📌 Snapshot
- Builds the entire classical-genetics scaffold of CUET Biology: Mendel's three laws, monohybrid (3:1) and dihybrid (9:3:3:1) ratios, and the Punnett-square machinery for predicting genotypes/phenotypes.
- Extends Mendelism beyond simple dominance into incomplete dominance (Snapdragon, 1:2:1 phenotype), co-dominance with multiple alleles (ABO blood groups, I^A I^B i), pleiotropy (phenylketonuria), and polygenic inheritance (human skin colour, height).
- Anchors genes to chromosomes through Sutton–Boveri's Chromosomal Theory (1902) and Morgan's Drosophila work on linkage and recombination, including Sturtevant's genetic mapping using recombination frequency.
- Covers all four major sex-determination systems — XO (grasshopper), XY (humans, Drosophila), ZW (birds), and haplodiploidy (honey bees) — plus mutation (point mutations, chromosomal aberrations) and pedigree analysis symbols.
- Closes with disease genetics: Mendelian disorders (haemophilia, sickle-cell anaemia, phenylketonuria, thalassaemia, colour blindness) and chromosomal disorders with karyotypes (Down's — trisomy 21; Klinefelter — 47, XXY; Turner — 45, XO). The single most-tested chapter in CUET Biology.
📖 Detailed Notes
2.1 Core concepts
- Genetics is the branch of biology dealing with inheritance (transmission of characters from parent to progeny) and variation (degree by which progeny differ from parents) (NCERT §4 intro, p. 54).
- Gregor Mendel conducted hybridisation experiments on garden peas for seven years (1856–1863); he used large samples and applied statistical/mathematical logic to biology for the first time and selected 14 true-breeding lines as 7 pairs with contrasting traits (NCERT §4.1, p. 54).
- The seven pairs of contrasting traits Mendel studied: stem height (tall/dwarf), flower colour (violet/white), flower position (axial/terminal), pod shape (inflated/constricted), pod colour (green/yellow), seed shape (round/wrinkled), seed colour (yellow/green) (NCERT §4.1 Table 4.1, p. 55).
- A monohybrid cross (Tt × Tt) gives an F2 phenotypic ratio of 3:1 (tall:dwarf) and a genotypic ratio of 1:2:1 (TT:Tt:tt); the F1 is all Tt and phenotypically tall (NCERT §4.2, pp. 56–58).
- The Punnett square, developed by Reginald C. Punnett, graphically represents probabilities of all possible offspring genotypes by writing gametes on the top row and left column (NCERT §4.2, p. 57).
- A test cross is a cross between an organism showing dominant phenotype (unknown genotype) and the homozygous recessive parent; it is used to determine whether the dominant individual is homozygous (TT) or heterozygous (Tt) (NCERT §4.2, p. 58).
- Law of Dominance (Mendel's First Law): characters are controlled by discrete units called factors that occur in pairs; in a dissimilar pair one factor dominates the other; explains the 3:1 F2 phenotypic ratio (NCERT §4.2.1, p. 59).
- Law of Segregation (Second Law): alleles do not blend; during gamete formation the two alleles segregate so that each gamete carries only one of them; a heterozygote produces two kinds of gametes in equal proportion (NCERT §4.2.2, p. 59).
- Incomplete dominance: in snapdragon (Antirrhinum) / Mirabilis-type cross RR (red) × rr (white) → F1 Rr is pink, F2 is 1 red : 2 pink : 1 white; phenotypic ratio (1:2:1) equals genotypic ratio because R is not fully dominant over r (NCERT §4.2.2.1, p. 60).
- Co-dominance: in human ABO blood groups, gene I has three alleles I^A, I^B and i; I^A and I^B are both dominant over i but co-dominant with each other, so I^A I^B genotype expresses both A and B sugars on RBCs (AB blood type); 6 genotypes give 4 phenotypes (A, B, AB, O) (NCERT §4.2.2.2 Table 4.2, p. 61).
- Multiple alleles: ABO blood grouping illustrates more than two alleles for one gene in a population, though any individual carries only two (NCERT §4.2.2.2, p. 62).
- Pleiotropy (incomplete-dominance angle): starch synthesis in pea seeds is controlled by one gene (B, b) — BB produces large round seeds, bb produces small wrinkled seeds, Bb produces round seeds with intermediate-sized starch grains; so the same gene shows complete dominance for shape but incomplete dominance for grain size (NCERT §4.2.2.2, p. 62).
