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

Showing the 12 most recent of 31 on this concept.

1. TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQ
   Q.Match the following List-1 List-2 A. linkage I. Common in cancer cells B. Recombination II. Parental gene combination C. Point mutation III. Non parental combination D. Chromosomal aberration IV. Sickle cell anemia V. Multiple phenotypes The correct answer is (A) A-II, B-V, C-IV, D-I (B) A-II, B-III, C-IV, D-I (C) A-I, B-IV, C-III, D-V (D) A-II, B-III, C-IV, D-V

   ›Reveal solutionSolution

   This question tests your understanding of key genetic terms and their real-world examples. The correct mapping is A-II, B-III, C-IV, D-I, which corresponds to option (B).

   Let’s build the intuition first. Each term in List-1 describes a specific genetic phenomenon, and List-2 gives either a cause, a consequence, or a classic example. The trick is to match each term with its most direct and well-known association — not just any possible connection, but the one that is textbook-standard for Indian exams.

   1. A. Linkage → II. Parental gene combination

      Linkage means genes located close together on the same chromosome tend to be inherited as a block. They do not assort independently, so the offspring often get the same combination of alleles that the parents had — hence “parental gene combination.” This is the classic definition: linkage preserves parental combinations unless crossing over breaks them.

   2. B. Recombination → III. Non-parental combination

      Recombination is the process that shuffles alleles between homologous chromosomes during meiosis (crossing over). The result is new combinations of genes that were not present together in either parent — these are called “non-parental” or “recombinant” combinations. So B matches III directly.

   3. C. Point mutation → IV. Sickle cell anemia

      A point mutation is a change in a single nucleotide base pair. Sickle cell anemia is the classic example: a single base substitution in the beta-globin gene (GAG → GTG) changes glutamic acid to valine, causing the disease. This is the go-to example in every syllabus.

   4. D. Chromosomal aberration → I. Common in cancer cells …

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2. TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQ
   Q.Study the following and choose the correct statements I. If a single gene influences more than one phenotypic trait, it is called pleiotropy II. Due to absence of antibodies anti A and anti B, the persons with AB blood group are described as universal recipients III. Inheritance of sex linked dominant traits follow cris cross inheritance IV. If sex index ratio is 0.33, then sexual phenotype of Drosophila is intersex (A) I, III (B) II, IV (C) III, IV (D) I, II

   ›Reveal solutionSolution

   Statements I and II are correct; III and IV are wrong, so the correct set is I, II — option (D).

   I – Correct. When a single gene affects several phenotypic traits, the effect is called pleiotropy (e.g. the sickle-cell gene affects RBC shape, anaemia and malaria resistance). This is the standard definition.

   II – Correct. Blood group AB carries both A and B antigens but no anti-A or anti-B antibodies in the plasma, so such persons can receive blood of any ABO type without agglutination — they are universal recipients.

   III – Incorrect. Criss-cross inheritance (father $\to$ daughter $\to$ grandson) is the pattern of X-linked recessive traits. X-linked dominant traits do not show criss-cross inheritance — an affected father passes the trait to all his daughters but to none of his sons. …

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3. TG EAPCET 2026Set ap-2026-05-05-FN1 markMCQ
   Q.Trisomy of $23^{\text{rd}}$ chromosome in male results in (A) Turner syndrome (B) Down syndrome (C) Edward syndrome (D) Klinefelter syndrome

   ›Reveal solutionSolution

   Trisomy of the 23rd chromosome in a male means an extra sex chromosome (XXY), which causes Klinefelter syndrome. The correct option is (D).

   The key here is understanding what "trisomy of the 23rd chromosome" means. Humans have 23 pairs of chromosomes — 22 pairs of autosomes and 1 pair of sex chromosomes (the 23rd pair). In a normal male, the sex chromosome pair is XY. Trisomy means having three copies of a particular chromosome instead of the usual two. So trisomy of the 23rd chromosome in a male means the sex chromosome constitution is XXY — two X chromosomes and one Y.

   Now, which syndrome does XXY produce? That is Klinefelter syndrome. Let’s walk through the options to see why the others are wrong.

   1. Turner syndrome results from monosomy of the X chromosome in females — a single X (45,X). That is a missing sex chromosome, not an extra one. So it cannot be trisomy of the 23rd.

   2. Down syndrome is trisomy of chromosome 21, not the 23rd. It is an autosomal trisomy, not a sex chromosome trisomy. So this is a common confusion — students sometimes mix up the chromosome numbers.

