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Question 15 Marks
Explain the Law of Dominance using a monohybrid cross.
Answer
Mendel's law of dominance states that a dominant allele expresses itself in a monohybrid cross and suppresses the expression of recessive allele. However, this recessive allele for a character is not lost and remains hidden or masked in the progenies of $F_1$ generation and reappears in the next generation. For example, when pea plants with round seeds (RR) are crossed with plants with wrinkled seeds ( rr ), all seeds in $\mathrm{F}_1$ generation were found to be round (Rr). When these round seeds were self-fertilized, both the round and wrinkled seeds appeared in $F_2$ generation in $3: 1$ ratio. Hence, in $\mathrm{F}_1$ generation, the dominant character (round seeds) appeared and the recessive character (wrinkled seeds) got suppressed, which reappeared in $F_2$ generation.
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Question 25 Marks
Briefly mention the contribution of T.H. Morgan in genetics.
Answer
TH Morgan's contribution in genetics are:
  • Morgan contributed to the understanding of sex-linked traits.
  • He carried out several experiments on fruit flies and his dihybrid crosses revealed that genes for some traits did not segregate independently and the F2 ratio was a lot different from 9 : 3 : 3 : 1.
  • He inferred that when two genes were situated on the same chromosome, the proportions of parental gene combinations were significantly higher than the non-parental type.
  • He established the principle of linkage crossing over, sex-linked inheritance and discovered the relation.
  • He established the technique of chromosome mapping.
  • He observed and worked on mutation.
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Question 35 Marks
A child has blood group O. If the father has blood group A and mother blood group B, work out the genotypes of the parents and the possible genotypes of the other offsprings.
Answer
The child with blood group 'O' will have homozygous recessive alleles. Therefore, both the parents should be heterozygous, i.e., the genotype of the father will be $l^iA$ and of mother will be $l^iB.$

Blood group: AB
The possible blood groups of other offsprings will be AB, A, B and O.
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Question 45 Marks
Mention any two autosomal genetic disorders with their symptoms.
Answer
Two autosomal genetic disorders are as follows:Sickle cell Anaemia:
It is an autosomal linked recessive disorder, which is caused by point mutation in the beta-globin chain of haemoglobin pigment of the blood. The disease is characterized by sickle shaped red blood cells, which are formed due to the mutant haemoglobin molecule. The disease is controlled by $\mathrm{Hb}^{\mathrm{A}}$ and $\mathrm{Hb}^{\mathrm{S}}$ allele. The homozygous individuals with genotype, $\mathrm{Hb}^{\mathrm{S}} \mathrm{Hb}^{\mathrm{S}}$, show the symptoms of this disease while the heterozygous individuals with genotype, $\mathrm{Hb}^A \mathrm{Hb}^{\mathrm{S}}$, are not affected. However, they act as carriers of the disease.
Symptoms:
Rapid heart rate, breathlessness, delayed growth and puberty, jaundice, weakness, fever, excessive thirst, chest pain, and decreased fertility are the major symptoms of sickle cell anaemia disease.
Down's syndrome:
It is an autosomal disorder that is caused by the trisomy of chromosome 21.
Symptoms:
The individual is short statured with round head, open mouth, protruding tongue, short neck, slanting eyes, and broad short hands. The individual also shows retarded mental and physical growth.
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Question 55 Marks
How is sex determined in human beings?
Answer
Humans have 47wQbNPTDJp9hMYdvogK2hAUiHsGeiybwaWe36bwtRQ3UTpYV7YuZ8FV5j9nauFCWwcjM6dTzpL5s2N79Rp5unwdMvc8ZKU and the remaining 1 pair is the sex chromosome that determines the sex of the child. In humans sex chromosome is of two types X and Y. The genetic composition of females is 44+XX whereas males have a genetic composition of 47wQbNPTDJp9hMYdvogK2hAUiHsGeiybwaWe36bwtRQ3UTpYV7YuZ8FV5j9nauFCWwcjM6dTzpL5s2N79Rp5unwdMvc8ZKUchromosomes from one parent and the other half from another the diploid state is restored. Now, the gametes formed by the female is all 22+X where as males produce two types of gametes, 22+X and 22+Y. During fertilization if the female gamete with 22+X chromosome is fertilized by the sperm bearing 22+X chromosome the resulting zygote has the genetic composition as 44+XX thus giving rise to a female child. But, if the female gamete with 22+X chromosome is fertilized by the sperm bearing 22+Y chromosome the resulting zygote has the genetic composition as 47wQbNPTDJp9hMYdvogK2hAUiHsGeiybwaWe36bwtRQ3UTpYV7YuZ8FV5j9nauFCWwcjM6dTzpL5s2N79Rp5unwdMvc8ZKUsperm that fertilizes the ovum and not the other way round.
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Question 65 Marks
Two heterozygous parents are crossed. If the two loci are linked what would be the distribution of phenotypic features in $F_1$ generation for a dibybrid cross?
Answer
If the two loci of characters are linked then the distance separating them will be very small and recombination will not be able to take place in between them. Thus no segregation can be observed. As the parent is heterozygous thus both dominant and recessive alleles are present on the two homozygous chromosomes and one chromosome is obtained by the gamete. Thus, in the $F_1$ generation for a di hybrid cross, if this gamete contains the dominant traits and is fertilized by either dominant or recessive alleles of the same trait from the other parent the dominant phenotypic character are observed. The recessive linked phenotype of the traits are only observable only when the both the gametes from the parents that are fertilized contains the recessive trait. Also due to no segregation between the two linked traits, they are always found together in both the cases.
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Question 75 Marks
  1. 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 the pea plant bearing axial flowers. Work out the cross to show the conclusions you arrive at.
  2. State the Mendel’s law of inheritance that is universally acceptable.
Answer
  1.  
  1. If the plants is homozygous for the dominant trait.
  1. 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.
  1. Law of Segregation, allelic pair segregate (separates) during gamete formation. (do not loose their identity)
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Question 85 Marks
  1. How are Mendelian inheritance, polygenic inheritance and pleiotropy different from each other?
  2. Explain polygenic inheritance pattern with the help of a suitable example.
Answer
 
  1. Pleiotropy: A single gene that has an effect on the expression of two or more phonotypic traits is said to have a pleiotropic effect on the traits. For example, testosterone controls the development of what are referred to as secondary sexual characteristics (e.g., a hon’s mane), but testosterone also relates behavioral traits like aggraession/Thus, a gene that controls the level of testosterone would have a pleiotropic iffect on the expression of many secondary sexual traitjs which are morphological, as well as behavioral. The concept of pleiotropy is intimately related to the concept of trade-off (Steams- 1976). Pleiotropy describes the proximate genetic gene controls the expression of two or more traits and those traits are related to a fitness trade-off, then we have identified the proximate source of the trade-off.
Polygenic: If two or more genes are responsible for a single trait, the phenotypic.
  1. If the gene for eye color is on the X chromosome, the red eyed male in the second cross will pass his red eyed X to only his daughters, who in turn received only a recessive white- carrying X from their mother. Thus all females had trend eyes like their father. Since the male fruit fly passes only the Y to his sons, their eye color is determined entirely by the single X chromosome they receive from their mother (in this case white). Thus all the males in the second cross were white eyed.
These experiments introduced the concept of sex-linkage, the occurrence of genes on that part of the X that lack a corresponding location on the Y. Sex- linked recessives (such as white eyes in fruit flies, hemophilia, baldness, and colorblindness in humans) occur more commonly in males, since there is no chance of them being heterozygous. Such condition is termedhemizygous.
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Question 95 Marks
  1. Explain Polygenic inheritance and Multiple allelism with the help of suitable examples.
  2. "Phenylketonuria is a good example that explains Pleiotropy." Justify.
Answer
  1. 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 $(I^A, I^B, i)$
  1. In pleiotropy a single gene can exhibit multiple phenotypic expressions , in phenyl ketonuria single mutated gene express mental retardation and reduction in hair and skin pigmentation.
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Question 105 Marks
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 $F_1-$population were tall with round seeds. However selfing among $F_1-$ population led to a 9:3:3:1 phenotypic ratio.
Answer
 
TR
Tr
tR
tr
TR
TTRR
Tall
Round
TTRr
Tall
Round
TtRR
Tall
Round
TtRr
Tall
Round
Tr
TTRr
Tall
Round
TTrr
Tall
Wrinkled
TtRr
Tall
Round
Ttrr
Tall
Wrinkled
tR
TtRR
Tall
Round
TtRr
Tall
Round
ttrr
Dwarf
round
ttRr
Dwarf
Round
tr
TtRr
Tall
Round
Ttrr
Tall
Wrinkled
ttRr
Dwarf
round
ttrr
Dwarf
round
Phenotypic ratio—
Tall Tall Dwarf Dwarf
Round Wrinkled Round Wrinkled
9 : 3 : 3 : 1
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 seperate 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 $F_2$ 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.
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Question 115 Marks
  1. Why are colourblindness and thalassemia categorised as Mendelian disorders? Write the symptoms of these diseases seen in people suffering from them.
  2. About 8% of human male population suffers from colourblindness whereas only about 0·4% of human female population suffers from this disease. Write an explanation to show how it is possible.
Answer
  1. Colour blindness and thalassaemia are categorised as Mendelian disorders because they occur by mutation in a single gene. Their mode of inheritance follows the principles of Mendelian genetics. Mendelian disorders can be
  • autosomal dominant (muscular dystrophy)
  • autosomal recessive (thalassaemia)
  • sex linked (colour blindness)
Symptoms of Thalassaemia:
  • Thalassaemia minor results only in mild anaemia, characterised by low haemoglobin level.
  • Thalassaemia major is also known as Cooley's anaemia. In this disease, affected infants are normal but as they reach 6 to 9 months of age, they develop severe anaemia, skeletal deformities, jaundice, fatigue, etc.
Symptoms of Colour Blindness:
  • Poor vision
  • Inability to distinguish between red and green colour
  • Inability to tolerate bright light.
  1. Colour blindness is a X-linked recessive disorder. Compared to females, males have higher chances of getting affected because females have two X chromosomes (XX) while males have only one X with Y chromosome. Thus, for a female to get affected by colour blindness, she has to have the mutant gene on both the X chromosomes while males can be affected if they carry it on the single X chromosome.
Conditions for colour blindness:
$XY; XX:$ Normal
$X^C Y :$ Colour-blind male
$X^C X^C: $ Colour-blind female
$X^C X : $ Carrier female
Let us assume that a carrier female $(X^C X)$ is married to a normal male.

