NEET Principles of Inheritance and Variation Previous Year Questions with Solutions

Down's syndrome (first described by Langdon Down in 1866) is an autosomal aneuploidy caused by non-disjunction leading to trisomy of chromosome 21 ($2n+1 = 47$). It does not involve sex chromosomes or the loss of an X-chromosome. Therefore, Statement I is definitively false.

• • Absence of an X chromosome causes Turner's syndrome, not Down's.

Turner's syndrome is a sex chromosomal monosomy caused by the absence of one X chromosome in females, resulting in the karyotype $44\text{A} + \text{XO}$ (45 chromosomes). Statement II asserts that it is caused by the "presence of an additional copy of the chromosomes", which describes Klinefelter's ($44\text{A} + \text{XXY}$) or Down's, not Turner's. Hence, Statement II is false.

• • Females are sterile with rudimentary ovaries and lack secondary sexual characters.

Because both statements swap and mischaracterize the underlying genetic lesions, both Statement I and Statement II are incorrect.

• • Matches official NTA final answer key Option B / (2).

ABO blood groups are controlled by the gene I, which has three alleles: I^A, I^B, and i. I^A and I^B are codominant with each other and completely dominant over the recessive allele i.

• • Any child with blood group O requires that both parents carry at least one recessive i allele.

Examine each given parental pair to check if both parents can produce an "i" gamete:

• • Possible offspring of (4) are only: I^A I^A (A), I^A i (A), I^A I^B (AB), or I^B i (B). None can be O.

When Mendel crossed true-breeding tall pea plants (TT) with true-breeding dwarf pea plants (tt), each parent produced only one type of gamete: T and t respectively. Fusion of these gametes produced F₁ progeny with uniform heterozygous genotype Tt, all exhibiting the tall phenotype.

• • Assertion (A) is completely true.

According to Mendel’s Law of Dominance: (i) Characters are controlled by discrete units called factors (genes). (ii) Factors occur in pairs. (iii) In a dissimilar pair of factors, one member of the pair dominates (dominant) the other (recessive). Because T completely dominates t in Tt, the dwarf trait is masked in F₁, perfectly explaining Assertion (A).

• • Recessive dwarf phenotype reappears only in F₂ in homozygous condition (tt).

Gregor Mendel studied traits that had distinct alternate forms (e.g. purple vs white flowers, yellow vs green seeds). However, in nature, many traits exhibit continuous gradation rather than distinct all-or-none phenotypes:

• • Traits that are generally controlled by three or more genes are called polygenic traits.
• • Classic examples include human skin colour (controlled by three genes A, B, C as explained in NCERT), human height, and intelligence.
• • Polygenic inheritance also takes into account the influence of the environment.

Evaluating inheritance classifications:

• • Mendelian inheritance is based on unit factors (genes) with dominant and recessive alleles yielding discrete qualitative phenotypic ratios (3:1 monohybrid, 9:3:3:1 dihybrid).
• • Polygenic inheritance involves multiple loci exhibiting additive or quantitative effects, where the phenotype reflects the total number of dominant/contributing alleles, resulting in a continuous Gaussian distribution.
• • Consequently, it represents an extension/deviation from classical Mendelian rules, categorized firmly as a Non-Mendelian inheritance pattern.

Polygenic traits display a Non-Mendelian pattern of inheritance.

Mendel crossed a pea plant homozygous for round yellow seeds with one having wrinkled green seeds:

• • P generation: $RRYY$ (Round, Yellow) $\times$ $rryy$ (Wrinkled, Green).
• • Gametes: $RY$ and $ry$.
• • F1 generation: All $RrYy$ individuals displaying the dominant phenotype (Round, Yellow seeds).

Because genes R and Y assort independently:

• • The F1 dihybrid ($RrYy$) produces 4 types of pollen grains and 4 types of eggs in equal proportions (25% each): $RY$, $Ry$, $rY$, $ry$.
• • Random union of these gametes yields a $4 \times 4 = 16$-box Punnett square in the F2 generation.

Combining monohybrid independent probabilities:

• • Round Yellow ($R\_Y\_$): $\frac{3}{4} \times \frac{3}{4} = \frac{9}{16}$.
• • Round Green ($R\_yy$): $\frac{3}{4} \times \frac{1}{4} = \frac{3}{16}$.
• • Wrinkled Yellow ($rrY\_$): $\frac{1}{4} \times \frac{3}{4} = \frac{3}{16}$.
• • Wrinkled Green ($rryy$): $\frac{1}{4} \times \frac{1}{4} = \frac{1}{16}$.