   3. Edward syndrome is trisomy of chromosome 18, another autosomal trisomy. Again, not the 23rd. …

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4. TG EAPCET 2026Set ap-2026-05-05-FN1 markMCQ
   Q.If the blood groups of parents are homozygous for A and homozygous for B blood groups, these blood groups are not expected in their children (A) A, AB, O (B) A, B, O (C) B, AB, O (D) A, B, AB

   ›Reveal solutionSolution

   When one parent is homozygous A (genotype $I^A I^A$) and the other is homozygous B ($I^B I^B$), all children inherit one $I^A$ and one $I^B$ allele, giving only blood group AB. So groups A, B, and O are not expected.

   The key here is understanding how blood groups are inherited. The ABO system is controlled by a single gene with three alleles: $I^A$, $I^B$, and $i$. Both $I^A$ and $I^B$ are dominant over $i$, but they are codominant with each other — meaning if both are present, both are expressed, giving the AB blood group.

   When a parent is homozygous for a blood group, it means they carry two copies of the same allele. So a homozygous A parent has genotype $I^A I^A$, and a homozygous B parent has genotype $I^B I^B$.

   Now let’s work through what happens in their children.

   1. Gamete formation

      The $I^A I^A$ parent can only produce gametes carrying the $I^A$ allele.

      The $I^B I^B$ parent can only produce gametes carrying the $I^B$ allele.

   2. Fertilization

      Every child receives one allele from each parent. So every child gets exactly one $I^A$ and one $I^B$ — the only possible genotype is $I^A I^B$.

   3. Resulting blood group

      Since $I^A$ and $I^B$ are codominant, the $I^A I^B$ genotype expresses as blood group AB.

      No other combination is possible.

   4. Which groups are not expected?

      The children cannot have: …

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5. TG EAPCET 2025Set ap-2025-04-29-AN1 markMCQ
   Q.In several fungi and plants, the bisexual condition is denoted by (A) Homothallic and monoecious (B) Heterothallic and monoecious (C) Homothallic and dioecious (D) Heterothallic and dioecious

   ›Reveal solutionSolution

   The question asks for the terms that describe the bisexual condition in fungi and plants. In fungi, “homothallic” means both sexes on the same thallus (bisexual); in plants, “monoecious” means male and female flowers on the same plant (bisexual). So the correct pairing is homothallic and monoecious.

   The key here is to match the biological terminology to the concept of bisexuality — having both male and female reproductive structures in the same organism. Different kingdoms use different words for the same idea, and this question tests your ability to translate across them.

   1. In fungi, the term “thallus” refers to the body of the fungus. If a single thallus produces both male and female gametes (or gametangia), it is called homothallic (“same thallus”). If two different thalli are needed — one male, one female — it is heterothallic. So the bisexual condition in fungi is homothallic.

   2. In plants, the term “monoecious” comes from Greek monos (single) and oikos (house) — meaning “one house.” A monoecious plant has separate male and female flowers on the same individual (e.g., maize, cucumber). That is the bisexual condition at the organism level. “Dioecious” (“two houses”) means male and female flowers are on separate plants (e.g., papaya, date palm), which is unisexual at the organism level. …

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6. TG EAPCET 2025Set ap-2025-04-29-AN1 markMCQ
   Q.The disease sickle cell anemia is caused by substitution of ‘A’ by ‘B’ of ‘C’ globin chain of hemoglobin molecule. Identify A, B and C respectively (A) A – Alanine, B – Glutamic acid, C – Beta (B) A – Glutamic acid, B – Valine, C – Beta (C) A – Valine, B – Glutamic acid, C – Alpha (D) A – Valine, B – Serine, C – Beta

   ›Reveal solutionSolution

   Sickle cell anemia results from a point mutation where glutamic acid is replaced by valine at the sixth position of the beta-globin chain of hemoglobin. The correct option is (B).

   Sickle cell anemia is a classic example of a genetic disorder caused by a single point mutation, leading to a change in a single amino acid in a protein. This seemingly small change has profound effects on the structure and function of hemoglobin, ultimately altering the shape of red blood cells and causing the disease's characteristic symptoms. Understanding this specific molecular change is key to grasping the pathology of sickle cell anemia.

   Here's a breakdown of the molecular basis of sickle cell anemia:

   1. Hemoglobin Structure: Hemoglobin is the oxygen-carrying protein in red blood cells. In adults, it is primarily composed of four polypeptide chains: two alpha ($\alpha$) globin chains and two beta ($\beta$) globin chains. Each chain contains a heme group that binds oxygen.

   2. The Genetic Defect: Sickle cell anemia arises from a mutation in the gene that codes for the beta-globin chain. This is a point mutation, meaning a change in a single nucleotide base pair in the DNA sequence.

