Principles of Inheritance and Variation Questions Class 12 Biology

ICSE Class 12 Biology

Mendelian Laws of Inheritance and Chromosomal
Theory of Inheritance

Very Short Answer Type Questions

Question. Write technical term used in human ABO blood groups for IA, IB and i 
Answer : IA, IB and i are the alleles of the gene

Question. How does a test cross help to determine the genotype of an individual ?
Answer : Individual of unknown genotype crossed with recessive parent.
All dominant in progeny—Homozygosity, dominant to recessive ratio 1 : 1 in progeny— Heterozygosity. 

Question. In a dihybrid cross carried out by T. H. Morgan in Drosophila the F2 ratio deviated from that of Mendel’s dihybrid F2 ratio. Give a reason.
Answer : Genes were linked/genes were on the same chromosome and closely associated

Question. Give an example of a plant where the F2 progeny of a monohybrid cross has same genotypic and phenotypic ratios.
Answer : Snapdragon/Antirrhinum majus/ Four O’ clock plants/Mirabilis jalapa. 

Question. If two genes are located far apart from each other on a chromosome, what will be its effect on the frequency of recombination ?
Answer : Frequency of recombination will be higher.

Question. Give an example of a polygenic trait in human
Answer : Skin colour/height in humans (any other suitable example)

Question. State a difference between a gene and an allele.
Answer : Gene : Contains information that is required to express a particular trait // unit of inheritance // segment of DNA called cistron // sequence of DNA coding for tRNA / rRNA / polypeptide / enzyme.
Allele : Genes which code for a pair of contrasting traits / different forms of the same gene / individual gene in a particular gene pair (for same character).

Question. A geneticist interested in studying variations and patterns of inheritance in living beings prefers to choose organisms for experiments with shorter life cycle. Provide a reason. 
Answer : Many generations can be obtained/variations can be exhibited/selected faster.

Question. List any two characters of Pea plants used by Mendel in his experiments other than height of the plant and the colour of the seed.
Answer : Flower colour / Flower position / Pod shape / pod colour / Seed shape 

Question. What are true breeding lines’ that are used to study inheritance pattern of traits in plants ?
Answer : Self pollination continuous for several generations / homozygous. 

Question. A garden pea plant produced axial white flowers. Another of the same species produced terminal violet flowers. Identify the dominant traits.
Answer : Axial, violet flower. 

Question. Name the stage of cell division where segregation of an independent pair of chromosomes occurs.
Answer : Anaphase-1 of Meiosis – 1.

Question. Mention any two contrasting traits with respect to seeds in pea plant that were studied by Mendel.
Answer : Round/Wrinkled, Yellow/Green.

Question. A garden pea plant (A) produced inflated yellow pod, and another plant (B) of the same species produced constricted green pods. Identify the dominant traits. 
Answer : Inflated, green pods.

Question. How many types of phenotypes would you expect in F2 generation in a monohybrid cross ?
Answer : Two types in the ratio of 3 : 1. 1

Question. What is the phenotype of the following ?
(i) IAi & (ii) ii 
Answer :
(i) Blood group A (Heterozygous) because the allele IA is dominant over allele i.
(ii) Blood group O (Homozygous condition)—because both alleles ii are recessive.

Question. Write the percentage of the pea plants that would be heterozygous tall in F2 generation when tall heterozygous F1 pea plants are selfed.
Answer :
50% heterozygous tall pea plants would be obtained in F2 generation when tall heterozygous pea plants are selfed. 

Question. In a dihybrid cross, when would the proportion of parental gene combinations be much higher than non-parental types, as experimentally shown by
Morgan and his group ?
Answer :
When the genes are linked i.e., when genes of dihybrid cross are closely situated on the same chromosome, the proportion of parental gene combinations will be much higher than nonparental types.

Question. Write the percentage of the pea plants that would be homozygous recessive in F2 generation when tall F1 heterozygous pea plants are selfed.
Answer : 25% homozygous recessive pea plants will be obtained in F2 generation when tall F1 heterozygous pea plants are selfed as follows :

Question. Mention two contrasting flower-related traits studied by Mendel in his pea plant experiments.
Answer :
Contrasting flower-related traits :
(i) Flower colour : violet/white or red/white.
(ii) Flower position : Axial/Terminal.

Question. AaBb was crossed with aabb. What would be the phenotypic ratio of the progeny ? Mention the term used to denote this kind of cross ?
Answer :
Phenotypic ratio would be 1:1:1:1. It is test cross.

