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. 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.
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. 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 :
Thus, the genotype of pea plant bearing purple coloured flowers in the school garden can be Pp or PP.
Purple : white : : 50% : 50% or 1 : 1
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. (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. 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. 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. 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.
Long Answer Type Questions
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).
Detailed Answer :
(i) Polygenic inheritance is the inheritance of traits that are produced by the combined effect of many genes. Polygenic trait is controlled by more than one pair of non-allelic genes and shows different types of phenotypes. For example, human skin colour is an example of polygenic inheritance. It is caused by a pigment called melanin due to three pairs of polygenes (A, B and C).
(ii) Mendelian inheritance refers to the expression of monogenic traits i.e. gene expression is controlled by one gene. In a pair of alleles, expression of the recessive gene is always masked by the expression of a dominant gene.
Pleiotropy is the ability of a gene to have multiple phenotypic effects because it influences several characters simultaneously.
Polygenic inheritance, on the other hand is a type of inheritance controlled by one or more genes in which the dominant alleles have a cumulative effect with each dominant allele expressing a
part or unit of the trait, the full being shown only when all the dominant alleles are present.
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. 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 IA, IB 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. 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. (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.
Detailed Answer :
(i) Explanation of Monohybrid cross : The monohybrid cross is a cross in which the inheritance of a single pair of contrasting characters is taken into consideration. A cross is made between true breeding yellow seed coat colour variety and green seed coat colour
variety of pea plants. All the hybrid offsprings obtained in F1 generation are with yellow seed coat colour. This means the yellow seed coat colour is dominant over green seed coat colour as only this trait appeared while the masked manifestation of other trait is totally marked.
When F1 offspring hybrids were allowed for selfing the yellow seed coat plants and green seed coat plants appeared in the ratio of 3 : 1.
The green seed coat plants are homozygous (yy) as they produced only green seed coat plants in successive generations when self pollinated. 1/3 of yellow seed coat plants also are homozygous (YY) because they produced only yellow seeded plants on selfing while 2/3 of yellow seeded plants are heterozygous (Yy) as they produced both yellow and green seeded plants on selfing. Thus the genotypic ratio is 1 : 2 : 1.
(ii) The following laws of inheritance can be derived from the monohybrid cross.
(a) Principle of dominance : During hybridization out of the two contrasting alleles / characters only one makes it appearance in the hybrid in F1 generation while the manifestation of the other is masked. Then the character which appears in the hybrid or the allele which expresses its effect in the individual / hybrid is called as dominant while the other, which does not show its effect in the hybrid / heterozygous individual is called as recessive. This is universal phenomenon in all types of crosses and is called as the Principle of Dominance.
(b) Law of Segregation : This is the second law of Mendel. This law states that if hybrids of F1 individuals are allowed for selfing the two expressions of a character separate out and appear in F2 generation in a definite proportion of 3 : 1. This is called as monohybrid ratio or segregation ratio. The two alleles for a character in the hybrid individual do not blend or influence each other but separate during gamete formation and therefore gamete receive only one of the two factors but never both. The gametes are pure for a given character. This is known as the Law of Purity of Gametes.
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. (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) 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. (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
(ii) If the plants is heterozygous for the dominant traitf flower)
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. 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
(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. 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 – IA I° or IAIA 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. 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. When a garden pea plant with violet flowers was crossed with another plant with white flowers, 50% of the progeny bore violet flowers.
(a) Work out the cross.
(b) Name the type of cross and mention its significance.
(c) How does the inheritance pattern of flower colour in snapdragon differ from the above?
Answer : (a) 50% of violet flower appear only if the parent is heterozygous for violet flower. The cross can be as follows :
The plants in F1 will bear purple and white coloured flowers in the ratio 1 : 1.
(b) The cross is test cross. A test cross is performed to find out the genotype of the unknown plant.
(c) In case of pea plant
The inheritance pattern of flower colour in garden pea plant is an example of complete dominance whereas inheritance pattern of flower colour in Antirrhinum is an example of incomplete dominance.
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. When a garden pea plant with green pods was cross pollinated with another plant with yellow pods, 50% of the progeny bore green pods.
(a) Work out the cross to illustrate this.
Answer : (a) In pea plant, green pod is dominant over yellow pod. 50% of dominant trait will only occur in progeny when parent is heterozygous for dominant trait. The cross can be illustrated as follows :
Question. You are given a red flower-bearing pea plant and a red flower bearing snapdragon plant.How would you find the genotypes of these two plants with respect to the colour of the flower?Explain with the help of crosses. Comment upon the pattern of inheritance seen in these two plants.
