Q.Name a human genetic disorder due to the following :

Showing the 12 most recent of 69 on this concept.

1. AP EAPCET 2026Set ap-2026-05-19-FN1 markMCQ
   Q.What is the percentage of Pink colour flowered plants in $F_2$ generation of snapdragon monohybrid cross (A) 25 (B) 50 (C) 75 (D) 100

   ›Reveal solutionSolution

   Snapdragon flower colour shows incomplete dominance, giving an F2 genotypic (and here, phenotypic) ratio of 1 red : 2 pink : 1 white, so pink = 50%.

   Concept and Intuition

   In cases of incomplete dominance, the heterozygote's phenotype is intermediate between the two homozygous parental phenotypes because neither allele is fully dominant — there is a partial/blended expression (e.g., due to partial enzyme/pigment production). In the snapdragon (Antirrhinum) monohybrid cross for flower colour, red (RR) crossed with white (rr) gives an F1 that is entirely pink (Rr), not red, showing the alleles are not fully dominant/recessive. Selfing the pink F1 (Rr × Rr) reproduces the classic Mendelian 1:2:1 genotypic ratio, and because genotype and phenotype track together here, the phenotypic ratio is also 1 red : 2 pink : 1 white.

   Step-by-Step Solution

   1. Set up the cross: RR (red) × rr (white) → F1 all Rr (pink), confirming incomplete dominance. …

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2. AP EAPCET 2026Set ap-2026-05-19-FN1 markMCQ
   Q.Study the following statements regarding genic balance theory and identify incorrect statements. A) The production of gametes with abnormal number of chromosomes is due to non-disjunction. B) In Drosophila 'XO' males produce motile sperms. C) Sex index < 0.5 indicates intersexes in Drosophila. D) 'Y' chromosome in Drosophila lacks testis determining factor. (A) A & C (B) A & D (C) B & C (D) B & D

   ›Reveal solutionSolution

   The genic balance theory (Bridges) uses the X:autosome ratio to fix sex in

   Drosophila; XO flies are sterile (not fertile) males, and ratio <0.5 gives

   super males, not intersexes — so B and C are the incorrect statements.

   Answer: (C).

   Concept and Intuition

   Bridges proposed that sex in Drosophila is decided not just by presence of Y,

   but by the ratio of X chromosomes to sets of autosomes (X:A, the "sex index"):

   • Sex index = 1.0 → normal female (XX)
   • Sex index = 0.5 → normal male (XY or XO)
   • 0.5 < index < 1.0 → intersex
   • index < 0.5 → super male (metamale)
   • index > 1.0 → super female (metafemale)

   Crucially, in Drosophila the Y chromosome does not determine maleness (that

   job is done by the X:A ratio) — but the Y chromosome IS still needed for

   fertility: XO males look and behave male but are sterile because normal

   sperm maturation requires Y-linked fertility genes.

   Step-by-Step Solution

   1. A: Non-disjunction → unequal chromosome distribution in gametes → gametes with abnormal chromosome numbers. This is a correct, well-established fact.
   2. D: Y chromosome in Drosophila lacks a testis-determining factor (sex is set by X:A ratio, not by Y) — correct statement.
   3. B: XO males — the X:A ratio (0.5) still makes them male, but they lack the Y chromosome's fertility genes, so they are sterile, not "produce motile …

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3. AP EAPCET 2026Set ap-2026-05-19-FN1 markMCQ
   Q.The blood group of mother is B and the progeny in the family is 25% A blood type, 25% AB and 50% B type. What are the genotypes of the parents. (A) $I^AI^A$ father and $I^BI^O$ mother (B) $I^AI^O$ father and $I^BI^O$ mother (C) $I^AI^B$ father and $I^BI^B$ mother (D) $I^AI^B$ father and $I^BI^O$ mother

   ›Reveal solutionSolution

   Testing each option against the observed 25% A : 25% AB : 50% B ratio, only

   father $I^AI^B$ × mother $I^BI^O$ reproduces it exactly. Answer: (D).