The F2 phenotypic ratio is 9 : 3 : 3 : 1, matching Option (3).

Analyzing the transmission pattern across generations:

• • In F0, an affected female (black circle, $X^a X^a$) mates with an unaffected male (empty square, $X^A Y$). All daughters inherit one paternal $X^A$ and one maternal $X^a$, making them obligate carrier females (circles with central dot, $X^A X^a$) in F1.
• • In F1, a carrier female ($X^A X^a$) mates with an unaffected male ($X^A Y$). They produce an affected son ($X^a Y$) in F2. This confirms X-linked recessive inheritance beyond doubt.
• • The consanguineous marriage in F2 involves a carrier female ($X^A X^a$) and an affected male ($X^a Y$).

Determining all possible genotypic and phenotypic outcomes:

• • Maternal gametes: $X^A$ ($50\%$) and $X^a$ ($50\%$).
• • Paternal gametes: $X^a$ ($50\%$) and $Y$ ($50\%$).
• • Cross outcomes ($4$ equally likely quadrants, each $p = 0.25$):
• • 1. $X^A X^a$: Phenotypically normal female, carries one mutated allele (Heterozygous carrier female).
• • 2. $X^a X^a$: Phenotypically affected female (Homozygous mutant).
• • 3. $X^A Y$: Phenotypically normal male (Hemizygous wild-type, carries zero mutant alleles, cannot be a carrier).
• • 4. $X^a Y$: Phenotypically affected male (Hemizygous mutant).

The question specifically requires the birth of a child (of either sex) having: (i) NO disease, AND (ii) being a carrier (bearing the disease mutation in one allele).

Out of all 4 possible zygotic combinations, exactly 1 corresponds to an unaffected carrier child ($X^A X^a$). The probability is 1/4.

Mendel stated the Law of Dominance using three core assertions: (i) Characters are controlled by discrete units called factors (Statement D). (ii) Factors occur in pairs (Statement C). (iii) In a dissimilar pair of factors, one member dominates the other (Statement A). It also explains the expression of only one parental character in F1 (Statement E) and the 3:1 ratio in F2.

Statement B asserts that alleles do not show blending and both characters are recovered as such in F2. This is the definition and evidence for the Law of Segregation (purity of gametes), not the Law of Dominance.

Statements A, C, D, and E belong to the Law of Dominance, making Option (4) the correct answer.

Flower colour in snapdragon exhibits incomplete dominance where the heterozygous condition (Rr) yields an intermediate pink phenotype. Homozygous dominant (RR) produces red flowers, and homozygous recessive (rr) produces white flowers.

Parent 1: Pink flowered = $Rr$ Parent 2: Red flowered = $RR$ Gametes produced: - Pink parent: $50\%\; R$, $50\%\; r$ - Red parent: $100\%\; R$

$$\text{Offspring Genotypes} = \frac{1}{2}\,RR + \frac{1}{2}\,Rr$$ - $RR$: Red flowered ($50\%$) - $Rr$: Pink flowered ($50\%$) No white ($rr$) plants can be formed. The progeny consists of red flowered as well as pink flowered plants.

In genetics, a test cross is the cross of an organism showing a dominant phenotype (whose genotype may be homozygous or heterozygous) with an organism that is homozygous recessive for the same trait.

If the black seed plant is homozygous dominant ($BB$): $$BB \times bb \rightarrow 100\%\; Bb\;(\text{all black})$$ If the black seed plant is heterozygous ($Bb$): $$Bb \times bb \rightarrow 50\%\; Bb\;(\text{black}) : 50\%\; bb\;(\text{white})$$ Because the recessive parent donates only recessive alleles ($b$), the phenotype of the progeny directly reveals the alleles donated by the unknown dominant parent.

The genotype crossed with the black seeded plant is homozygous recessive ($bb$), which is Option (4).

Alleles are slightly different forms of the same gene occupying identical loci on homologous chromosomes (A -> III). Ploidy refers to the number of complete sets of chromosomes in a cell (e.g., haploidy, diploidy) (D -> II).

A back cross is any cross between an F1 individual and either of its parents (dominant or recessive) (C -> I). A test cross is specifically a cross between an F1 individual and its homozygous recessive parent, used to determine whether the F1 is homozygous or heterozygous (B -> IV).

The resulting match is A-III, B-IV, C-I, D-II, which corresponds to Option (1).

Down’s syndrome is an aneuploid condition caused by the presence of an additional copy of chromosome number 21 (trisomy 21) (A -> III). Klinefelter’s syndrome is caused due to the presence of an additional copy of the X-chromosome resulting into a karyotype of 47, XXY (D -> II).

