Genetics and Evolution constitute the core of modern biological sciences. Genetics explains how traits are preserved and transmitted across generations (heredity) and why offspring display unique characteristics (variation). Evolution explains the macro-historical consequences of these genetic processes: how genetic variations, accumulated over billions of years, led to the diversification of life from a single common ancestor.
This unit integrates three fundamental areas of study:
NEET Relevance: This unit is one of the highest-yielding domains in the NEET UG biology paper, accounting for 8–12 questions annually. Rote learning is insufficient; questions are increasingly conceptual, demanding mathematical problem-solving (Hardy-Weinberg, probability of crosses), analysis of molecular biology experiments, and pedigree decoding.
By the end of this unit, you should be able to:
Before starting, verify you are comfortable with:
NEET Priority: Critical
Gregor Mendel's success with the garden pea (Pisum sativum) was due to his quantitative approach, keeping track of individual traits through successive generations. He analyzed seven characters: stem height, flower colour, flower position, pod shape, pod colour, seed shape, and seed colour.
For any cross involving \( n \) heterozygous gene pairs showing complete dominance:
NEET Priority: Very High
The F₁ hybrid exhibits a phenotype intermediate between the dominant and recessive phenotypes.
Both alleles in a heterozygote are expressed fully and simultaneously.
NEET Priority: Critical
Pedigree analysis is the study of an inherited trait in a family over multiple generations. Use this step-by-step algorithm to determine the mode of inheritance:
Thomas Hunt Morgan working on Drosophila melanogaster observed that genes on the same chromosome do not assort independently.
NEET Priority: Critical
DNA replication is semi-conservative (proven by Meselson and Stahl using \(^{15}\text{N}\) and \(^{14}\text{N}\) in E. coli) and semi-discontinuous.
| Enzyme | Function | Directionality |
|---|---|---|
| Helicase | Unwinds the double helix by breaking hydrogen bonds | Moves \( 5' \rightarrow 3' \) or \( 3' \rightarrow 5' \) |
| Topoisomerase (Gyrase) | Relieves torsional strain and supercoiling ahead of the replication fork | Cuts and reseals phosphodiester bonds |
| SSB Proteins | Keep unwound single strands stable and prevent re-annealing | Non-directional binding |
| Primase (RNA Polymerase) | Synthesizes a short RNA primer (\( \sim 10\ \text{nucleotides} \)) to provide a free 3'-OH group | Synthesizes \( 5' \rightarrow 3' \) |
| DNA Polymerase III | Primary replicative enzyme. Adds dNTPs complementary to the template strand | Synthesizes \( 5' \rightarrow 3' \) only |
| DNA Polymerase I | Removes RNA primers (5'\( \rightarrow \)3' exonuclease) and replaces them with DNA | Synthesizes \( 5' \rightarrow 3' \) |
| DNA Ligase | Seals nicks between Okazaki fragments by forming phosphodiester bonds | Active on lagging strand |
NEET Priority: Critical
A transcription unit consists of a Promoter, a Structural Gene, and a Terminator.
Unlike prokaryotic transcription, eukaryotic transcription produces a precursor heterogeneous nuclear RNA (hnRNA) that must undergo processing inside the nucleus:
The genetic code is a triplet codon system (61 coding codons, 3 stop codons: UAA, UAG, UGA).
NEET Priority: Critical
The Lac Operon in E. coli is an inducible operon that controls the transport and metabolism of lactose.
NEET Priority: Very High
Chemical evolution posits that life arose from non-living organic molecules through a series of progressive chemical reactions:
| Class | Anatomical Basis | Evolutionary Pattern | Exhaustive NCERT Examples |
|---|---|---|---|
| Homologous Organs | Same internal skeletal structure and origin; different functions | Divergent Evolution (adaptation to different niches from a common ancestor) | 1. Forelimbs of whale, bat, cheetah, and human.<br>2. Thorns of Bougainvillea and tendrils of Cucurbita (both are axillary bud modifications).<br>3. Vertebrate hearts and brains. |
| Analogous Organs | Different internal skeletal structure and origin; same function | Convergent Evolution (adaptation to similar ecological pressures in unrelated groups) | 1. Sweet potato (root modification) and potato (stem modification) storing starch.<br>2. Wings of butterflies (chitinous) and birds (feathered/bony).<br>3. Eyes of octopuses and mammals.<br>4. Flippers of penguins (birds) and dolphins (mammals). |
NEET Priority: Very High
Natural selection acts on phenotypic variations in three distinct ways:
In a large, panmictic (randomly mating) population where evolutionary forces are absent, allele and genotype frequencies remain constant.