   3. Identifying the Globin Chain (C): The specific mutation responsible for sickle cell anemia occurs in the beta ($\beta$) globin chain. This is why the disease is often referred to as a "beta-thalassemia" in some contexts, though sickle cell anemia has its own distinct pathology.

   4. Identifying the Amino Acid Substitution (A and B):

      • At the sixth position of the beta-globin chain, the normal amino acid is glutamic acid (A). Glutamic acid is a hydrophilic (water-loving) amino acid.
      • Due to the point mutation (a change from GAG to GTG in the mRNA codon), valine (B) is substituted in place of glutamic acid at this sixth position. Valine is a hydrophobic (water-fearing) amino acid. …

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7. TG EAPCET 2025Set ap-2025-04-29-AN1 markMCQ
   Q.Study the following and identify the correct combinations

   	Phenomenon	Phenotypic ratio	Genotypic ratio
   I	Co-dominance	$1:2:1$	$1:2:1$
   II	Incomplete dominance	$3:1$	$1:2:1$
   III	Monohybrid test cross	$1:1$	$1:1$
   IV	Dihybrid test cross	$9:3:3:1$	$1:1:1:1$

   (A) I & III (B) II & III (C) II & IV (D) I & II

   ›Reveal solutionSolution

   Only I (co-dominance: $1:2:1$ and $1:2:1$) and III (monohybrid test cross: $1:1$ and $1:1$) are stated correctly. Incomplete dominance gives a $1:2:1$ phenotypic ratio (not $3:1$), and a dihybrid test cross gives $1:1:1:1$ phenotypes (not $9:3:3:1$). Hence the correct combination is I & III — option (A).

   The concept first

   Two different things are being asked in every row:

   • The genotypic ratio comes only from which gametes combine — it is pure Punnett-square bookkeeping and is the same whatever the dominance relationship is.
   • The phenotypic ratio depends on how the alleles interact:
     • Complete dominance — the heterozygote looks like the dominant homozygote, so two genotype classes fuse and $1:2:1$ collapses to $3:1$.
     • Incomplete dominance — the heterozygote is intermediate (e.g. pink Mirabilis), so nothing fuses and the phenotypic ratio equals the genotypic ratio, $1:2:1$.
     • Co-dominance — the heterozygote shows both phenotypes together (e.g. blood group $AB$), so again nothing fuses and the phenotypic ratio equals the genotypic ratio, $1:2:1$.

   So the memorable rule is: in both incomplete dominance and co-dominance, phenotypic ratio $=$ genotypic ratio $= 1:2:1$.

   Step-by-step

   Step 1 — Row I: Co-dominance.

   Cross $I^A I^B \times I^A I^B$ (or any co-dominant heterozygote $\times$ itself). Gametes from each parent: $\tfrac12 I^A,\ \tfrac12 I^B$.

   $\text{Offspring} = \tfrac14 I^A I^A : \tfrac12 I^A I^B : \tfrac14 I^B I^B \;=\; 1:2:1 \text{ (genotypic)}$

   Because the heterozygote expresses both alleles, it is its own distinct phenotype, so the phenotypic ratio is also $1:2:1$. The row says $1:2:1$ and $1:2:1$ — correct.

   Step 2 — Row II: Incomplete dominance.

   $Rr \times Rr$ gives genotypes $RR : Rr : rr = 1:2:1$ — the row's genotypic ratio is right. But the heterozygote $Rr$ is pink, distinct from red $RR$ and white $rr$, so the phenotypes are $1$ red $:2$ pink $:1$ white $= 1:2:1$. The row prints $3:1$, which is the complete-dominance result. Row II is incorrect.

   Step 3 — Row III: Monohybrid test cross.

   A test cross is heterozygote $\times$ recessive homozygote: $Aa \times aa$. Gametes: $\tfrac12 A, \tfrac12 a$ from one side; only $a$ from the other.

   $\text{Offspring} = \tfrac12 Aa : \tfrac12 aa = 1:1$ …

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8. TG EAPCET 2025Set ap-2025-04-29-AN1 markMCQ
   Q.Consider the following statements Assertion (A): In human beings, Y-linked genes of non homologous part are called holandric genes. Reason (R): Y-chromosome is present in males only The correct answer is (A) Both (A) and (R) are true, (R) is the correct explanation of (A) (B) Both (A) and (R) are true, (R) is not the correct explanation of (A) (C) (A) is true, but (R) is false (D) (A) is false, but (R) is true

   ›Reveal solutionSolution

   Holandric genes are Y-linked genes found only in males since the Y chromosome is male-specific; the reason correctly explains why these genes show this pattern of inheritance. The correct option is (A).