Answer : A test cross is required to find out the genotype of both the plants.
Case I In garden pea :
From the above crosses, it is clear that if the F1 generation contains all red coloured flowers, then the genotype of the given plant is RR (homozygous dominant); whereas if the F1 generation contains red and white flowers in the ratio of 1 : 1, then the genotype of the given plant is Rr (heterozygous dominant).
Case II In snapdragon:
From the above crosses, it is clear that if the F1 generation contains all pink coloured flowers, then the genotype of the given plant is RR (homozygous dominant); whereas if the F1 generation contains both pink and white flowers, then the genotype of the given plant is Rr.
The inheritance pattern of flower colour in garden pea plant is an example of complete dominance whereas inheritance pattern of flower colour in snapdragon is an example of incomplete dominance.
Question. (a) List the three different allelic forms of gene ‘I’ in humans. Explain the different phenotypic expressions, controlled by these three forms.
(b) A woman with blood group ‘A’ marries a man with blood group ‘O’. Discuss the possibilities of the inheritance of the blood groups in the following starting with “yes” or “no” for each :
(i) They produce children with blood group “A” only.
(ii) They produce children some with “O” blood group and some with “A” blood group.
Answer : (a) In humans, ABO system of blood group is a case of multiple allelism, codominance and dominance. In the ABO system, there are four blood groups A, B, AB and O. ABO blood groups are controlled by gene I. The gene I has three alleles IA, IB and i. This phenomenon is known as multiple allelism. IA and IB are completely dominant over i.
This is known as dominance. When IA and IB are present together they both express themselves and produce blood group AB. This phenomenon is known as codominance. The blood groups and their possible genotypes are given below in the table :
(b) (i) Yes, there may be a possibility of progeny having blood group ‘A’ only. It can be explained as follows:
(ii) Yes, this may be the possibility when woman is heterozygote for blood group ‘A’. It can be explained as follows:
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. 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.
Mutation
Very Short Answer Type Questions
Question. Name the event during cell divison cycle that results in the gain or loss of chromosome.
Answer. Mutation
Short Answer Type Questions
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.
Genetic Disorders
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) HbAHbS
(b) HbSHbS
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. 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.
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. 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).
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.
Short Answer Type Questions
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. (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. 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. 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. 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. 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. 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. (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.
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 showing the pattern of blood group inheritance in a family.
(a) Give the genotype of the following :
(i) parents
(ii) the individual ‘X’ in second generation
(b) State the possible blood groups of the individual ‘Y’ in third generation.
(c) How does the inheritance of this blood group explain co-dominance?
Answer.
(a) (i) Parents : IAIO and IBIO
(ii) Since the offspring of X is a individual with blood group A, so the genotype of individual X can be IOIO or IOIA or IAIA. It can be explained as follows:
(b) The possible blood groups of individual ‘Y’ in third generation may be A or O. This can be illustrated as given below:
Case I : When an individual with blood group O (IOIO) marries with an individual with blood group A (IAIA) : The offspring will have blood group A with genotype IO IA.
Case II :When an individual with blood group O (IOIO) marries a person with blood group A (IOIA)
The offspring can either have blood group O with genotype IOIO or A with genotype IAIO.
(c) The pedigree shown occurrence of blood group AB in the family. Blood group AB is an example of co-dominance. Both the alleles IA and IB are codominant as both of them are able to express themselves in the presence of each other in blood group AB by forming antigens A and B.
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. 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 Xh
5 – XhY
6 – X Y
(b) The probability of their first child being a haemophilic male is 25%.
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.
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 – IOIB
Individual 2 – IOIA
(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 – IBIO
Individual 4 – IAIO
Question. Study the following pedigree chart of a family starting with mother with AB blood group and father with O blood group.
(a) Mention the blood group as well as its genotype of the offspring numbered 1 in generation II.
(b) Write the possible blood groups as well as their genotypes of the offsprings numbered 2 and 3 in generation III.
Answer.(a) Offspring numbered 1 has blood group B with genotype IOIB.
(b) For offspring numbered 2, the possible blood groups as well as their genotype would be O (IOIO) or A (IOIA) as shown:
There can be two cases for offspring numbered 3 to know the possible blood groups as well as their genotypes.
Long Answer Type Questions
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. 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. (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 su ers 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. (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.