From the given table, it can be concluded that females have very less probability of getting this disease as compared to males. Females will be colour-blind only when either both parents are affected or male is affected and female is carrier, while males can be colour-blind even if female is carrier and male is normal.
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Question 125 Marks
Work out a typical Mendelian dihybrid cross and state the law that he derived from it.
Answer

Law of Independent Assortment- It states that when two pairs of traits are combgined
in a hybrid, seggregation of one pair of character is independent of the other pair of characters.
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Question 135 Marks
  1. Explain a monohybrid cross taking seed coat colour as a trait in Pisum sativum. Work out the cross upto $F_2$ generation.
  2. State the laws of inheritance that can be derived from such a cross.
  3. How is the phenotypic ratio of $F_2$ generation different in a dihybrid cross?
Answer
$F_2-$ Phenotypic ratio = 3:1

Genotypic ratio = 1:2:1
  1. 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.
  2. Phenotypic ratio of $F_2$ in monohybrid cross is 3:1 whereas in a dihybrid cross the phenotypic ratio is 9:3:3:1.
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Question 145 Marks
  1. Why is haemophilia generally observed in human males? Explain the conditions under which a human female can be haemophilic.
  2. A pregnant human female was advised to undergo M.T.P. It was diagnosed by her doctor that the foetus she is carrying has developed from a zygote formed by an XX-egg fertilised by Y-carrying sperm. Why was she advised to undergo M.T.P.?
Answer
  1. Haemophilia is caused due to the recessive, gene on X chromosome, Y has no allele for this/if a male is $X^hY$ then he is haemophilic/if male inherits $X^h$ from the mother he is haemophilic (with the genotype $X^hY$)
    If female inherits $X^hX^h$, one from the carrier mother and one from her haemophilic father (then she can be haemophilic).
  2. Embryohas (trisomy of sex chromosome) XXY/Klinefelter’s syndrome
Advised MTP since the child will have the following problems:

Male with feminine traits/like gynaecomastia/under developed testes/sterile.
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Question 155 Marks
Describe the mechanism of pattern of inheritance of ABO blood groups in humans.
Answer
Human blood group is determined by glycoprotein/antigen A, and glycoprotein/antigen B The alleles are $I^A, I^B$ and i -Hence referred to asmultiple allelism. The individual inherits any two of them as given below
$I^A I^A, I^A i$ A group
$I^BI^B, I^Bi$ B group
$I^AI^B$ AB group
ii O group

In the case of A, B and O Law of dominance is the pattern of inheritance as $I^A/ I^B$ dominant over i.
In AB group both the alleles $I^A$ and $I^B$ express It is the case of Co-dominance.
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Question 165 Marks
  1. State the law of independent assortment.
  2. Using Punnett Square demonstrate the law of independent assortment in a dihybrid cross involving two heterozygous parents.
Answer
  1. When two pairs of traits are combined in a hybrid, segregation of one pair of character is independent of the other pair of characters.
  2. 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.
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Question 175 Marks
A particular garden pea plant produces only violet flowers.
  1. Is it homozygous dominant for the trait or heterozygous?
  2. How would you ensure its genotype? Explain with the help of crosses.
Answer
  1. It could be homozygous dominant, heterozygous dominant.
  2. By test cross.
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Question 185 Marks
Thalassemia and Haemophilia are both Mendelian disorders related to blood. Write the symptoms of the diseases. Explain with the help of crosses the difference in the inheritance pattern of the two diseases.
Answer
Thalassemia- Anaemia.
Haemophilia- Non stop bleeding.
Haemophilia- Sex linked recessive disorder, is generally passed on from (carrier) mother to some of her sons/from affected father to daughter. (carrier)


Thalassemia- Autosome linked recessive blood disease , inheritance is like Mendelian inheritance pattern.

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Question 195 Marks
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 were very 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.
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Question 205 Marks
  1. State the cause and symptoms of colour-blindness in humans.
  2. Statistical data has shown that 8% of the human males are colour-blind whereas only 0.4% of females are colour-blind. Explain giving reasons how is it so.
Answer
  1. Cause- sex-linked recessive disorder.
Symptoms- failure to discriminate between red and green colour.
  1. Since males have only one X chromosome gene for colour blindness, if present in any one parent will always be expressed, whereas in female it will be expressed only if it is present on both the X chromosomes, when both parents are carrying gene for colour blindness.
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Question 215 Marks
State and explain the “law of independent assortment” in a typical Mendelian dihybrid cross.
Answer
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 and crossed them with homozygous pea plant producing green and wrinkled seeds/shown in a flow chart of a dihybrid cross given.
Phenotypes
Yellow
round
:
Yellow
wrinkled
:
Green
round
:
Green
wrinkled
Phenotype ratio
9
:
3
:
3
:
1
(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)
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Question 225 Marks
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.
  1. What is Thalassemia?
  2. How would you counsel the family not to blame the mother for delivering a child suffering from this disease? Explain.
  3. List the values your counselling can propagate in the families.
Answer
  1. Thalassemia is a group of autosomal recessive blood disorders characterised by sever anaemia as a result of production of faulty haemoglobin chains. Mutations in the genes coding for the alpha, beta or delta chains constituting haemoglobin lead to the synthesis of improperly folded haemoglobin that is incapable of transporting oxygen efficiently.
  2. Thalassemia is an autosomal recessive disease, which means the mutation is carried on one of the autosomes, so the carrier can be any one of two parents. It has an equal probability of coming from the mother or the father, so to just blame the mother for the child’s abnormality is unjustified.
  3. The values counselling can propagate in the families are-
  1. Give healthy diet to the child.
  2. Accepting their child with all his/her positives and negatives.
  3. Neither of the parents is responsible for giving birth to a sick baby.
  4. The defect is caused by a random change in the genes of child.
  5. Encouraging the child to follow his/her treatment regularly and lead a happy and normal Life.
  6. Support the child emotionally by taking about fear, anxiety and depression or stress the child feels.
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Question 235 Marks
What is the inheritance pattern observed in the size of starch grains and seed shape of Pisum sativum? Workout 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 grains and the seed shape.(Deviation fromMendelian Law of Dominance): The trait of size of starch grain shows incomplete dominance. Hence in heterozygous condition the starch grain are of intermediate size. The trait of seed shape follows Law of Dominance and the hybrid will show only dominant trait.
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Question 245 Marks
  1. Write the scientific name of the organism Thomas Hunt Morgan and his colleagues worked with for their experiments. Explain the correlation between linkage and recombination with respect to genes as studied by them.
  2. How did Sturtevant explain gene mapping while working with Morgan?
Answer
  1. Drosophila melanogaster:
Morgan carried out several dihybrid crosses in Drosophila to study gens that were sex-lined. Morgan and his group knew that the genes were located on the X chromosome and saw quickly that when the two genes in a dihybrid cross were situated on the same chromosome, the proportion of parental gene combinations were much higher than the non-parental type. Morgan attributed this due to the physical association or linkage of the two genes and coined the term linkage to describe this physical association of genes on a chromosome and the term recombination to describe the generation of non-parental gene combination. Morgan and his group also found that even when genes were grouped on the same chromosome, some genes were very tightly linked (showed very low recombination) while others were loosely linked.
  1. Morgans student Alfred Sturtevant 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. Today genetic maps are extensively used as a starting point in the sequencing of whole genomes.
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Question 255 Marks
Differentiate between incomplete dominance and co-dominance. Substantiate your answer with one example of each.
Answer
Co-dominance is the phenomenon in which both the alleles of a contrasting character are expressed in the heterozygous condition. Both the alleles of a gene are equally dominant. ABO blood group in human beings is an example of co-dominance. The blood group character is controlled by three sets of alleles, namely, $I^A, I^B$, and i. The alleles, $I^A$ and $I^B$, are equally dominant and are said to be co­­-dominant as they are expressed in the AB blood group. Both these alleles do not interfere with the expression of each other and produce their respective antigens. Hence, the AB blood group is an example of co-dominance.
Allele from Parent 1
Allele from Parent 2
Genotype of offspring
Blood type of offspring
$I^A$
$I^A$
$I^A I^A$
A
$I^A$
$I^B$
$I^AI^B$
AB
$I^A$
i
$I^A i$
A
$I^B$
$I^A$
$I^A I^B$
AB
$I^B$
$I^B$
$I^BI^B$
B
$I^B$
i
$I^B i$
B
i
i
i i
O
Incomplete dominance is a phenomenon in which one allele shows incomplete dominance over the other member of the allelic pair for a character. For example, a monohybrid cross between the plants having red flowers and white flowers in Antirrhinum species will result in all pink flower plants in the $F_1$ generation. The progeny obtained in $F_1$ generation does not resemble either of the parents and exhibits intermediate characteristics. This is because the dominant allele, R, is partially dominant over the other allele, r. Therefore, the recessive allele, r, also gets expressed in the $F_1$ generation resulting in the production of intermediate pink flowering progenies with Rr genotype.
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Question 265 Marks
Sunny was rushed to a nearby hospital after an accident, which had caused too much blood loss. The hospital failed to supply 'O' negative blood for transfusion. Ahmed, who was attending a patient there, heard about the situation and agreed to donate blood as he had the same blood group. Sunny's mother initially refused but was later convinced by her daughter's view.
  1. Why cannot 'O' positive blood be transfused into Sunny's body?
  2. What is the genetic basis of blood group inheritance?
  3. Human blood grouping is an example of?
  4. What values do you find in Sunny's sister and Ahmed?
Answer
  1. $O^+$ blood group contains antigen for Rh factor. $Rh^-$ blood lacks the antigen, so if the two blood mixes, it will cause clotting resulting in death of the patient.
  2. Blood groups are inherited from both parents. The ABO blood type is controlled by a single gene (the ABO gene) with three types of alleles inferred from classical genetics: $i, I^A$, and $I^B$. The I designation stands for isoagglutinogen, another term for antigen.
  3. Multiple alleles.
  4. Sunny's sister shows understanding of science and Ahmed believes in helping a person in need without thinking about which religion or caste he belongs to.
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Question 275 Marks
Can a child have blood group O if his parents have blood group ‘A’ and ‘B’. Explain.
Answer
If parents are heterozygous then child can have O blood group but if parents are homozygous then child cannot have O blood group.
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Question 285 Marks
What is Down’s syndrome? Give its symptoms and cause. Why is it that the chances of having a child with Down’s syndrome increases if the age of the mother exceeds forty years?
Answer
Down’s syndrome is a chromosomal disorder. A person suffering from Down’s syndrome shows following symptoms:
  • Short stature.
  • Small round head.
  • Furrowed tongue and partially open mouth.
  • Palm is broad with characteristic palm crease.
  • Physical, psychomotor and mental development is retarded in the person.
This condition happens because of an extra copy of chromosome 21. As per various research reports, chances of having a child with Down’s syndrome increases with advancing maternal age because ova are present in females right from their birth. More is the age of the mother, cells will be more older. Hence, chances of chromosomal non-disjunction will be more because of various physico-chemical exposures during the mother’s life-time.
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Question 295 Marks
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.
  1. What is thalassemia?
  2. How would you counsel the family not to blame the mother for delivering a child suffering from this disease? Explain.
  3. List the values your counselling can propagate in the families.
Answer
  1. Thalassemia is a blood disorder passed down through families (inherited) in which the body makes an abnormal form or inadequate amount of hemoglobin. Hemoglobin is the protein in red blood cells that carries oxygen.
  2. Mother cannot be blamed for the disease because it is an autosomal recessive blood disorder. The genes for the synthesis of globin chains are present on one of the autosomes. So, the carrier can be any one of the two parents. It has equal probability of coming from the mother or father. So, to blame the mother for the child's abnormality is unjustified.
  3. Values that can be propogated in families are:
  1. One should be aware of such autosomal diseases and should get themselves checked for the same before marriage.
  2. Give a healthy diet plan to child.
  3. Neither of the parents is responsible for giving birth to a sick baby.
  4. Accepting their child with all his/ her positives and negatives.
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Question 305 Marks
In a court of law, two couples X and Y, claim a child Z of blood group O belongs to them. Couple X has man with blood group O and woman with blood group AB. Couple Y has man with blood group A and woman with blood group B. So, a geneticists helped in solving the problem but the judge ordered the= use of more recent and advance techniques and then the child was handed over ultimately to couple Y.
  1. Is it a correct judgement regarding the true/ biological parents of the child? Justify with your knowledge of genetics.
  2. Name the other technique that the judge has ordered for solving this problem.
  3. Indicate the value(s) exhibited by the judge.
Answer
  1. Yes, the judgment is correct. Couple X has blood groups O (man) and AB (woman).
Their genotypes and the progeny are as follows:



Thus, couple-X cannot have a child of O blood group. Couple Y has blood group A (man) and B (woman); since, the child is O group (ii), both these parents must be heterozygous, i.e. man $(I^Ai)$ and woman $(I^Bi)$.

