10 interactive concept widgets for Principles of Inheritance and Variation. Drag any slider, change any number, and watch the formula and the answer update live. Built so you understand how each NEET problem actually works, not just the final number.
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Pick parent genotypes and see all offspring with ratios. Includes preset crosses for monohybrid F1, F2, and the test cross.
Pick the genotypes of two parents (T = tall, dominant; t = dwarf, recessive). The Punnett square shows all possible offspring genotypes with their ratios.
Parent 1 (mother): Tt
Heterozygous (carrier)
Parent 2 (father): Tt
Heterozygous (carrier)
Tt × Tt
TT
Tall
Tt
Tall
Tt
Tall
tt
Dwarf
Genotypic ratio
1 TT : 2 Tt : 1 tt
(out of 4)
Phenotypic ratio
3 Tall : 1 Dwarf
(out of 4)
Try these classic crosses
NEET key facts
!
F1 of TT x tt = all Tt (heterozygous, all show DOMINANT phenotype).
!
F2 of Tt x Tt = 1 TT : 2 Tt : 1 tt (genotypic) and 3 tall : 1 dwarf (phenotypic) = 3:1.
!
TEST CROSS (Tt x tt): if heterozygous → 1:1 ratio. If homozygous (TT x tt) → all tall.
!
Mendel's laws: Dominance + Segregation (from monohybrid). Independent Assortment (from dihybrid).
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Mendel's classic dihybrid cross with full 4x4 Punnett square. Visualise round/wrinkled and yellow/green seed combinations.
Mendel's classic dihybrid cross with pea (round/wrinkled, yellow/green seeds). The F2 generation from RrYy x RrYy gives the famous 9:3:3:1 phenotypic ratio.
F1 self-cross: RrYy × RrYy
R = round (dominant), r = wrinkled; Y = yellow (dominant), y = green
RRYY
Round
Yellow
RRYy
Round
Yellow
RrYY
Round
Yellow
RrYy
Round
Yellow
RRyY
Round
Yellow
RRyy
Round
Green
RryY
Round
Yellow
Rryy
Round
Green
rRYY
Round
Yellow
rRYy
Round
Yellow
rrYY
Wrinkled
Yellow
rrYy
Wrinkled
Yellow
rRyY
Round
Yellow
rRyy
Round
Green
rryY
Wrinkled
Yellow
rryy
Wrinkled
Green
Phenotypic ratio (the famous 9:3:3:1)
Round Yellow
9/16
Both dominant; parental type
Round Green
3/16
Round dom + green rec; recombinant
Wrinkled Yellow
3/16
Wrinkled rec + yellow dom; recombinant
Wrinkled Green
1/16
Both recessive; parental type
9 : 3 : 3 : 1
(Round-Yellow : Round-Green : Wrinkled-Yellow : Wrinkled-Green)
Why 9:3:3:1?
The 9:3:3:1 ratio is the PRODUCT of two independent 3:1 ratios (one for each gene). Mathematically:
This works ONLY when the two genes are on DIFFERENT chromosomes (independent assortment). For LINKED genes, the ratio is different.
NEET key facts
!
Dihybrid F2 phenotypic ratio = 9:3:3:1.
!
Genotypic ratio = 1:2:1:2:4:2:1:2:1 (9 genotype classes - more complex).
!
Mendel's 3rd law: INDEPENDENT ASSORTMENT - alleles of different genes assort independently into gametes.
!
Test cross of RrYy x rryy gives 1:1:1:1 phenotypic ratio (4 gamete types revealed).
!
For LINKED genes, the 9:3:3:1 ratio fails. Linked genes show parental types more than recombinants.
Try this
Four major inheritance patterns: complete dominance, incomplete dominance, codominance, multiple alleles. Compare heterozygote phenotypes and F2 ratios.
Four major inheritance patterns: complete dominance, incomplete dominance, codominance, and multiple alleles. Click each to compare heterozygotes, F2 ratios, and examples.
Complete Dominance
Heterozygote looks like the dominant parent
When two contrasting alleles are present, only the DOMINANT allele is expressed. The recessive allele is masked. Heterozygote is phenotypically identical to the homozygous dominant. Mendel's discoveries are based on this.
Heterozygote phenotype:
Same as dominant homozygote
F2 ratio:
F2 = 3:1 phenotypic; 1:2:1 genotypic
Classic examples:
• Mendel's pea: tall x dwarf → tall heterozygote
• Mendel's 7 traits in pea
• Most simple Mendelian crosses
Quick comparison
NEET key facts
!
Complete dominance: heterozygote = dominant homozygote. Mendel's pea (3:1).
!
Incomplete dominance: heterozygote = INTERMEDIATE (e.g., pink Mirabilis). 1:2:1 ratio.
!
Codominance: heterozygote = BOTH phenotypes (e.g., AB blood, roan coat). 1:2:1 ratio.
!
Multiple alleles: more than 2 alleles in population. ABO has 3 (IA, IB, i).
!
IA and IB are CODOMINANT to each other. Both are DOMINANT over i. So O blood = ii (recessive).