   Concept and Intuition

   ABO blood grouping is governed by multiple alleles ($I^A$, $I^B$, $I^O$) at a

   single locus, where $I^A$ and $I^B$ are co-dominant to each other and both

   dominant to $I^O$. To find parental genotypes from an observed progeny ratio,

   we can test each candidate cross by Punnett-square logic and check whether it

   reproduces the given proportions.

   Step-by-Step Solution

   1. Mother's phenotype is B, so her genotype must be $I^BI^B$ or $I^BI^O$ — this alone doesn't decide between the options, so test the crosses.
   2. Try option (D): father $I^AI^B$ (gametes $I^A$, $I^B$, each 1/2), mother $I^BI^O$ (gametes $I^B$, $I^O$, each 1/2).
   3. Combine gametes:
      • $I^A \times I^B \to I^AI^B$ (AB) — 1/4
      • $I^A \times I^O \to I^AI^O$ (A) — 1/4
      • $I^B \times I^B \to I^BI^B$ (B) — 1/4
      • $I^B \times I^O \to I^BI^O$ (B) — 1/4 …

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4. AP EAPCET 2026Set ap-2026-05-19-AN1 markMCQ
   Q.Mendelian genetic disorder controlled by a single gene on chromosome 11 of each parent is (A) Phenylketonuria (B) Cystic fibrosis (C) Sickle-cell anaemia (D) Cooley's Anaemia

   ›Reveal solutionSolution

   Cooley's anaemia (beta-thalassemia) is controlled by the single HBB gene on chromosome 11 of each parent — option (D).

   Concept and Intuition

   NCERT distinguishes the thalassemias: alpha-thalassemia is controlled by two closely linked genes (HBA1, HBA2) on chromosome 16 of each parent, while beta-thalassemia — Cooley's anaemia — is controlled by a single gene, HBB, on chromosome 11 of each parent. The wording of the stem matches this exact statement.

   Step-by-Step Solution

   1. Phenylketonuria: PAH gene on chromosome 12 — not chromosome 11.
   2. Cystic fibrosis: CFTR gene on chromosome 7 — not chromosome 11.
   3. Sickle-cell anaemia also involves HBB on chromosome 11, but NCERT does not use the 'single gene ... chromosome 11 of each parent' descriptor for it. …

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5. AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQ
   Q.Assertion (A): Though the parents contain two alleles during gamete formation, the alleles of a pair segregate from each other. Reason (R): Segregation is a universal phenomenon in all organisms showing sexual method of reproduction. (A) Both (A) and (R) are correct and (R) is the correct explanation to (A) (B) Both (A) and (R) are correct but (R) is not correct explanation for (A) (C) (A) is correct (R) is wrong (D) (A) is wrong (R) is correct

   ›Reveal solutionSolution

   Mendel's Law of Segregation: alleles separate during gamete formation, and this is universally true across sexually reproducing organisms because it is a direct consequence of meiosis. Both statements true, R explains A.

   Concept and Intuition

   A diploid organism carries two alleles for every gene (one from each parent). During meiosis, homologous chromosomes — and with them, the two alleles of each gene — separate into different gametes, so any single gamete carries only one allele per gene. This isn't a quirk of Mendel's pea plants; it is a mechanical outcome of the meiotic process itself, which every sexually reproducing organism undergoes to produce haploid gametes.

   Step-by-Step Solution

   1. Assertion: alleles of a pair segregate during gamete formation — this is exactly Mendel's Law of Segregation, verified true by countless organisms since. …

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6. AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQ
   Q.Expression of more than one phenotypic trait by a single gene is known as (A) Pleiotropy (B) Polygenetic inheritance (C) Multiple allelism (D) Lyonisation

   ›Reveal solutionSolution

   A single gene producing effects on more than one phenotypic trait is the definition of pleiotropy.

   Concept and Intuition

   Most genes are studied for one visible trait, but many gene products (often enzymes early in a biochemical pathway) influence several downstream processes at once. When a single gene's mutation therefore shows up as changes in multiple, often unrelated, characteristics simultaneously, geneticists call this pleiotropy. This is distinct from polygenic inheritance (many genes controlling one trait), multiple allelism (many alternate forms of one gene, e.g., ABO blood groups), and lyonisation (X-chromosome inactivation in females).