The cross is shown below:



It is possible that child with O blood group is produced by couple Y.
  1. DNA-fingerprinting is another technique to solve the paternity cases.
  2. The judge wanted to be very sure of the parents before pronouncing his judgment; he made his self-conscience satisfied.
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Question 315 Marks
Discuss why Drosophila has been used extensively for genetical studies.
Answer
Morgan worked with the tiny fruit files, Drosophila melanogaster, which were found very suitable for such studies, as:
  1. They could be grown on simple synthetic medium (ripe banana) in the laboratory.
  2. They complete their life cycle in about two weeks.
  3. A single mating could produce a large number of progeny flies.
  4. There was a clear differentiation of the sexes – the male and female flies are easily distinguisable.
  5. It has many types of hereditary variations that can be seen with low power microscopes.
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Question 325 Marks
  1. Explain the phenomenon of multiple allelism and co-dominance taking ABO blood group as an example,
  2. What is the phenotype of the following:
  1. $I^Ai$
  2. $ii$
Answer
  1. Co-dominance: The alleles which are able to express themselves independently when present together are called co-dominant alleles and this phenomenon is termed as co-dominance. For example AB blood group, in which both the alleles IA (For blood group A) and I (For blood group B) are co-dominant, hence when they come together in an individual, they form blood group AB.
Multiple Allelism: It refers to the occurrence of more than two alleles associated with a single locus although no individual carries more than two.
  • ABO blood groups inheritance in man is example of this phenomenon.
  • The gene for blood group has three alleles$-I^A, I^B$ and i. (The letter I is related to their isoagglutination).
  • An individual contains two out of these three alleles.
  • Glycoprotein A is produced by allele $I^A$ and this glycoprotein is present on the surface of red blood cells.
  • Glycoprotein B on the surface of RBCs is produced by allele Io.
  • No glycoprotein is produced by allele i.
  • The alleles 14 and 15 are dominant over i.
  • Alleles I and I when together express themselves in combination as both are equally dominant and both the glycoproteins A and B are produced. The blood group is decided by the presence of the glycoproteins i.e., group A contains glycoprotein A, group B contains glycoprotein B, group AB contains both the glycoproteins A and B and group O does not contain any glycoprotein.
  • The blood groups and their possible genotype are shown by following table:
S. No.
Blood group
Possible genotypes
i.
A
$I^AI^A$ or $I^A i$
ii.
B
$I^BI^B$ or $I^B i$
iii.
AB
$I^A I^B$
iv.
O
ii
  • Mendelian pattern of Inheritance is followed by the inheritance of blood group character.
  1.  
  1. A blood group.
  2. O blood group.
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Question 335 Marks
Work out a monohybrid cross upto F, generation, between two pea plants and two Antirrhinum plants, both having contrasting traits with respect to colour of flower. Comment on the patterns of inheritance in the crosses carried above.
Answer
The plants must be crossed with a white flowered (homozygous recessive) plant, i.e. a test cross is conducted.
Garden Pea:
  1. If the plant is homozygous dominant, all the plants in the progeny would bear red flowers (heterozygous) and if the plant is heterozygous, the progeny would consist of red-flowered plants and white-flowered plants in the ratio of 1 : 1.
  1. Homozygous Red:
  1. Heterozygous Red:

Red flowered : White flowered, 1 : 1.
Snapdragon:
  1. In snapdragon, when a red-flowered plant is crossed to a white-flowered plant, the progeny would consist of pink-flowered plants.
  2. So, red-flowered plants are homozygous dominant, while heterozygous condition produces pink-flowered plants.
  3. The cross is as follows:
  1. It is because of incomplete dominance, where neither of the two alleles is completely dominant over the other and the hybrid is intermediate between the two i.e. pink-flowered.
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Question 345 Marks
Inheritance pattern of flower colour in garden pea plant and snapdragon differs. Why is this difference observed? Explain showing the crosses up to $F_2$ generation.
Answer
Inheritance pattern of flower colour in garden pea follows principle of dominance whereas inheritance in snapdragon shows incomplete dominance.

Phenotypic ratio–3 : 1 Genotypic ratio–1 : 2 : 1. Inheritance of flower colour in snapdragon:

Phenotypic ratio–1 : 2 : 1 Genotypic ratio–1 : 2 : 1.
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Question 355 Marks
Answer the following questions:
Why is haemophilia generally observed in human males? Explain the conditions under which a human female can be haemophilic.
Answer
Haemophilia is caused due to the recessive gene on X chromosome. Y chromosome has no allele for this.
If a male is $X^h Y$, then he is haemophilic.
If male inherits Xh from the mother, he will be haemophilic (with the genotype $\mathrm{X}^h \mathrm{Y}$ ). If female inherits $\mathrm{X}^h \mathrm{X}^h$, one from the carrier mother and one from her haemophilic father, then she can be haemophilic.
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Question 365 Marks
State the law of independent assortment.
Answer
According to this law the two factors of each character assort or separate out independent of the factors of other characters, at the time of gamete formation and get randomly rearranged in the offsprings, producing both parental and new combinations of characters.
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Question 375 Marks
Answer the following questions:
During a cross involving true breeding red flowered and true breeding white flowered snapdragon plants, the $F_1$ progeny did not show any of the parental traits, while they reappeared in $F_2$ progenies. Explain the mechanism using Punnett Square.
Answer
Incomplete Dominance: It is a phenomenon in which the $F_1$ hybrid exhibits characters intermediate of the parental genes. Here, the phenotypic ratio deviates from the Mendel’s monohybrid ratio. It is seen in flower colours of Mirabilis jalapa (4 o’ clock plant) and Antirrhinum majus (snapdragon), where red colour is due to gene RR, white colour is due to gene rr and pink colour is due to gene Rr.
Monohybrid cross in snapdragon, where one allele is incompletely dominant over the other allele.
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Question 385 Marks
Explain the chromosomal theory of inheritance.
Answer
Chromosomal Theory of Inheritance:
The chromosomal theory of inheritance was proposed independently by Walter Sutton and Theodore Boveri in 1902. According to this theory,
  1. Since the sperm and egg cells provide the only bridge from one generation to the other, all hereditary characters must be carried in them. ii. The hereditary factors are carried in the nucleus.
  2. Like the Mendelian alleles, chromosomes are also found in pairs.
  3. The sperm and egg having haploid sets of chromosomes fuse to re-establish the diploid state.
  4. The genes are located on the chromosomes in a linear order. As there are two chromosomes of each kind in somatic (diploid) cell there must be two genes of each kind, one in each of the two homologous chromosomes.
  5. Homologous chromosomes synapse during meiosis and get separated to pass into different cells. This forms the basis for segregation and independent assortment. A gamete receives only one chromosome of each type and thus has only one gene for a trait. The paired condition is restored by fusion of gametes.
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Question 395 Marks
Thalassemia and haemophilia are both Mendelian disorders related to blood. Write the symptoms of the diseases. Explain with the help of crosses the difference in the inheritance pattern of the two diseases.
OR
Why are thalassemia and hemophilia categorised as Mendelian disorders? Write the symptoms of these diseases. Explain their pattern of inheritance in humans. Write the genotypes of the normal parents producing a haemophilic son.
Answer
Both are caused due to alteration or mutation, in a single gene and follow Mendelian pattern of inheritance.
Symptoms:
Thalassemia: Anaemia (caused due to defective/ abnormal Hb).

Haemophilia: Non–stop bleeding even in minor injury.
Pattern of inheritance:

Thalassemia: Autosomal recessive inheritance pattern inherited from eterozygous/ parent carrier.

Haemophilia: X–linked recessive inheritance inherited from a haemophilic father/carrier mother (females are rarely haemophilic).



Thalassemia is an autosome–linked recessive blood disease. Its inheritance is like Mendelian inheritance pattern.
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Question 405 Marks
State the Mendel’s law of inheritance that is universally acceptable.
Answer
Law of Segregation is universally accepted. It states that allelic pair segregate (separates) during gamete formation.
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Question 415 Marks
Using Punnett square demonstrate the law of independent assortment in a dihybrid cross involving two heterozygous parents.
Answer

