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Principles of Inheritance and Variation carries some of the most reliable marks in NEET Biology, because its questions follow patterns you can practise. These notes cover Mendel's monohybrid and dihybrid crosses, the exceptions to his laws, linkage, sex determination, pedigree analysis and genetic disorders, with the problems worked step by step.
The shortcut that solves most cross problems
Treat each gene separately, find its probability, then multiply. For $Aa \times Aa$, the chance of $aa$ is $\tfrac{1}{4}$; for two independent genes, the chance of $aabb$ is $\tfrac{1}{4} \times \tfrac{1}{4} = \tfrac{1}{16}$. You rarely need to draw a 16-box Punnett square.
1. Terms you must be exact about
| Term | Meaning |
|---|---|
| Gene (factor) | The unit of inheritance that controls a trait |
| Allele | One of the alternative forms of a gene, such as $T$ and $t$ |
| Homozygous | Both alleles the same: $TT$ or $tt$ |
| Heterozygous | Two different alleles: $Tt$ |
| Genotype / phenotype | The genetic make-up / the observable trait |
| True-breeding line | One that shows the same trait over several generations of self-pollination |
| Test cross | Crossing an organism of unknown genotype with a homozygous recessive |
2. Mendel and the garden pea
Gregor Mendel ran his hybridisation experiments on garden pea from 1856 to 1863. The pea worked well because it had clearly contrasting traits, true-breeding lines, and could be self- or cross-pollinated. Mendel also used large samples and statistics — new for biology at the time. He studied seven pairs of contrasting traits:
| Character | Dominant | Recessive |
|---|---|---|
| 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 |
Note the trap: green is dominant for pod colour but recessive for seed colour.
3. Monohybrid cross
A true-breeding tall plant ($TT$) is crossed with a dwarf plant ($tt$). Every F1 plant is $Tt$ and tall. When F1 plants self-pollinate, the F2 generation looks like this:
| $T$ | $t$ | |
|---|---|---|
| $T$ | $TT$ — tall | $Tt$ — tall |
| $t$ | $Tt$ — tall | $tt$ — dwarf |
- Phenotypic ratio — tall : dwarf $= 3 : 1$
- Genotypic ratio — $TT : Tt : tt = 1 : 2 : 1$
- Law of dominance: in a heterozygote, one allele (dominant) expresses itself and the other (recessive) does not.
- Law of segregation: the two alleles of a gene separate during gamete formation, so each gamete receives only one.
- Test cross: $Tt \times tt$ gives tall : dwarf $= 1 : 1$, while $TT \times tt$ gives all tall. That is how a test cross reveals the unknown genotype.
4. When Mendel's ratios change
| Pattern | What happens | Example | F2 ratio |
|---|---|---|---|
| Incomplete dominance | The heterozygote shows an intermediate phenotype | Snapdragon: red $RR$ × white $rr$ gives pink $Rr$ | $1 : 2 : 1$ for both phenotype and genotype |
| Codominance | Both alleles are fully expressed together | Blood group AB, with genotype $I^AI^B$ | — |
| Multiple alleles | A gene has more than two alleles in the population, though each person carries only two | ABO blood groups: $I^A$, $I^B$, $i$ | — |
| Pleiotropy | One gene affects several traits | Phenylketonuria | — |
| Polygenic inheritance | Several genes add up to control one trait | Human skin colour | A continuous range of phenotypes |
| Genotype | Blood group |
|---|---|
| $I^AI^A$ or $I^Ai$ | A |
| $I^BI^B$ or $I^Bi$ | B |
| $I^AI^B$ | AB (codominance) |
| $ii$ | O |
Three alleles give six genotypes but only four phenotypes.
Worked example: blood groups
A father with blood group A (genotype $I^Ai$) and a mother with blood group B ($I^Bi$) have children. What blood groups are possible?
Father's gametes: $I^A$ or $i$. Mother's gametes: $I^B$ or $i$.
Children: $I^AI^B$ (AB), $I^Ai$ (A), $I^Bi$ (B), $ii$ (O) — each with probability $\tfrac{1}{4}$. All four blood groups are possible.
5. Dihybrid cross
Crossing a round yellow pea ($RRYY$) with a wrinkled green pea ($rryy$) gives F1 plants that are all $RrYy$, round and yellow. In F2:
$$\text{round yellow} : \text{round green} : \text{wrinkled yellow} : \text{wrinkled green} = 9 : 3 : 3 : 1$$Law of independent assortment: when two pairs of traits are combined in a hybrid, the segregation of one pair is independent of the other.
Worked examples: probability shortcuts
1. In an $RrYy \times RrYy$ cross, what fraction of the offspring is round and green?
$P(\text{round}) = \tfrac{3}{4}$ and $P(\text{green}) = \tfrac{1}{4}$, so $\tfrac{3}{4} \times \tfrac{1}{4} = \tfrac{3}{16}$.
2. What does the dihybrid test cross $RrYy \times rryy$ give?
Four phenotypes in the ratio $1 : 1 : 1 : 1$.
3. For $AaBbCc \times AaBbCc$: each parent makes $2^3 = 8$ kinds of gametes, and the chance of an $aabbcc$ offspring is $\left(\tfrac{1}{4}\right)^3 = \tfrac{1}{64}$.