$$ p + q = 1 \quad \text{(Allele Frequencies)} $$ $$ p^2 + 2pq + q^2 = 1 \quad \text{(Genotype Frequencies)} $$Forces that disrupt Hardy-Weinberg Equilibrium:
| Hominid | Age (MYA) | Brain Capacity (cc) | Key Morphological & Cultural Features | Diet |
|---|---|---|---|---|
| Dryopithecus | \( \sim 15 \) | N/A | Ape-like, hairy, knuckle-walker, arms and legs of equal length | Herbivorous (fruits and leaves) |
| Ramapithecus | \( \sim 15 \) | N/A | More man-like, walked more erect, teeth/jaw man-like | Herbivorous (nuts and seeds) |
| Australopithecus | \( 3\text{–}2 \) | \( 400\text{–}500 \) | First bipedal hominid. Lived in East African grasslands (e.g., "Lucy"). Used stone weapons | Herbivorous (primarily fruits) |
| Homo habilis | \( 2.5\text{–}1.5 \) | \( 650\text{–}800 \) | First human-like hominid. First tool maker ("Handy Man"). Handedness | Herbivorous (did not eat meat) |
| Homo erectus | \( 1.5 \) | \( 900 \) | Fully upright posture. Java Man, Peking Man. First to use fire | Omnivorous (meat-eater) |
| Homo neanderthalensis | \( 0.1\text{–}0.04 \) | \( 1400 \) | Lived in Near East and Central Asia. Buried their dead, wore animal hides to protect bodies. Developed family structures | Omnivorous |
| Homo sapiens | \( 0.075\text{–}0.01 \) | \( 1350\text{–}1400 \) | Evolved in Africa. Cave art (developed \( 18,000\ \text{years ago} \)), agriculture (\( 10,000\ \text{years ago} \)), and settlements | Omnivorous |
Where \( n \) is the number of heterozygous gene pairs.
The haploid human genome contains \( 3.3 \times 10^9\ \text{bp} \). Therefore, a diploid human cell contains:
$$ \text{Diploid Genome Size (N)} = 2 \times 3.3 \times 10^9\ \text{bp} = 6.6 \times 10^9\ \text{bp} $$The distance between two consecutive base pairs in the B-DNA helix is \( 0.34\ \text{nm} \) or \( 0.34 \times 10^{-9}\ \text{meters} \).
The total length of DNA (\( L \)) is derived by multiplying the total number of base pairs by the distance between them:
$$ L = \text{Number of Base Pairs} \times \text{Distance between base pairs} $$ $$ L = (6.6 \times 10^9\ \text{bp}) \times (0.34 \times 10^{-9}\ \text{m/bp}) $$ $$ L = 6.6 \times 0.34\ \text{meters} $$ $$ L \approx 2.244\ \text{meters} $$Conclusion: The total length of DNA in a single human diploid cell is approximately \( 2.2\ \text{meters} \), packaged within a nucleus measuring only about \( 6\ \mu\text{m} \) in diameter.