Results of a dihybrid cross where the two parents differed in two pairs of contrasting traits: seed colour and seed shape.
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Question 425 Marks
State the laws of inheritance that can be derived from such a cross.
Answer
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 separate from each other such that gamete receives only one of the two factors.
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Question 435 Marks
List the characteristics of the disorders mentioned above that help in their diagnosis.
Answer
Down’s syndrome: The individuals have overall masculine development but they express feminine development like development of east, i.e., gynaecomastia. They are sterile.
Klinefelter’s syndrome: The females are sterile as ovaries are rudimentary. Other secondary sexual characters are also lacking.
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Question 445 Marks
Write the symptoms of haemophilia and sickle-cell anaemia in humans. Explain how the inheritance pattern of the two diseases differ from each other.
Answer
  1. Symptoms of Haemophilia:
  • Since a protein necessary for blood clotting is not formed, the blood does not clot and there is non-stop bleeding in case of an injury in the affected individual.
  1. Symptoms of sickle-cell anaemia:
  • The RBCs become sickle-shaped.
  • The oxygen transport to the tissues is impaired.
  1. Differences:
S.No
Haemophilia
Sickle-cell anaemia
1.
It is due to a defective recessive allele present on the X-chromosome, i.e. it is a sex-linked disorder.
It is due to point mutation, i.e. a single base pair change leading to a change in an amino acid, i.e. it is an autosomal disorder.
2.
More males than females are affected.
Both males and females are affected equally.
3.
The female parent passes on the disorder to male progeny, but father never passes it on to the male progeny.
The female parent passes on the disorder to male or female progeny in equal frequency and father also passes on the disorder to male and female progeny.
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Question 455 Marks
Explain the mechanism of sex–determination in humans.
Answer
Sex Determination in Humans:
  • Humans show XY type of sex determining mechanism.
  • Out of 23 pair of chromosomes, 22 are autosomes (same in both males and females).
  • Females have a pair of X–chromosomes.
  • Males have an X and a Y chromosome.
  • During spermatogenesis males produce two types of gametes with equal probability – sperm carrying either X or Y chromosome.
  • During oogenesis females produce only one types of gamete – having X chromosome.
  • An ovum fertilised by the sperm carrying X–chromosome develops into a female (XX) and an ovum fertilised by the sperm carrying Y–chromosome develops into a male (XY).
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Question 465 Marks
Explain with an example, where a gene, which carries a major disadvantage in homozygous state, confers an advantage in heterozygous condition.
Answer
An example for this condition is sickle-cell anaemia. The person with homozygous condition $(Hb^sHb^s)$ for sickle-cell anaemia does not survive beyond the age of 20 years. The individual with normal gene (HbHb) does not have sickle-cell anaemia but they are sensitive towards malaria.
The person with heterozygous condition $(HbHb^s)$ for sickle-cell anaemia does not develop both sickle-cell anaemia and malaria.
Due to abnormal haemoglobin, the malarial parasite will not harm the person. Hence, in this case, the heterozygous condition has more survival chances, an advantage with respect to the homozygous individuals.
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Question 475 Marks
Explain a monohybrid cross taking seed coat colour as a trait in Pisum sativum. Work out the cross up to $F_2$ generation.
Answer


$F_2$ Phenotypic ratio = 3 : 1; $F_2$ Genotypic ratio = 1 : 2 : 1
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Question 485 Marks
A true breeding pea plant homozygous for axial violet flowers is crossed with another pea plant with terminal white flowers (aavv).
What would be the phenotype and genotype of $F_1$ and $F_2$ generations?
Answer


Phenotype of $F_1$ generation–All axial, Violet flowers.
Genotype of $F_1$ generation–AaVv.
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Question 495 Marks
With the help of one example each, provide genetic explanation for the following observations:
$F_1-$generation resembles both the parents.
Answer
$F_1$ generation resembles both the parents: This happens in the case of co-dominance where both alleles express themselves fully in heterozygous condition. For example: Different types of red blood cells determine ABO blood grouping in human beings.
Co-dominance:
  • The alleles which are able to express themselves independently, even when present together are called co-dominant alleles and this biological phenomenon is called co–dominance. For example, ABO blood grouping in humans.
  • ABO blood groups are controlled by gene I. Gene I has three alleles $I^A, I^B$ and $I^O/i.$
  • $I^A$ and $I^B$ produce RBC surface antigens sugar polymer A and B, respectively, whereas i does not produce any antigen.
  • $I^A$ and $I^B$ are dominant over i hence $I^A$ and $I^B$ are dominant alleles and i is recessive allele as in $I^Ai$ and $I^Bi.$
  • When $I^A$ and $I^B$ are present together, both express equally and produce the surface antigens A and B, hence show co-dominance.
  • Since humans are diploid, each person possesses any two of the three ‘I’ gene alleles, resulting into six different genotypic combinations and four phenotypic expressions.
Table showing the genetic basis of blood groups in human population:
Allele from Parent 1
Allele from Parent 2
Genotype of offspring
Blood groups of offspring
$I^A$
$I^A$
$I^AI^A$
A
$I^A$
B
$I^AI^B$
AB
$I^A$
i
$I^Ai$
A
$I^B$
$I^B$
$I^BI^B$
B
$I^B$
i
$I^Bi$
B
i
i
ii
O
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Question 505 Marks
How are genetic disorders broadly classified? Differentiate between them. Explain by giving one example for each.
Answer
These are disorders or illness caused by one or more abnormlities in the autosomes or sex chromosomes of the person. Thus, referred to as autosomal disorders or sex-linked disorders, respectively.
The generic disorders are divided into:

Differentiate between Mendelian and Chormosomal disorders are:
S.No.
Mendelian disorders
Chormosomal disorders
(i)
The disorders are caused by allelic abnormalities.
The disorders are caused by chromosomal abnormalities
(ii)
Allelic abnormalities develop due to mutations.
Chromosomal abnormalities develop due to defective synapsis and disjunction.
(iii)
The defect can be predicted through pedigree analysis.
The defect can be known through amniocentesis.
(iv)
The disorder is transmitted to the progeny.
The disorder is rarely transmitted.
(v)
Examples: Haemophilia, Cystic fibrosis, Colour blindness, Sickle-cell anaemia, Thalassemia, Phenylketonuria, etc.
Examples: Down's syndrome, Turner's syndrome, Klinefelter's syndrome, etc.
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Question 515 Marks
State and explain with the help of a cross, the law of segregation as proposed by Mendel.
Answer
Law of Segregation:
  1. law of segregation states that the members of the allelic pair that remained together in the parent/ hybrid, segregate during gamete formation and enter different gametes.
  2. As a result, gametes have only one allele for a trait and are pure for a character.
  1. The phenotypic ratio is 3 tall plants : 1 dwarf plant.
  2. The genotypic ratio is 1TT : 2Tt : 1tt
  3. In this case, tallness is dominant and dwarfness is recessive.
  4. The F, hybrid is tall (dominant character).
  5. The recessive character, dwarfness, remains hidden in the F, but reappeares in the F, generation without any change.
  6. This is because the factors Tandt remained together in the hybrid, but segregated during gamete formation and entered different gametes.
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Question 525 Marks
State the three principles of Mendel’s law of inheritance.
Answer
Mendel’s Laws of Inheritance:
  • Based on his hybridisation experiments, Mendel proposed the laws of inheritance.
  • His theory was rediscovered by Hugo de Vries of Holland, Carl Correns of Germany and Eric von Tschermak of Austria in 1901.
  1. Law of dominance:
  • This law states that when two alternative forms of a trait or character (genes or alleles) are present in an organism, only one factor expresses itself in $F_1$ progeny and is called dominant while the other that remains masked is called recessive.
  1. Law of segregation:
  • This law states that the factors or alleles of a pair segregate from each other during gamete formation, such that a gamete receives only one of the two factors. They do not show any blending.
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Question 535 Marks
  1. A true breeding pea plant, homozygous for inflated green pods is crossed with another pea plant with constricted yellow pods (ffgg). What would be the phenotype and genotype of $F_1$ and $F_2$ generations? Give the phenotype ratio of generation.
  2. State the generalisation proposed by Mendel on the basis of the above–Mentioned cross.
Answer
  1. Mendel’s Law of Independent Assortment: This law states that the different factors or allelomorphic pair in gametes assort themselves and segregate independently of one another.
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Question 545 Marks
  1. Work out a cross upto F, generation between two pure-bred pea plants, one bearing violet flowers and the other white flowers.
  2.  
  1. Name this type of cross.
  2. State the different laws of Mendel that can be derived from such a cross.
Answer
  • The phenotypic ratio of violet flowers : white flowers is 3 : 1.
  1.  
  1. It is a monohybrid cross.
  2.  
  • Law of dominance states that dominance is a phenomenon, in which one allele of a gene expresses itself and suppresses the expression of the other allele of the same gene, when they are present together in a hybrid.
  • Law of segregation states that the two factors of a trait that remained together in the hybrid segregate during gametogenesis and enter different gametes.
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Question 555 Marks
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
  1. A test cross is required to find out the genotype of both the plants.


If the $F_1$ generation plants have all red flowers, the genotype of the parent plant will be homozygous dominant and if the $F_1$ generation plants have red and white flowers in the ratio of 1 : 1, then the genotype of the parent plant is heterozygous dominant. This inheritance follows the Mendelian law of dominance.
  1.  


The parent plant will be homozygous for flower colour because a heterozygous plant will have pink flowers due to the phenomenon of incomplete inheritance.
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Question 565 Marks
A true breeding pea plant homozygous for axial violet flowers is crossed with another pea plant with terminal white flowers (aavv).
List the Mendel’s generalisations that can be derived from the above cross.
Answer
Law of Independent Assortment: This law states that the different factors or allelomorphic pair in gametes and zygotes assort themselves and segregate independently of one another.
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Question 575 Marks
  1. List the three different allelic forms of gene ‘I’ in humans. Explain the different phenotypic expressions, controlled by these three forms.
  2. 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:
  1. They produce children with blood group ‘A’ only.
  2. They produce children, some with 'O' blood group and some with 'A' blood group.
Answer
  1. Inheritance of human blood-group character:
  • The gene for blood group character exists in three allelic forms, $I^A, I^B$ and i; it is a case of multiple allelism.
  • The phenotypes and genotypes of these alleles are as follows:
Blood group
Genotype(s)
A
$I^AI^A$ or $I^Ai$
B
$I^BI^B$ or $I^Bi$
AB
$I^AI^B$
O
ii
  • The alleles $I^A$ and $I^B$ are codominant and express themselves as blood group AB, when they are together.
  • Blood group O is homozygous recessive.
  1.  
  1. Yes, when the woman is homozygous $(I^AI^A)$ for blood group A, the children can be with only blood group A,

