Mendel's laws, dominance patterns, dihybrid cross, linkage, polygenic inheritance, pleiotropy, sex determination, mutation and genetic disorders — with Punnett squares, SVG diagrams, MCQs, A-R and PYQs.
Genetics deals with inheritance and variation. Inheritance is the process by which characters pass from parent to progeny (basis of heredity). Variation is the degree by which progeny differ from their parents.
Humans knew from about 8000–1000 B.C. that sexual reproduction is a cause of variation and used it for selective breeding (e.g. Sahiwal cows of Punjab).
| S.No. | Character | Dominant / Recessive |
|---|---|---|
| 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 |
Crossing an organism of dominant phenotype (unknown genotype) with the recessive parent to determine its genotype.
| Law | Statement |
|---|---|
| I. Dominance | (i) Characters are controlled by discrete factors; (ii) factors occur in pairs; (iii) in a dissimilar pair one factor (dominant) masks the other (recessive). Explains F₁ expression and 3:1 in F₂. |
| II. Segregation | Alleles do not blend; during gamete formation (meiosis) the two alleles of a pair separate so each gamete gets only one. Homozygous → one gamete type; heterozygous → two types in equal proportion. (Law of purity of gametes.) |
The unmodified allele makes a normal enzyme; the modified (recessive) allele makes a non-functional enzyme or none. The phenotype then depends on the functioning allele, which is dominant.
F₁ is intermediate between parents. Example: snapdragon (Antirrhinum) flower colour — RR (red) × rr (white) → Rr pink; F₂ = 1 Red : 2 Pink : 1 White. Genotypic and phenotypic ratios are both 1:2:1.
| Genotype | Blood group |
|---|---|
| IAIA, IAi | A |
| IBIB, IBi | B |
| IAIB | AB |
| ii | O |
RRYY (round, yellow) × rryy (wrinkled, green) → F₁ RrYy (all round, yellow). F₁ selfed gives F₂ phenotypes in 9 : 3 : 3 : 1.
= 9 Round-Yellow : 3 Wrinkled-Yellow : 3 Round-Green : 1 Wrinkled-Green. F₂ has 9 genotypes (ratio 1:2:1:2:4:2:1:2:1) and 4 phenotypes.
When two pairs of traits are combined in a hybrid, segregation of one pair is independent of the other pair. F₁ makes 4 gamete types — RY, Ry, rY, ry — each ¼.
Why Mendel was ignored until 1900: poor communication; the idea of discrete, non-blending factors did not fit continuous variation; use of mathematics was new; no physical proof of what factors were.
A single gene shows multiple phenotypic effects (acts on metabolic pathways). Example: phenylketonuria (PKU) — mutation in the gene for phenylalanine hydroxylase causes mental retardation and reduced hair and skin pigmentation.
Henking (1891) found a nuclear structure (X body) in 50% of sperms in insects — later called the X chromosome. Chromosomes that differ between sexes are sex chromosomes; the rest are autosomes.
| Type | Male | Female | Example |
|---|---|---|---|
| XO (male heterogamety) | Autosomes + X (one X) | Autosomes + XX | Grasshopper |
| XY (male heterogamety) | Autosomes + XY | Autosomes + XX | Humans, Drosophila |
| ZW (female heterogamety) | ZZ | ZW | Birds |
| Haplodiploid | Haploid (16) | Diploid (32) | Honey bee |
Mutation is an alteration in DNA sequence that changes the genotype and phenotype. With recombination, it is a source of variation.
Analysis of a trait over several generations of a family, drawn as a family tree. It shows whether a trait is dominant or recessive, autosomal or sex-linked, since controlled crosses are not possible in humans.
Due to alteration in a single gene; follow Mendelian inheritance. Examples: haemophilia, cystic fibrosis, sickle-cell anaemia, colour blindness, phenylketonuria, thalassemia.
| Disorder | Inheritance | Key facts |
|---|---|---|
| Colour blindness | X-linked recessive | Red-green discrimination fails; defective red/green cone genes on X. About 8% of males, 0.4% of females. Carrier mother → 50% sons affected. |
| Haemophilia | X-linked recessive | A protein in the clotting cascade is affected → non-stop bleeding. Carrier female transmits to sons; affected female is extremely rare. Queen Victoria was a carrier. |
| Sickle-cell anaemia | Autosomal recessive | Alleles HbA, HbS; only HbSHbS is diseased; HbAHbS is a carrier. Point mutation: Glu → Val at 6th position of β-globin. Two carriers → 50% chance of passing the gene. |
| Phenylketonuria | Autosomal recessive | Lack of phenylalanine hydroxylase; mental retardation (pleiotropy). |
Caused by absence, excess or abnormal arrangement of chromosomes. Aneuploidy = gain/loss of one chromosome (failure of chromatid segregation) — trisomy or monosomy. Polyploidy = extra whole set(s) of chromosomes (failure of cytokinesis after telophase) — common in plants. Normal human cell: 46 (23 pairs; 22 autosomal + 1 sex pair).
| Syndrome | Cause | Features |
|---|---|---|
| Down's | Trisomy of chromosome 21 (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 and mental development. |
| Klinefelter's | Extra X — 47, XXY | Overall masculine, but gynaecomastia (breast development); sterile. |
| Turner's | Loss of one X — 45, XO (females) | Rudimentary ovaries, sterile, lack of secondary sexual characters, short stature. |
Q1. A diploid organism is heterozygous for 4 loci. How many types of gametes can it produce?
Each heterozygous locus gives 2 allele types, assorting independently: gamete types = 2n.
n = 4 → 24 = 16.
Q2. TtYy (tall, yellow) × Ttyy (tall, green). What proportion of offspring is (a) tall and green, (b) dwarf and green?
Tt × Tt → ¾ tall : ¼ dwarf. Yy × yy → ½ yellow : ½ green.
¾ × ½ = 3/8.
¼ × ½ = 1/8.
Q3. A child has blood group O. Father has group A, mother group B. Give the parents' genotypes and the possible genotypes of other children.
Child is ii, so each parent gave one i. Father = IAi, mother = IBi.
Father: IA, i. Mother: IB, i.
IAIB (AB), IAi (A), IBi (B), ii (O) — each ¼.
(A) Both true, R explains A. (B) Both true, R does NOT explain A. (C) A true, R false. (D) A false, R true.
Q. Mention the advantages of selecting pea plant for Mendel's experiments.
Many clear contrasting traits (7 pairs); easy to grow, short life cycle, many seeds per plant; bisexual flowers that normally self-pollinate, yet can be artificially cross-pollinated; true-breeding lines available.
Q. Define and design a test cross.
Cross of an individual of dominant phenotype with the homozygous recessive parent. Tt × tt → ½ Tt (tall) : ½ tt (dwarf) = 1 : 1; TT × tt → all Tt (tall).
Q. Differentiate between incomplete dominance and co-dominance with examples.
Incomplete dominance: F₁ is intermediate (snapdragon: Rr pink; F₂ 1:2:1). Co-dominance: F₁ shows both parental traits (ABO group: IAIB expresses both A and B sugars).
Q. How is sex determined in human beings?
XY type: female 44 + XX, male 44 + XY. Sperms are 50% X and 50% Y; ova all X. Fertilisation by X-sperm → girl (XX); by Y-sperm → boy (XY). So the sperm determines the sex.
Q. What is pedigree analysis? How is it useful?
Study of inheritance of a trait over several generations of a family using a family tree. It shows whether a trait is dominant/recessive and autosomal/sex-linked, and helps trace carriers of genetic disorders.