🔀 Epistasis, Polygenic Inheritance & Cytoplasmic Inheritance
Learn epistasis types — complementary (9:7), supplementary (9:3:4), duplicate (15:1) for CUET Agriculture. Polygenic and cytoplasmic inheritance.
Non-Allelic Gene Interactions (Epistasis)
Non-allelic gene interactions occur between genes located on different loci (different chromosomes or far apart on the same chromosome). These interactions modify the standard 9:3:3:1 dihybrid ratio into various modified ratios. Understanding these modifications is one of the most frequently tested topics in CUET genetics.
Epistasis: When one non-allelic gene suppresses the effect of another gene, it is called epistasis. Think of it as "dominance between different genes" rather than between alleles of the same gene.
- The gene that suppresses is called the Epistatic gene (the one in control)
- The gene that gets suppressed is called the Hypostatic gene (the one being masked)
- This is essentially dominance expressed between non-allelic genes
TIP
To identify which type of epistasis is occurring, always add up the numbers in the modified ratio — they should total 16 (since it is a modification of the 9:3:3:1 dihybrid ratio where 9+3+3+1 = 16).
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Non-Allelic Gene Interactions (Epistasis)
Non-allelic gene interactions occur between genes located on different loci (different chromosomes or far apart on the same chromosome). These interactions modify the standard 9:3:3:1 dihybrid ratio into various modified ratios. Understanding these modifications is one of the most frequently tested topics in CUET genetics.
Epistasis: When one non-allelic gene suppresses the effect of another gene, it is called epistasis. Think of it as "dominance between different genes" rather than between alleles of the same gene.
- The gene that suppresses is called the Epistatic gene (the one in control)
- The gene that gets suppressed is called the Hypostatic gene (the one being masked)
- This is essentially dominance expressed between non-allelic genes
TIP
To identify which type of epistasis is occurring, always add up the numbers in the modified ratio — they should total 16 (since it is a modification of the 9:3:3:1 dihybrid ratio where 9+3+3+1 = 16).
1. Complementary Genes (9:7)
Two genes at different loci, both needed in dominant form for the expression of a character. If either gene is homozygous recessive, the character is not expressed. The two genes "complement" each other — both must be present for the trait to appear.
Example: Flower Colour in Sweet Pea (Lathyrus)
Both gene C and gene P must be present in dominant form for coloured flowers. Each gene alone is not sufficient.
Parents: CCPP (Coloured) × ccpp (Colourless)
F1: CcPp (All Coloured)
F2: (Self-pollination)
F2 Punnett Square (4 x 4):
| CP | Cp | cP | cp | |
|---|---|---|---|---|
| CP | CCPP (Coloured) | CCPp (Coloured) | CcPP (Coloured) | CcPp (Coloured) |
| Cp | CCPp (Coloured) | CCpp (Colourless) | CcPp (Coloured) | Ccpp (Colourless) |
| cP | CcPP (Coloured) | CcPp (Coloured) | ccPP (Colourless) | ccPp (Colourless) |
| cp | CcPp (Coloured) | Ccpp (Colourless) | ccPp (Colourless) | ccpp (Colourless) |
F2 Ratio: Coloured : Colourless = 9 : 7
- At least one dominant allele of both genes (C_ and P_) is required for colour
- Any combination with cc__ or __pp gives colourless flowers
- The 9:7 ratio comes from: 9 (C_P_) are coloured; 3 (C_pp) + 3 (ccP_) + 1 (ccpp) = 7 are colourless
NOTE
In complementary gene interaction, two separate biochemical pathways must both be functional. Gene C might produce an enzyme for one step, and gene P for another. If either enzyme is missing (homozygous recessive), the end product (colour) is not formed.
2. Dominant Epistasis (12:3:1)
One dominant gene (epistatic) masks the expression of another gene at a different locus, regardless of the other gene's alleles. The epistatic gene has a visible phenotype of its own.
