Genetics and Molecular Inheritance — NEET Biology

Genetics is the single highest-yield block in NEET Biology, worth 9–11 questions a year, and it is unusually solvable — the monohybrid and dihybrid ratios, the exceptions to dominance, the base-pairing and replication rules, and the steps of the central dogma are all fixed facts you can bank. This chapter builds classical (Mendelian) genetics first, then the molecular machinery of DNA, in the ordered, value-rich form the exam quotes almost verbatim.

Part A — Principles of Inheritance and Variation

1. Mendel and the monohybrid cross

Gregor Mendel worked on the garden pea (Pisum sativum) with seven contrasting traits. A monohybrid cross (one trait) of a tall (TT) × dwarf (tt) plant gives:

• F₁: all Tall (Tt) — the tall allele is dominant.
• F₂: on selfing F₁ → 3 Tall : 1 dwarf (phenotypic ratio 3:1), and the genotypic ratio 1 TT : 2 Tt : 1 tt (1:2:1).

This gives the Law of Dominance (one allele masks the other) and the Law of Segregation (the two alleles of a gene separate during gamete formation, each gamete getting one).

Worked example 1.1. In a monohybrid cross, why is the F₂ phenotypic ratio 3:1 but the genotypic ratio 1:2:1? The three tall plants of the 3:1 phenotype are not identical: one is homozygous TT and two are heterozygous Tt, all looking tall because T is dominant. Adding the one tt dwarf gives genotypes in 1 : 2 : 1, which collapse to the 3 tall : 1 dwarf phenotype because TT and Tt look alike.

2. The dihybrid cross and independent assortment

A dihybrid cross follows two traits (e.g. seed shape and colour): round-yellow (RRYY) × wrinkled-green (rryy).

• F₁: all round-yellow (RrYy).
• F₂: 9 round-yellow : 3 round-green : 3 wrinkled-yellow : 1 wrinkled-green — the 9:3:3:1 ratio.

This gives the Law of Independent Assortment: the alleles of one gene segregate independently of another (when the genes are on different chromosomes). A test cross (F₁ × homozygous recessive) reveals the genotype: a dihybrid test cross gives 1:1:1:1.

Worked example 2.1. How many types of gametes does a dihybrid RrYy produce, and in what proportion? Four types — RY, Ry, rY, ry — in equal (1:1:1:1) proportion. Because R/r and Y/y assort independently, each combination is equally likely, which is why a dihybrid self-cross gives the 9:3:3:1 F₂ ratio and a test cross gives 1:1:1:1.

3. Exceptions to Mendelian dominance

Real inheritance is often more subtle:

• Incomplete dominance: the heterozygote is intermediate — Mirabilis (4 o'clock): red (RR) × white (rr) → pink (Rr); F₂ ratio 1 red : 2 pink : 1 white (1:2:1 for both phenotype and genotype).
• Codominance: both alleles express fully — human ABO blood group (I^A and I^B are codominant → AB blood).
• Multiple alleles: more than two alleles in a population — ABO has three alleles (I^A, I^B, i).
• Pleiotropy: one gene affects many traits (e.g. phenylketonuria, sickle-cell).

Worked example 3.1. A cross of a red and a white Mirabilis gives all pink offspring. What is this called and what will the F₂ ratio be? This is incomplete dominance — neither allele is fully dominant, so the Rr heterozygote is pink (intermediate). Selfing the pink F₁ gives an F₂ of 1 red : 2 pink : 1 white, in which the phenotypic ratio equals the genotypic ratio (1:2:1) because each genotype has its own appearance.

4. Chromosomal theory and sex determination

The chromosomal theory of inheritance (Sutton and Boveri) states that genes are located on chromosomes, whose behaviour in meiosis explains Mendel's laws.

Sex determination:

• Humans (XX–XY): females XX, males XY. The sperm decides the sex (X-bearing → girl, Y-bearing → boy).
• Birds (ZZ–ZW): males ZZ, females ZW.
• Grasshopper (XX–XO): females XX, males XO.

Humans have 46 chromosomes (23 pairs) — 22 pairs of autosomes + 1 pair of sex chromosomes.

5. Linkage, recombination and pedigree analysis

Linked genes on the same chromosome tend to be inherited together (Morgan, in Drosophila), reducing recombinants; crossing over separates them, and the recombination frequency measures the distance between genes.

