Q.(a)

Showing the 12 most recent of 48 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.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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3. AP EAPCET 2026Set ap-2026-05-19-AN1 markMCQ
   Q.Gene 'y' in the Lac operon synthesizes the following protein/enzyme. (A) Transacetylase (B) Permease (C) $\beta$-galactosidase (D) Repressor

   ›Reveal solutionSolution

   In the lac operon's z-y-a gene cluster, gene y (lacY) encodes permease, the membrane transporter that brings lactose into the cell. Answer: (B) Permease.

   Concept and Intuition

   The lac operon (Jacob and Monod's model of inducible gene regulation) has a single promoter-operator controlling three structural genes transcribed as one polycistronic mRNA: lacZ ($\beta$-galactosidase, hydrolyses lactose into glucose + galactose and also converts some lactose to allolactose, the real inducer), lacY (permease, transports lactose across the cell membrane into the cytoplasm), and lacA (transacetylase, whose exact metabolic role is less central but is transcribed along with z and y). The repressor protein, coded by the separate lacI gene (with its own independent promoter), binds the operator to block transcription in the absence of lactose.

   Step-by-Step Solution

   1. Identify the gene naming convention: lacZ, lacY, lacA correspond to $\beta$-galactosidase, permease, and transacetylase respectively.
   2. Gene 'y' = lacY $\rightarrow$ codes for permease. …

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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-AN1 markMCQ
   Q.During the expression of Lac operon, the repressor protein binds to (A) Promoter (B) Operator (C) Inducer (D) Terminator

   ›Reveal solutionSolution

   This tests the basic mechanism of negative regulation in the lac operon: the

   repressor's binding site is the operator, not the promoter, inducer, or terminator.

   Concept and Intuition

   The lac operon is a classic example of negative inducible regulation. The lacI

   gene constitutively produces a repressor protein. In the absence of lactose (the

   inducer), this repressor binds the operator — a DNA sequence that overlaps/lies

   next to the promoter — and this bound repressor sterically blocks RNA polymerase

   from moving from the promoter into the structural genes, so transcription is

   switched off. When lactose (via allolactose) is present, it binds the repressor,

   changes its shape, and the repressor can no longer bind the operator, so RNA

   polymerase transcribes lacZ, lacY, lacA freely.

   Step-by-Step Solution

   1. Identify the four choices as operon elements: Promoter (where RNA polymerase binds), Operator (the repressor's binding site), Inducer (lactose/allolactose, which binds the repressor, not DNA), Terminator (where transcription ends).
   2. Recall the definition of the operator: a DNA sequence whose occupation by the repressor blocks transcription initiation.
   3. The repressor protein's job, by definition, is to bind the operator — this is the textbook mechanism of lac operon repression. …

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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 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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11. AP EAPCET 2025Set ap-2025-05-19-FN1 markMCQ
    Q.In which of the following crosses, both phenotypic and genotypic ratios at F$_2$ generation are 1 : 2 : 1 I) Red flowered plant crossed with white flowered plant in snapdragon II) Tall plant crossed with dwarf plant in garden pea III) Plant with dotted seed coat is crossed with spotted seed coat plant in lentil IV) A homozygous plant is crossed with a heterozygous plant The correct answer is (A) I, II (B) III, IV (C) I, III (D) II, III

    ›Reveal solutionSolution

    Only incomplete dominance (snapdragon flower colour) and codominance (lentil seed-coat pattern) give matching 1:2:1 phenotypic and genotypic ratios at F2 — complete dominance (pea height) gives 3:1 phenotypically despite 1:2:1 genotypically, and a homozygous x heterozygous cross gives 1:1. Answer: (C), I and III.

    Concept and Intuition

    In a standard monohybrid self-cross (Aa x Aa to F2), the genotypic ratio is always 1 AA : 2 Aa : 1 aa. Whether the phenotypic ratio also comes out 1:2:1 depends entirely on whether the heterozygote (Aa) looks different from both homozygotes:

    • If dominance is complete (as in pea height, Tall dominant over dwarf), Aa looks identical to AA, so the two dominant genotypic classes merge phenotypically — giving the familiar 3:1 phenotypic ratio, even though the underlying genotypic ratio is still 1:2:1. Phenotypic and genotypic ratio do not match here.
    • If dominance is incomplete (as in snapdragon flower colour: red x white gives pink F1), the heterozygote has its own distinct, intermediate phenotype — so phenotypic classes align exactly with genotypic classes, giving 1 red : 2 pink : 1 white, matching the 1:2:1 genotypic ratio.
    • If the trait is codominant (as in lentil seed-coat pattern: dotted x spotted gives a distinct dotted-and-spotted heterozygote pattern in which both parental patterns are simultaneously visible), the same logic applies — the heterozygote's phenotype is unique, so phenotypic and genotypic ratios both come out 1:2:1.
    • A cross between a homozygous and a heterozygous individual (e.g., AA x Aa, or Aa x aa) is a test/back-cross type, not a self-cross producing an "F2" generation in the usual sense, and produces a 1:1 ratio of genotypes/phenotypes (not 1:2:1).

    Step-by-Step Solution

    1. (I) Snapdragon red x white: incomplete dominance to F1 pink; F2 = 1 red : 2 pink : 1 white (phenotypic) and 1 RR : 2 Rr : 1 rr (genotypic) — both 1:2:1. QUALIFIES. …

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12. AP EAPCET 2025Set ap-2025-05-20-AN1 markMCQ
    Q.If father has blood group A (heterozygous) and mother has blood group B (homozygous), these blood groups are not expected in their children (A) AB, O (B) AB, B (C) A, O (D) B, O

    ›Reveal solutionSolution

    A father who is $I^Ai$ crossed with a mother who is $I^BI^B$ can only produce children

    of blood group AB ($I^AI^B$) or B ($I^Bi$) — groups A and O are impossible from this

    particular cross. Answer: (C).

    Concept and Intuition

    Human ABO blood groups are controlled by three alleles at one locus: $I^A$ and $I^B$ are

    co-dominant to each other and both dominant over $i$; genotype $I^AI^A$ or $I^Ai$ gives

    blood group A, $I^BI^B$ or $I^Bi$ gives blood group B, $I^AI^B$ gives blood group AB, and

    $ii$ gives blood group O. To determine which blood groups are possible in offspring, we

    need the parents' gamete contributions, not just their phenotypes.

    Step-by-Step Solution

    1. Father's genotype: heterozygous A = $I^A i$. His gametes: $I^A$ or $i$ (each with 50% probability).
    2. Mother's genotype: homozygous B = $I^B I^B$. Her gametes: only $I^B$ (100%).
    3. Combine gametes: $I^A$ (from father) + $I^B$ (from mother) = $I^A I^B$ → blood group AB.
    4. Combine gametes: $i$ (from father) + $I^B$ (from mother) = $I^B i$ → blood group B.
    5. These are the only two possible offspring genotypes/phenotypes: AB and B. Blood …

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