type eye lid fold (epicanthus). The victim has little intelligence.
(b) Inheritance pattern of Thalassemia :
Thalassemia is an autosomal, recessively inherited disorder, transmitted to the offspring when both the parents are heterozygous/carriers for the disease. Alpha thalassemia is controlled by two closely linked genes, HBA1 and HBA2, located on chromosome 16. Beta thalassemia is controlled by a single gene, HBB, located on chromosome 11.
Inheritance pattern of Haemophilia : Haemophilia is a sex-linked, recessively inherited disorder, whose gene is present on the X chromosome. It affects more males than females, because a male has only one X chromosome and the female has two X chromosomes and has to be homozygous recessive for the disease to develop.

Question. With the help of one example each provide genetic explanation for the following observations :
(a) F₁ – generation resembles one of two parents.
(b) F₁ – generation resembles both the parents.
Answer: (a) In dominance, F₁ resembles one of the two parents, i.e., it resembles dominant trait of the parent.
Example : When violet flower coloured pea plant is crossed with white flower coloured pea plant, the F₁ is violet coloured pea plant. This can be illustrated as given below :

Genetic explanation : When two individuals of a species, differing in a pair of contrasting forms of a trait are crossed, the form of the trait that appears in the F₁ hybrid is dominant i.e., it resemble one of the two parents.
(b) In codominance both the alleles express themselves independently. Hence, in case of codominance, F₁ resembles both the parent. E.g., roan coat colour in cattle. The cross can be illustrated as given below :

Genetic explanation : Codominance is a phenomenon in which alleles which do not show dominance-recessive relationship are able to express themselves independently when present together.

Question. (i) Write the conclusions Mendel arrived at on dominance of traits on the basis of monohybrid crosses that he carried out in pea plants.
(ii) Explain why a recessive allele is unable to express itself in a heterozygous state.
Answer: (i)
(a) Characters are controlled by discrete unit called factors.
(b) Factors occur in pairs.
(c) In a dissimilar pair of factors one member of the pair dominates / only one of the parental character is expressed in a monohybrid cross in the F₁ and both are expressed in the F₂
(ii) Due to non-functional enzyme / less efficient enzyme / no enzyme at all.
Detailed Answer :
(i) On the basis of monohybrid crosses carried in pea plants, Mendel concluded that :
(a) The characters are controlled by discrete units called factors now known as genes or alleles.
(b) These factors/genes occur in pairs.
(c) When the factors of a pair are dissimilar only one expresses itself in F₁ generation of a monohybrid cross while the manifestation of the other is masked. The character which finds its expression in F₁ is called as dominant and the latter as recessive factor. On this basis Mendel formulated the Law of Dominance.
(ii) The recessive allele is unable to express itself in heterozygous state because this does not code for its product probably due to non-functional or less active enzymes.

Question. (a) How does a chromosomal disorder differ from a Mendelian disorder?
(b) Name any two chromosomal aberration associated disorders.
(c) List the characteristics of the disorders mentioned above that help in their diagnosis.
Answer: (a)

(b) Two chromosomal aberration-associated disorders are Down’s syndrome and Klinefelter’s syndrome.
(c) (i) Down’s syndrome: The individuals have overall masculine development but they express feminine development like development of breast, i.e., gynaecomastia. They are sterile.
(ii) Klinefelter’s syndrome: The females are sterile as ovaries are rudimentary. Other secondary sexual characters are also lacking.

 Question. Haemophilia is sex linked recessive disorder of humans. The pedigree chart given below shows the inheritance of haemophilia in one family. Study the pattern of inheritance and answer the questions given.

(a) Give all the possible genotypes of the members 4, 5 and 6 in the pedigree chart.
(b) A blood test shows that the individual 14 is a carrier of haemophilia. The member numbered 15 has recently married the member numbered 14. What is the probability that their first child will be a haemophilic male?
Answer: (a) 4 – X X^h
5 – X^hY
6 – X Y
(b) The probability of their first child being a haemophilic male is 25%.

Question. A particular garden pea plant produces only violet flowers
(i) Is it homozygous dominant for the trait or heterozygous ?
(ii) How would you ensure its genotype ? Explain with the help of crosses.
Answer: (i) The plant should be homozygous dominant as it produces violet flowers only.
(ii) Its genotype can be ensured by performing the test cross. In this cross if all the F₁ plants obtained (100%) are with violet flowers then it is homozygous & if the violet & white flowers appear in 1:1 ratio then the plant is heterozygous (Vv). Test crosses can be shown as follows :

Question. (a) Work out a cross upto F2 generation between two pure breed pea plants, one bearing violet flowers and the other white flowers.
(b) (i) Name this type of cross.
(ii) State the different laws of Mendel that can be derived from such a cross.
Answer: In pea plant purple colour is dominant over the white colour. The cross between the two can be shown as below :
(a) In pea plant purple colour is dominant over the white colour. The cross between the two can be shown as below :

Mendel’s law of dominance and law of seggregation can be derived from this cross. Law of dominance states that when individuals differing in a pair of contrasting characters are crossed, the character that appears in the F1 hybrid is dominant over the alternate form that remain hidden. Principle 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”, i.e, allele pairs segregate during gamete formation and the paired condition is restored by random fusion of gametes during fertilisation.
(b) (i) Monohybrid cross
(ii) Two laws of inheritance can be derived from such a cross. These are given below:
Law of dominance: According to this law, characters are controlled by discrete units called factors, which occur in pairs with one member of the pair dominating over the other dissimilar pair. This law explains expression of only one of the parental character in F1 generation.
Law of segregation : Principle 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.