   Step-by-Step Solution

   1. Read the definition carefully: 'expression of more than one phenotypic trait by a single gene.'
   2. Match this to the standard genetics term: pleiotropy is defined exactly this way. …

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7. AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQ
   Q.If blood group of father is A (homozygous) and that of mother is O, these blood groups are not expected in their children. (A) B, AB, A (B) B, AB, O (C) A, O, AB (D) A, B, O

   ›Reveal solutionSolution

   With father $I^AI^A$ and mother $ii$, every child is genotype $I^Ai$ (blood group A) — so B, AB and O are all impossible.

   Concept and Intuition

   ABO blood group is controlled by multiple alleles $I^A$, $I^B$, $i$, with $I^A$ and $I^B$ co-dominant and both dominant over $i$. A homozygous $I^AI^A$ father can only pass on the $I^A$ allele (he has no other allele to give), and an $ii$ mother can only pass on $i$. Every offspring therefore receives exactly one $I^A$ and one $i$, giving genotype $I^Ai$, phenotype blood group A, with no variation possible.

   Step-by-Step Solution

   1. Father's genotype: $I^AI^A$ (homozygous A) — gametes are all $I^A$.
   2. Mother's genotype: $ii$ (O) — gametes are all $i$.
   3. Cross: $I^AI^A \times ii \Rightarrow$ all offspring $I^Ai$ — phenotype A, with 100% certainty. …

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8. AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQ
   Q.If sex index ratio of a Drosophila is 0.66, its sexual phenotype is (A) Intersex (B) Meta male (C) Meta female (D) Triploid male

   ›Reveal solutionSolution

   A sex index of 0.66 lies strictly between the male (0.5) and female (1.0) values, which by Bridges' genic balance theory in Drosophila defines an intersex.

   Concept and Intuition

   Calvin Bridges' genic balance theory explains Drosophila sex determination by the ratio of X chromosomes to sets of autosomes (X:A ratio, or 'sex index'). A ratio of 1.0 (2X : 2 autosome-sets) gives a normal female; 0.5 (1X : 2 autosome-sets) gives a normal male. Ratios above 1.0 give metafemales (superfemales), ratios below 0.5 give metamales (supermales), and ratios strictly between 0.5 and 1.0 give intersexes — flies with an ambiguous mix of male and female characteristics because the genic balance is intermediate.

   Step-by-Step Solution

   1. Recall the reference points: female index = 1.0, male index = 0.5.
   2. Given index = 0.66 — this lies between 0.5 and 1.0. …

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9. AP EAPCET 2026Set ap-2026-05-20-AN1 markMCQ
   Q.If blood group of mother is B (homozygous) and that of father is A (homozygous), these blood groups are absent in their children. (A) A, AB, B (B) A, AB, O (C) B, O, A (D) AB, O, B

   ›Reveal solutionSolution

   This is a classic ABO blood group cross using co-dominant alleles $I^A$ and $I^B$.

   Concept and Intuition

   The ABO blood group system involves three alleles at one locus: $I^A$ and $I^B$ are co-dominant to each other and both dominant over $i$ (the recessive allele for O). A homozygous B individual has genotype $I^B I^B$ and can only contribute the $I^B$ allele to offspring. A homozygous A individual has genotype $I^A I^A$ and can only contribute the $I^A$ allele. Since neither parent carries the recessive $i$ allele, no child can be blood group O; and since every child receives one $I^A$ and one $I^B$, every child is genotype $I^A I^B$ = blood group AB, so no child can be pure A or pure B either.

