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AITS 03Biology · Chapter 5 of 6

Biology · last-minute revision

Principles of Inheritance and Variation

Mendel's laws and ratios, deviations from dominance, linkage, sex determination, mutation, pedigree analysis and human genetic disorders — every ratio checked by enumeration.

9 : 3 : 3 : 1dihybrid F₂: two genes assorting independently

Mendel's experiments and lawsMust

Mendel crossed garden pea (Pisum sativum) from 1856 to 1863 using true-breeding lines for seven pairs of contrasting traits, and published in 1865. Pea suited him: clear contrasting forms, natural self-pollination, easy artificial crossing, many seeds.

CharacterDominantRecessive
Stem heightTallDwarf
Flower colourVioletWhite
Flower positionAxialTerminal
Pod shapeInflatedConstricted
Pod colourGreenYellow
Seed shapeRoundWrinkled
Seed colourYellowGreen
Pod colour versus seed colour

Green is dominant for pod colour; yellow is dominant for seed colour. A favourite statement-question swap.

Monohybrid Punnett square for Tt crossed with Tt giving 3 tall to 1 dwarfF₁ selfed: Tt × TtTTttgametesTTtallTttallTttallttdwarfPhenotype 3 tall : 1 dwarfGenotype 1 TT : 2 Tt : 1 ttTest cross Tt × tt → 1 tall : 1 dwarfTT × tt → all tallSegregation: one allele per gamete
Two heterozygous parents: alleles separate into gametes and recombine at random. Phenotype 3 : 1, genotype 1 : 2 : 1.
Law of Dominance
Factors (genes) come in pairs; in a dissimilar pair one (dominant) masks the other (recessive).
Law of Segregation
The two alleles do not blend; they separate at gamete formation, so each gamete gets only one.
Test cross
Cross with homozygous recessive. All dominant → parent homozygous; 1 : 1 → heterozygous.
Terms
Alleles: alternative forms of a gene. Homozygous TT or tt; heterozygous Tt. Genotype = genetic make-up; phenotype = observable form.

Beyond simple dominanceMust

Incomplete dominance — snapdragon (Antirrhinum)

RR red × rr white → F₁ Rr pink. F₂ is 1 red : 2 pink : 1 white — phenotypic ratio equals genotypic ratio.

Co-dominance and multiple alleles — ABO blood groups

Gene I has three alleles, IA, IB and i. IA and IB are co-dominant (both expressed together); both are dominant over i. An individual carries only two alleles, so multiple alleles show up only in population studies.

GenotypeBlood groupGenotypeBlood group
IAIA or IAiAIAIBAB
IBIB or IBiBiiO

6 genotypes, 4 phenotypes. Antigens on the RBC surface are sugar polymers made by the alleles' products; i makes none.

Dominance depends on the trait you look at

Pea gene B controls starch synthesis. BB: large starch grains, round seeds. bb: less starch, wrinkled seeds. Bb: seeds round (dominance) but starch grains of intermediate size (incomplete dominance). Dominance is not an autonomous property of the gene.

Pleiotropy
One gene, several phenotypic effects. Phenylketonuria: one mutant gene → mental retardation and reduced hair and skin pigmentation.
Polygenic inheritance
Three or more genes control one trait, with environmental influence. Human skin colour: AABBCC darkest, aabbcc lightest, AaBbCc intermediate.

Two genes: independent assortmentMust

Dihybrid Punnett square with the four phenotype classes 9 3 3 1 highlighted in turnRYRYRyRyrYrYryryRRYYRRYyRrYYRrYyRRYyRRyyRrYyRryyRrYYRrYyrrYYrrYyRrYyRryyrrYyrryyRRYY × rryy → F₁ RrYy, selfedRound yellow9Round green3Wrinkled yellow3Wrinkled green1Parental combinations: 10/16New combinations: 6/16Test cross RrYy × rryy → 1 : 1 : 1 : 1
Each phenotype class lights up in turn. Two gene pairs that sit on different chromosomes sort independently, giving 9 : 3 : 3 : 1.
Law of Independent Assortment
When two pairs of traits combine in a hybrid, segregation of one pair is independent of the other.
Dihybrid F₂
Phenotypes 9 : 3 : 3 : 1 · 9 genotypes · 16 squares
Dihybrid test cross
1 : 1 : 1 : 1
n heterozygous gene pairs
gamete types 2n · F₂ genotypes 3n · F₂ phenotypes 2n (complete dominance) · Punnett squares 4n
Probability shortcut
Multiply single-gene probabilities: P(RRyy) = ¼ × ¼ = 1/16; P(homozygous for both genes) = ½ × ½ = ¼