Or
No, When the woman is heterozygous $(I^Ai)$ for blood group A, children will be of A group and O group.
  1. Yes, when the woman is heterozygous $(I^Ai)$ for blood group A, children can be with either blood group A or O (as in the cross above).
    Or
    No, when the woman is homozygous $(I^AI^A)$ for blood group A, the children can be with only blood group A (as in the first cross of (i) of this answer).
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Question 585 Marks
  1. How does a chromosomal disorder differ from a Mendelian disorder?
  2. Name any two chromosomal aberration associated disorders.
  3. List the characteristics of the disorders mentioned above that help in their diagnosis.
Answer
  1. Mendelian disorders are mainly determined by the alteration or mutation in the single gene. These mutations are transmitted to the offspring. On the other hand, Chromosomal disorders are caused due to absence or excess of one or more chromosomes.
  2. The two chromosomal aberrations that are most commonly seen are Down's syndrome (trisomy of 21st chromosome) and Turner's syndrome (absence of one of the X chromosome i.e. 45 + X0).
  3. The characteristics of the two chromosomal disorders are as follows:
  1. Characteristics of Down's syndrome are furrowed tongue, partially opened mouth, broad palm with characteristic palm creases, short-statured, small round head physical, psychomotor and mental development is retarded.
  2. Characteristics of Turner's syndrome are sterile or rudimentary ovary, lack of secondary sexual characteristics. It occurs only in females.
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Question 595 Marks
How many different types of gametes could result from each of the following genotypes? What will be their genotypes?
  1. Aa.
  2. AABB.
  3. AaBb.
  4. EeCc.
  5. FFIIJj.
Answer
The number of different types of gametes and their genotypes are:
  1. Aa : 2 (A, a).
  2. AABB : 1 (AB).
  3. AaBb : 4 (AB, Ab, aB, ab).
  4. EeCc : 4 (EC, Ec, eC, ec).
  5. FFIIJJ : 2 (FIJ, FIj).
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Question 605 Marks
Explain why a recessive allele is unable to express itself in a heterozygous state.
Answer
The alleles are present on homologous chromosomes.
The recessive allele does not code for its product or codes for a defective product.
The other allele remains normal and thus expresses itself.
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Question 615 Marks
Answer the following questions:
How did Morgan show the deviation in inheritance pattern in Drosophila with respect to this law?
Answer
Linkage and Recombination:
  • T. H. Morgan carried out several dihybrid crosses in Drosophila to study the genes that are sex–linked. He observed that when the two genes in a dihybrid cross are located on the same chromosome, the proportion of parental gene combinations in the progeny was much higher than the non-parental or recombination of genes.
  • Morgan and his group found that when genes are grouped on the same chromosome, some genes are tightly linked or associated and show little recombination.
  • Linkage: Results of two dihybrid crosses conducted by Morgan. Cross A shows crossing between genes y and w; Cross B shows crossing between genes w and m. Here dominant wild type alleles are represented with (+) sign in superscript.
  • When the genes are loosely linked they show higher percentage of recombination.
  • Morgan hybridised yellow bodied and white eyed females with brown bodied and red eyed males (wild type) (cross–A) and inter–crossed their $F_1$ progeny.
  • Alfred Sturtevant determined that genes of Drosophila are arranged in a linear order. He measured the distance between genes and prepared chromosome maps with the position of genes on the chromosomes based on percentage of recombinants. These are also called genetic maps.
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Question 625 Marks
Explain co–dominance taking an example of human blood groups in the population.
Answer
Co–dominance:
  • The alleles which are able to express themselves independently, even when present together are called co–dominant alleles and this biological phenomenon is called co–dominance. For example, ABO blood grouping in humans.
  • ABO blood groups are controlled by gene I. Gene I has three alleles $I^A,I^B$ and $I^O/i.$
  • $I^A$ and $I^B$ produce RBC surface antigens sugar polymer A and B, respectively, whereas i does not produce any antigen.
  • $I^A$ and $I^B$ are dominant over i hence $I^A$ and $I^B$ are dominant alleles and i is recessive allele as in $I^Ai$ and $I^Bi.$
  • When $I^A$ and $I^B$  are present together, both express equally and produce the surface antigens A and B, hence show co–dominance.
  • Since humans are diploid, each person possesses any two of the three ‘I’ gene alleles, resulting into six different genotypic combinations and four phenotypic expressions.
Table showing the genetic basis of blood groups in human population.
Allele fromParent 1 Allele fromParent 2 Genotype of offspring Blood groups of offspring
$I^A$ $I^A$ $I^AI^A$ A
$I^A$ $I^B$ $I^AI^B$ AB
$I^A$ i $I^Ai$ A
$I^B$ $I^B$ $I^BI^B$ B
$I^B$ i $I^Bi$ B
i i ii O
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Question 635 Marks
Identify ‘a’, ‘b’, ‘c’, ‘d’, ‘e’ and ‘f’ in the table given below:
S. No.
Syndrome
Cause
Characteristics of affected individuals
SexMale/ Female/ Both
1.
Down’s
Trisomy of 21
‘a’-
(i)
(ii)
‘b’
2.
‘c’
XXY
Overall masculinedevelopment.
‘d’
3.
Turner’s
45 with XO
‘e’-
(i)
(ii)
‘f’
Answer
  1. Short stature/ small round head/ furrowed tongue/ partially open mouth/ mental development retarded.
  2. Both.
  3. Klinefelter’s syndrome.
  4. Male.
    1. Sterile ovaries;
    2. Lack of secondary sexual characters.
  5. Female.
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Question 645 Marks
  1. You are given tall pea plants with yellow seeds, whose genotypes are unknown. How would you find the genotype of these plants? Explain with the help of cross.
  2. Identify A, B and C in the table given below:
Pattern of Inheritance
Monohybrid $F_1$ Phenotypic Expression
Codominance
‘A’
‘B’
Progeny resembled only one of the parents.
Incomplete dominance.
‘C’
Answer
  1. The given tall pea plant with yellow seeds need to be crossed with a dwarf plant with green seeds:
  1. The dominant traits are tallness and yellow colour of seeds. The recessive traits are dwarfness and green colour of seeds.
  2. Cross between tall plant/ yellow seeds and dwarf plant/green seeds. Cross showing heterozygous nature for both traits.


In this cross, the F-generation shows four phenotypes in the ratio of 1 : 1 : 1 : 1. So, the given plant is heterozygous for both the traits.
  1.  
A - Dominant trait.

B - Dominance.

C - Phenotype obtained in between of two parental traits.
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Question 655 Marks
Write the blood group of people with genotype $I^AI^B$. Give reasons in support of your answer.
Answer
Blood group AB. Both the alleles $I^A$ and $I^B$ are co–dominant and express themselves completely.
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Question 665 Marks
A true breeding homozygous pea plant with green pods and axial flowers as dominant characters, is crossed with a recessive homozygous pea plant with yellow pods and terminal flowers. Work out the cross up to $F_2$ generation giving the phenotypic ratios of $F_1$ and $F_2$ generation respectively.
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Question 675 Marks
Match the sex-chromosome complements in Column I with the male/ female organisms in Column II.
S.No
Column I
S.No
Column II
A.
XO
1.
Human male
B.
XX
2.
Male bird
C.
XY
3.
Male grasshopper
D.
ZZ
4.
Female Drosophila
 
 
5.
Female bird
Answer
S.No
Column I
S.No
Column II
A.
XO
3.
Male grasshopper
B.
XX
4.
Female Drosophila
C.
XY
1.
Human male
D.
ZZ
4.
Female bird
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Question 685 Marks
Explain polygenic inheritance pattern with the help of a suitable example.
Answer
Human height or skin colour are examples of polygenic inheritance.
Height trait is controlled by at least three gene pairs.
Additive effect allele contributes to the phenotypic expression of the trait.
The dominant alleles more are, more pronounced is the phenotypic expression or more in other word.
The recessive alleles less pronounced is the phenotypic expression.
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Question 695 Marks
“Phenylketonuria is a good example that explains Pleiotropy.” Justify.
Answer
In pleiotropy a single gene can exhibit multiple phenotypic expressions. In phenylketonuria single mutated gene express multiple phenotypic expression like mental retardation and reduction in hair and skin pigmentation.
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Question 705 Marks
Provide genetic explanation for the observation in which the flower colour in $F_1$ generation of snapdragon did not resemble either of the two parents. However, the parental characters reappeared when $F_1$ progenies were selfed.
Answer
This is an exception to Mendel’s principle of dominance and can be explained by the phenomenon of ‘Incomplete dominance’.
It is a phenomenon where none of the two contrasting alleles or factors are dominant.
The expression of the character in a hybrid or $F_1$ individual is intermediate or a fine mixture of expression of the two factors (pink flowers in this case from two parents with red and white flowers).
This may be considered as an example of quantitative inheritance where only a single gene pair is involved.
$F_2$ phenotypic ratio is 1 : 2 : 1, similar to the genotypic ratio, in which the parental characters also reappear.
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Question 715 Marks
Why is thalassemia categorised as a Mendelian disorder? State the condition when an individual will suffer from the disease.
Answer
Thalassemia is categorised as Mendelian disorder because these are caused due to alteration or mutation in single gene and follow the Mendel’s principles of inheritance.
It occurs due to either mutation or deletion resulting in reduced rate of synthesis of one of globin chains of haemoglobin.
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Question 725 Marks
Why are colour blindness and thalassemia categorised as Mendelian disorders? Write the symptoms of these diseases seen in people suffering from them.
Answer
Both are caused due to mutation or alteration in a single gene, and follow Mendelian inheritance, therefore, they are called Mendelian disorders.
Symptoms of colour blindness: Unable to discriminate between red and green colours.
Symptoms of thalassemia: Formation of abnormal haemoglobin resulting in Anaemia.
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Question 735 Marks
Answer the following questions:
Explain Polygenic inheritance and Multiple allelism with the help of suitable examples.
Answer
Co–dominance:
  • The alleles which are able to express themselves independently, even when present together are called co–dominant alleles and this biological phenomenon is called co–dominance. For example, ABO blood grouping in humans.
  • ABO blood groups are controlled by gene I. Gene I has three alleles $I^A,I^B$ and $I^O/i.$
  • $I^A$ and $I^B$ produce RBC surface antigens sugar polymer A and B, respectively, whereas i does not produce any antigen.
  • IA and IB are dominant over i hence $I^A$ and $I^B$ are dominant alleles and i is recessive allele as in $I^Ai$ and $I^Bi.$ • When IA and IB are present together, both express equally and produce the surface antigens A and B, hence show co–dominance.
  • Since humans are diploid, each person possesses any two of the three ‘I’ gene alleles, resulting into six different genotypic combinations and four phenotypic expressions.
Table showing the genetic basis of blood groups in human population.
Allele from Parent 1
Allele from Parent 2
Genotype of offspring
Blood groups of offspring
$I^A$
$I^A$
$I^AI^A$
A
$I^A$
$I^B$
$I^AI^B$
AB
$I^A$
i
$I^Ai$
A
$I^B$
$I^B$
$I^BI^B$
B
IB
i
IBi
B
i
i
ii
O
Polygenic Inheritance:
  • It is a type of inheritance, in which traits are controlled by three or more genes. Such traits are called polygenic traits.
  • The phenotype reflects contribution of each allele and is also influenced by the environment.
  • For example, human skin colour. Suppose three genes A, B and C control skin colour with A, B, C being the dominant alleles and a, b, c being the recessive alleles. Then,
  • The $F_2$ generation will have varied skin tones, with each type of allele in the genotype determining the darkness or lightness of the skin.
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Question 745 Marks
Answer the following questions:
You are given tall pea plants with yellow seeds whose genotypes are unknown. How would you find the genotype of these plants? Explain with the help of cross.
Answer
Test cross will be performed to know the genotype of these plants.