| Metric/Constant | Value | Physical Significance |
|---|---|---|
| Base pair distance | \( 0.34\ \text{nm} \) | Linear rise per nucleotide step |
| Map unit (mu/cM) | \( 1\% \) recombination | Relative distance between loci on a chromosome |
| Hardy-Weinberg variables | \( p, q \in [0, 1] \) | Proportions of alleles in a gene pool |
| Brain capacity (cc) | \( 400\text{–}1650\ \text{cm}^3 \) | Cranial volume reflecting evolution of hominids |
| Meselson-Stahl timing | \( 20\ \text{minutes} \) | Generation time of E. coli at \( 37^\circ\text{C} \) |
Example 1 (Hardy-Weinberg): In a randomly mating population of wildflowers, the frequency of the recessive allele for white flower colour is \( 0.4 \). If the population is in Hardy-Weinberg equilibrium, what is the frequency of pink flowers, assuming the dominant allele codes for red flowers and the heterozygotes are pink (incomplete dominance)? Solution:
Example 2 (Gene Mapping): In a test cross of a dihybrid Drosophila female, the following progeny classes were obtained:
Calculate the distance between the two genes on the chromosome. Solution:
Example 3 (Chargaff's Rule): A double-stranded DNA molecule contains \( 20\% \) Adenine. What is the percentage of Cytosine in this DNA molecule? Solution:
Example 4 (DNA Packaging): How many nucleosomes are present in a diploid human cell containing \( 6.6 \times 10^9\ \text{bp} \) of DNA? Solution:
Example 5 (Hardy-Weinberg [NEET 2020]): In a gene pool, the frequency of a recessive allele is \( 0.3 \). What is the frequency of homozygous dominant individuals? Solution:
Example 6 (DNA Replication): If E. coli replicates its genome every 20 minutes, what will be the ratio of hybrid (\(^{15}\text{N}/^{14}\text{N}\)) to light (\(^{14}\text{N}/^{14}\text{N}\)) DNA molecules after 60 minutes of growth in a \(^{14}\text{N}\)-containing medium, starting from a pure \(^{15}\text{N}\) culture? Solution:
Example 7 (Probability of Crosses): In a cross between individuals with genotypes \( \text{AaBb} \times \text{AaBb} \), what is the probability of obtaining an offspring with the genotype \( \text{AABb} \)? Solution:
Example 8 (Chargaff's Rule [AIPMT 2012]): In a sample of double-stranded DNA, the ratio of \( \frac{\text{A+T}}{\text{G+C}} = 1.2 \). If the total number of base pairs is 120, find the number of Guanine bases. Solution:
This means \( \text{G} + \text{C} = 109 \) bases, and \( \text{A} + \text{T} = 131 \) bases.
Example 9 (Pedigree Probability): A man who is heterozygous for the autosomal recessive condition Albinism marries a normal woman whose father was albino. What is the probability that their first child will be albino? Solution:
Example 10 (Hominid Evolution): Arrange the following hominid ancestors in increasing order of their cranial capacity: Homo erectus, Homo habilis, Homo neanderthalensis, Australopithecus. Solution:
Example 11 (Hardy-Weinberg [NEET 2017]): In a population of 2000 individuals, 800 belong to genotype AA, 800 to Aa, and 400 to aa. Find the allele frequency of A. Solution:
Example 12 (Transcription Coding): Write the sequence of mRNA transcribed from a DNA template strand with the sequence \( 3'\text{-TACGGCCTAATG}\text{-}5' \). Solution:
How many different types of gametes can be produced by an individual with the genotype \( \text{AaBbCcDd} \), assuming independent assortment?
** The number of gamete types is calculated using the formula \( 2^n \), where \( n \) is the number of heterozygous gene pairs. The genotype \( \text{AaBbCcDd} \) has 4 heterozygous pairs (\( n=4 \)).
$$ 2^4 = 16\ \text{types of gametes} $$If a double-stranded DNA has \( 20\% \) Cytosine, what is the percentage of Adenine in it?
** By Chargaff's rule, in double-stranded DNA, the percentage of Cytosine is equal to Guanine, and Adenine is equal to Thymine.
If \( \text{C} = 20\% \), then \( \text{G} = 20\% \).
The remaining percentage belongs to Adenine and Thymine:
$$ 100\% - 40\% = 60\% $$Since \( \text{A} = \text{T} \), divide the remaining percentage by 2:
$$ \text{A} = \frac{60\%}{2} = 30\% $$Which of the following did Mendel NOT use in his hybridization experiments?
** Mendel used pod shape and pod colour, but he did NOT use pod position as one of the seven character pairs. He did use flower position (axial vs terminal).
A recombination frequency of \( 50\% \) between two genes indicates that:
** A recombination frequency of \( 50\% \) is the maximum possible and is equivalent to the outcome of independent assortment. It indicates that the genes are either on different chromosomes or located very far apart on the same chromosome, allowing crossing over to occur on every chromatid.
Which codon acts as both a start codon and codes for methionine?