Chromosomes, linkage and recombinationMust

Mendel's work was rediscovered in 1900 by de Vries, Correns and von Tschermak. Sutton and Boveri (1902) noticed that chromosomes pair and separate exactly like Mendel's factors; Sutton united chromosomal segregation with Mendel's laws as the chromosomal theory of inheritance.

Morgan used Drosophila melanogaster — grows on simple synthetic medium, completes its life cycle in about two weeks, gives many progeny per mating, has clear male–female differences and many visible variations. For two genes on the same X chromosome, parental combinations were far more common than 9 : 3 : 3 : 1 predicts.

Crossing over between non-sister chromatids and Morgan's recombination frequenciesABabhomologous pair · 4 chromatidsAfter crossoverABparentalAbrecombinantaBrecombinantabparentalMorgan's Drosophila crossesbody colour (y) – eye colour (w)1.3%eye colour (w) – wing size (m)37.2%Tightly linked → few recombinantsFar apart on chromosome → manySturtevant: recombination % → gene maps
Exchange between non-sister chromatids creates Ab and aB. The farther apart two genes are, the more often a crossover falls between them.
Linkage
physical association of genes on one chromosome (Morgan's term)
Recombination
generation of non-parental gene combinations
Genetic map
Sturtevant used recombination frequency as a measure of distance between genes
ScientistContribution
MendelLaws of inheritance from pea (1856–63; published 1865)
de Vries, Correns, von TschermakRediscovered Mendel's work, 1900
Sutton and BoveriChromosomal theory of inheritance, 1902
T. H. MorganLinkage and recombination in Drosophila
A. H. SturtevantGenetic (linkage) maps
HenkingX body in insect spermatogenesis, 1891
PunnettPunnett square
Langdon DownDescribed Down's syndrome, 1866

Sex determinationMust

Henking (1891) saw a nuclear structure passed to only half the sperm of certain insects and called it the X body; it was later identified as the X chromosome. Chromosomes involved in sex determination are sex chromosomes; the rest are autosomes. Humans have 23 pairs: 22 pairs of autosomes plus XX or XY.

Sex determination by X or Y sperm in humans, with XO, ZW and haplodiploid systemsHumans: XX female · XY maleXegg (always X)XYXX → girlXY → boySperm decides sex · 50 : 50 at every birthMale is heterogametic (X- or Y-bearing sperm)Grasshopper (XO type)♀ XX · ♂ XO (single X, no partner)sperm with X or without Xmale has one chromosome fewerBirds (ZW type)♂ ZZ · ♀ ZWfemale heterogametic → egg decidesHoney bee (haplodiploid)fertilised egg → ♀, diploid 2n = 32unfertilised egg → ♂, haploid n = 16
Every human egg carries X; the sperm carries X or Y with equal probability, so the father's gamete decides the sex. In birds it is the other way round.
SystemFemaleMaleHeterogametic sexExample
XX–XOXXXOMaleGrasshopper
XX–XYXXXYMaleHumans, Drosophila
ZW–ZZZWZZFemaleBirds
Haplodiploid2n = 32n = 16—Honey bee
Gap content

Honey-bee haplodiploidy is listed in the NTA syllabus but is absent from some rationalised NCERT printings — revise it anyway. Males develop from unfertilised eggs by parthenogenesis and make sperm by mitosis; so a drone has no father and cannot have sons, but has a grandfather and can have grandsons.