If all the plants of $F_1$ generation are tall with yellow seeds, then the phenotype of the parent is homozygous dominant (case i). If the plants in $F_1$ generation are in the ratio of 1 : 1 : 1 : 1, then the parent plant is heterozygous dominant
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Question 755 Marks
Answer the following questions:
Explain Mendel’s law of independent assortment by taking a suitable example.
Answer
According to this law, the two factors of each character assort or separate out independent of the factors of other characters at the time of gamete formation and get randomly rearranged in the offsprings producing both parental and new combinations of characters.
The Punnett square can be effectively used to understand the independent segregation of the two pairs of genes during meiosis and the production of eggs and pollen in the $F_1$ (RrYy) plant.
Consider the segregation of one pair of genes R and r. Fifty per cent of the gametes have the gene R and the other 50 per cent have gene r.
Now besides each gamete having either R or r, it should also have the allele Y or y.
The important thing to remember here is that segregation of 50 per cent R and 50 per cent r is independent from the segregation of 50 per cent Y and 50 per cent y.
Therefore, 50 per cent of the r bearing gamete has Y and the ther 50 per cent has y.
Similarly, 50 per cent of the R bearing gamete has Y and the other 50 per cent has y.
Thus there are four genotypes of gametes (four types of pollen and four types of eggs).
The four types are RY, Ry, rY and ry each with a frequency of 25 per cent of 1/4th of the total gametes produced.
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Question 765 Marks
Explain polygenic inheritance with the help of an example.
Answer
Polygenic Inheritance:
  • It is a type of inheritance, in which traits are controlled by three or more genes. Such traits are called polygenic traits.
  • The phenotype reflects contribution of each allele and is also influenced by the environment.
  • For example, human skin colour. Suppose 3 genes A, B and C control skin colour with A, B, C being the dominant alleles and a, b, c being the recessive alleles. Then,
  • The $F_2$ generation will have varied skin tones, with each type of allele in the genotype determining the darkness or lightness of the skin.
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Question 775 Marks
Differentiate between male heterogamety and female heterogamety with the help of an example of each.
Answer
S. No.
Male heterogamety
Female heterogamety
(i)
Males produce two types of gametes.
Females produce two types of gametes
(ii)
Example, male grasshopper produce gametes of two types–X and O.
Example, female birds produce gametes of two types–Z and W.
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Question 785 Marks
Answer the following questions:
State and explain the law of dominance as proposed by Mendel.
Answer
This law states that when two alternative forms of a trait or character (genes or alleles) are present in an organism, only one factor expresses itself in $F_1$ progeny and is called dominant while the other that remains masked is called recessive. The characters are controlled by discrete units called factors. These factors occur in pairs.
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Question 795 Marks
  1. Work out a dihybrid cross upto F, generation between pea plants bearing violet-coloured axial flowers and white-coloured terminal flowers. Give their phenotypic ratio.
  2. State the Mendel's law of inheritance that was derived from such a cross.
Answer
  1. Law of Independent Assortment: This law states that when two pairs of traits are combined in hybrid, the factors of every character segregate independently of the factors of other pair of characters.
  2.  
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Question 805 Marks
Answer the following questions:
Four children with four different blood groups are born to parents where the mother has blood group ‘A’ and the father has blood group ‘B’. Work out the cross to show the genotypes of the parents and all four children.
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Question 815 Marks
About 8% of human male population suffers from colourblindness whereas only about 0.4% of human female population suffers from this disease. Write an explanation to show how it is possible.
Answer
  • Thalassemia:
  1. It is an autosome–linked recessive disease.
  2. It occurs due to either mutation or deletion resulting in reduced rate of synthesis of one of globin chains of haemoglobin.
  3. Anaemia is the characteristic of this disease.
  4. Thalassemia is classified into two types:
    1. α–thalassemia–Production of α-globin chain is affected. It is controlled by the closely linked genes $\mathrm{Hb}^{\mathrm{A} 1}$ and $\mathrm{Hb}^{\mathrm{A} 2}$ on chromosome 16. It occurs due to mutation or deletion of one or more of the four genes.
    2. β–thalassemia–Production of β–globin chain is affected. It occurs due to mutation of one or both $\mathrm{Hb^B}$ genes on chromosome 11.
  • Colour blindness:
  1. It is a sex–linked recessive disorder.
  2. It results in defect in either red or/ and green cone of eye, resulting in failure to discriminate between red and green colour.
  3. The gene for colour blindness is present on X chromosome.
  4. It is observed more in males $(X^cY)$ because of presence of only one X chromosome as compared to two chromosomes of females.
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Question 825 Marks
List the three different allelic forms of gene “I” in humans. Explain the different phenotypic expressions, controlled by these three forms.
Answer
The three different allelic forms are: $I^A, I^B, I^O/i.$Co–dominance:
  • The alleles which are able to express themselves independently, even when present together are called co–dominant alleles and this biological phenomenon is called co–dominance. For example, ABO blood grouping in humans.
  • ABO blood groups are controlled by gene I. Gene I has three alleles $I^A, I^B$ and $I^O/i.$
  • $I^A$ and $I^B$ produce RBC surface antigens sugar polymer A and B, respectively, whereas i does not produce any antigen.
  • $I^A$ and $I^B$are dominant over i hence $I^A$ and $I^B$ are dominant alleles and i is recessive allele as in $I^Ai$ and $I^Bi$.
  • When $I^A$ and $I^B$ are present together, both express equally and produce the surface antigens A and B, hence show co–dominance.
  • Since humans are diploid, each person possesses any two of the three ‘I’ gene alleles, resulting into six different genotypic combinations and four phenotypic expressions.
Table showing the genetic basis of blood groups in human population:
Allele from Parent 1
Allele from Parent 2
Genotype of offspring
Blood groups of offspring
$I^A$
$I^A$
$I^AI^A$
A
$I^A$
$I^B$
$I^A​​​​​​​I^B$
AB
$I^A$​​​​​​​
i
$I^A​​​​​​​i$
A
$I^B$ $I^B$
$I^BI^B$
B
$I^B$
i
$I^Bi$
B
i
i
ii
O
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Question 835 Marks
A pregnant human female was advised to undergo M.T.P. It was diagnosed by her doctor that the foetus she is carrying has developed from a zygote formed by an XX egg fertilised by Y–carrying sperms. Why was she advised to undergo M.T.P.?
Answer
Embryo has (trisomy of sex chromosome) XXY karyotype or Klinefelter’s syndrome.
She was advised to undergo MTP since the child will have the following problems:
  1. Male with feminine traits.
  2. Gynaecomastia.
  3. Underdeveloped testes.
  4. Sterile.
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Question 845 Marks
Describe the mechanism of inheritance of the ABO system of blood group, highlighting the principle of genetics involved in it.
OR
Explain the genetic basis of blood grouping in human population.
Answer
Co-dominance:
The alleles which are able to express themselves independently, even when present together are called co-dominant alleles and this biological phenomenon is called co-dominance. For example, ABO blood grouping in humans. ABO blood groups are controlled by gene I. Gene I has three alleles $I^A, I^B$ and $I^O / \mathrm{i}$. $I^A$ and $I^B$ produce $R B C$ surface antigens sugar polymer $A$ and $B$, respectively, whereas $i$ does not produce any antigen. $I^A$ and $I^B$ are dominant over $i$ hence $I^A$ and $I^B$ are dominant alleles and $i$ is recessive allele as in $I^{A}i$ and $I^{B}i$. When $I^A$ and $I^B$ are present together, both express equally and produce the surface antigens $A$ and $B$, hence show co-dominance.
Since humans are diploid, each person possesses any two of the three ‘I’ gene alleles, resulting into six different genotypic combinations and four phenotypic expressions.
Table showing the genetic basis of blood groups in human population:
Allele from Parent 1
Allele from Parent 2
Genotype of offspring
Blood groups of offspring
$I^A$
$I^A$
$I^AI^A$
A
$I^A$ $I^B$
$I^AI^B$
AB
$I^A$
i
$I^Ai$
A
$I^B$
$I^B$
$I^BI^B$
B
$I^B$
i
$I^Bi$
B
i
i
ii
O
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Question 855 Marks
State the Mendelian principle which can be derived from such a cross and not from monohybrid cross.
Answer
From the above 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 character is independent of the other pair of characters.
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Question 865 Marks
Identify a, b and c in the table given below:
S. No.
Pattern of inheritance
Monohybrid $F_1$ phenotypic expression
(i)
Co–dominance
a
(ii)
b
The progeny resembled only one of the parents.
(iii)
Incomplete dominance
c
Answer
  1. Both the forms of a trait are equally expressed in $F_1$ generation.
  2. Dominance.
  3. Phenotypic expression of $F_1$ generation is somewhat intermediate between the two parental forms of a trait.
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Question 875 Marks
A cross was carried out between a pea plant heterozygous for round and yellow seeds with a pea plant having wrinkled and green seeds.
  1. Show the cross in a Punnett square.
  2. Write the phenotype of the progeny of this cross.
  3. What is this cross known as? State the purpose of conducting such a cross.
Answer
  1. Both the phenotypic and genotypic ratio are same, i.e., 1: 1 : 1 : 1.
  2. This cross is known as test cross.
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Question 885 Marks
Haemophilia is a sex–linked recessive disease. Study the pedigree analysis given below showing the inheritance of the disease in a family and answer the questions that follow.
  1. Give the evidence from the above analysis which suggests that the disease is:
    1. Sex–linked and,
    2. Caused by a recessive allele.
  2. Write the possible genotypes of the individuals ‘2’ and ‘5’.
Answer
    1. In all the generations, only the males are affected with the disease.
    2. The parents of the affected individuals are not affected which implies that they are carriers and the gene is recessive.
  1. Possible genotype of ‘2’:$X^d\ X$
Possible genotype of ‘5’: $X^d\ Y$
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Question 895 Marks
Differentiate between dominance and co–dominance.
Answer
Dominance: It is a phenomenon in which when two contrasting alleles are present together, only one expresses itself and is called dominant whereas the other which does not express itself is called recessive.
Co–dominance: It is a phenomenon in which when two contrasting alleles are present together, both of the alleles express themselves.
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Question 905 Marks
Match the items in Column I with those in Column II.
S.No Column I S.No Column II
A. ABO blood group in humans 1. Polygenic inheritance
B. Flower colour in snapdragon 2. Mendelian genetic disorder
C. Human skin colour 3. Sex-linked Mendelian disorder
D. Phenylketonuria 4. Incomplete dominance
    5. Multiple allelism
Answer
S.No Column I S.No Column II
A. ABO blood group in humans 5. Multiple allelism
B. Flower colour in snapdragon 4. Incomplete dominance
C. Human skin colour 1. Polygenic inheritance
D. Phenylketonuria 2. Mendelian genetic disorder
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Question 915 Marks
A true breeding pea plant homozygous for axial violet flowers is crossed with another pea plant with terminal white flowers (aavv).
Give the phenotypic ratio of $F_2$ generation.
Answer
Phenotypic ratio of $F_2$ generation:
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Question 925 Marks
In the case of snapdragon (Antirrhinum majus) a plant with red flowers was crossed with another plant with white flowers. Trace the inheritance of flower colour up to $F_2$ generation indicating the genotype and phenotype at each level. What special feature do you notice in the genotype and phenotype ratio in $F_2$ generation?
Answer


Comment: This is a case of Mendelian deviation and that shows incomplete dominance as red and white both are not expressed but produce pink trait in $F_1$. Here, both the genotypic and phenotypic ratio are 1 : 2 : 1.
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Question 935 Marks
How would phenotypes of monohybrid $F_1$ and $F_2$ progeny showing incomplete dominance in Snapdragon and co-dominance in human blood group be different from Mendelian monohybrid $F_1$ and $F_2$ progeny? Explain.
Answer
 