- A dihybrid cross RrYy × RrYy gives an F2 phenotypic ratio of 9 round-yellow : 3 wrinkled-yellow : 3 round-green : 1 wrinkled-green (NCERT §4.3 and §4.3.1, pp. 63–64).
- Law of Independent Assortment: when two pairs of traits are combined in a hybrid, segregation of one pair is independent of the other; gametes formed by RrYy are RY, Ry, rY, ry each at 25% (NCERT §4.3.1, p. 64).
- Chromosomal Theory of Inheritance: Walter Sutton and Theodore Boveri (1902) noted that the behaviour of chromosomes during meiosis parallels the behaviour of genes — both occur in pairs, both segregate at gamete formation, and independent pairs assort independently (NCERT §4.3.2 Table 4.3, p. 66).
- Morgan & Drosophila: T.H. Morgan used the fruit fly Drosophila melanogaster (simple medium, ~2-week life cycle, large progeny, distinguishable sexes) to experimentally verify the chromosomal theory (NCERT §4.3.2, p. 67).
- Linkage and recombination: when two genes lie on the same chromosome, parental combinations exceed non-parental (recombinant) combinations; Morgan coined "linkage" for physical association and "recombination" for generation of non-parental combinations. y–w showed 1.3% recombination (tightly linked) while w–m showed 37.2% (loosely linked); Sturtevant used recombination frequency to map gene positions (NCERT §4.3.3, pp. 67–68).
- Polygenic inheritance: traits like human skin colour and height are controlled by three or more genes with additive allelic effects and environmental influence; e.g., AABBCC = darkest, aabbcc = lightest skin (NCERT §4.4, p. 69).
- Pleiotropy: a single gene producing multiple phenotypic effects — phenylketonuria is caused by mutation in the gene coding for phenylalanine hydroxylase and manifests as mental retardation plus reduced hair and skin pigmentation (NCERT §4.5, p. 69).
- Sex determination — XO type (grasshopper, insects): males have only one X-chromosome (besides autosomes); females have two X-chromosomes; male-heterogamety (NCERT §4.6, p. 70).
- XY type (humans, Drosophila): males XY, females XX; male-heterogamety; both sexes have same total chromosome number; humans have 22 pairs of autosomes plus one pair of sex chromosomes (NCERT §4.6 / §4.6.1, p. 70–71).
- ZW type (birds): females ZW, males ZZ; female-heterogamety (NCERT §4.6, p. 71).
- Haplodiploidy (honey bees): females (queen/worker) develop from fertilised eggs and are diploid (32 chromosomes); males (drones) develop from unfertilised eggs by parthenogenesis and are haploid (16 chromosomes); drones have no father, cannot have sons, but have a grandfather and grandsons (NCERT §4.6.2, p. 71).
- Mutation: alteration of DNA sequence resulting in change in genotype and phenotype; arises from deletions/insertions/duplications of DNA segments (chromosomal aberrations, common in cancer cells) or single-base-pair changes (point mutations, e.g., sickle-cell anaemia); UV radiation is a mutagen (NCERT §4.7, p. 72).
- Pedigree analysis: study of inheritance of a trait across generations in a family tree, since controlled crosses are impossible in humans; uses standard symbols (NCERT §4.8.1, p. 72).
- Colour blindness: X-linked recessive defect in red or green cone; ~8% of males and ~0.4% of females are affected because the gene is on X chromosome; mother is unaffected carrier if heterozygous (NCERT §4.8.2, p. 73).
- Haemophilia: X-linked recessive; one of the clotting-cascade proteins is defective causing non-stop bleeding from a simple cut; heterozygous female (carrier) transmits to sons; female homozygotes are extremely rare because the father has to be haemophilic; Queen Victoria's family pedigree shows multiple haemophilic descendants (NCERT §4.8.2, p. 74).
- Sickle-cell anaemia: autosomal recessive; controlled by alleles Hb^A and Hb^S; only Hb^SHb^S shows the disease, Hb^AHb^S are carriers; caused by substitution of Glutamic acid by Valine at the sixth position of the β-globin chain (single base substitution GAG → GUG at the sixth codon); mutant haemoglobin polymerises under low oxygen tension changing RBC shape from biconcave to sickle-like (NCERT §4.8.2, pp. 74–75).
- Phenylketonuria: autosomal recessive inborn error of metabolism; affected individual lacks the enzyme that converts phenylalanine to tyrosine; accumulation of phenylalanine and phenylpyruvic acid in brain causes mental retardation; excreted via urine (NCERT §4.8.2, p. 75).