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3 alleles (IA, IB, i) → 4 blood groups (A, B, AB, O). Pick parent genotypes and see possible children blood groups with reference table.
The ABO system has 3 alleles (IA, IB, i) giving 4 blood groups (A, B, AB, O). IA and IB are codominant; i is recessive to both. Pick parents to see possible offspring blood groups.
The 3 alleles
IA
A antigen
IB
B antigen
i
No antigen
IA and IB are CODOMINANT to each other. Both DOMINANT over i.
Parent 1 (mother): IAi → Group
Parent 2 (father): IBi → Group
IAi × IBi
IAIB
Group AB
IAi
Group A
iIB
Group B
ii
Group O
Offspring blood group probabilities
A
1/4
= 25%
B
1/4
= 25%
AB
1/4
= 25%
O
1/4
= 25%
ABO blood group reference
NEET key facts
!
ABO has 3 alleles: IA, IB, i. Multiple alleles example.
!
IA and IB CODOMINANT (both expressed in IA IB → AB blood). Both DOMINANT over i.
!
Genotype to phenotype: IA IA or IA i = A. IB IB or IB i = B. IA IB = AB. ii = O.
!
O is universal donor (no A or B antigens to react with recipient antibodies).
!
AB is universal recipient (no antibodies; can accept any blood type).
!
AB man + O woman → only A or B children possible. NEET classic problem.
Try this
Four systems across organisms: XY (humans), XO (grasshoppers), ZW (birds), ZO (some moths). Click each to see who is heterogametic.
Four major chromosomal sex determination systems. The HETEROGAMETIC sex (which produces 2 types of gametes) determines the offspring sex. Click each to see who is heterogametic.
XY system (humans)
The classic mammal system. The Y chromosome carries male-determining genes (SRY in humans). Male is heterogametic. Father determines sex of children (X-bearing sperm = girl, Y-bearing = boy). 22 pairs of autosomes + XX or XY = 46 chromosomes in humans.
FEMALE
XX
MALE
XY
Heterogametic sex (determines offspring sex):
MALE (produces X and Y sperm)
Homogametic sex:
FEMALE (all eggs are X)
Examples:
• Humans
• Most mammals
• Drosophila (fruit fly)
NEET key facts
!
XY system (humans, Drosophila): male XY (heterogametic), female XX. SRY gene on Y determines maleness.
!
XO system (grasshopper): male XO (only ONE X, no Y), female XX. Males produce X-bearing and no-chromosome sperm.
!
ZW system (birds, butterflies): FEMALE ZW (heterogametic), male ZZ. Reverse of XY.
!
ZO system (some moths): female ZO, male ZZ.
!
In humans, the FATHER determines the child's sex (he provides X or Y). Probability of son = 50%.
!
NEET memory: XY = male heterogametic, ZW = female heterogametic.
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Why haemophilia and colour blindness are more common in males. Pick parent genotypes and see daughter / son outcomes.
X-linked recessive traits like haemophilia and colour blindness are more common in males. Pick parents and see offspring outcomes for each sex.
X = normal allele. Xc = recessive disease allele (e.g., haemophilia). Y has no allele for this trait.
Mother: XcX (Carrier (heterozygous))
Father: XY (Normal)
XXc
CARRIER daughter
XcY
AFFECTED son
XX
Normal daughter
XY
Normal son
Why are males more affected by X-linked recessive traits?
Males have only ONE X chromosome (XY). If they inherit the recessive allele Xc, there is NO second X to mask it. So a single recessive allele is enough to cause the disease. Males are HEMIZYGOUS for X-linked genes.
Females have TWO X chromosomes (XX). They need TWO recessive alleles (XcXc) to express the disease. With one recessive allele (XcX), they are CARRIERS - they have the allele but do not show the disease.
NEET key facts
!
X-linked recessive: haemophilia, colour blindness, Duchenne muscular dystrophy (DMD).
!
Affected MALES have only ONE X (XcY) - hemizygous - single allele expresses.
!
CARRIER females are XcX (heterozygous) - phenotypically normal but pass the allele.
!
A father's X-linked allele is passed to ALL daughters (never to sons; sons get Y from father).
!
A carrier mother (XcX) x normal father (XY): 50% sons affected, 50% daughters carriers.
Try this
Recombination frequency calculator. RF = recombinants / total × 100. Visualise linkage strength on a 0 to 50% scale.
When two genes are on the same chromosome, they tend to be inherited together. Crossing over creates recombinants. The recombination frequency tells you how close the genes are.
Out of 100 test cross offspring, how many are PARENTAL (non-recombinant) types?
85
So 15 are recombinant types (new combinations).
Recombination frequency
15.0%
= 15.0 cM (centimorgans)
MEDIUM linkage
The formula
1 cM (centimorgan) = 1% recombination frequency.
MAXIMUM possible RF = 50% (genes effectively unlinked, on different chromosomes or very far apart).