Question. Write the percentage of F2 homozygous & heterozygous populations in a typical monohybrid cross. 
Answer :
50% homozygous and 50% heterozygous. Out of 50% homozygous population 25% is homozygous dominant and 25% homozygous recessive. 

Question. Name the respective pattern of inheritance where F1 phenotype.
(i) does not resemble either of the two parents and is in between the two.
(ii) resemble only one of the parents.
Answer :
(i) Incomplete dominance
(ii) Dominance.

Question. Mention the type of allele that expresses itself only in homozygous state in an organism.
Answer :
Recessive allele.

Short Answer Type Questions – l

Question. In Snapdragon, a cross between true breeding red flower (RR) plants and true breeding white flower
(rr) plants showed a progeny of plants with all pink flowers.
(i) The appearance of pink flowers is not known as blending. Why ?
(ii) What is the phenomenon known as ?
Answer :
 (i) R (dominant allele red colour) is not completely dominant over r (recessive allele white colour) / r maintains its originality and reappear in F2 generation.
(ii) Incomplete dominance. 

Question. State the Mendelian principle which can be derived from a dihybrid cross and not from monohybrid cross.
Answer : From the dihybrid cross, law of independent assortment can be derived which states that when two pairs of traits are combined in a hybrid, segregation of one pair of characters is independent of the other pair of characters.

Question. Write the scientific name of the fruit-fly. Why did Morgan prefer to work with fruit-flies for his experiments ? State any three reasons.
Answer : Drosophila melanogaster Grown in simple synthetic medium, complete the life cycle in two weeks / short life cycle, single mating produce more progeny, dimorphism, many heritable variations / easy to handle.

Question. How does the gene I control ABO blood groups in humans ? Write the effect the gene has on the structure of red blood cells.
Answer : (i) Gene I has three different alleles IA, IB, i. ½
(ii) IA produces A type of sugar / Antigen → A group IB produces B type of sugar / Antigen → B group 
(iii) i – No sugar/Antigen – O group.
(iv) Structure – sugar polymers protrude from the surface of plasma membrane of RBCs.

Question. Linkage or crossing-over of genes are alternatives of each other. Justify with the help of an example.
Answer : In Drosophila, a yellow bodied white eyed female was crossed with brown bodied red eyed male, F1 progeny produced and intercrossed. The F2 phenotypic ratio of Drosophila deviated significantly from Mendel’s 9 : 3 : 3 : 1, the genes for eye colour & body colour are closely located on the ‘X’ chromosome showing linkage & therefore inherited together, recombinants were formed due to crossing over but at low percentage.

Question. Explain pleiotropy with the help of an example.
Answer : Effect of single gene on multiple phenotypic expressions. e.g. size of the starch grains produced and shape of the seeds in pea plant are controlled by a single gene // Phenylketonuria characterised by mental retardation and reduction in hair and skin pigmentation. .

Question. A cross was carried out between two pea plants showing the contrasting traits of height of the plant. The result of the cross showed 50% of parental characters.
(i) Work out the cross with the help of a Punnett square.
(ii) Name the type of the cross carried out.
Answer :
(i) Tt × tt.

(ii) Test cross

Question. Study the figures given below and answer the question.

Identify in which of the crosses the strength of linkage between the genes is higher. Give reasons in support of your answer.
Answer :
Cross A, because they are tightly linked / due to close physical association / they are closely located.

Question. In a cross between two tall pea plants, some of the offsprings produced were dwarf. Show with the help of the Punnett Square how this is possible.
Answer : In a cross between two tall pea plants, some of the offsprings produced were dwarf. It indicates that parent pea plants were heterozygous for tallness (Tt) i.e., they contain a recessive gene (t) for dwarfness from each of the parent plant.

25% plants were dwarf. Phenotypic ratio – Tall : Dwarf :: 3 : 1 Genotypic ratio – 1 : 2 : 1 Punnett square : Punnett square is a graphic representation of the probabilities of all the possible genotypes and phenotypes of offsprings in a cross.

Question. Differentiate between multiple allelism and pleiotropy with the help of an example each.
Answer : Multiple alleles : More than two alternate forms of a gene present on the same locus of a homologous pair of chromosomes in a population are called multiple alleles. They control the single trait. For example, ABO blood group in human Pleiotropic genes : The gene having a multiple phenotypic effect because of its ability to control the expression of a number of characters is called pleiotropic gene. For example, skin pigmentation and phenylketonuria.