Example: Fruit Colour in Summer Squash (Cucurbita pepo)
- Y = dominant allele for yellow fruit colour
- y = recessive allele (allows green)
- W = epistatic gene — when dominant (W_), it masks Y/y and produces white fruit
- w = recessive (no epistatic effect)
Parents: WWYY (White) × wwyy (Green)
F1: WwYy (All White)
F2:
F2 Punnett Square:
| WY | Wy | wY | wy | |
|---|---|---|---|---|
| WY | WWYY (White) | WWYy (White) | WwYY (White) | WwYy (White) |
| Wy | WWYy (White) | WWyy (White) | WwYy (White) | Wwyy (White) |
| wY | WwYY (White) | WwYy (White) | wwYY (Yellow) | wwYy (Yellow) |
| wy | WwYy (White) | Wwyy (White) | wwYy (Yellow) | wwyy (Green) |
F2 Phenotypic Ratio: White : Yellow : Green = 12 : 3 : 1
The logic: When W is present (dominant), the fruit is always white regardless of the Y gene (9 W_Y_ + 3 W_yy = 12 white). When ww and Y_ are present, the fruit is yellow (3). When both are recessive (wwyy), the fruit is green (1).
3. Recessive Epistasis (Supplementary Gene Interaction) (9:3:4)
A recessive allele at one locus, when homozygous, masks the expression of alleles at another locus. The recessive epistatic gene must be homozygous (cc) to suppress the other gene.
Example: Coat Colour in Mice
- C = gene for coat colour (C_ allows colour expression — it acts as a "switch")
- c = recessive (cc = no colour = albino, regardless of A gene)
- A = Agouti (brown) colour gene
- a = non-agouti (black)
Parents: CCaa (Black) × ccAA (Albino)
F1: CcAa (All Agouti)
F2:
F2 Punnett Square:
| CA | Ca | cA | ca | |
|---|---|---|---|---|
| CA | CCAA (Agouti) | CCAa (Agouti) | CcAA (Agouti) | CcAa (Agouti) |
| Ca | CCAa (Agouti) | CCaa (Black) | CcAa (Agouti) | Ccaa (Black) |
| cA | CcAA (Agouti) | CcAa (Agouti) | ccAA (Albino) | ccAa (Albino) |
| ca | CcAa (Agouti) | Ccaa (Black) | ccAa (Albino) | ccaa (Albino) |
F2 Phenotypic Ratio: Agouti : Black : Albino = 9 : 3 : 4
- The cc genotype is epistatic over the A/a gene — when cc is present, the animal is always albino regardless of A or a (because the colour "switch" is off)
- The 9:3:4 comes from: 9 (C_A_) Agouti + 3 (C_aa) Black + 3 (ccA_) Albino + 1 (ccaa) Albino = 9:3:4
4. Duplicate Genes (15:1)
Two genes at different loci produce the same phenotypic effect. Either gene alone (in dominant form) can produce the trait. Only the double recessive (both genes homozygous recessive) shows the alternative phenotype.
Example: Kernel Colour in Wheat
Two genes (A and B) both contribute to kernel colour:
- Any dominant allele of A or B → Coloured kernel
- Only aabb → White kernel
F2 Ratio: Coloured : White = 15 : 1
This is a "backup system" — either gene can do the job alone, so only when both fail (double recessive) does the alternative phenotype appear.
5. Inhibitory Genes (13:3)
One dominant gene inhibits the expression of another gene. The inhibitor gene itself has no visible phenotypic effect — it only prevents the other gene from expressing.