A pedigree traces a trait through a family. Autosomal recessive traits can skip generations; X-linked recessive traits (haemophilia, colour blindness) appear far more in males (who have one X).

6. Human genetic disorders

• Mendelian disorders:
  • Sickle-cell anaemia — autosomal recessive; a single base change (GAG→GTG) puts valine for glutamic acid in β-globin (HbS).
  • Haemophilia — X-linked recessive; blood fails to clot.
  • Colour blindness — X-linked recessive.
  • Phenylketonuria — autosomal recessive enzyme defect.
  • Thalassaemia — autosomal recessive; reduced globin synthesis.
• Chromosomal disorders:
  • Down's syndrome — trisomy 21 (47 chromosomes).
  • Klinefelter's — XXY (47; male, sterile).
  • Turner's — XO (45; female, sterile).

Worked example 6.1. Why is haemophilia far more common in males than females? Haemophilia is X-linked recessive. A male has only one X, so a single defective allele expresses the disease. A female has two Xs and would need the defective allele on both to be affected — much rarer; usually she is a carrier. Hence the strong male bias.

Part B — Molecular Basis of Inheritance

7. DNA as the genetic material

DNA is the genetic material, shown by Griffith's transforming principle, Avery–MacLeod–McCarty (identified it as DNA), and the Hershey–Chase experiment (bacteriophage — DNA, not protein, enters the host). In some viruses RNA is the genetic material.

Structure (Watson & Crick, 1953): a double helix of two antiparallel strands (5′→3′ and 3′→5′), a sugar–phosphate backbone, with bases pairing by hydrogen bonds:

• A = T (2 H-bonds), G ≡ C (3 H-bonds) — Chargaff's rule: A = T and G = C.
• One helical turn = 3.4 nm, 10 base pairs per turn, so adjacent bases are 0.34 nm apart. Diameter 2 nm.

Worked example 7.1. If a DNA sample is 30% adenine, what percentage is guanine? By Chargaff's rule A = T and G = C. If A = 30%, then T = 30% (together 60%). The remaining 40% is shared equally by G and C, so G = 20% (and C = 20%). Always A + G = T + C = 50%.

8. Packaging of DNA — the nucleosome

A human cell's ~2 m of DNA is packed into the nucleus by winding around histone proteins. Negatively charged DNA wraps around a positively charged histone octamer (2 each of H2A, H2B, H3, H4) to form a nucleosome (~200 bp). Nucleosomes form the "beads-on-a-string" chromatin, coiling further into chromosomes. Euchromatin is loosely packed and active; heterochromatin is densely packed and inactive.

9. DNA replication — semiconservative

Meselson and Stahl (1958, using ¹⁵N in E. coli) proved replication is semiconservative — each daughter DNA has one old (parental) and one new strand.

Key features:

• Enzyme DNA polymerase synthesises the new strand 5′→3′, using the parental strand as template; helicase unwinds the helix.
• Synthesis is continuous on the leading strand and discontinuous (short Okazaki fragments joined by DNA ligase) on the lagging strand — because polymerase works only 5′→3′.
• Replication is highly accurate (proofreading).

Worked example 9.1. Why is one new DNA strand made continuously and the other in fragments? Because DNA polymerase can add nucleotides only in the 5′→3′ direction. On the strand whose template runs 3′→5′ (leading), synthesis is continuous; on the opposite (lagging) strand the template runs the "wrong" way, so synthesis proceeds in short Okazaki fragments (each 5′→3′) that DNA ligase later joins.

10. Transcription — DNA to RNA

Transcription copies one strand of DNA into RNA by RNA polymerase. Only the template strand (3′→5′) is read; the RNA made is identical to the coding strand except U replaces T.

In prokaryotes RNA is used directly; in eukaryotes the primary transcript (hnRNA) is processed — introns removed (splicing), exons joined, a 5′ cap and a 3′ poly-A tail added — before it leaves the nucleus as mature mRNA.

Three RNAs: mRNA (message), tRNA (adaptor, carries amino acids), rRNA (ribosome).

11. The genetic code

The genetic code reads mRNA in triplets (codons) — 3 bases per amino acid. Its features:

• 64 codons (4³) for 20 amino acids — so the code is degenerate (most amino acids have more than one codon).
• AUG = start codon (and codes methionine); UAA, UAG, UGA = stop codons (no amino acid).
• The code is universal (nearly the same in all life), non-overlapping and comma-less, read in a fixed frame.