Question. ABO blood groups is a good example of codominance. Justify.
Answer :
(i) ABO blood group in humans is contributed by gene ‘I’ that has 3 alleles ‘IA’ ‘IB’ and ‘i.’
(ii) Because human beings are diploid and each person has two of the three alleles.
(iii) IA and IB produce two different types of sugar while allele i does not produce sugar on the plasma membrane of RBC.
(iv) When IA and IB present they produce their own type of sugar-this is called co-dominance.

Question. List all three different allelic forms of gene I in human Explain different phenotypic expressions controlled by these three forms.
Answer :
The three different forms of gene I are IA, IB and i. 
ABO blood grouping is controlled by these three alleles, hence it is an example of multiple allelism. Each person possesses any two of the three I gene alleles. IA and IB are completely dominant over allele (i) while IA and IB are co-dominant because both of them express themselves equally and independently when present together. As there are three alleles of gene I there would be following six types of allelic combinations of genotypes resulting in four phenotypic expressions as follows :

Question. Explain co-dominance with the help of one example.
Answer :
When the dominant alleles of the same gene which are contributed by both parents are expressed (called co-dominance) // F1 generation resembles both the parents :
In human blood group :
Parents   IA IA IB IB
Gametes IA      IB
F1 –          IAIB
In human red blood cells, alleles IA and IB of gene I are both dominant, when IA & IB are present together in an individual both are expressed as IA IB, (AB blood group).

Question. Tallness of pea plants is a dominant trait, while dwarfness is the alternate recessive trait. When a pure-line tall is crossed with pure-line dwarf, what fraction of tall plants in F2 shall be heterozygous ? Give reasons.
Answer :
Pure line tall is crossed with pure line dwarf.

Phenotypic ratio = Tall : dwarf
                              3  : 1
Genotypic ratio = TT : Tt : tt
                           1  : 2  : 1
Two third of tall progenies are heterozygous
because gene for tallness (T) is dominant and
expresses itself in heterozygous condition.
A Punnet square is used to understand a typical
monohybrid cross between tall and dwarf plants.

Long Answer Type Question

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.
(ixi) Single gene controls multiple phenotypic expression (Pleiotropy), one gene controls one phenotypic expression (Mendelian)

Question. Give a genetic explanation for the following cross. When a tall pea plant with round seeds was crossed with a dwarf pea plant with wrinkled seeds then all the individual of F1-populations were tall with round seeds. However selfing among F1-population led to a 9 : 3 : 3 : 1 phenotypic ratio.
Answer :

Law of dominance : In a dissimilar pair of factors, one member of the pair is dominant and the other is recessive. In the given cross tall and round are dominant where as dwarf and wrinkled are recessive (explain with or without a cross).
Law of Segregation : Allelic pairs separate or segregate during gamete formation and the paired condition is restored during fertilisation (explain with or without a cross).
Law of Independent Assortment : The new combination seen in F2 generation (Tall wrinkled) (Dwarf round) is only possible when the two gene pairs for height and seed shape assort independently of each other during gamete formation. The law states that when two pairs of traits are combined in a hybrid, segregation of one pair of characters is independent of the other pair of characters.

Question. How do ‘Pleiotropy’, ‘incomplete dominance’, ‘co-dominance’ and ‘polygenic inheritance’ deviate from the observation made by Mendel ? Explain with the help of example for each.
Answer :

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

Question. A pea plant producing yellow coloured and round seeds is given with unknown genotypes. Explain how you would find the correct genotypes of the plants with respect to the two traits mentioned.
Work out the cross and name it.
Answer : Test Cross 1
Case 1 :

If all phenotypes is yellow and round then the genotype of parents is YYRR. (Homozygous)
Case 2 :

Phenotypes : 1 Yellow round and 1 Green round
Then the genotype of the parent is YyRR (Heterozygous for yellowness)
Case 3

Phenotype : 1 Yellow round and 1 Yellow wrinkled.
Then the genotype of parent is YYRr
(Heterozygous for shape of seed)

Phenotype : 1 Yellow round , 1 Yellow wrinkled ,
1 Green round and 1 Green wrinkled.
Then parent is YyRr

Question. A tall pea plant bearing violet flowers is given with its unknown genotypes. Explain by working out the crosses how would you find the correct genotypes with respect to the two traits mentioned only by ‘selfing’ the given plants.
Answer :
Tall plant = TT/Tt
Violet flowers = WW/Ww
Genotype of given plant could be any of the assumed four : TTWW, TTWw, TtWW, TtWw. 1
Case 1 :

Tall violet = 9
Tall white = 3
Dwarf violet = 3
Dwarf white = 1
If phenotypic ratio 9 : 3 : 3 : 1, then parent is TtWw

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. (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 ?