Mendelian monohybrid cross
Incomplete dominance
Co–dominance
$F_1$
All members resemble the parent with dominant trait.
All members do not resemble either of the two parents but show an intermediate trait.
Blood groups of all members resemble combination of dominant traits of both the parents.
$F_2$
Both the parental traits reappear.
Both the parental traits and an intermediate trait appear.
Both the parental traits as well as the co–dominant trait appear.
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Question 945 Marks
A normal visioned woman, whose father is colour blind, marries a normal visioned man. What would be the probability of her (a) sons (b) daughters to be colour blind? Explain with the help of pedigree chart.
Answer

All daughters are normal visioned and 50% of sons are likely to be colour blind.
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Question 955 Marks
Name and state the law that can be derived from this cross and not from a monohybrid cross.
Answer
Law of Independent Assortment: When two pairs of traits are combined in a hybrid, segregation of one pair of characters is independent of the other pair of character.
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Question 965 Marks
Given below is a table showing the genotypes and the phenotypes of blood groups in the human population.
Genotype
Phenotype
W
Y
$I^BI^O$
Z
$I^AI^B$
O
X
O
  1. Identify the genotype Wand X, and the phenotype Y and Z.
  2. Name the pattern of inheritance exhibited by the phenotypes Y and Z in the table.
Answer
  1. Blood group of given geneotype and phenotype.
$\begin{matrix} \text{W}&=& \text{I}^{\text{A}}\text{I}^{\text{O}}\\ \text{X}&=& \text{I}^{\text{O}}\text{I}^{\text{O}}\\ \end{matrix}\Bigg\}\text{Genotype}$

$\begin{matrix} \text{Y}&=& \text{B}\ -\text{blood graup}\ \ \ \ \ \ \ \ \ \ \\ \text{Z}&=&\text{AB}\ -\text{AB}\ \text{blood group}\\ \end{matrix}\Bigg\}\text{Phenotype} $
  1. Pattern of inheritance by B-blood group (Y) and Ab blood group (Z) when $I^B$ and i are present, $I^B$ expresses because it is dominant. When $I^A$ and $I^B$ both are present, they both express their own types of sugars. So, AB blood group expresses this due to codominance.
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Question 975 Marks
$F_1$–generation does not resemble either of the parents.
Answer
$F_1$–generation does not resemble either of the parents: In incomplete dominance, a heterozygous organism carrying two alleles wherein one is dominant and the other one is recessive, (e.g., Rr). Hence, the heterozygote (Rr) will have an intermediate phenotype and will not resemble any parent.Incomplete Dominance:
It is a phenomenon in which the $F_1$ hybrid exhibits characters intermediate of the parental genes. Here, the phenotypic ratio deviates from the Mendel’s monohybrid ratio. It is seen in flower colours of Mirabilis jalapa (4 O'clock plant) and Antirrhinum majus (snapdragon), where red colour is due to gene RR, white colour is due to gene rr and pink colour is due to gene Rr.
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Question 985 Marks
Name any two chromosomal aberration associated disorders.
Answer
Two chromosomal aberration–associated disorders are Down’s syndrome and Klinefelter’s syndrome.
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Question 995 Marks
Answer the following questions:
Write the Mendelian $F_2$ phenotypic ratio in a dihybrid cross. State the law that he proposed on the basis of this ratio. How is this law different from the law of segregation?
Answer
The $F_2$ phenotypic ratio is 9:3:3:1. On the basis of this ratio Mendel proposed Law of Independent Assortment.
Law of independent assortment:
  • According to this law the two factors of each character assort or separate out independent of the factors of other characters at the time of gamete formation and get randomly rearranged in the offsprings producing both parental and new combinations of characters.
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Question 1005 Marks
Let us assume in a given plant the genotype symbol “Y” stands for dominant yellow seed colour and “y” for recessive green seed colour; symbol “R” for round seed shape and “r” for wrinkled seeds.
Two homozygous parents (plants) with genotypes “RRYY” and “rryy” are crossed and their $F_1\_$generation progeny is then selfed. What shall be the:
  1. Phenotype of $\mathrm{F}_1$-progeny.
  2. Genotype of $\mathrm{F}_1$-progeny.
  3. Gamete genotypes of $\mathrm{F}_1$-progeny.
  4. Phenotypic ratio of $\mathrm{F}_2$ population.
  5. Phenotypic ratio of yellow seed to green seed and round seed to wrinkled seed in $F_2$ population.
Answer

Results of a dihybrid cross where the two parents differed in two pairs of contrasting traits: seed colour and seed shape.
  1. Phenotype of $F_1\_$progeny: Round seeds that are yellow in colour.
  2. Genotype of $F_1\_$ progeny: RrYy.
  3. Gamete genotypes of $F_1\_$progeny: RY, Ry, rY and ry.
  4. Phenotypic ratio of $F_2$ population: 9 : 3 : 3 : 1.
Nine round–yellow seeds; three round–green seeds; three wrinkled–yellow seeds; one wrinkled–green seed.
  1. Phenotypic ratio of yellow seed to green seed and round seed to wrinkled seed in $F_2$ population:
Yellow seed to green seed = 3 : 1
Round seed to wrinkled seed = 3 : 1
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Question 1015 Marks
Describe the nature of inheritance of the ABO type of blood group in humans. In which ways does this inheritance differ from that of height of the plant in garden pea?
Answer
ABO blood group system in human:Co–dominance:
  • The alleles which are able to express themselves independently, even when present together are called co–dominant alleles and this biological phenomenon is called co–dominance. For example, ABO blood grouping in humans.
  • ABO blood groups are controlled by gene I. Gene I has three alleles $I^A, I^B$ and IO/i.
  • $I^A$ and $I^B$ produce $R B C$ surface antigens sugar polymer $A$ and $B$, respectively, whereas $i$ does not produce any antigen.
  • $I^A$ and $I^B$ are dominant over $i$ hence $I^A$ and $I^B$ are dominant alleles and $i$ is recessive allele as in $I^A i$ and $I^B i$.
  • When $I^A$ and $I^B$ are present together, both express equally and produce the surface antigens $A$ and $B$, hence show co-dominance.
  • Since humans are diploid, each person possesses any two of the three ‘I’ gene alleles, resulting into six different genotypic combinations and four phenotypic expressions.
Table showing the genetic basis of blood groups in human population.
Allele from Parent 1
Allele from Parent 2
Genotype of offspring
Blood groups of offspring
$I^A$
$I^A$
$I^AI^A$
A
$I^A$
$I^B$
$I^AI^B$
AB
$I^A$
i
$I^Ai$
A
$I^B$
$I^B$
$I^BI^B$
B
$I^B$
i
$I^Bi$
B
i
i
ii
O
S. No.
Blood group in man
Height of plant in garden pea
(i)
The gene I responsible for the blood group exists in three allelic alternative forms $I^A$, IBand i.
The gene for this trait exists in two allelic forms T and t.
(ii)
$I^A$ and $I^B$ are dominant over i and $I^A$ and $I^B$ are co-dominant.
Allele T is dominant over t.
(iii)
They exhibit four phenotypes with six possible genotypes.
They exhibit two phenotypes with three possible genotype.
(iv)
Blood group in man exhibit phenomena of multiple allelism.
Height in garden pea plant do not exhibit multiple allelism.
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Question 1025 Marks
A pregnant woman who have had amniocentesistest was advised to undergo MTP. It was diagnosed by her doctor that the embryo she is carrying has an extra barr body.
  1. Name the syndrome which is likely to be associated with the embryo.
  2. Why was she advised to undergo MTP?
  3. Do you justify the statutory ban on amniocentesis in our country Give reasons.
Answer
  1. Klinefelter's syndrome It occurs due to trisomy of sex chromosome.
  2. She was advised to undergo MTP since, the child will have the following problems.
  1. Male with feminine traits, i.e. gynaecomastia.
  2. Underdeveloped testes leading to sterility.
  1. Yes, this test is actually meant to determine the genetic defects or metabolic disorders in foetus by doing a chromosomal analysis. In such extreme cases that could be incurable, a decision to abort the foetus could be taken. But, amniocentesis is misused now a days.
It is used to determine the sex of the foetus and in many cases it leads to female foeticide. It has became so serious that it disturbed the male-female ratio, that can have a negative impact on society. So, it should be banned.
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Question 1035 Marks
Write the conclusions Mendel arrived at on dominance of traits on the basis of monohybrid crosses that he carried out in pea plants.
Answer
Mendel concluded that:
  1. Characters are controlled by discrete units called factors.
  2. Factors occur in pair.
  3. 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 $F_1$ and both are expressed in the $F_2$.
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Question 1045 Marks
In a plant tallness is dominant over dwarfness and red flower is dominant over white. Starting with the parents work out a dihybrid cross. What is standard dihybrid ratio? Do you think the values would deviate if the two genes in question are interacting with each other?
Answer