- Thalassaemia: autosomal recessive; reduced synthesis of one of the globin chains; α-thalassaemia controlled by two closely linked genes HBA1 and HBA2 on chromosome 16; β-thalassaemia controlled by single gene HBB on chromosome 11; differs from sickle-cell anaemia in being a quantitative (too few globin molecules) rather than qualitative defect (NCERT §4.8.2, p. 75).
- Chromosomal disorders: caused by absence/excess/abnormal arrangement of chromosomes; aneuploidy from failure of segregation of chromatids during cell division (gain/loss of chromosome); polyploidy from failure of cytokinesis (whole set added) (NCERT §4.8.3, p. 75).
- Down's syndrome: trisomy of chromosome 21 (extra copy of chromosome 21, total 47); first described by Langdon Down (1866); short stature, small round head, furrowed tongue, partially open mouth, broad palm with characteristic crease, retarded physical/psychomotor/mental development (NCERT §4.8.3, p. 76).
- Klinefelter's syndrome: karyotype 47, XXY (extra X); overall masculine development but gynaecomastia (breast development); sterile (NCERT §4.8.3, p. 76).
- Turner's syndrome: karyotype 45, X0 (one X chromosome missing); sterile females with rudimentary ovaries; lack secondary sexual characters (NCERT §4.8.3, p. 76).
2.2 Definitions to memorise
| Term | Definition | Page |
|---|---|---|
| Inheritance | Process by which characters are passed on from parent to progeny; basis of heredity | 54 |
| Variation | Degree by which progeny differ from their parents | 54 |
| True-breeding line | Line that, after continuous self-pollination, shows stable trait inheritance for several generations | 54 |
| Allele | Slightly different forms of the same gene; code for a pair of contrasting traits | 56 |
| Genotype / Phenotype | Genetic constitution (TT/Tt/tt) vs descriptive appearance (tall/dwarf) | 56 |
| Homozygous / Heterozygous | Identical allelic pair (TT or tt) vs dissimilar (Tt) | 56 |
| Dominant / Recessive | In a dissimilar pair, the factor that dominates vs the one whose expression is masked | 56 |
| Monohybrid cross | Cross between two plants differing in one character (e.g., TT × tt) | 57 |
| Punnett square | Graphical tabular device by R.C. Punnett to calculate probabilities of offspring genotypes | 57 |
| Test cross | Cross of an organism showing dominant phenotype with the homozygous recessive parent to determine its genotype | 58 |
| Incomplete dominance | F1 phenotype is intermediate between the two parents (e.g., pink Rr) | 60 |
| Co-dominance | F1 expresses both parental phenotypes (e.g., AB blood from I^AI^B) | 60–61 |
| Multiple alleles | More than two alleles of a gene in a population (e.g., I^A, I^B, i) | 62 |
| Pleiotropy | A single gene producing multiple phenotypic effects (e.g., phenylketonuria) | 69 |
| Dihybrid cross | Cross between plants differing in two traits (RrYy × RrYy → 9:3:3:1) | 64 |
| Law of Independent Assortment | When two pairs of traits are combined in a hybrid, segregation of one pair is independent of the other | 64 |
| Chromosomal Theory of Inheritance | Sutton-Boveri synthesis: chromosomes carry the genes; their pairing/separation explains Mendel's laws | 67 |
| Linkage | Physical association of two or more genes on the same chromosome | 67 |
| Recombination | Generation of non-parental gene combinations through crossing over | 67 |
| Polygenic trait | Trait controlled by three or more genes, showing additive allelic effects and continuous variation | 69 |
| Point mutation | Change in a single base pair of DNA (e.g., GAG → GUG in sickle-cell anaemia) | 72 |
| Mutagen | Chemical/physical factor that induces mutation; UV radiation is a mutagen | 72 |
| Pedigree analysis | Study of inheritance of a particular trait in a family across generations | 72 |
| Aneuploidy | Gain or loss of a chromosome due to failure of segregation of chromatids | 75 |
| Polyploidy | Increase in a whole set of chromosomes due to failure of cytokinesis | 75 |
| Trisomy / Monosomy | Presence of an extra chromosome (2n+1) / absence of one chromosome (2n−1) | 76 |
2.3 Diagrams / processes to remember
- Figure 4.1: Seven pairs of contrasting traits Mendel studied in pea (stem height, flower colour, flower position, pod shape, pod colour, seed shape, seed colour), p. 54.
- Figure 4.2: Steps in making a cross in pea — emasculation, dusting pollen, bagging, p. 55.