   Understanding Sex Chromosomes and Gene Inheritance

   In human genetics, sex chromosomes (X and Y) determine biological sex and carry genes beyond just sex determination. The X and Y chromosomes have both homologous regions (where they can pair during meiosis) and non-homologous regions (unique to each chromosome).

   The key concept here involves understanding what happens with genes located in the non-homologous portion of the Y chromosome—the part that has no corresponding region on the X chromosome.

   Evaluating the Assertion

   Assertion (A): "In human beings, Y-linked genes of non homologous part are called holandric genes."

   Let's break this down:

   1. Holandric genes are indeed genes located on the Y chromosome, specifically in the non-homologous region (the region unique to Y that doesn't pair with X).

   2. These genes are passed exclusively from father to son, creating a strict patrilineal inheritance pattern.

   3. Examples include genes for testis development (SRY gene), some spermatogenesis factors, and certain genes controlling male-specific traits.

   This assertion is TRUE.

   Evaluating the Reason

   Reason (R): "Y-chromosome is present in males only"

   1. In humans, the typical sex chromosome composition is:

      • Males: XY (one X, one Y)
      • Females: XX (two X chromosomes, no Y)

   2. Therefore, the Y chromosome is indeed present only in males (barring rare chromosomal abnormalities).

   This reason is TRUE.

   Determining the Relationship

   Now the critical question: Does (R) correctly explain (A)? …

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9. TG EAPCET 2025Set ap-2025-04-29-FN1 markMCQ
   Q.Consider the following statements Assertion (A): In hymenopteran insects, sperms are produced by mitosis Reason (R): Male hymenopteran insects are haploid The correct answer is (A) Both (A) and (R) are true, (R) is the correct explanation of (A) (B) Both (A) and (R) are true, (R) is not the correct explanation of (A) (C) (A) is true, but (R) is false (D) (A) is false, but (R) is true

   ›Reveal solutionSolution

   In hymenopteran insects (ants, bees, wasps), males are haploid and produce sperm by mitosis, not meiosis. Both Assertion and Reason are true, and the Reason correctly explains the Assertion.

   The key here is understanding haploidy and sex determination in hymenopterans. In these insects, females develop from fertilized (diploid) eggs, while males develop from unfertilized (haploid) eggs through a process called arrhenotokous parthenogenesis. This means males have only one set of chromosomes — they are haploid.

   Now, think about what happens when a haploid organism needs to produce gametes. Normal meiosis would halve the chromosome number, but a haploid cell cannot undergo reduction division — it would produce inviable gametes with zero chromosomes. So nature has a clever workaround: male hymenopterans produce sperm by mitosis instead. Each sperm is an exact copy of the male's haploid genome, which is exactly what's needed to fertilize an egg and restore diploidy in the female offspring.

   Let's break it down step by step.

   1. Why the Reason (R) is true

      Male hymenopterans (drones in honeybees, for example) develop from unfertilized eggs. Since no sperm contributed genetic material, the egg's nucleus alone forms the embryo — it is haploid ($n$). This is a well-established biological fact.

   2. Why the Assertion (A) is true

      Because the male is haploid, it cannot undergo meiosis to produce sperm — meiosis would reduce the chromosome number to $n/2$, which is not viable. Instead, the cells in the testes divide by mitosis, producing haploid sperm that are genetically identical to the male parent. This is a direct adaptation to haploidy.

   3. How (R) explains (A) …

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10. TG EAPCET 2025Set ap-2025-04-29-FN1 markMCQ
    Q.Study the following and identify the correct statements: I. The alleles $I^A$ and $I^B$ regarding blood groups are dominant over $I^O$, and $I^A$ and $I^B$ are co-dominant II. In some fishes, reptiles and birds, females are heterogametic (ZW) and males are homogametic (ZZ). III. Genic balance theory states that Y-chromosome is essential for determination of sex in male Drosophila IV. Person with O blood group are called universal donors because their RBC contain both antigens A and B. (A) I, II (B) III, IV (C) I, III (D) II, IV

    ›Reveal solutionSolution

    The question tests four statements on genetics — blood group inheritance, sex determination in ZW systems, the genic balance theory in Drosophila, and the basis of universal donation. Statements I and II are correct; III and IV are false. The correct option is (A).

    Let’s examine each statement carefully, one by one.

    1. Statement I: The alleles $I^A$ and $I^B$ regarding blood groups are dominant over $I^O$, and $I^A$ and $I^B$ are co-dominant.