Mutation: the sickle-cell exampleHigh yield

Mutation
change in DNA sequence → change in genotype and phenotype
Chromosomal aberrations
deletion, insertion or duplication of DNA segments; common in cancer cells
Point mutation
single base-pair change — sickle-cell anaemia
Frameshift mutation
insertion or deletion of base pairs shifts the reading frame
Mutagens
chemical and physical agents, e.g. UV radiation
Single base substitution GAG to GUG replacing glutamic acid with valine and the red cell sicklingNormal β-globin mRNA (HbA)GUGValCACHisCUGLeuACUThrCCUProGAGGluGAGGluSickle-cell β-globin mRNA (HbS)GUGValCACHisCUGLeuACUThrCCUProGUGValGAGGlu6th codon · 6th amino acidbiconcave RBC (HbA)HbS under low O₂ → sickleGAG → GUGGlu → Val
One base changes at codon 6: GAG → GUG. Glutamic acid becomes valine; HbS polymerises at low oxygen tension and the RBC turns from a biconcave disc into a sickle.
Trap · DNA or mRNA?

NCERT writes the change as GAG → GUG, which is the mRNA codon. On the DNA coding strand it is GAG → GTG. Check which molecule the question names.

Pedigree analysis and Mendelian disordersMust

Controlled crosses are impossible in humans, so inheritance is traced through family trees. Symbols: square = male, circle = female, shaded = affected, horizontal line = mating, double line = consanguineous mating, vertical line down to a sibship line = offspring.

Pedigree of a carrier mother and normal father showing carrier daughter and haemophilic soncarrier mothernormal fatherXhXXYXXnormalXhXcarrierXYnormalXhYaffectedX-linked recessiveCarrier mother × normal father:½ sons affected · ½ daughters carriersAffected father × normal mother:all daughters carriers · no son affectedAffected daughter needs an affectedfather and a carrier motherSame pattern: colour blindness
Sons receive their only X from the mother, so a carrier mother passes the recessive allele to half her sons (affected) and half her daughters (carriers).
DisorderInheritanceDefect and key fact
HaemophiliaX-linked recessiveOne clotting-cascade protein affected; non-stop bleeding from a small cut. Female haemophiliacs extremely rare. Queen Victoria was a carrier.
Colour blindnessX-linked recessiveDefect in red or green cone; about 8% of males, 0.4% of females.
Sickle-cell anaemiaAutosomal recessiveGlu → Val at 6th position of β-globin. Only HbSHbS is diseased; HbAHbS carriers look unaffected.
PhenylketonuriaAutosomal recessiveNo enzyme to convert phenylalanine → tyrosine; phenylpyruvic acid builds up → mental retardation; excreted in urine.
ThalassemiaAutosomal recessiveReduced synthesis of α or β globin. α: HBA1 and HBA2 on chromosome 16. β: HBB on chromosome 11. Quantitative defect (too little globin) — sickle-cell is qualitative (faulty globin).
Myotonic dystrophyAutosomal dominantAppears in NCERT's pedigree figure; affected parent in every generation.

Chromosomal disordersMust

Normal chromosome separation compared with non-disjunction producing n plus 1 and n minus 1 gametesNormal separationnn× n → normal 2nNon-disjunctionn + 1n − 1× n → 2n + 1 · trisomy (Down: 47)× n → 2n − 1 · monosomy (Turner: 45, XO)Failed segregation of chromatids → aneuploidy · failed cytokinesis → polyploidy (common in plants)
If a chromosome pair fails to separate in meiosis, gametes get one chromosome extra or one missing; fertilisation then gives 2n + 1 or 2n − 1.
DisorderKaryotypeFeatures
Down's syndrome47; trisomy 21Short stature, small round head, furrowed tongue, partially open mouth, broad palm with characteristic palm crease; retarded physical, psychomotor and mental development.
Klinefelter's syndrome47, XXYMale with overall masculine development plus gynaecomastia (breast development); sterile.
Turner's syndrome45, XOFemale; sterile with rudimentary ovaries; secondary sexual characters lacking.

