Chapter 4: Principles of Inheritance and Variation Application Questions | biology Class 12 CBSE

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Chapter 4: Principles of Inheritance and Variation Application Questions | biology Class 12 CBSE | ByteNotes.in

Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 1

Applied Concept

A farmer notices that crossing two varieties of wheat that are true-breeding for different grain colour always produces offspring with an intermediate grain colour. With reference to patterns of inheritance, explain this observation and predict the phenotypic ratio expected in the next generation if these intermediate plants are self-pollinated.

This is an example of incomplete dominance, where neither allele is completely dominant over the other. When true-breeding red (RR) and white (rr) plants are crossed, F1 plants are Rr and display an intermediate pink/intermediate colour because neither allele fully suppresses the other's phenotypic effect.
If the F1 intermediate plants (Rr) are self-pollinated, the F2 generation will produce offspring in the ratio 1 RR : 2 Rr : 1 rr. The phenotypic ratio will also be 1:2:1 (1 original red : 2 intermediate : 1 original white), since genotypic and phenotypic ratios coincide in incomplete dominance.
This differs from complete dominance (where F2 phenotypic ratio is 3:1) because in incomplete dominance heterozygotes can be distinguished from both homozygotes. The farmer can therefore identify homozygous varieties by screening F2 plants, which is useful for maintaining true-breeding lines.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 2

Applied Concept

A child is born with blood group O. Her father has blood group A and her mother has blood group B. With reference to ABO blood grouping, determine the genotypes of the parents and list the possible blood groups of other children born to these parents.

The child has blood group O (genotype ii). She must have received one i allele from each parent. Therefore, the father (blood group A) must be heterozygous: genotype IAi. The mother (blood group B) must also be heterozygous: genotype IBi.
Cross: IAi x IBi produces four possible offspring genotypes — IAIB (blood group AB), IAi (blood group A), IBi (blood group B), ii (blood group O) — each with a 25% probability. Thus, all four ABO blood groups are possible in this family.
This cross also illustrates co-dominance (IAIB gives AB phenotype), simple dominance (IA and IB each dominant over i), and multiple alleles (three alleles IA, IB, i control the trait). It demonstrates that blood group O children can be born to parents of different non-O blood groups if both parents are carriers of the i allele.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 3

Applied Concept

A woman who is a carrier of haemophilia marries a normal man. With reference to sex-linked inheritance, predict the probability of haemophilia among their sons and daughters, and explain why haemophilic daughters are extremely rare.

The carrier mother has genotype XHXh and the normal father has genotype XHY. Cross: XHXh x XHY gives four genotypes — XHXH (normal female, 25%), XHXh (carrier female, 25%), XHY (normal male, 25%), XhY (haemophilic male, 25%). Thus, 50% of sons will be haemophilic and 50% of daughters will be carriers.
Haemophilic daughters (XhXh) require two recessive alleles — one from each parent. For this, the father must be haemophilic (XhY), which is the precondition. However, haemophilic males often do not survive to reproductive age, making this combination extremely rare.
This pattern illustrates X-linked recessive inheritance: males are hemizygous (only one X chromosome) so a single recessive allele causes the disease, while females need two copies. Carrier females are phenotypically normal but serve as transmitters of the disease to future generations.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 4

Applied Concept

Two tall pea plants when crossed produce both tall and dwarf offspring. When a tall plant from this cross is mated with a dwarf plant, all progeny are tall. Using appropriate crosses, explain the genotypes of all the plants involved.

Two tall plants producing both tall and dwarf offspring in a 3:1 ratio indicates both parents are heterozygous (Tt x Tt). From this cross, offspring are 1/4 TT, 1/2 Tt, 1/4 tt.
This is a test cross scenario: when the tall plant (of unknown genotype) is crossed with a dwarf (tt) and all progeny are tall, it indicates the tall plant is homozygous dominant (TT). Cross: TT x tt produces all Tt progeny, which are all phenotypically tall.
Had the tall plant been heterozygous (Tt), a cross with dwarf (tt) would have produced 50% tall (Tt) and 50% dwarf (tt). The observation of all-tall progeny thus unambiguously identifies the tested plant as TT, demonstrating the utility of the test cross in determining genotype.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 5

Applied Concept

A geneticist performs a dihybrid cross between tall yellow-seeded (TtYy) and dwarf green-seeded (ttyy) pea plants. Predict the phenotypic ratio in the offspring and explain which law of Mendel is demonstrated by this cross.