Thalassemia is an autosome-linked recessive blood disease. α-Thalassemia is controlled by two closely linked genes HBA1 and HBA2 on chromosome 16 of each parent (B -> IV). β-Thalassemia is controlled by a single gene HBB on chromosome 11 of each parent (C -> I).

The resulting mapping is A-III, B-IV, C-I, D-II, corresponding to Option (1).

ABO blood groups are controlled by the gene $I$, which has three alleles: $I^A$, $I^B$, and $i$. The alleles $I^A$ and $I^B$ are codominant with each other and both are completely dominant over the recessive allele $i$.

A person with blood group O exhibits the phenotype of the recessive allele and must have the homozygous genotype: $$\text{Child} = ii$$

For the child to receive one $i$ allele from the father and one $i$ allele from the mother, both parents must possess at least one $i$ allele: - Father (Blood group B): Can be $I^B I^B$ or $I^B i$. Since he must donate $i$, his genotype is $I^B i$. - Mother (Blood group A): Can be $I^A I^A$ or $I^A i$. Since she must donate $i$, her genotype is $I^A i$. Respective genotypes (Father / Mother / Child): $$I^B i \;/\; I^A i \;/\; ii$$

Only Statement A correctly represents $I^B i \;/\; I^A i \;/\; ii$. Thus, the answer is "A only", corresponding to Option (3).

Alfred Sturtevant (1913), a student of Thomas Hunt Morgan, reasoned that the frequency of recombination between two linked gene pairs is proportional to the physical distance separating them on the chromosome:

Key genetic milestones in NCERT include:

Sturtevant's genetic mapping principle served as the foundation for modern genome sequencing projects, including the Human Genome Project.

Pleiotropy is the phenomenon wherein a single gene influences multiple unrelated phenotypic traits:

The underlying mechanism in most cases is the effect of the gene product on a central metabolic pathway that contributes to different development systems:

NCERT defines pleiotropy: "There are instances where a single gene can exhibit multiple phenotypic expressions. Such a gene is called a pleiotropic gene."

Klinefelter's syndrome is caused by the presence of an additional copy of the X-chromosome resulting in a $47,\, XXY$ karyotype:

Analyzing the canonical textbook descriptions: - "Such an individual has overall masculine development, however, the feminine development (development of breast, i.e., Gynaecomastia) is also expressed." (Statement B is TRUE) - "Such individuals are sterile." (Statement E is TRUE)

Statements A, C, and D describe Down's syndrome (trisomy 21): - First described by Langdon Down (1866) (Down's syndrome). - Short statured with small round head (Down's syndrome). - Physical, psychomotor, and mental development is retarded (Down's syndrome).

Pedigree charts use standardized geometric symbols to represent familial relationships and phenotypic statuses:

Under "Symbols used in the Human Pedigree Analysis", NCERT explicitly diagrams a square joined to a circle by two parallel horizontal bars labeled: "Mating between relatives (consanguineous mating)".

Option (1) depicts shaded symbols (affected individuals). Option (2) depicts a single line (non-consanguineous mating). Option (4) depicts parents with an affected male offspring.

First described by Langdon Down in 1866, Down's syndrome is an autosomal aneuploidy caused by nondisjunction resulting in trisomy of chromosome 21 ($2n + 1 = 47$):

NCERT provides four unmistakable physical signs: 1. Short statured with small round head. 2. Furrowed tongue and partially open mouth. 3. Palm is broad with characteristic palm crease (Simian crease). 4. Physical, psychomotor, and mental development is retarded.

Thalassemia affects globin chain synthesis. Turner's syndrome ($45, XO$) presents with short stature and webbed neck. Klinefelter's ($47, XXY$) presents with tall stature and gynaecomastia.

The production of gametes by parents, the formation of zygotes, and the resulting $F_1$ and $F_2$ generations can be systematically mapped using a checkerboard graphical representation:

All possible types of gametes produced by the male parent are listed across the top, while gametes produced by the female parent are listed along the left side. Each internal cell represents the union of two haploid gametes, visually illustrating the resulting diploid zygotic genotypes and genotypic ratios.

Sickle-cell anaemia is an autosomal recessive Mendelian disorder controlled by a single pair of alleles, $Hb^A$ (normal $\beta$-globin allele) and $Hb^S$ (mutant $\beta$-globin allele with a GAG $\to$ GUG transversion at codon 6, substituting glutamic acid with valine). - Normal homozygous ($Hb^A Hb^A$): Completely unaffected (100% normal biconcave erythrocytes). - Heterozygous carriers ($Hb^A Hb^S$): Phenotypically normal under routine atmospheric oxygen tensions; carry sickle-cell trait but do not suffer from anaemia. - Homozygous diseased ($Hb^S Hb^S$): Express the full clinical syndrome of sickle cell anaemia with microvascular vaso-occlusion, acute sickle crises, and haemolytic anaemia.