Standard question patternsMust

Probability of one genotype from a dihybrid cross
Split into single genes and multiply: RrYy × RrYy → P(rryy) = ¼ × ¼ = 1/16.
Gamete types, genotypes, phenotypes for n heterozygous genes
2n, 3n, 2n.
F₂ ratio 1 : 2 : 1 for phenotypes
Incomplete dominance (snapdragon) — not a Mendelian 3 : 1 case.
Parents of blood groups A and B have an O child
Parents must be IAi and IBi; all four groups possible, ¼ each.
Unaffected parents, affected child
Recessive trait. Mostly sons affected through carrier mothers → X-linked recessive.
Karyotype match
Down 47 (+21) · Klinefelter 47, XXY · Turner 45, XO.
Recombination percentages compared
Higher percentage → genes farther apart (1.3% tightly linked; 37.2% loosely linked).
Statement set on ABO or sex determination
Check each statement against the table: codominance, 6 genotypes, 4 phenotypes; birds female heterogametic.

Traps that cost marksMust

Trap · Ratio reversal

F₂ phenotypes 3 : 1 but genotypes 1 : 2 : 1. Dihybrid F₂ 9 : 3 : 3 : 1 but dihybrid test cross 1 : 1 : 1 : 1. Read whether the question asks for the F₂ or the test cross.

Trap · Who is heterogametic

Humans and grasshopper: male. Birds: female (ZW). Swapping these is a standard wrong option.

Trap · Attribution

Morgan: linkage and recombination. Sturtevant: gene maps. Sutton and Boveri: chromosomal theory. Henking: X body. Do not credit maps to Morgan.

Trap · Right content, wrong arrangement

In match-the-column items Klinefelter (XXY, male, sterile) and Turner (XO, female, sterile) are swapped deliberately. Tick each row separately before you look at the options.

Trap · Assertion–reason

'Sickle-cell anaemia is a qualitative defect' is true; the reason 'it is caused by a frameshift mutation' is false — it is a point mutation. Judge A and R separately, then ask whether R explains A.

Ten-question checkMust

NEET Biology now leans on statements, match-the-column and assertion–reason. Mark each statement true or false before you read the options. Aim for 9/10 in 9 minutes.

Q1.In the F₂ of a dihybrid cross RrYy × RrYy, the fraction of plants homozygous for both gene pairs is

  1. (A)1/16
  2. (B)9/16
  3. (C)1/8
  4. (D)1/4
Show answer

Answer (D). Each gene gives ½ homozygous (RR or rr). ½ × ½ = ¼, i.e. 4 of 16 squares: RRYY, RRyy, rrYY, rryy. 1/16 counts only rryy.

Q2.A woman with blood group A and a man with blood group B have a child with blood group O. Blood groups possible among their children are

  1. (A)A and B only
  2. (B)A, B and AB only
  3. (C)A, B, AB and O
  4. (D)AB and O only
Show answer

Answer (C). An O child means both parents carry i: IAi × IBi. Offspring IAIB, IAi, IBi and ii — AB, A, B and O, each ¼.

Q3.Consider the statements about human ABO blood groups:
a. IA and IB are co-dominant.
b. Three alleles give six possible genotypes.
c. Allele i is dominant over IA.
d. There are four possible phenotypes.
Choose the correct option.

  1. (A)a and c only
  2. (B)b, c and d only
  3. (C)a, b, c and d
  4. (D)a, b and d only
Show answer

Answer (D). Statement c is false: IA and IB are both dominant over i.

Q4.Match Column I with Column II:
A. Down's syndrome   B. Klinefelter's syndrome   C. Turner's syndrome   D. Sickle-cell anaemia
i. 47, XXY   ii. Glu → Val at 6th position of β-globin   iii. Trisomy of chromosome 21   iv. 45, XO

  1. (A)A-iii, B-i, C-ii, D-iv
  2. (B)A-i, B-iii, C-iv, D-ii
  3. (C)A-iv, B-i, C-iii, D-ii
  4. (D)A-iii, B-i, C-iv, D-ii
Show answer

Answer (D). Down = trisomy 21; Klinefelter = XXY; Turner = XO; sickle-cell = Glu → Val. The first distractor swaps only rows C and D.

Q5.Assertion (A): A haemophilic man married to a normal (non-carrier) woman has no haemophilic sons.
Reason (R): Sons receive their X chromosome from their mother.