** AUG is the universal start (initiation) codon. It codes for methionine in eukaryotes and N-formylmethionine in prokaryotes.
The Miller-Urey experiment used which of the following mixtures of gases in the spark discharge flask?
** The Miller-Urey experiment simulated primitive Earth's reducing atmosphere by sealing water vapour (\( \text{H}_2\text{O} \point{)} \), methane (\( \text{CH}_4 \)), ammonia (\( \text{NH}_3 \)), and hydrogen gas (\( \text{H}_2 \)) in a glass apparatus. Oxygen (\( \text{O}_2 \)) was strictly absent.
Thorns of Bougainvillea and tendrils of Cucurbita are examples of:
** Thorns of Bougainvillea and tendrils of Cucurbita both develop from the axillary buds, demonstrating similar structural origin (homology) but have adapted to different functions (protection vs climbing), representing divergent evolution.
Which hominid first used fire?
** Homo erectus (cranial capacity \( \sim 900\ \text{cc} \)) is historically credited with the discovery and control of fire, which allowed them to cook meat and migrate out of Africa.
If the frequency of a recessive allele \( \text{a} \) is \( 0.2 \), what is the frequency of heterozygotes in a population at Hardy-Weinberg equilibrium?
** Let \( q = \text{freq(a)} = 0.2 \).
$$ p = \text{freq(A)} = 1 - 0.2 = 0.8 $$The frequency of heterozygotes is represented by \( 2pq \):
$$ 2pq = 2 \times 0.8 \times 0.2 = 0.32 $$The average size of a human gene is:
** According to the Human Genome Project, the average human gene consists of \( 3,000\ \text{bases} \), although sizes vary greatly (with the largest being dystrophin at \( 2.4\ \text{million bp} \)).
Assertion (A): The human male is heterogametic.
Reason (R): Human males produce two types of sperms, half containing X chromosomes and half containing Y chromosomes.
** Human males have XY sex chromosomes, making them heterogametic. During spermatogenesis, they produce two types of sperms: 50% carrying the X chromosome and 50% carrying the Y chromosome, which determines the sex of the offspring.
Assertion (A): Splicing is an essential processing step for hnRNA in prokaryotes.
Reason (R): Prokaryotic structural genes are polycistronic and split, containing both exons and introns.
** Both statements are false. Splicing is absent in prokaryotes because their structural genes are continuous (no introns). Eukaryotes have split genes containing introns and exons.
Assertion (A): In the presence of glucose, transcription of the lac operon is very low even if lactose is present.
Reason (R): High glucose levels lead to low intracellular cAMP levels, preventing the formation of the active CAP-cAMP complex required for high transcription rates.
** Glucose is the preferred energy source for E. coli. High glucose levels inhibit adenylyl cyclase, lowering cAMP. Without cAMP, CAP cannot bind to the lac promoter, and RNA polymerase cannot initiate high-level transcription, ensuring the cell uses glucose first.
Assertion (A): Sweet potato and potato are homologous organs.
Reason (R): Both organs have the same anatomical origin and structure but perform different functions.
** Both statements are false. Sweet potato (adventitious root modification) and potato (stem tuber modification) are analogous organs, not homologous. They have different anatomical origins but perform the same function of food storage.
Assertion (A): Genetic drift has a significant evolutionary effect only in large populations.
Reason (R): Large populations are more susceptible to random environmental changes that shift allele frequencies.
** Both statements are false. Genetic drift (random fluctuations in allele frequencies) is highly significant in small populations, where chance events can easily eliminate alleles. Large populations are buffered against such sampling errors.
Statement I: DNA replication takes place in the S-phase of the cell cycle.
Statement II: A failure in cell division after DNA replication results in polyploidy.
** DNA replication occurs during the S (Synthesis) phase of interphase. If the chromosome replicates but cytokinesis (cell division) fails, the cell ends up with multiple sets of chromosomes, leading to polyploidy (common in plants).
Statement I: RNA was the first genetic material to evolve.
Statement II: DNA evolved from RNA with chemical modifications that made it more stable.
** RNA is believed to have been the first genetic material because it can both store genetic information and catalyze reactions (ribozymes). DNA evolved later as a more stable alternative, replacing the reactive 2'-OH group of ribose with hydrogen (deoxyribose) and using thymine instead of uracil.