- Figure 4.3: Monohybrid cross diagrammatic representation (TT × tt → F1 all Tt tall → F2 3:1 phenotype, 1:2:1 genotype), p. 56.
- Figure 4.4: Punnett square for monohybrid cross between true-breeding tall and dwarf plants, p. 57.
- Figure 4.5: Test cross of violet (V) × white (v) showing 1:1 ratio, p. 59.
- Figure 4.6: Monohybrid cross in Snapdragon with incomplete dominance — F2 ratio 1 red : 2 pink : 1 white, p. 60.
- Figure 4.7: Dihybrid cross with seed colour and seed shape — F2 9:3:3:1 grid (16 squares), p. 63.
- Figure 4.8: Meiosis and germ-cell formation in a four-chromosome cell, p. 65.
- Figure 4.9: Independent assortment of chromosomes — two possibilities of metaphase alignment, p. 66.
- Figure 4.10: Drosophila melanogaster male and female, p. 67.
- Figure 4.11: Morgan's two dihybrid crosses showing tight linkage (y–w: 1.3% recombination) vs loose linkage (w–m: 37.2%), p. 68.
- Figure 4.12: Sex determination — (a) human XX/XY (b) Drosophila XX/XY (c) bird ZZ/ZW, p. 70.
- Figure 4.13: Sex determination in honey bee (haplodiploid; diploid females 32, haploid males 16), p. 71.
- Figure 4.13 (also): Symbols used in human pedigree analysis (male = square, female = circle, affected = shaded, mating = horizontal line, sibship = vertical descent), p. 72.
- Figure 4.14: Pedigrees of (a) autosomal dominant trait — myotonic dystrophy and (b) autosomal recessive trait — sickle-cell anaemia, p. 73.
- Figure 4.15: Micrograph of normal vs sickle-cell RBCs and the amino-acid composition of β-chain showing Glu→Val substitution at position 6, p. 74.
- Figure 4.16: Down's syndrome — flat back of head, broad flat face, big wrinkled tongue, palm crease, congenital heart disease; with trisomy-21 karyotype, p. 76.
- Figure 4.17: (a) Klinefelter's syndrome (tall, feminised character, gynaecomastia, 47 XXY); (b) Turner's syndrome (short stature, underdeveloped feminine character, 45 X0), p. 76.
2.4 Common confusions / NTA trap points
- Phenotypic vs genotypic ratios. In a normal monohybrid Tt × Tt cross, phenotypic is 3:1 but genotypic is 1:2:1; in incomplete dominance (Snapdragon RR × rr), both phenotypic and genotypic ratios are 1:2:1. NTA loves swapping these.
- Co-dominance vs incomplete dominance. Co-dominance — both alleles express their distinct phenotypes simultaneously (AB blood). Incomplete dominance — the heterozygote shows an intermediate/blended phenotype (pink Snapdragon).
- Multiple alleles vs polygenic inheritance. Multiple alleles = more than two alleles of the same gene (I^A, I^B, i for ABO). Polygenic = many different genes acting additively on the same trait (skin colour, height).
- X-linked recessive disorders sex bias. Colour blindness and haemophilia are far more common in males because they have only one X — a single recessive allele expresses. NTA distractors will swap the % frequencies (8% male / 0.4% female for colour blindness).
- Sickle-cell anaemia substitution direction. The mutation is GAG → GUG (codon change) and Glu → Val (amino-acid change) at the sixth position of the β-globin chain. Wrong-position (4th, 7th) or wrong-chain (α-chain) distractors are classic NTA traps.
- Down's syndrome karyotype. Trisomy of chromosome 21 → 47 chromosomes total (NOT 47, XX or 47, XY notation alone — the autosomal 21 is the extra one). Don't confuse with Klinefelter (47, XXY — sex-chromosome trisomy).
- Turner vs Klinefelter karyotypes. Turner = 45, X0 (female, missing X). Klinefelter = 47, XXY (male, extra X). Both are sterile, but one has a chromosome missing and the other has an extra one.
- Honey bee haplodiploidy. Males are haploid (16, from unfertilised egg), females are diploid (32). "Drones produce sperm by mitosis" and "drones have no father" are favourite assertion-reason hooks.
- Linkage strength vs recombination frequency. Tightly linked genes show low recombination frequency (y-w: 1.3%), loosely linked show high (w-m: 37.2%). Distractors will invert this relationship.