  1. (A)Both A and R are true, and R is the correct explanation of A
  2. (B)Both A and R are true, but R is not the correct explanation of A
  3. (C)A is true, but R is false
  4. (D)A is false, but R is true
Show answer

Answer (A). Both true. The father gives sons his Y, and their X comes from the non-carrier mother — that is exactly why no son is affected.

Q6.Assertion (A): In snapdragon the F₂ phenotypic ratio is the same as the genotypic ratio.
Reason (R): The heterozygote Rr is phenotypically different from both homozygotes.

  1. (A)Both A and R are true, and R is the correct explanation of A
  2. (B)Both A and R are true, but R is not the correct explanation of A
  3. (C)A is true, but R is false
  4. (D)A is false, but R is true
Show answer

Answer (A). Both true, and because pink Rr is distinguishable from red RR and white rr, phenotypes map one-to-one on to genotypes (1 : 2 : 1).

Q7.Assertion (A): Individuals with Turner's syndrome are sterile females.
Reason (R): Turner's syndrome is caused by an additional copy of the X chromosome.

  1. (A)Both A and R are true, and R is the correct explanation of A
  2. (B)Both A and R are true, but R is not the correct explanation of A
  3. (C)A is true, but R is false
  4. (D)A is false, but R is true
Show answer

Answer (C). A is true. R is false: Turner's is 45, XO — one X is absent. An additional X in a male is Klinefelter's (XXY).

Q8.In Morgan's crosses, the recombination frequency between the genes for white eye and miniature wing in Drosophila was

  1. (A)37.2%
  2. (B)1.3%
  3. (C)50%
  4. (D)62.8%
Show answer

Answer (A). White–miniature: 37.2% (loosely linked). 1.3% is yellow body–white eye (tightly linked).

Q9.In birds, the sex of an offspring is decided by

  1. (A)the sperm, because the male is ZW
  2. (B)the egg, because the female is ZW
  3. (C)the sperm, because the male is XY
  4. (D)the egg, because the female is XO
Show answer

Answer (B). Birds show female heterogamety: ZW female, ZZ male. Eggs carry Z or W; all sperm carry Z.

Q10.Which statement correctly distinguishes thalassemia from sickle-cell anaemia?

  1. (A)Both are X-linked recessive disorders
  2. (B)Thalassemia is a quantitative defect in globin synthesis; sickle-cell anaemia is a qualitative defect
  3. (C)β-thalassemia is controlled by HBA1 and HBA2 on chromosome 16
  4. (D)Sickle-cell anaemia is caused by a frameshift deletion
Show answer

Answer (B). Both are autosomal recessive. HBA1/HBA2 on chromosome 16 control α-globin; β is HBB on chromosome 11. Sickle-cell is a point mutation.

60-second recap before the paperMust

  1. Pea, 7 traits; pod green dominant but seed yellow dominant.
  2. Mono F₂ 3 : 1 (1 : 2 : 1); test cross 1 : 1. Di F₂ 9 : 3 : 3 : 1; test cross 1 : 1 : 1 : 1.
  3. Snapdragon pink = incomplete dominance, 1 : 2 : 1. ABO = co-dominance + multiple alleles; 6 genotypes, 4 phenotypes.
  4. Pleiotropy: PKU. Polygenic: skin colour. Starch grains: dominance depends on the level of phenotype.
  5. Sutton–Boveri theory; Morgan linkage/recombination (1.3% vs 37.2%); Sturtevant maps.
  6. Henking X body. XO grasshopper, XY human (male heterogametic), ZW birds (female heterogametic), bee 2n = 32 / n = 16.
  7. Sickle-cell: GAG → GUG, Glu → Val at 6; point mutation; qualitative. Thalassemia quantitative (α: ch 16; β: ch 11).
  8. X-linked recessive: haemophilia, colour blindness (8% ♂, 0.4% ♀). Autosomal recessive: sickle-cell, PKU, thalassemia.
  9. Down 47 (+21) · Klinefelter 47 XXY · Turner 45 XO. Aneuploidy vs polyploidy.