       This is textbook ABO blood group genetics. The $I^A$ and $I^B$ alleles each produce a specific antigen (A and B respectively), while $I^O$ produces no functional antigen. Both $I^A$ and $I^B$ are dominant over $I^O$, so a person with genotype $I^A I^O$ has blood group A, and $I^B I^O$ gives group B. When both $I^A$ and $I^B$ are present together, both antigens are expressed equally — that is co-dominance, giving blood group AB.

       Statement I is correct.

    2. Statement II: In some fishes, reptiles and birds, females are heterogametic (ZW) and males are homogametic (ZZ).

       In the ZW sex-determination system, the female has two different sex chromosomes (ZW) and the male has two identical ones (ZZ). This is indeed found in birds, many reptiles, and some fishes. The opposite (XX/XY) system is typical in mammals and Drosophila.

       Statement II is correct.

    3. Statement III: Genic balance theory states that Y-chromosome is essential for determination of sex in male Drosophila.

       This is a classic trap. In Drosophila melanogaster, sex is determined not by the presence of a Y chromosome, but by the ratio of X chromosomes to autosomes (the X:A ratio). The Y chromosome in Drosophila is required only for male fertility, not for male determination. The genic balance theory, proposed by Calvin Bridges, explicitly states that the Y chromosome is not essential for maleness — a fly with XXY is female, and a fly with XO (no Y) is male but sterile.

       Statement III is false. …

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11. TG EAPCET 2025Set ap-2025-04-30-FN1 markMCQ
    Q.Consider the following statements Assertion (A): Skin colour in human beings is a polygenetic trait. Reason (R): Human skin colour is controlled by cumulative effect of genes. The correct answer is (A) Both (A) and (R) are true, (R) is the correct explanation of (A) (B) Both (A) and (R) are true, (R) is not the correct explanation of (A) (C) (A) is true, but (R) is false (D) (A) is false, but (R) is true

    ›Reveal solutionSolution

    Skin colour is polygenic because multiple genes act cumulatively/additively; R explains A.

    Assertion: Skin colour in human beings is a polygenic trait - a phenotype governed by more than one gene, showing a continuous gradation of expression rather than discrete classes. …

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12. TG EAPCET 2024Set ap-2024-05-07-AN1 markMCQ
    Q.Observe the phenotypic ratios given below A. $9:3:3:1$ B. $1:2:1$ C. $1:1$ Arrange the ratios in the order of monohybrid test cross; incomplete dominance; and dihybrid cross (A) B-C-A (B) C-B-A (C) B-A-C (D) A-B-C

    ›Reveal solutionSolution

    A monohybrid test cross gives $1:1$ (ratio C), incomplete dominance gives $1:2:1$ (ratio B) and a dihybrid cross gives $9:3:3:1$ (ratio A). In the order asked, that is C – B – A, i.e. option (B).

    The concept first

    Ratios in genetics are not things to be memorised as trivia — each one is generated by a specific set of gametes, and if you can build the Punnett square you never have to remember anything.

    Two questions decide every ratio:

    1. How many gene pairs are segregating? One → monohybrid arithmetic. Two → multiply the two monohybrid outcomes (Law of Independent Assortment).
    2. Is the heterozygote phenotypically identical to the dominant homozygote, or does it look different? If identical → complete dominance, and genotypes collapse into fewer phenotypes. If different → incomplete dominance, and no collapsing happens.

    Step-by-step

    1. Monohybrid test cross → ratio C ($1:1$). A test cross means crossing an individual of unknown genotype with the homozygous recessive. For a heterozygote:

    $Tt \times tt$

    Gametes: $T$ and $t$ from the first parent; only $t$ from the second.

    Offspring: $\tfrac12\,Tt$ (tall) and $\tfrac12\,tt$ (dwarf).

    $\therefore \text{phenotypic ratio} = 1:1 \;\Rightarrow\; \mathbf{C}$

    This is exactly why a test cross works — a heterozygote betrays itself by producing recessive offspring.

    2. Incomplete dominance → ratio B ($1:2:1$). In Mirabilis jalapa (four o'clock plant), $RR$ = red, $rr$ = white, and $Rr$ is pink — its own distinct phenotype. Selfing the $F_1$:

    $Rr \times Rr \;\rightarrow\; 1\,RR : 2\,Rr : 1\,rr \;=\; 1\ \text{red} : 2\ \text{pink} : 1\ \text{white}$

    Because the heterozygote is visibly different, the $2\,Rr$ class does not merge with the $1\,RR$ class. So the phenotypic ratio equals the genotypic ratio:

    $1:2:1 \;\Rightarrow\; \mathbf{B}$ …

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