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 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 : (i)

(ii) Round and yellow – 25%
Wrinkled and yellow – 25%
Round and green – 25%
Wrinkled and green – 25%
(iii) Test cross, to identify the genotype of unknown if it is homozygous dominant or heterozygous dominant

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. Mendel published his work on inheritance of characters in 1865, but it remained unrecognized till 1900. Give three reasons for the delay in accepting his    work.
Although Mendel published his work on inheritance of characters in 1865 but for several reasons, it remained unrecognised till 1900.
Explain giving three reasons why it took so long ?

Answer : Work of Mendel could not be widely publicised due to poor communication / his concept of genes or factors as discrete units and which did not blend was not accepted due to continuous variation seen in nature/no proof of existence of factors and what they were made of /Mendel used mathematics to explain biological phenomenon which was new and unacceptable.

Question. Describe the mechanism of pattern of inheritance of ABO blood groups in human
Answer : Human blood group is determined by glycoprotein / antigen A and glycoprotein / antigen B.
The alleles are IA, IB and i – Hence referred to as multiple allelism.
The individual inherits any two of them as given below
IA IA, IA i — A group
IB IB, IB i — B group
IA IB — AB group
i i — O group 
In the case of A, B and O — Law of dominance is the pattern of inheritance as IA / IB dominant over i.
In AB group both the alleles IA and IB express — It is the case of Co-dominance.

Question. (i) Explain Mendel’s Law of Independent Assortment by taking a suitable example.
(ii) How did Morgan show the deviation in inheritance pattern in Drosophila with respect to this law ?
Answer : (i)

Phenotypic ratio : round yellow : round green :
                                   9                   3
wrinkled yellow : wrinkled green
          3                    1
Law of Independent Assortment : It states that when two pairs of contrasting traits are combined in a hybrid, segregation of contrasting one pair of character is independent of the other pair of characters.
(ii) Both parental type and recombinant types are observed to show that genes for the colour and genes for the shape of seeds segregate independently during gamete formation.

Question. State and explain the ”law of independent assortment” in a typical Mendelian dihybrid cross.
Work out a typical Mendelian dihybrid cross and state the law that he derived from it.
(i) State the Law of Independent Assortment.
(ii) Using Punnett Square demonstrate the law of independent assortment in a dihybrid cross involving two heterozygous parents.
Answer : (i) Law of Independent Assortment : When two pair of traits are combined in a hybrid, inheritance of one pair of characters is independent of the other pair of characters / when two pairs of contrasting characters or genes or traits are inherited together in a dihybrid cross (in a pea plant) the inheritance of one pair of character is independent of inheritance of the other character in the progeny.
Explanation : Mendel took homozygous pea plant producing yellow and round seeds, crossed them with homozygous pea plant producing green and wrinkled seeds / shown in a flow chart of a dihybrid cross given.

Phenotypes – Yellow : Yellow : Green : Green
                      round wrinkled round wrinkled
Phenotype ratio – 9    :   3    :    3    :    1
(Four different types of phenotypes in correct ratio)
(Formation of new phenotypes along with parental phenotypes is possible because inheritance of two pairs of contrasting traits or genes in the progeny is independent of each other)

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.

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) Explain Polygenic and Multiple allelism with the help of suitable examples.
(ii) ”Phenylketonuria is a good example that explains Pleiotropy.” Justify.
Answer :
(i) Traits that are generally controlled by three or more genes, the phenotype reflects the contribution of each allele / effect of each allele is additive.
eg. Human skin colour, controlled by three genes (A, B, C).
In multiple allelism more than two alleles, govern the same character / phenotype.
eg. Human blood group (ABO system), controlled by three different alleles (IA, IB, i)
(ii) In pleiotropy a single gene can exhibit multiple phenotypic expressions, in phenylketonuria single mutated gene express mental retardation and reduction in hair and skin pigmentation.

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 : (i)

(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. Skin colour in humans does not have distinct alternate forms but shows a whole range of possible variations in skin colour. Explain the pattern of inheritance of such a triait. What is this type of inheritance known as ? Provide another example of exhibiting such an inheritance pattern.
Answer :
Skin colour is controlled by three genes; A, B, C dominant genes and a, b, c the recessive genes; the effect of each type of allele is additive; more dominant allele, darker the skin colour; more the recessive allele, lighter the skin colour; when three dominant alleles and three recessive alleles are present in an individual the skin colour is intermediate.
(i) Polygenic inheritance.
(ii) Human Height / or any other correct example.