Linkage strength scale
T.H. Morgan and Drosophila
Thomas Hunt Morgan and his students (Sturtevant, Bridges, Muller) discovered linkage in Drosophila melanogaster (fruit fly). They found that some genes did NOT assort independently as Mendel had predicted - they tended to be inherited together. They concluded these genes are on the SAME chromosome (linked). Crossing over during meiosis I separates linked genes, creating recombinants. Morgan won the Nobel Prize in 1933 for this work.
NEET key facts
!
Linked genes: located on the SAME chromosome. Tend to be inherited together.
!
Crossing over during meiosis I creates RECOMBINANT gametes.
!
Recombination frequency (RF) = (recombinants / total) × 100. Maximum = 50%.
!
1 cM (centimorgan) = 1% RF. Used to map gene positions on chromosomes.
!
Mendel's law of independent assortment WORKS only for genes on different chromosomes.
!
Morgan studied this in Drosophila. Sturtevant created the first genetic map.
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Five major aneuploidies: Down's, Klinefelter's, Turner's, Edward's, Patau. All caused by non-disjunction.
Five major chromosomal disorders. All caused by ANEUPLOIDY (abnormal chromosome number from non-disjunction). Click each to see karyotype, symptoms, and discovery.
Down's syndrome
Karyotype
Trisomy 21 (47, +21)
Total chromosomes
47
Type
Autosomal
Cause:
Extra copy of chromosome 21 (3 copies instead of 2). Caused by NON-DISJUNCTION of chromosome 21 during meiosis I or II.
Key symptoms / features:
• Short stature
• Characteristic facial features (small round face, flat nasal bridge, slanted eyes)
• Intellectual disability (mild to moderate)
• Congenital heart defects (~50% of cases)
• Single deep transverse palm crease (simian crease)
• Increased risk with maternal age over 35
Discovered: Langdon Down (1866) - clinical description; cause identified by Lejeune in 1959
What is non-disjunction?
NON-DISJUNCTION is the failure of homologous chromosomes (in meiosis I) or sister chromatids (in meiosis II) to separate properly. Result: one daughter cell gets an EXTRA chromosome (n+1 gamete) and another gets ONE FEWER (n-1 gamete). When these abnormal gametes fertilise normal ones, the zygote has extra (trisomy) or missing (monosomy) chromosomes.
ANEUPLOIDY = abnormal chromosome number. POLYPLOIDY = extra full chromosome SETS (common in plants, e.g., 4n).
NEET key facts
!
Down's syndrome = TRISOMY 21 (47 chromosomes). Autosomal aneuploidy. Risk increases with maternal age.
!
Klinefelter's syndrome = 47, XXY. Extra X in MALE. Sterile, tall, gynecomastia.
!
Turner's syndrome = 45, XO. MISSING X in female. Short, sterile, no secondary sexual characteristics.
!
Edward's syndrome = trisomy 18. Patau syndrome = trisomy 13. Both severe, usually fatal.
!
All caused by NON-DISJUNCTION (failure of chromosomes to separate during meiosis).
!
ANEUPLOIDY = abnormal number. POLYPLOIDY = extra full sets (4n, 6n in plants).
Try this
Standard pedigree symbols and decision tree. Identify autosomal vs X-linked, dominant vs recessive from family trees.
A pedigree is a family tree showing inheritance of a trait. Identifying the pattern (autosomal dominant, recessive, X-linked dominant, or recessive) helps determine genotypes and predict offspring.
Pedigree symbols
□
Unaffected male
■
Affected male
○
Unaffected female
●
Affected female
⊡
Carrier male (rare)
⊙
Carrier female
─
Marriage line
|
Parent-offspring line
Autosomal Recessive
Pedigree features:
• May SKIP generations (carriers do not show the trait)
• Often appears in offspring of unaffected parents (both carriers)
• Both sexes equally affected
• Increased frequency with consanguineous marriage
Inheritance rules:
• Two unaffected carriers (Aa x Aa) have ~25% affected children
• Two affected parents (aa x aa) have ALL affected children
• Horizontal pattern in pedigree (siblings affected, parents not)
Examples:
• Sickle cell anaemia
• Cystic fibrosis
• Phenylketonuria (PKU)
• Albinism
• Tay-Sachs disease
How to identify the pattern from a pedigree
NEET key facts
!
Pedigree symbols: square = male, circle = female, filled = affected, half-filled = carrier.
!
Autosomal recessive: skips generations, two carriers can have affected child. Sickle cell, PKU, albinism.
!
Autosomal dominant: appears in every generation. Marfan, Huntington's, polydactyly.
!
X-linked recessive: more in MALES, skips through carrier mothers. Haemophilia, colour blindness, DMD.
!
NEET trick: father-to-son transmission RULES OUT X-linked (sons get Y from father, not X).
Try this
12-question scored quiz covering Mendel, monohybrid / dihybrid crosses, ABO blood, sex determination, linkage, and chromosomal disorders.
One question at a time. Pick an option, see the explanation, then move to the next.
Mendel chose Pisum sativum (garden pea) primarily because:
A. Pea is the only plant with contrasting traits
B. Pea has true-breeding lines, contrasting traits, controllable pollination, and large progeny
C. Pea grows only in temperate climates
D. Pea has many chromosomes
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