   Step-by-Step Solution

   1. Mother: $I^B I^B$ (homozygous B) → gametes are all $I^B$.
   2. Father: $I^A I^A$ (homozygous A) → gametes are all $I^A$.
   3. Cross: every offspring gets $I^A$ from father and $I^B$ from mother → genotype $I^A I^B$ → phenotype AB, with 100% probability. …

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10. AP EAPCET 2026Set ap-2026-05-20-AN1 markMCQ
    Q.Sex determination in butter flies is (A) XX-XY type (B) XX-XO type (C) ZZ-ZW type (D) ZZ-ZO type

    ›Reveal solutionSolution

    Sex determination in butterflies (Lepidoptera) follows the ZW system, where the female is heterogametic (ZW) and the male is homogametic (ZZ).

    Concept and Intuition

    Sex-chromosome systems aren't universal across animals. In the familiar XX–XY system (mammals, and also Drosophila), the male is the heterogametic sex. But several groups have the opposite arrangement, called the ZW system, where the female is heterogametic (ZW) and the male is homogametic (ZZ). Birds are the textbook example of ZW; Lepidoptera (moths and butterflies) also use this system.

    Step-by-Step Solution

    1. Identify the taxonomic group: butterflies belong to order Lepidoptera.
    2. Recall the sex-determination systems taught: XX–XY (e.g., humans, Drosophila, male heterogametic), XX–XO (e.g., grasshoppers, male has only one X, no Y), ZZ–ZW (birds, most Lepidoptera; female heterogametic), and ZZ–ZO (some other insects).
    3. Lepidoptera (which includes butterflies) is one of the classic examples, alongside birds, of the ZZ–ZW system. …

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11. AP EAPCET 2025Set ap-2025-05-19-AN1 markMCQ
    Q.If two heterozygous tall garden pea plants are crossed, the expected genotypic ratio in their off spring is (A) 3 : 1 (B) 1 : 1 (C) 1 : 2 : 1 (D) 1 : 0

    ›Reveal solutionSolution

    This tests the classic Mendelian monohybrid cross genotypic ratio. The answer is (C) 1 : 2 : 1.

    Concept and Intuition

    When two heterozygotes for a single gene are crossed (Tt × Tt), each parent contributes either the dominant (T) or recessive (t) allele with equal probability (1/2 each) to the gametes. Combining gametes via a Punnett square yields four equally likely combinations: TT, Tt, Tt, tt — i.e., genotypes in the ratio 1 TT : 2 Tt : 1 tt. This is distinct from the phenotypic ratio, which collapses TT and Tt into the same "tall" phenotype, giving 3 tall : 1 dwarf (3:1).

    Step-by-Step Solution

    1. Set up the cross: Tt (tall, heterozygous) × Tt (tall, heterozygous).
    2. Gametes from each parent: T or t, each with probability 1/2. …

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12. AP EAPCET 2025Set ap-2025-05-19-AN1 markMCQ
    Q.Match the following Method of sex determination - Example A) XX - XY I) Grasshoppers B) XX - XO II) Birds C) ZW - ZZ III) Human beings D) ZZ - ZO IV) Moths V) Honey bees Identify the correct match (A) A-II, B-III, C-IV, D-V (B) A-III, B-IV, C-I, D-II (C) A-V, B-II, C-III, D-IV (D) A-III, B-I, C-II, D-IV

    ›Reveal solutionSolution

    This tests recall of chromosomal sex-determination systems across different animal groups. The answer is (D) A-III, B-I, C-II, D-IV.

    Concept and Intuition

    Different animal groups use different combinations of sex chromosomes to determine sex, and whether the male or the female is the heterogametic sex varies by group:

    • XX-XY: the female is XX (homogametic), the male is XY (heterogametic) — the system found in human beings and most mammals.
    • XX-XO: the female is XX, the male has only a single X and no partner (XO) — classically illustrated by grasshoppers.
    • ZW-ZZ: here it is the female that is heterogametic (ZW), while the male is ZZ (homogametic) — found in birds.
    • ZZ-ZO: a rarer variant where the female is ZO (lacking a chromosome partner) and the male is ZZ — found in certain moths.

    Step-by-Step Solution

    1. A) XX-XY → matches Human beings (III), the standard mammalian system.
    2. B) XX-XO → matches Grasshoppers (I), the textbook example of this system. …

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