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 trai

(ii) If the plants is heterozygous 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. Describe the dihybrid cross carried on Drosophila melanogaster by Morgan and his group. How did they explain linkage, recombination and gene mapping on the basis of their observations ?
Answer :
According to Morgan and his group, if genes are tightly linked they showed very low recombination.
(shown in cross A)
If genes were loosely linked they showed very high recombination.
(shown in cross B)
The group used the frequency of recombination between gene pairs on the same chromosome as a measure of distance between genes and ‘mapped‘ their position on the chromosome.

Question. (a) Write the scientific name of the organism
Thomas Hunt Morgan and his colleagues worked with for their experiments the correlation between linkage and recombination with respect to genes as studied by them.
(b) How did Sturtevant explain gene mapping while working with Morgan
Answer :
(a) Drosophila melanogaster
They observed that two genes (located closely on a chromosome) did not segregate independently of each other (F2 ratio deviated significantly from 9 : 3 : 3 : 1). =½ Tightly linked genes tend to show fewer (lesser) recombinant frequency of parental traits / show higher (more) frequency of parental type Loosely linked genes show higher percentage (more) of recombinant frequency of parental traits / lower frequency percentage of parental type Genes present on same chromosome are said to be linked and the recombinant frequency depends on 
their relative distance on the chromosome.
(b) He used the frequency of recombination between gene pairs on the same chromosome, as a measure of the distance between genes and mapped their position on the chromosome. of mapping, which are used today for genome sequencing projects as in Human Genome Project.

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 :

Sex Determination and Chromosomal Disorder

Question. Give an example of a human disorder that is caused due to a single gene mutation.
Answer : Sickle cell anaemia/Thalassemia/Phenylketonuria.

Question. Mention two causes of frame-shift Mutation.
Answer : Insertion, deletion of three bases/one codon or multiple of three bases/multiple codon (hence one or more amino acid) (reading frame remains unaltered from that point onwards) 

Question. What is point mutation? Give one example.
Answer : Arising due to change in a single base pair of DNA, sickle cell anaemia.

Question. What is a Mutagen? Name a physical factor that can be a Mutagen. 
Answer : All the physical and chemical factors that induce mutation, UV radiation/X rays. 

Question. Write the chromosomal defect in individuals affected with Klinefelter’s syndrome.
Answer : Male – Additional copy of X chromosome / XXY.

Question. Give an example of a sex-linked recessive disorder in humAnswer :
Answer : Colour blindness. 

Question. Why do normal red blood cells become elongated sickle shaped structures in a person suffering from sickle cell anaemia?
Answer : The mutant haemoglobin molecule (substitution of Glutamic acid by valine) undergoes polymerization, under low oxygen tension causing the change. 

Question. A colour blind boy is born to a couple with a normal colour vision. Write the genotype of the parents.
Answer : Father – XY, Mother – XXC 

Question. Give an example of an organism that exhibits haplodiploid sex-determination system.
Answer : Honey bees.

Question. Give one example of organism exhibiting female heterogamety.
Answer : In many birds (ZZ / ZW) Male / female/heterogamety 

Question. Indiscriminate diagnostic practices using X-rays etc., should be avoided. Give one reason.
Answer : Indiscriminate diagnostic practices using X-rays,
gamma rays etc. are ionizing radiations which usually produce breaks in the chromosomes and chromatids and abnormal mitosis in the irradiated cells. They cause abnormal functioning of the cells, mutations resulting in the development of various types of cancers specially blood cancer or leukemia etc.

Question. State the chromosomal defects in individuals with Turner’s syndrome. 
Answer : Monosomy of sex chromosome/XO condition/ Absence of one X chromosome (in female).

Question. Write the chromosomal basis of sex determination in birds.
Answer : (Male) ZZ + (Female)ZW/Heterogamety

Question. On what basis is the skin colour in humans considered polygenic ?
Answer : Controlled by more than one gene, cumulative and additive effect of genes.

Question. Name the disorder caused due to the absence of one of the X-chromosomes in a human female.
Answer : Turner’s syndrome

Question. Give an example of a chromosomal disorder caused due to non-disjunction of autosomes.
Answer : Down’s syndrome.

Question. A male honeybee has 16 chromosomes whereas its female has 32 chromosomes. Give one reason.
Answer : Male honey bee develops from unfertilized female gamete/unfertilised egg/ Parthenogenesis of female gamete (16 chromosomes), female develops by fertilization/fertilised egg. (32 chromosomes).