Worked example 11.1. The genetic code is called "degenerate". What does this mean, and give the numbers behind it? Degenerate means one amino acid can be specified by more than one codon. There are 64 codons but only 20 amino acids (3 of the 64 are stop codons, leaving 61 coding codons), so on average each amino acid has about three codons — the surplus is the degeneracy, which buffers against some mutations.

12. Translation and the lac operon

Translation builds a protein on the ribosome. tRNAs bring amino acids matching each mRNA codon (by anticodon pairing); peptide bonds link them from the start (AUG) to a stop codon. The ribosome moves along the mRNA 5′→3′.

Gene regulation — the lac operon (Jacob & Monod, in E. coli) is the classic example of prokaryotic regulation:

• The operon has a promoter, an operator and three structural genes (z, y, a) for lactose metabolism.
• No lactose: a repressor binds the operator → transcription off.
• Lactose present: lactose (inducer) binds the repressor → it leaves the operator → transcription on (an inducible operon).

Worked example 12.1. In the lac operon, what happens when lactose is added to the medium? Lactose acts as an inducer: it binds the repressor protein, changing its shape so it can no longer sit on the operator. RNA polymerase is then free to transcribe the structural genes (z, y, a), and the enzymes for lactose breakdown (e.g. β-galactosidase) are made — the operon is switched on.

13. Common traps NEET sets here

• Monohybrid F₂: 3:1 phenotype, 1:2:1 genotype; dihybrid F₂: 9:3:3:1; dihybrid test cross 1:1:1:1.
• Incomplete dominance 1:2:1 (phenotype = genotype); codominance = both alleles show (AB blood).
• ABO: three alleles (I^A, I^B, i); I^A & I^B codominant, i recessive.
• Humans XX/XY — sperm decides sex; 46 chromosomes (22 pairs autosomes + 1 pair sex).
• Haemophilia & colour blindness = X-linked recessive (more in males).
• Down = trisomy 21 (47); Klinefelter XXY (47); Turner XO (45).
• A=T (2 H-bonds), G≡C (3 H-bonds); 10 bp/turn, 3.4 nm/turn, 0.34 nm/bp.
• Replication semiconservative (Meselson–Stahl); polymerase 5′→3′; lagging strand = Okazaki fragments + ligase.
• Transcription: template strand read; RNA has U for T. Splicing removes introns in eukaryotes.
• 64 codons, degenerate; AUG start; UAA/UAG/UGA stop.
• lac operon inducible: lactose binds repressor → operon ON.

14. Memory aids

• "3:1 you see, 1:2:1 it be" — monohybrid phenotype vs genotype.
• "Nine-three-three-one" — the dihybrid F₂.
• "Incomplete = in-between (pink)" — incomplete dominance.
• "A-T two, G-C three" — hydrogen bonds per base pair.
• "Semi = half old, half new" — semiconservative replication.
• "Okazaki lags" — the discontinuous lagging strand.
• "AUG starts, U-A-A/A-G/G-A stops" — start and stop codons.
• "Lactose OFF the repressor, operon ON" — the lac operon logic.

15. Exam protocol

1. Mendel: monohybrid (3:1 / 1:2:1), dihybrid (9:3:3:1), test cross (1:1 or 1:1:1:1); laws of dominance, segregation, independent assortment.
2. Exceptions: incomplete dominance (1:2:1), codominance (ABO/AB), multiple alleles, pleiotropy.
3. Sex determination (XX-XY, ZZ-ZW, XX-XO); 46 human chromosomes; linkage & recombination.
4. Disorders: sickle-cell/haemophilia/colour blindness/PKU/thalassaemia; Down (trisomy 21), Klinefelter (XXY), Turner (XO).
5. DNA proof (Griffith, Hershey–Chase); structure (double helix, A=T/G≡C, 10 bp/turn, 3.4 nm); nucleosome (histone octamer).
6. Replication (semiconservative, Meselson–Stahl; 5′→3′; Okazaki/ligase).
7. Transcription (template strand, U for T, splicing); genetic code (64, degenerate, AUG/stop); translation; lac operon (inducible).