TtYy x ttyy is a dihybrid test cross. TtYy produces four gamete types: TY, Ty, tY, ty (25% each). The ttyy parent produces only ty gametes. Combining these gives offspring in four phenotypic classes.
Expected offspring: 1 tall yellow (TtYy) : 1 tall green (Ttyy) : 1 dwarf yellow (ttYy) : 1 dwarf green (ttyy) — a 1:1:1:1 phenotypic ratio. Tall and dwarf each appear in equal frequency because Tt x tt segregates 1:1; yellow and green appear equally because Yy x yy segregates 1:1.
This demonstrates the Law of Independent Assortment — the inheritance of seed colour (Y/y) is independent of plant height (T/t). The 1:1:1:1 ratio is specific to dihybrid test crosses (one parent heterozygous for both loci x double recessive) and is evidence that the two gene pairs segregate independently.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 6

Applied Concept

A couple is worried because the wife's brother has sickle-cell anaemia. The wife appears healthy, and her husband has no family history of the disease. With reference to the inheritance of sickle-cell anaemia, explain the risk to their children.

Sickle-cell anaemia is autosomal recessive, controlled by alleles HbA (normal) and HbS (sickle). The wife's brother is affected (HbSHbS). For both parents of the wife to have an affected child, both must be carriers (HbAHbS). Therefore, the wife has a 2/3 probability of being a carrier (HbAHbS) — given she is unaffected but has an affected sibling.
The husband has no family history; without testing, we assume he is HbAHbA, but he could also be a carrier. If both the wife and husband are carriers (HbAHbS x HbAHbS), the risk of each child being affected (HbSHbS) is 25%.
Genetic counselling and molecular testing of both partners for the HbS allele would accurately determine their carrier status and calculate the actual risk. If only the wife is a carrier and the husband is normal, their children cannot be affected (only carriers at most), illustrating the importance of carrier testing before making predictions.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 7

Applied Concept

Morgan hybridised yellow-bodied, white-eyed female Drosophila with brown-bodied, red-eyed males. The F2 offspring did not follow the expected 9:3:3:1 ratio. With reference to linkage and recombination, explain this observation.

In Morgan's cross, both genes (body colour and eye colour) are located on the X chromosome — they are sex-linked. Because they are physically on the same chromosome, they do not assort independently, violating the assumption underlying the 9:3:3:1 ratio. The parental combinations (yellow-white and brown-red) appear at much higher frequency than the non-parental combinations.
This deviation is explained by linkage — the physical association of genes on the same chromosome. Tightly linked genes are usually inherited together, so parental gene combinations predominate in F2. Non-parental (recombinant) types arise only through crossing over (recombination) during meiosis.
Morgan coined the term recombination to describe the generation of non-parental gene combinations. The frequency of recombinant offspring is used to estimate the map distance between genes — tightly linked genes have low recombination frequency, while loosely linked genes have higher frequency, as used by Sturtevant to build the first genetic map.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 8

Applied Concept

During a karyotyping procedure, a 15-year-old girl is found to have 45 chromosomes with only one X chromosome (XO). What is the likely condition? Describe the clinical features and origin of this chromosomal abnormality.