🎯 Practice MCQs
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Q1. For how many years did Gregor Mendel conduct his hybridisation experiments on garden pea, and during which period?
▸ Show answer & explanation
Answer: B
The NCERT explicitly states that Mendel conducted hybridisation experiments on garden peas for seven years (1856–1863) before proposing his laws of inheritance.
Q2. Which of the following pairs of contrasting traits was **NOT** among the seven studied by Mendel in pea plants?
▸ Show answer & explanation
Answer: C
Mendel's seven contrasting traits were stem height, flower colour, flower position, pod shape, pod colour, seed shape and seed colour. Leaf shape is not in Table 4.1.
Q3. In a monohybrid cross between true-breeding tall (TT) and dwarf (tt) pea plants, the F2 progeny show the following ratios:
▸ Show answer & explanation
Answer: B
In a Tt × Tt cross the F2 shows a 3:1 phenotypic ratio (tall:dwarf) and a 1:2:1 genotypic ratio (TT:Tt:tt). Option A inverts these.
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Q4. A test cross in pea was performed between a tall plant of unknown genotype and a dwarf plant. The progeny showed tall and dwarf plants in a 1:1 ratio. The genotype of the tall parent was:
▸ Show answer & explanation
Answer: B
A test cross uses the homozygous recessive parent (tt). If progeny is 1:1 tall:dwarf the unknown plant must be Tt (heterozygous); a TT parent would give 100% tall progeny.
Q5. The Punnett square as a graphical device to calculate probabilities of offspring genotypes was developed by:
▸ Show answer & explanation
Answer: C
The Punnett square was developed by Reginald C. Punnett, a British geneticist. Sutton and Boveri are associated with the chromosomal theory of inheritance, and Morgan with linkage and recombination.
Q6. **Assertion (A):** In a cross between a true-breeding red-flowered Snapdragon (RR) and a true-breeding white-flowered plant (rr), the F2 shows a phenotypic ratio of 1:2:1. **Reason (R):** In incomplete dominance, R is not completely dominant over r, so the heterozygote Rr is phenotypically distinct (pink) from both homozygotes.
▸ Show answer & explanation
Answer: A
The F2 phenotypic ratio in Snapdragon is 1 red : 2 pink : 1 white precisely because R is incompletely dominant — Rr shows a phenotype distinct from both RR and rr.
Q7. Match the following ABO blood-group genotypes (Column I) with their phenotypes (Column II): | Column I (Genotype) | Column II (Blood type) | |---|---| | P. I^A i | i. AB | | Q. I^A I^B | ii. O | | R. I^B I^B | iii. A | | S. ii | iv. B |
▸ Show answer & explanation
Answer: A
I^A is dominant over i (so I^A i = A); I^A I^B are co-dominant (so I^A I^B = AB); I^B I^B = B; ii = O. The remaining options shuffle these incorrectly.
Q8. The starch grain size in pea seeds is controlled by the gene B/b. BB seeds are round with large starch grains, bb seeds are wrinkled with small starch grains, and Bb seeds are round but with intermediate-sized starch grains. This best illustrates that:
▸ Show answer & explanation
Answer: B
NCERT explicitly states that dominance is not an autonomous feature of a gene — for seed shape Bb is round (dominance) but for starch-grain size Bb is intermediate (incomplete dominance). Option C is wrong because co-dominance is not involved.
Q9. A cross is made between two pea plants of genotype RrYy × RrYy. What fraction of the F2 progeny will be phenotypically round and green?
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Answer: B
In the 9:3:3:1 dihybrid F2, the four phenotypic classes are 9 round-yellow, 3 wrinkled-yellow, 3 round-green and 1 wrinkled-green. Round-green therefore = 3/16.
Q10. When a heterozygous tall plant with yellow seeds (TtYy) is crossed with a tall plant with green seeds (Ttyy), what is the expected proportion of "dwarf and green" offspring?
▸ Show answer & explanation
Answer: B
Probability of dwarf (tt) from Tt × Tt = 1/4; probability of green (yy) from Yy × yy = 1/2. Combined probability = 1/4 × 1/2 = 1/8.
Q11. The Chromosomal Theory of Inheritance, which united Mendelian principles with the behaviour of chromosomes during meiosis, was proposed by:
▸ Show answer & explanation
Answer: B
Walter Sutton and Theodore Boveri (1902) noted the parallel between chromosome behaviour and gene behaviour and called the synthesis the Chromosomal Theory of Inheritance.