Question. The son of a haemophilic man does not get this genetic disorder. Mention the reason. 
Answer : The son gets Y chromosome from the father and X chromosome from the mother. Therefore, as the gene for haemophilia is located on X chromosome, a son cannot get the disease from his father. 

Question. What is the cause of Down’s syndrome in humans ?
Answer : This syndrome develops due to trisomy of chromosome 21. The non-disjunction of the 21 chromosome during meiosis causes the trisomy of 21st chromosome and results in Down’s syndrome.

Question. How many chromosomes do drones of honeybee possess ? Name the type of cell division involved in the production of sperms by them.
Answer : 16, Mitosis.

Question. Mention the combination (s) of sex chromosomes in a male and a female bird.
Answer : Male – ZZ, Female – ZW 

Question. Write the sex of a human having XXY chromosomes with 22 pairs of autosomes. Name the disorder this human suffers from.
Answer : Male, Klinefelter’s syndrome

Question. Name one autosomal dominant and one autosomal recessive Mendelian disorder in human :
Answer : Autosomal dominant — Myotonic dystrophy Autosomal recessive — Phenylketonuria / sickle cell anaemia / cystic fibrosis / Thalassaemia.

Question. Observe the pedigree chart and answer the following questions :

i) Identify whether the trait is sex-linked or autosomal.
(ii) Give an example of a disease in human beings which shows such a pattern of inheritance.
Answer :
(i) Sex-linked.
(ii) Haemophilia/colour blindness.

Question. The prophase I stage of meiosis plays a vital role in r-DNA formation. Justify with reason.
Answer : The prophase I stage of meiosis plays vital role in r-DNA formation because crossing over occurs at this stage, which helps in recombination.

Question. What is the difference in the amino-acid sequence in the B-chain of haemoglobin in a normal person and a sickle-cell anaemia person ?
Answer : In a person suffering from sickle-cell anaemia, the amino acid glutamine present at the sixth position, is replaced by the valine.

Question. Write the genotype of (i) an individual who is carrier of sickle cell anaemia gene but apparently unaffected and (ii) an individual affected with the disease.
Answer : (i) HbA, HbS
(ii) HbS HbS

Short Answer Type Questions – l

Question. What happens when chromatids fail to segregate during cell division cycle ? Explain your answer with an example.
Answer : Failure of segregation of chromatids during cell division cycle results in the gain or loss of chromosome / called aneuploidy e.g., Downs’ syndrome results in the gain of extra copy of chromosome 21 / Turner’s syndrome results due to loss of an X-chromosome in human female.

Question. Which chromosomes carry the mutant genes causing thalassaemia in humans ? What are the problems caused by these mutant genes ?
Answer :
11th and 16th chromosomes carry the mutant gene causing thalassaemia.
These cause the formation of abnormal haemoglobin molecules resulting into anaemia.

Question. Differentiate between ‘ZZ’ and ‘XY’ type of sexdetermination mechanisms. 
Answer : ZZ type of sex determination mechanism is found in birds, reptiles and fishes. In this type, the females have heteromorphic sex chromosomes (ZW),
while males have homomorphic sex-chromosomes (ZZ). Females are heterogametic i.e. produce two dissimilar types of eggs while males produce only
one type of sperms. The egg determines the sex of the individual. XY type of sex determination mechanism is found in human beings. In this type, the male individuals have heteromorphic sex chromosomes (XY) and are therefore heterogametic i.e. producing two types of sperms are with X and the other carrying Y
chromosome. The females have homomorphic sex chromosome (XX) and homogametic i.e. produce only one type of eggs. The sex of the offspring is determined by type of sperm taking part in fertilization.

Question. Is haemophilia in humans a sex linked or autosomal disorder ? Work out a cross in support of your answer.
Answer : Haemophilia in humans is a sex linked disorder

(any one incorrect genotype or phenotype-no marks)
Cross showing haemophilic father and normal mother

Question. Why is the possibility of human female suffering from haemophilia rare ? Explain.
Answer :

Rare because mother should be atleast carrier and father haemophilic (non viable at later stage).

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 : In the given pedigree of a family, a normal female is crossed with a haemophilic male. The offspring generated are carrier females as haemophilia is an X-linked disease and female has two X chromosomes. These carrier females when crossed with normal males produced carrier females and haemophilic males.

Question. A haemophilic father can never pass the gene for haemophilia to his son. Explain.
Answer :
It is a sex linked recessive disorder in which X-chromosome has the haemophilic gene. Son inherits a Y chromosome from father and gene for haemophilia is not present on Y chromosome.