The girl has Turner's syndrome — a chromosomal disorder caused by monosomy of the sex chromosome (karyotype 45, XO). This arises due to non-disjunction during meiosis in one of the parents, resulting in a gamete lacking a sex chromosome; when this gamete is fertilised by a normal X-bearing gamete, the resulting zygote is XO.
Clinical features include: short stature, rudimentary (non-functional) ovaries leading to primary amenorrhoea, infertility/sterility, lack of secondary sexual characters (underdeveloped breasts, sparse pubic/axillary hair), and sometimes a webbed neck and broad chest.
This is a numerical chromosomal disorder (aneuploidy). Turner's syndrome is detectable by karyotype analysis showing 45 chromosomes. Early diagnosis allows management with hormone replacement therapy to promote secondary sexual development, though fertility cannot be restored. Understanding this disorder illustrates the critical role of the second X chromosome in normal female development.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 9

Applied Concept

A pedigree chart of a family shows that the grandfather was colour blind, his daughter has normal vision, and her son is colour blind. With reference to X-linked inheritance, trace the transmission of this trait across three generations.

Colour blindness is X-linked recessive. The grandfather (XcY) passes his Xc to his daughter. The daughter inherits one Xc from her father and one XC from her normal mother, making her a carrier (XCXc) — she has normal vision due to the dominant normal allele.
The carrier daughter (XCXc) marries a normal male (XCY). She passes Xc to 50% of her sons. The colour-blind grandson (XcY) inherited the Xc from his carrier mother and Y from his normal father — he is hemizygous for the recessive allele and therefore expresses colour blindness.
This three-generation pedigree illustrates the classic pattern of X-linked recessive inheritance: grandfather affected → daughter unaffected carrier → grandson affected. The trait appears to skip a generation and disproportionately affects males. No father-to-son transmission occurs because sons receive Y from the father.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 10

Applied Concept

A diploid organism is heterozygous for 4 independent gene loci. How many types of gametes can it produce? Explain using the principle of independent assortment.

For each heterozygous locus, an organism produces two types of gametes — one with the dominant allele and one with the recessive allele. The number of gamete types for n heterozygous loci is 2n.
For 4 heterozygous loci (e.g., AaBbCcDd), the number of gamete types = 24 = 16. Each gamete will carry one allele from each locus independently, producing combinations like ABCD, ABCd, ABcD, ABcd, AbCD, etc.
This is a direct application of the Law of Independent Assortment — the segregation of alleles at each locus is independent of the segregation at all other loci. This principle explains how sexual reproduction generates enormous genetic diversity in offspring.

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Chapter 4: Principles of Inheritance and Variation | Application Questions

A child is born with blood group O. Her father has blood group A and her mother has blood group B. With reference to ABO blood grouping, determine the genotypes of the parents and list the possible blood groups of other children born to these parents.

A woman who is a carrier of haemophilia marries a normal man. With reference to sex-linked inheritance, predict the probability of haemophilia among their sons and daughters, and explain why haemophilic daughters are extremely rare.

Two tall pea plants when crossed produce both tall and dwarf offspring. When a tall plant from this cross is mated with a dwarf plant, all progeny are tall. Using appropriate crosses, explain the genotypes of all the plants involved.

A geneticist performs a dihybrid cross between tall yellow-seeded (TtYy) and dwarf green-seeded (ttyy) pea plants. Predict the phenotypic ratio in the offspring and explain which law of Mendel is demonstrated by this cross.

A couple is worried because the wife's brother has sickle-cell anaemia. The wife appears healthy, and her husband has no family history of the disease. With reference to the inheritance of sickle-cell anaemia, explain the risk to their children.

Morgan hybridised yellow-bodied, white-eyed female Drosophila with brown-bodied, red-eyed males. The F2 offspring did not follow the expected 9:3:3:1 ratio. With reference to linkage and recombination, explain this observation.

During a karyotyping procedure, a 15-year-old girl is found to have 45 chromosomes with only one X chromosome (XO). What is the likely condition? Describe the clinical features and origin of this chromosomal abnormality.

A pedigree chart of a family shows that the grandfather was colour blind, his daughter has normal vision, and her son is colour blind. With reference to X-linked inheritance, trace the transmission of this trait across three generations.

A diploid organism is heterozygous for 4 independent gene loci. How many types of gametes can it produce? Explain using the principle of independent assortment.

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