6. Chromosomal theory, linkage and recombination
- Walter Sutton and Theodor Boveri proposed the chromosomal theory of inheritance: genes sit on chromosomes, and chromosomes pairing and separating at meiosis explains Mendel's laws.
- Thomas Hunt Morgan worked with Drosophila and found that genes on the same chromosome tend to be inherited together. He called this linkage, and the new combinations produced by crossing over recombination.
- The body-colour and eye-colour genes in Drosophila showed only 1.3% recombination, meaning they are tightly linked. The eye-colour and wing-size genes showed 37.2%, meaning they are loosely linked.
- Alfred Sturtevant used recombination frequency to map the positions of genes on a chromosome. 1% recombination is taken as 1 map unit.
7. Sex determination
| System | Female | Male | Example |
|---|---|---|---|
| XX–XO | XX | XO (one fewer chromosome) | Grasshopper |
| XX–XY | XX | XY — male heterogametic | Humans, Drosophila |
| ZW–ZZ | ZW — female heterogametic | ZZ | Birds |
| Haplodiploidy | Diploid, 32 chromosomes, from a fertilised egg | Haploid, 16 chromosomes, from an unfertilised egg | Honey bee |
In humans the sperm decides the sex of the child: an X-bearing sperm gives a girl, a Y-bearing sperm gives a boy. The chance of either is 50% in every pregnancy.
8. Reading a pedigree
Squares are males, circles are females, shaded symbols are affected individuals, and a line joining a male and a female is a mating. Use these clues:
- Two unaffected parents have an affected child: the trait is recessive.
- The trait appears in every generation, and every affected child has an affected parent: it is probably dominant.
- Mostly males are affected, and the trait passes to them through unaffected mothers: it is probably X-linked recessive.
9. Mendelian disorders
| Disorder | Inheritance | Key fact |
|---|---|---|
| Haemophilia | X-linked recessive | A protein in the blood-clotting cascade is affected. Females are affected only if the mother is at least a carrier and the father is haemophilic. Queen Victoria's family is the classic pedigree. |
| Colour blindness | X-linked recessive | Defect in red or green cones. Affects about 8% of males and 0.4% of females. |
| Sickle-cell anaemia | Autosomal recessive | The codon GAG changes to GUG, so glutamic acid is replaced by valine at the 6th position of the β-globin chain. |
| Phenylketonuria | Autosomal recessive | The enzyme that converts phenylalanine to tyrosine is missing, so phenylalanine builds up. Also an example of pleiotropy. |
| Thalassaemia | Autosomal recessive | Too little globin is made — a quantitative defect. α-thalassaemia genes are on chromosome 16; β-thalassaemia on chromosome 11. |
| Myotonic dystrophy | Autosomal dominant | A dominant disorder: one copy of the allele is enough to cause it. |
Worked example: colour blindness
A woman with normal vision whose father was colour blind marries a man with normal vision. What can their children inherit?
She must be a carrier, $X^CX^c$. He is $X^CY$.
| $X^C$ (father) | $Y$ (father) | |
|---|---|---|
| $X^C$ (mother) | $X^CX^C$ — normal daughter | $X^CY$ — normal son |
| $X^c$ (mother) | $X^CX^c$ — carrier daughter | $X^cY$ — colour-blind son |
No daughter is colour blind, but half the daughters are carriers. Half the sons are colour blind. Across all children, the chance of a colour-blind child is $\tfrac{1}{4}$.
10. Chromosomal disorders
| Disorder | Chromosomes | Key features |
|---|---|---|
| Down syndrome | Trisomy 21 (47 chromosomes) | First described by Langdon Down (1866). Short stature, small round head, furrowed tongue, partly open mouth, broad palm with a characteristic crease, delayed physical and mental development. |
| Klinefelter syndrome | 47, XXY | Male with an extra X. Overall masculine development, but may have gynaecomastia (breast development) and is sterile. |
| Turner syndrome | 45, XO | Female missing one X. Sterile, with rudimentary ovaries and lack of other secondary sexual characters. |
Mistakes that cost marks
- Giving the genotypic ratio $1 : 2 : 1$ when the question asks for the phenotypic ratio $3 : 1$.
- Mixing up incomplete dominance (a blend, as in pink snapdragons) with codominance (both expressed, as in blood group AB).
- Doing a test cross with a heterozygote. The tester is always homozygous recessive.
- Assuming the male is heterogametic in every species. In birds it is the female (ZW).
- Saying haemophilia can never affect females. It is rare, not impossible.
- Taking the male honey bee as diploid. Drones are haploid.
- Calling sickle-cell anaemia X-linked. It is autosomal recessive.
60-second recap
- Monohybrid F2: $3 : 1$ phenotype, $1 : 2 : 1$ genotype. Dihybrid F2: $9 : 3 : 3 : 1$.
- Test cross ratios: $1 : 1$ monohybrid, $1 : 1 : 1 : 1$ dihybrid.
- ABO: three alleles, six genotypes, four phenotypes; AB shows codominance.
- Low recombination frequency means tight linkage; 1% recombination is 1 map unit.
- Birds: female ZW. Honey bee: male haploid.
- Haemophilia and colour blindness: X-linked recessive. Sickle-cell, PKU, thalassaemia: autosomal recessive.
- Down: trisomy 21. Klinefelter: XXY. Turner: XO.