Question. Explain the cause of chromosomal disorders. 
Describe the effect of such disorders with the help of an example each involving (i) autosomes, and
(ii) sex chromosomes.
Answer :
Gain or loss of a chromosome (due to non disjunction)
(i) Down Syndrome – Additional copy of 21st chromosome / trisomy of 21. ½ + ½
(ii) Klinefelter’s Syndrome – presence of an additional copy of X chromosome leading to XXY. Turner’s Syndrome – absence of one of the X chromosome i.e., 44 with XO 

Question. During a cytological study conducted on the chromosomes of the insects, it was observed that only 50% of the sperms had a specific structure after spermatogenesis. Name the struture and write its significance in sex determination of insects. 
Answer :
X body / X factor / X chromosome.
In insects the sex chromosome consists of XX female; XO-Males 

Question. A non-haemophilic couple was informed by their doctor that there is possibility of a haemophilic child be born to them. Draw a checker board and find out the percentage of possibility of such child in the progeny
Answer :

Phenotypes : 50% daughter normal (XX)
50% daughter carrier (XXh)
50% son normal (XY)
50% son haemophilic (XhY)

Question. (i) Why are grasshopper and Drosophila said to show male heterogamety ? Explain.
(ii) Explain female heterogamety with the help of an example.
Answer :
(i) In grasshopper, males have one X only (XO type), in Drosophila males have one X and one Y (XY type) – Males in both cases produce different kinds of gametes so heterogametic.
(ii) In birds female has ZW, produce two kinds of gametes and so heterogametic. 
(i) Male heterogamety, Grasshopper. 
(ii) Female heterogamety, Birds. 

Question. Why is pedigree analysis done in the study of human genetics ? State the conclusions that can be drawn from it.
Answer : (
i) Control crosses are not possible in case of humans beings.
(ii) Analysis of traits in several generations of family / To trace pattern of inheritance / whether the trait is dominant or recessive / sex linked or not.

Short Answer Type Questions – ll

Question. Give an example of an autosomal recessive trait in human Explain its pattern of inheritance with the help of a cross.
Answer :
Sickle cell anaemia is an example of autosomal recessive trait in human

The disease is controlled by a single pair of allele HbA and HbS. The disease is only expressed if both the copies are defective i.e. only when the autosomal recessive genes are present in homozygous condition (HbSHbS). People with a single defective copy of the gene are clinically normal, however they act as carrier and can pass on the defective gene to their next generations in the ratio of 1 : 2 : 1 as shown in the above cross.

Question. One of the twins born to parents having normal colour vision was colour blind whereas the other twin had normal vision. Work out the cross. Give two reasons how it is possible
Answer :

Genes that lead to colour blindness located on X-chromosome.
Gene is recessive and is suppressed in heterozygous mother (female) but expressed in male in single dose.

Question. Write the type and location of the gene causing thalassaemia in human State the cause and symptoms of the disease. How is sickle cell anaemia different from this disease ?
Answer :
(i) Autosomal, recessive gene, gene for alpha thalassemia is on chromosome 16, for Beta thalassemia it is on chromosome 11. 1
(ii) Cause of symptoms—Mutation or deletion of the gene / genes, resulting in reduced rate of synthesis of one of the globin chains / alpha or beta chains).
(iii) Thalassaemia is a quantitative problem of too few globin molecules of haemoglobin, while sickle-cell is a qualitative problem of synthesizing an incorrectly functioning globin.

Question. A couple with normal vision bear a colour blind child. Workout a cross to show how it is possible and mention the sex of the affected child.
Answer :

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

Answer : (a) Short statured / small round head / furrowed tongue / partially open mouth / palm is broad / physical development retarded / psychomotor development retarded / mental development retarded.
(b) both / male and female
(c) Klinefelter’s syndrome
(d) male 
(e) sterile ovaries / rudimentary ovaries, lack of secondary sexual characters.

Question. Explain with the help of an example each– Male and female heterogamety mechanisms of sex determination. 
Answer :
Male produces 2 different types of gametes
XO – e.g. grasshopper.
XY – e.g. human, it is the type of sperm fertilising the egg that determine the sex of the offspring.
Female produces 2 different types of gametes.
ZW – eg. : Birds, it is the type of egg getting fertilised with the sperm that determine the sex of the chick.