Question. A snapdragon plant homozygous for red flower when crossed with a white owered plant of the same species produced pink owers in F1 generation.
(a) What is this phenotypic expression called?
(b) Work out the cross to show the F₂ generation when F1 was self – pollinated. Give the phenotypic and genotypic ratios of F₂ generation.
(c) How do you compare the F₂ phenotypic and genotypic ratios with those of Mendelian monohybrid F₂ ratios? 
Answer: (a) This phenotypic expression is called incomplete dominance.
(b) In this neither of the two alleles of a gene is completely dominant over the other, hence the phenomenon is known as incomplete dominance.

(c) Mendelian phenotypic F₂ monohybrid ratio = 3 : 1
Mendelian genotypic F₂ monohybrid ratio = 1 : 2 : 1

 Question. Give an example of an autosomal recessive trait in humans. Explain its pattern of inheritance with the help of a cross.
Answer: Autosomal recessive traits are expressed only when autosomal recessive genes are present in homozygous condition. For example, sickle cell anaemia in humans is an autosomal recessive trait/ disorder. In this disorder, the erythrocytes become sickle shaped under oxygen deficiency. This occurs due to formation of abnormal haemoglobin-S (Hbs). It is formed when the glutamic acid present at 6th amino acid in b-chain of normal Hb-A, is replaced by valine.
Sickle cell anaemia can be transmitted from parents to the offspring when both the parents are carrier for the gene or are heterozygous. When two sickle celled heterozygotes marry they may give birth to three types of children— homozygous normal, heterozygous carrier and homozygous sickle celled in the ratio of 1 : 2 : 1. However, homozygous sicklecelled individuals (HbSHbS) die in childhood (before reproductive age) due to acute anaemia. This can be shown by the given cross:

 Question. Name a blood related autosomal Mendelian disorder. Why is it called Mendelian disorder?
How is the disorder tansmitted from parents to offsprings?
Answer: Sickle-cell anaemia is a blood related autosomal Mendelian disorder. It is called Mendelian disorder because it is transmitted to the offspring as per Mendelian principles. The gene for sickle-celled erythrocytes is represented by Hbs while that of normal erythrocytes is written as HbA. The homozygotes for the two types are Hbs Hbs and HbA HbA.The heterozygotes are written as HbA HbS. When two sickle cell heterozygotes marry they produce three types of children–homozygous normal, heterozygous carrier and homozygous sickle celled in the ratio of 1 : 2 :1. However, homozygous sickle-celled individuals (Hbs Hbs) die in childhood (before reproductive age) due to acute anaemia. Therefore, a ratio of one normal to two carriers is obtained.

Question. (i) Explain a monohybrid cross taking seed coat colour as a trait in Pisum sativum. Work out the cross upto F₂ generation.
(ii) State the laws of inheritance that can be derived from such a cross.
(iii) How is the phenotypic ratio of F₂ generation different in a dihybrid cross ?
Answer: (i)

F₂– Phenotypic ratio = 3 : 1
Genotypic ratio = 1 : 2 : 1
(ii) Law of Dominance – In a contrasting pair of factors one member of the pair dominates (dominant) the other (recessive). 
Law of Segregation : Factors or allele of pair segregate from each other so that a gamete receives only one of the two factors. 
(iii) Phenotypic ratio of F₂ in monohybrid cross is 3: 1 whereas in a dihybrid cross the phenotypic ratio is 9 : 3 : 3 : 1.

Question. A non-haemophilic couple was informed by their doctor that there is possibility of a haemophilic child be born to them. Explain the basis on which the doctor conveyed this information. Give the genotypes and the phenotypes of all possible children who could have born to them. 
Answer: The doctor must have used pedigree analysis which refers to the analysis of distribution and movement of traits in a series of generations of a family.
Since the non-haemophilic parents may give rise to a haemophilic child, the genotypes of them should be:
Father : XY (normal)
Mother : XXh (carrier/heterozygous, nonhaemophilic)

 Question. (a) Sickle celled anaemia in humans is a result of point mutation. Explain.
(b) Write the genotypes of both the parents who have produced a sickle celled anaemic offspring. 
Answer: (a) Sickle cell anaemia is caused by the formation of an abnormal haemoglobin called haemoglobin-S. Haemoglobin-S differs from normal haemoglobin-A in only one amino acid – 6thamino acid of b-chain. Here, glutamic acid is replaced by valine due to substitution (trans-version) of T by A in the second position of the triplet codon (CTC) which is changed to CAC. The gene is situated on chromosome 11. The codon CTC is transcribed into GAG (coding for glutamic acid) but due to substitutions of T by A the new codon CAC is transcribed into GUG that codes for valine. Hence it is a result of point mutation.
Sickle-cell anaemia is a blood related autosomal Mendelian disorder. It is called Mendelian disorder because it is transmitted to the offspring as per Mendelian principles. The gene for sickle-celled erythrocytes is represented by Hbs while that of normal erythrocytes is written as HbA. The homozygotes for the two types are Hbs Hbs and HbA HbA. The heterozygotes are written as HbA HbS. When two sickle cell heterozygotes marry they produce three types of children–homozygous normal, heterozygous carrier and homozygous sickle celled in the ratio of 1 : 2 :1. However, homozygous sickle-celled individuals (Hbs Hbs) die in childhood (before reproductive age) due to acute anaemia. Therefore, a ratio of one normal to two carriers is obtained.