Q12. In Morgan's Drosophila experiments, the genes "white" (w) and "yellow" (y) showed 1.3% recombination, while "white" (w) and "miniature" (m) showed 37.2% recombination. This indicates that:
▸ Show answer & explanation
Answer: B
Recombination frequency reflects the distance between linked genes. Low frequency (1.3% for y–w) means tight linkage (close together); higher frequency (37.2% for w–m) means looser linkage (farther apart). Sturtevant used this to map genes.
Q13. Which of the following statements about polygenic inheritance is **NOT** correct?
▸ Show answer & explanation
Answer: C
NCERT specifies that polygenic inheritance also takes into account the influence of environment, in addition to multiple genes with additive effects. Options A, B and D are correct.
Q14. Phenylketonuria is caused by mutation in the gene coding for phenylalanine hydroxylase, leading to mental retardation along with reduced hair and skin pigmentation. This best exemplifies:
▸ Show answer & explanation
Answer: C
Pleiotropy refers to a single gene producing multiple phenotypic expressions, often via metabolic-pathway effects. Phenylketonuria is the NCERT-cited example of a pleiotropic gene.
Q15. Match the organism (Column I) with its sex-determination system (Column II): | Column I | Column II | |---|---| | P. Grasshopper | i. ZZ / ZW | | Q. Drosophila | ii. XX / XO | | R. Bird | iii. Haplodiploid | | S. Honey bee | iv. XX / XY |
▸ Show answer & explanation
Answer: A
Grasshopper shows XO type (male has one X, female has XX); Drosophila has XX/XY; birds have ZZ/ZW; honey bees show haplodiploid sex determination (diploid females, haploid males).
Q16. **Assertion (A):** In honey bees, male drones do not have a father but do have a grandfather. **Reason (R):** Drones develop from unfertilised eggs by parthenogenesis and produce sperm by mitosis, so they are haploid with 16 chromosomes.
▸ Show answer & explanation
Answer: A
Because drones develop parthenogenetically from unfertilised (haploid) eggs, they have no father; the queen who produced their egg had a father, who is the drone's grandfather. R correctly explains A.
Q17. The sickle-cell anaemia mutation is caused by:
▸ Show answer & explanation
Answer: B
NCERT explicitly states that sickle-cell anaemia is due to substitution of Glutamic acid (Glu) by Valine (Val) at the sixth position of the **β-globin** chain, owing to a single base change GAG → GUG. Options A, C and D mis-state the chain, direction or codon.
Q18. A phenotypically normal couple has a child with sickle-cell anaemia. The genotypes of the parents must therefore be:
▸ Show answer & explanation
Answer: C
Sickle-cell anaemia (Hb^SHb^S) appears only in homozygous recessive offspring. Both phenotypically normal parents must therefore be heterozygous carriers (Hb^A Hb^S × Hb^A Hb^S) for there to be a 1/4 chance of an affected child.
Q19. Colour blindness shows the following pattern of occurrence in humans:
▸ Show answer & explanation
Answer: B
NCERT states that red-green colour blindness occurs in about 8% of males and only about 0.4% of females, because the gene is recessive and X-linked — males with one X show the trait whenever the allele is present.
Q20. A phenotypically normal woman whose father was haemophilic marries a normal man. The probability that their son is haemophilic is:
▸ Show answer & explanation
Answer: C
A woman whose father was haemophilic is an obligate carrier (X^H X^h). Crossing with a normal X^H Y father, sons receive either X^H (normal) or X^h (haemophilic) from her with equal probability — 50% chance of being haemophilic.
Q21. Match the genetic disorder (Column I) with its karyotype (Column II): | Column I | Column II | |---|---| | P. Down's syndrome | i. 45, X0 | | Q. Klinefelter's syndrome | ii. Trisomy of chromosome 21 (47) | | R. Turner's syndrome | iii. 47, XXY |
▸ Show answer & explanation
Answer: A
Down's syndrome = trisomy 21 (47 chromosomes); Klinefelter's = 47, XXY (extra X in a male, leading to gynaecomastia and sterility); Turner's = 45, X0 (missing X, sterile female).
Q22. Which of the following statements about thalassaemia is **CORRECT**?
▸ Show answer & explanation
Answer: B
Thalassaemia is autosome-linked recessive and results from a quantitative deficit in globin synthesis. α-thalassaemia is controlled by HBA1 and HBA2 on chromosome 16, while β-thalassaemia is controlled by HBB on chromosome 11 — option C swaps these. Sickle-cell anaemia is a qualitative defect (incorrectly functioning globin) — option D is wrong.
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