Question. Explain the mechanism of ‘sex determination’ in birds. How does it differ from that of human beings?
Answer :.

types and (W) type of gametes
Humans : Male heterogamety / male produces (X) types and (Y) types of gametes

Question. Why are human females rarely haemophilic ?
Explain. How do haemophilic patients suffer ?
Answer : Haemophilia is a X-linked genetic disorder, which means that it shows criss cross inheritance. Like most recessive sex-linked X chromosome disorders, haemophilia is more likely to occur in males than females. This is because females have two X chromosomes (XX), while males have only one (XY), so the defective gene is guaranteed to manifest in any male who carries it. A female having two defective copies of the gene is very rare. 
Haemophilia impairs the body’s ability to control blood clotting and coagulation.

Question. If there is a history of haemophilia in the family, the chances of male members becoming haemophilic are more than that of the female.
(i) Why is it so ?
(ii) Write the symptoms of the disease.
Answer : (i) Defective gene is on X chromosome, in case the carrier female (mother) passes Xh to the son he suffers, if she passes Xh to the daughter, she has the other X (from father) to make it heterozygous so the daughters escape as carriers.
(ii) The blood does not clot in the affected person after an injury or a small cut. 

Question. (i) Sickle cell anaemia in human is a result of point mutation. Explain
(ii) Write the genotypes of both the parents who have produced a sickle celled anaemic offspring.
Answer : (i) Mutation arising due to change in single base pair of DNA, the defect is caused by the substitution of Glutamic acid (Glu) by Valine
(Val) at the sixth position of the beta globin chain of the haemoglobin molecule. 
(ii) Father – HbA HbS, Mother – HbA HbS  (Both parents are heterozygous)

Question. (i) Name the genetic disorder in a human female having 44 + XO karyotype. Mention the diagnostic features of the disorder.
(ii) Explain the cause of such chromosomal disorder.
Answer : (i) (a) Turner’s syndrome.
(b) 44 with XO chromosomes — such females are sterile as ovaries are rudimentary. Other features include lack of other secondary sexual characters, short stature and under developed feminine characters.
(ii) Such a disorder is caused due to the absence of one of X chromosomes.

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 cells.

(i) Is this representation indicating a normal human or a sufferer from certain genetic disease ? Give reason in support of your answer.
(ii) What difference would be noticed in the phenotype of the normal and the sufferer related to this gene ?
(iii) 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 : (i) This representation (HbA peptide) indicates a normal human, because the glutamic acid in the sixth position is not substituted by Valine.
(ii) The sufferer’s RBCs become elongated and sickle shaped as compared to the normal biconcave RBCs.
(ii) Both males and females are likely to suffer from the disease equally, as this is not a sex linked disease. It is an autosomal linked recessive trait.

Question. Study the given pedigree chart and answer the questions that follow:

(a) Is the trait recessive or dominant?
(b) Is the trait sex-linked or autosomal?
(c) Give the genotypes of the parents shown in generation I and their third child shown in  generation II and the first grandchild shown in generation III. 
Answer :
(a) Dominant
(b) Autosomal.
(c) Genotype of parents in generation I – Female – aa and Male – Aa.
Genotype of third child in generation II – Aa.
Genotype of first grandchild in generation III – Aa.

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.
(i) Represent this cross in the form of a pedigree chart. Give the genotypes of the daughter and her husband.
(ii) Write the conclusion you draw from the inheritance pattern of this disease. 
Answer :

Question. Haemophilia is a sex linked recessive disorder of human 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? Show with the help of Punnett square.
Answer : (a) Genotypes of member 4 – XX or XXh
Genotypes of member 5 – XhY
Genotypes of member 6 – XY

(b) The probability of first child to be a haemophilic male is 25%.
1 mark for punnetts square + ½ for probability

Question. Why do normal RBCs become elongated sickle shaped structures in a person suffering from sickle shaped anaemia.
Answer : Sickle cell anaemia is caused due to point mutation because of which in β-globulin chain of haemoglobin molecule, the glutamic acid (Glu) is replaced by valine. This results in oxygen stress. Under this condition the RBCs loose their circular shape, polymerise and become elongated and sickle shaped.

Question. A haemophilic son was born to normal parents.
Give the genotypes of parents and son.
Answer : Father (A + XY)
Mother (A + XXh)
Son (A + XhY)
(A = Autosomes which are 44 in number &
XY = Sex chromosomes).

Question. Why is that the father never passes on the genes for haemophilia to his son ?
Answer : Haemophilia is a sex linked trait. The gene for it is located on X chromosome only. Since, father contributes only Y chromosome to the son, he never passes haemophilic gene to his son.

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