Cross: Heterozygous male ($Hb^A Hb^S$) $\times$ Heterozygous female ($Hb^A Hb^S$): $$\begin{array}{c|c|c} & Hb^A & Hb^S \\ \hline Hb^A & Hb^A Hb^A \, (25\%) & Hb^A Hb^S \, (25\%) \\ \hline Hb^S & Hb^A Hb^S \, (25\%) & Hb^S Hb^S \, (25\%) \\ \end{array}$$ - Normal phenotype (disease-free): $Hb^A Hb^A$ (25%) + $Hb^A Hb^S$ (50%) = 75% - Diseased phenotype: $Hb^S Hb^S$ = $1/4 = 25\%$.

The human ABO blood group system is governed by a single polymorphic gene, gene $I$, located on chromosome 9: - The plasma membrane of red blood cells contains protruding glycoprotein/glycolipid polymers based on a core H-antigen. - Three Alleles: Gene $I$ exists as three distinct alleles in human populations: $I^A$, $I^B$, and $i$. - Allele $I^A$: Codes for $\alpha$-1,3-N-acetylgalactosaminyltransferase, which adds N-acetylgalactosamine (A-sugar antigen). - Allele $I^B$: Codes for $\alpha$-1,3-galactosyltransferase, which adds galactose (B-sugar antigen, a distinctly different sugar). - Allele $i$: Contains a single base deletion causing a frameshift, producing a truncated non-functional enzyme that does not produce any sugar.

Statement (3) asserts that "When $I^A$ and $I^B$ are present together, they express same type of sugar." In reality, they express different types of sugars because of codominance. Therefore, statement (3) is false and represents the correct choice.

- 1. Haemophilia: A sex-linked (X-linked) recessive blood coagulation disorder involving Factor VIII (Haemophilia A) or Factor IX (Haemophilia B) deficiency. Because it is X-linked, it exhibits criss-cross inheritance and predominantly affects males. It is never Y-linked. - 2. Phenylketonuria (PKU): An autosomal recessive inborn error of phenylalanine metabolism caused by mutation in the gene encoding phenylalanine hydroxylase on chromosome 12. It is not an autosomal dominant trait. - 3. Sickle Cell Anaemia: An autosomal recessive haematological disorder caused by a single nucleotide transversion (GAG $\to$ GUG) at the sixth codon of the $\beta$-globin gene located on autosomal chromosome 11, replacing glutamic acid with hydrophobic valine. - 4. Thalassemia: An autosomal recessive quantitative hemoglobinopathy caused by defective synthesis of $\alpha$-globin chains (genes *HBA1* and *HBA2* on autosome chromosome 16) or $\beta$-globin chains (gene *HBB* on autosome chromosome 11). It is completely autosomal, not sex-linked.

Only pair (3) matches both the mode of inheritance (autosomal recessive) and the chromosomal assignment (chromosome-11) correctly.

NCERT states verbatim: "Mendel selected 14 true-breeding pea plant varieties, as pairs which were similar except for one character with contrasting traits."

The 7 pairs of contrasting traits included: 1. Stem height (Tall / Dwarf) 2. Flower colour (Violet / White) 3. Flower position (Axial / Terminal) 4. Pod shape (Inflated / Constricted) 5. Pod colour (Green / Yellow) 6. Seed shape (Round / Wrinkled) 7. Seed colour (Yellow / Green).

Number of characters = 7. Number of pairs of contrasting traits = 7. Total number of true-breeding plant varieties = 14.

The question specifically asks for the number of true-breeding varieties, which is 14.

In 1902, Walter Sutton and Theodore Boveri noted that the behavior of chromosomes was parallel to the behavior of genes and used chromosome movement during meiosis to explain Mendel's laws, proposing the Chromosomal Theory of Inheritance.

NCERT states verbatim: "Experimental verification of the chromosomal theory of inheritance by Thomas Hunt Morgan and his colleagues, led to discovering the basis for the variation that sexual reproduction produced."

Morgan worked with the tiny fruit fly, Drosophila melanogaster, because they could be grown on simple synthetic medium, completed their life cycle in two weeks, had high fecundity, clear sex dimorphism, and easily visible hereditary variations under low power microscopes.

Thomas Hunt Morgan (known as the Father of Experimental Genetics) uniquely performed the experimental verification.