Yes, the ratio will deviate if the two genes are interacting with each other. In this condition the genes do not independently assort with each other, so, ratio will be changed from 9 : 3 : 3 : 1.
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Question 1055 Marks
Differentiate between the following:
  1. Polygenic inheritance and Pleiotropy.
  2. Dominance, Codominance and Incomplete dominance.
Answer
  1. Pleiotropy:
  • In this, a single gene controls more than one phenotype.
  • The F, resembles one of the parents, i.e., dominant parent.
  • There is no gradation in the phenotype.
  1. Dominance:
  • It is the phenomenon in which one allele of a gene expresses itself and suppresses the expression of the other (recessive) allele of the same gene, when they are present together in a hybrid
  • The hybrid resembles one of the parents.
  1. Codominance:
  • It is the phenomenon in which two alleles of a gene are equally dominant and express themselves in the presence of the other.
  • The hybrid shows characters of both the parents.
  1. Incomplete dominance:
  • It is the phenomenon in which neither of the two alleles of a gene is completely dominant over the other.
  • The hybrid is intermediate between the two parents.
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Question 1065 Marks
A garden pea plant bearing terminal, violet flowers, when crossed with another pea plant bearing axial, violet flowers, produced axial, violet flower and axial, white flowers in the ratio of 3 : 1. Work out the cross showing the genotypes of the parent pea plants and their progeny.
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Question 1075 Marks
A tall pea plant with yellow seeds (heterozygous for both the traits) is crossed with a dwarf pea plant with green seeds. Using a Punnett square work out the cross to show the phenotypes and the genotypes of $F_1$ generation.
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Question 1085 Marks
How are Mendelian inheritance, polygenic inheritance and pleiotropy different from each other?
Answer
Mendelian Inheritance
Polygenic inheritance
Pleiotropy
One gene controls one trait/ character/ phenotype.
Two or more genes influence the expression of one trait/ character/ phenotype.
One genes controls the expression of more than one trait/character/ phenotype.
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Question 1095 Marks
Write the type and location of the gene causing thalassemia in humans. State the cause and symptoms of the disease. How is sickle cell anaemia different from this disease?
Answer
Sickle–cell anaemia:
  1. It is an autosome–linked recessive trait.
  2. The disease is controlled by a single pair of allele $\mathrm{Hb}^{\mathrm{A}}$ and $\mathrm{Hb}^{\mathrm{S}}$.
  3. Only the homozygous individuals for $\mathrm{Hb}^{\mathrm{S}}$, i.e., $\mathrm{Hb}^{\mathrm{S}} \mathrm{Hb}^{\mathrm{S}}$ show the diseased phenotype.
  4. The heterozygous individuals are carriers $\left(\mathrm{Hb}^{\mathrm{A}} \mathrm{Hb}^{\mathrm{S}}\right)$.
  5. Due to point mutation, glutamic acid (Glu) is replaced by valine (Val) at the sixth position of β–globin chain of haemoglobin molecule.
  6. HbS behaves as normal haemoglobin except under oxygen stress where erythrocytes lose their circular shape and become sickle–shaped. As a result, the cells cannot pass through narrow capillaries. Blood capillaries are clogged and thus affect blood supply to different organs.
Thalassemia:
  1. It is an autosome-linked recessive disease.
  2. It occurs due to either mutation or deletion resulting in reduced rate of synthesis of one of globin chains of haemoglobin.
  3. Anaemia is the characteristic of this disease.
  4. Thalassemia is classified into two types:
  • α-thalassemia–Production of α-globin chain is affected. It is controlled by the closely linked genes $\mathrm {Hb^{A1}}$ and $\mathrm{Hb^{A2}}$ on chromosome 16. It occurs due to mutation or deletion of one or more of the four genes.
  • β-thalassemia–Production of β-globin chain is affected. It occurs due to mutation of one or both $\mathrm {Hb^B}$ genes on chromosome 11.
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Question 1105 Marks
How does a chromosomal disorder differ from a Mendelian disorder?
Answer
S. No.
Mendelian disorder
Chromosomal disorder
(i)
This disorder is mainly due to alteration or mutation in the single gene.
This disorder is caused due to absence or excess or abnormal arrangement of one or more chromosomes.
(ii)
This follows Mendel’s principles of inheritance.
This does not follow Mendel’s principles ofinheritance.
(iii)
This may be recessive or dominant in nature.
This is always dominant in nature.
(iv)
For example, haemophilia, sickle–cell anaemia.
For example, Turner’s syndrome.
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Question 1115 Marks
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:
  1. They produce children with blood group “A” only.
  2. They produce children some with “O” blood group and some with “A” blood group.
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Question 1125 Marks
Explain the causes, inheritance pattern, symptoms of two Mendelian genetic disorders.
Answer
Two mendelian disorder are:
  1. Sickle cell disease is a genetic disease which is caused due to molecular mutation of gene Hb on chromosome 11 which produces the beta chain of adult hemoglobin. The mutated gene Hbs produces sickle cell haemoglobin.The 6th amino acid in the beta chain of normal haemoglobin is glutamic acid, in sickle cell haemoglobin this amino acid is replaced by valine. The children that are homozygous produce rigid chains. When oxygen level of the blood drops below a certain level, RBC undergoes sickling. Such cells do not transport oxygen efficiently, they are removed by spleen causing severe anaemia. Sickle cell is an autosomal linked disease in which both male and female progenies can be equally affected.
  2. Haemophilia is a sex-linked disorder. In haemophilia, the blood fails to clot when exposed to air and even a small skin injury results in continuous bleeding and can lead to death from the loss of blood. The recessive X linked gene for haemophilia shows characteristic criss-cross inheritance. It's a single gene in male results and disease haemophilia, where is a woman needs two such genes for the same.
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Question 1135 Marks
How is the phenotypic ratio of $F_2$ generation different in a dihybrid cross?
Answer
Phenotypic ratio of $F_2$ in monohybrid cross is 3 : 1 whereas in a dihybrid cross the phenotypic ratio is $9 : 3 : 3 : 1.$
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Question 1145 Marks
In one family, the four children each have a different blood group. Their mother has blood group A and their father has blood group B. Work out a cross to explain how it is possible.
Answer
A cross is carried out between heterozygous father (for blood group B) and heterozygous mother (of blood group A) to get four children with different blood groups.

All the four blood groups are controlled by three allelic genes $I^A, I^B$, i and thus it shows phenomena of multiple allelism. Both $I^A$ and $I^B$ are dominant over i. However, when together, both are dominant and show the phenomena of co–dominance forming the blood group AB. Six genotypes are possible with combination of these three alleles.
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Question 1155 Marks
Answer the following questions:
State and explain the law of segregation as proposed by Mendel in a monohybrid cross.
Answer
Mendel’s Observations:
  1. $F_1$ progenies always resembled one of the parents and trait of other parent was not seen.
  2. $F_2$ stage expressed both the parental traits in the proportion 3 : 1.
  3. The contrasting traits did not show any blending at either $F_1$ or $F_2$ stage.
  4. In dihybrid cross, he got identical results as in monohybrid cross.
  5. He found that the phenotypes in $F_2$ generation appeared in the ratio 9 : 3 : 3 : 1.
Law of segregation:
  • This law states that the factors or alleles of a pair segregate from each other during gamete formation, such that a gamete receives only one of the two factors. They do not show any blending.

Monohybrid cross of true–breeding pea plant.
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Question 1165 Marks
Explain the contribution of Alfred Sturtevant in ‘Chromosome mapping’.
Answer
Alfred Sturtevant 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.
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Question 1175 Marks
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 the pea plant bearing axial flowers. Work out the cross to show the conclusions you arrive at.
Answer
If the plant is homozygous for the dominant trait.

Conclusion: If all progeny show axial flowers (dominant) the plant is homozygous (AA), If 50% of progeny show axial flower (Dominant) and 50% terminal flower (Recessive) the plant is heterozygous.
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Question 1185 Marks
  1. A garden pea plant bearing terminal, violet flowers, when crossed with another pea plant bearing axial, violet flowers produced axial, violet flowers and axial, white flowers in the ratio of 3 : 1. Work out the cross showing the genotypes of the parent pea plants and their progeny.
  2. Name and state the law that can be derived from this cross and not from a monohybrid cross.
Answer
  1. Dihybrid cross:
  • Since white flowers have appeared in the progeny, both the parents must be heterozygous for flower colour, i.e. Vv.
  • Since all plants in the progeny have axial flowers, the parent with axial flowers, must be homozygous for the trait, i.e. AA.
  • Genotypes of the parents are as follows:
Terminal, violet flowers : aaVv
Axial, violet flowers : AAVv

The phenotypic ratio:
3 Axial, violet flower: 1 axial, white flowers is justified
  1. Law of independent assortment: It states that when two pairs of traits are combined in a hybrid, segregation of one pair of traits is independent of the other pair of trait.
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Question 1195 Marks
Define aneuploidy. How is it different from polyploidy? Describe the individuals having following chromosomal abnormalities.
  1. Trisomy of 21st Chromosome.
  2. XXY.
  3. XO.
Answer
Failure of segregation of chromatids during cell division cycle results in the gain or loss of a chromosome(s), called aneuploidy. – Difference between aneuploidy and polyploidy
  1. Failure of segregation of chromatids during cell division cycle results in the gain or loss of a chromosome(s), called aneuploidy. Failure of cytokinesis after telophase stage of cell division results in an increase in a whole set of chromosomes in an organism and,this phenomenon is known as polyploidy.
  2. Polyploidy occurs due to altering set of chromosome number such as 2n, 3n, 5n, whereas aneuploidy occurs due to altering particular chromosome or part of a chromosome such as 2n + 1 (trisomic) and 2n – 1 (monosomic).
  3. Aneuploidy can be seen in human as genetic disorders; for example, Tuner syndrome, Klinefelter syndrome and Down syndrome, whereas polyploidy is common in plants.
Down’s Syndrome (Mongolism),
  • The cause of this genetic disorder-is the presence of an additional copy of the chromosome number 21 (trisomy of 21) due to non-disjunction of chromosomes during sperm or ova formation.
  • The affected individual is short statured with small round head, furrowed tongue and partially open mouth. Palm is broad with characteristic palm crease.
  • Physical, psychomotor and mental development is retarded.
Klinefelter’s Syndrome,
  • This genetic disorder is also caused due to the presence of an additional copy of X-chromosome resulting into a karyotype of 47, XXY.
  • Such an individual has overall masculine development, however, the feminine development (development of breast, i.e., Gynaecomastia) is also expressed. Such individuals are sterile male.
Turner’s Syndrome,
Such a disorder is caused due to the absence of one of the X chromosomes, i.e., 45 with XO. Such females are sterile as ovaries are rudimentary besides other features including lack of other secondary sexual characters.
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Question 1205 Marks
  1. In humans, males are heterogametic and females are homogametic. Explain. Are there any examples where males are homogametic and females heterogametic?
  2. Also describe as to, who determines the sex of an unborn child? Mention whether temperature has a role in sex determination.
Answer
  1. In humans, the 23rd pair of chromosome contains and X chromosome and a Y chromosome. Hence, males are called heterogametic. Females, on the other hand, have XX chromosomes in the 23rd pair. Hence, females are called homogametic. But in bird’s females have ZW chromosome and male have ZZ Chromosome. So, in some cases, males can be homogametic and females can be heterogametic.
  2. In case of humans, sex is determined by X and Y-chromosomes. Out of the 23 pairs of chromosomes in human beings, the 23rd pair is called sex chromosome while the remaining 22 pairs are called autosomes. All males have X and Y-chromosomes in the 23rd pair while females have XX-chromosomes in the 23rd pair. Thus, a sperm can have either X or Y chromosome, while all the eggs will have X-chromosome. When a sperm with X chromosome fertilizes the ovum; the zygote will result in development of a girl child. If a sperm with Y chromosome fertilizes the ovum; the zygote will result in development of a male child.
Temperature dependent sex determination is found in many animals, e.g. in crocodiles. When eggs are incubated at higher temperature, it results in birth of male crocodiles.
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Question 1215 Marks
Discuss in detail the contributions of Morgan and Sturvant in the area of genetics.
Answer
Morgan and his group conducted various experiments in the field of genetics.
Sturtevant was a student of Morgan. Some of the contributions by them are as follows:
Morgan carried out several dihybrid crosses of Drosophila. He observed that the phenotypic ratio was not similar to the standard phenotypic ratio as observed by Mendel. Morgan and his team were aware that the genes were located on X chromosome. Thy inferred that when the genes were situated on the same chromosome, they did not segregate independently of each other.
When the genes are situated on the same chromosome, the chances of parental combination are much higher than non-parental combination. The physical association of genes on the same chromosome was termed as linkage; by Morgan. Morgan also coined the term recombination to describe generation of non-parental combination.
Sturtevant came out with the finding that relative distance between two genes on the same chromosome was an important factor in recombination or lack of recombination. If the genes were tightly linked, they did not show recombination. But if the genes were far apart then chances of recombination were higher. Today’s genetic mapping could be developed because of contributions made by Morgan and his team.
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