Statement I: Natural selection is the only force that can change allele frequencies in a population.
Statement II: Genetic drift changes allele frequencies purely by chance, primarily in small populations.
** Statement I is incorrect. Besides natural selection, allele frequencies can be altered by mutation, gene flow, non-random mating, and genetic drift. Statement II is correct; genetic drift is a stochastic process that is most pronounced in small populations.
Statement I: Codominance is seen in Snapdragon flowers where red and white crosses yield pink flowers.
Statement II: The genotypic and phenotypic ratios in Snapdragon incomplete dominance crosses are both 1:2:1.
** Statement I is incorrect. Snapdragon crosses show incomplete dominance (pink flowers), not codominance. Statement II is correct; the monohybrid F₂ phenotypic and genotypic ratios for incomplete dominance are both 1:2:1.
Statement I: Eukaryotic transcription requires three different RNA Polymerases.
Statement II: RNA Polymerase II is responsible for transcribing tRNA and 5S rRNA.
** Statement I is correct. Eukaryotes have RNA Pol I, II, and III. Statement II is incorrect; RNA Polymerase III transcribes tRNA and 5S rRNA, while RNA Polymerase II transcribes mRNA (hnRNA).
Match the Mendelian disorder in Column I with its characteristic feature in Column II:
| Column I | Column II |
|---|---|
| A. Sickle-cell anaemia | I. Deficiency of phenylalanine hydroxylase |
| B. Phenylketonuria | II. Point mutation in beta-globin chain (Glu \( \rightarrow \) Val) |
| C. Haemophilia | III. Reduced synthesis of alpha or beta globin chains |
| D. Thalassemia | IV. X-linked recessive clotting disorder |
** Sickle-cell is a point mutation in the beta-globin chain (Glu\( \rightarrow \)Val). PKU is a deficiency of phenylalanine hydroxylase. Haemophilia is an X-linked recessive clotting factor VIII deficiency. Thalassemia is characterized by reduced globin chain synthesis.
Match the evolutionary ancestor in Column I with its cranial capacity in Column II:
| Column I | Column II |
|---|---|
| A. Australopithecus | I. \( 1400\ \text{cc} \) |
| B. Homo habilis | II. \( 900\ \text{cc} \) |
| C. Homo erectus | III. \( 650\text{–}800\ \text{cc} \) |
| D. Homo neanderthalensis | IV. \( 400\text{–}500\ \text{cc} \) |
** The cranial capacities are: Australopithecus (\( 400\text{–}500\ \text{cc} \)), Homo habilis (\( 650\text{–}800\ \text{cc} \)), Homo erectus (\( 900\ \text{cc} \point{)} \), and Homo neanderthalensis (\( 1400\ \text{cc} \)).
Study the pedigree chart below showing the inheritance of an autosomal dominant trait (e.g., Myotonic dystrophy):
If individual B is an affected male who marries an unaffected female, what is the probability that their first child will be affected?
** Because Myotonic dystrophy is autosomal dominant, an affected individual (B) is likely heterozygous (\( \text{Mm} \)) if one parent is unaffected. Crossing with an unaffected female (\( \text{mm} \)) yields:
$$ \text{Mm} \times \text{mm} \rightarrow 50\%\ \text{Mm (affected)} \quad \text{and} \quad 50\%\ \text{mm (unaffected)} $$The diagram below represents a nucleosome. Identify the labels A, B, and C:
** A represents the double-stranded DNA wrapped around the core. B represents the linker histone H1. C represents the histone octamer core.
In the replication fork diagram below, identify the strands A and B:
** The DNA polymerase synthesizes new strands only in the \( 5' \rightarrow 3' \) direction. Strand A is synthesised continuously toward the replication fork (leading strand). Strand B is synthesised discontinuously away from the fork (lagging strand).
In a transcription unit, if the coding strand sequence is \( 5'\text{-ATGCATGC}\text{-}3' \), the sequence of the transcribed RNA will be:
** The transcribed RNA sequence is identical to the coding strand sequence, running in the same \( 5' \rightarrow 3' \) direction, except that Thymine (T) is replaced by Uracil (U).
A population has two alleles for a gene, T and t. The frequency of allele T is \( 0.8 \). What is the frequency of homozygous dominant individuals in this population?
** Let \( p = \text{freq(T)} = 0.8 \). The frequency of homozygous dominant individuals (\( \text{TT} \)) is \( p^2 \):
$$ p^2 = (0.8)^2 = 0.64 $$If a mutation changes a codon from UGG to UGA, what will be the effect on translation?
** UGG codes for tryptophan. UGA is a stop codon (nonsense codon). A mutation changing a coding codon to a stop codon is a nonsense mutation, causing translation to terminate prematurely.
What was the chemical condition of the primitive Earth's atmosphere?
** The primitive Earth's atmosphere was highly reducing because free oxygen was absent. Lighter gases like hydrogen reacted with other elements to form methane, ammonia, and water vapour.
The complexes that remove introns from eukaryotic pre-mRNA are:
** Spliceosomes, which are complexes of proteins and small nuclear RNAs (snRNPs), recognize intron boundary sequences and catalyze the excision of introns and ligation of exons.
Inheritance is governed by Mendel's laws: Dominance, Segregation, and Independent Assortment. Deviations include incomplete dominance (blended phenotype), codominance (both alleles expressed, like AB blood group), multiple allelism, pleiotropy (one gene affecting multiple traits), and polygenic inheritance (multiple genes affecting one trait, like skin colour). Linkage and crossing over occur on the same chromosome. Genetic mapping utilizes recombination frequencies. Autosomal and sex-linked disorders present distinct inheritance patterns in pedigrees.
DNA is the genetic material. Watson-Crick B-DNA parameters outline a double helix packaged via histone octamers into nucleosomes. DNA replication is semi-conservative and catalyzed by DNA Polymerases. Transcription converts DNA into RNA, which undergoes capping, tailing, and splicing in eukaryotes. Translation synthesizes proteins on ribosomes, guided by the triplet genetic code. The Lac Operon regulates lactose metabolism through negative (induction) and positive (CAP-cAMP) control.
Chemical evolution suggests life originated from inorganic precursors, verified by the Miller-Urey experiment. Comparative anatomy supports evolution through homologous (divergent) and analogous (convergent) structures. Natural selection can be stabilizing, directional, or disruptive. The Hardy-Weinberg equilibrium (\(p^2 + 2pq + q^2 = 1\)) is maintained unless evolutionary forces act. Human evolution exhibits a clear path from arboreal apes to modern technological humans, marked by expanding brain capacity.
| Genetic Concept | Key Fact / Formula | Common Exam Trap |
|---|---|---|
| Dihybrid F₂ Ratio | Phenotypic = 9:3:3:1; Genotypic = 1:2:1:2:4:2:1:2:1 | Assumes independent assortment; fails if linkage exists |
| Incomplete Dominance | Phenotypic and Genotypic ratios are both 1:2:1 | Do not confuse with codominance (where both traits show simultaneously) |
| Chargaff's Rules | \( \text{A} + \text{G} = \text{T} + \text{C} \) | Only applies to double-stranded DNA; invalid for single-stranded DNA/RNA |
| Replication Direction | DNA synthesis always proceeds \( 5' \rightarrow 3' \) | The template strand is read in the \( 3' \rightarrow 5' \) direction |
| Stop Codons | UAA, UAG, UGA | Do not code for any amino acid; trigger binding of release factors |
| Inducer of Lac Operon | Allolactose (isomer of lactose) | Lactose itself does not bind the repressor; allolactose is the active inducer |
| Hardy-Weinberg | \( p^2 + 2pq + q^2 = 1 \) | \( p^2 \) is the frequency of the homozygous dominant genotype, not the allele \( p \) |
| Homologous Organs | Same origin, divergent evolution (e.g., thorn/tendril) | Sweet potato (root) and potato (stem) are ANALOGOUS, not homologous |
| Tool Maker Hominid | Homo habilis (\( 650\text{–}800\ \text{cc} \)) | Australopithecus used stones but did not manufacture tools |
| First Hominid with Fire | Homo erectus (\( 900\ \text{cc} \)) | Neanderthals buried dead and used hides, but H. erectus first controlled fire |
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