Example: Feather Colour in Poultry (Fowl)
- C = gene for coloured feathers
- I = inhibitor gene (dominant I_ inhibits colour expression)
F2 Ratio: White : Coloured = 13 : 3
Breaking down the ratio:
- 9 (C_I_) = White (colour gene present but inhibited by I)
- 3 (ccI_) = White (no colour gene present)
- 1 (ccii) = White (no colour gene, no inhibitor)
- 3 (C_ii) = Coloured (colour gene expressed, no inhibitor to stop it)
Total White = 9 + 3 + 1 = 13; Coloured = 3
WARNING
Do not confuse inhibitory genes (13:3) with dominant epistasis (12:3:1). In inhibitory genes, the inhibitor itself produces no phenotype — it only blocks the other gene. In dominant epistasis, the epistatic gene has its own visible phenotype (like white fruit colour).
Complete Summary Table of Modified Dihybrid Ratios
This table is one of the most important tables for CUET genetics. Memorize the ratios and their classic examples:
| Type of Interaction | Modified F2 Ratio | Standard 9:3:3:1 Modification | Classic Example |
|---|---|---|---|
| Normal dihybrid | 9:3:3:1 | No modification | Seed shape and colour in pea |
| Complementary genes | 9:7 | 9:(3+3+1) | Flower colour in sweet pea (Lathyrus) |
| Dominant epistasis | 12:3:1 | (9+3):3:1 | Fruit colour in summer squash |
| Recessive epistasis | 9:3:4 | 9:3:(3+1) | Coat colour in mice |
| Duplicate genes | 15:1 | (9+3+3):1 | Kernel colour in wheat |
| Inhibitory genes | 13:3 | (9+3+1):3 | Feather colour in poultry |
| Duplicate recessive epistasis | 9:7 | 9:(3+3+1) | Flower colour in Lathyrus |
Memory trick for modified ratios
All modified dihybrid ratios add up to **16**. Here is a quick way to remember them:- 9:7 = Complementary (both needed) — 9 + 7 = 16
- 12:3:1 = Dominant epistasis (one dominates everything) — 12 + 3 + 1 = 16
- 9:3:4 = Recessive epistasis (recessive masks) — 9 + 3 + 4 = 16
- 15:1 = Duplicate (either gene works) — 15 + 1 = 16
- 13:3 = Inhibitory (blocks expression) — 13 + 3 = 16
Polygenic Inheritance (Quantitative Inheritance)
Mendel studied qualitative characters (clearly distinguishable categories like tall vs dwarf). However, many traits in nature show continuous variation — these are controlled by multiple genes (polygenes), each contributing a small additive effect.
Key Features of Polygenic Inheritance
- Many genes together control a single trait — each has a small, additive effect
- The trait shows continuous variation (no distinct categories — think of human height, which varies continuously)
- Each gene has a contributing/dominant allele (adds to the trait) and a non-contributing/recessive allele (adds nothing)
- The number of phenotypic classes = (2n + 1), where n = number of gene pairs
- Intermediate values are most common; extremes are rare
- Produces a bell-shaped (normal distribution) curve in F2 — most individuals cluster around the mean
Example 1: Kernel Colour in Wheat
Studied by Nilsson-Ehle. Two gene pairs (A and B) control kernel colour. Each dominant allele contributes equally to colour intensity.
Parents: AABB (Dark Red) × aabb (White)
F1: AaBb (Intermediate Red)
F2 Phenotype Distribution:
| Phenotype | No. of Dominant Alleles | Genotypes | F2 Frequency |
|---|---|---|---|
| Full Red | 4 | AABB | 1 |
| Light Red | 3 | AABb (2), AaBB (2) | 4 |
| Intermediate Red | 2 | AAbb (1), AaBb (4), aaBB (1) | 6 |
| Very Light Red | 1 | Aabb (2), aaBb (2) | 4 |
| White | 0 | aabb | 1 |
F2 Ratio: 1 : 4 : 6 : 4 : 1
Number of phenotypes = 2n + 1 = 2(2) + 1 = 5 phenotypes
NOTE
The 1:4:6:4:1 ratio follows the binomial distribution — the same mathematical pattern as Pascal's triangle. The most common phenotype class (6/16) is the intermediate, and the extremes (1/16 each) are the rarest.
Example 2: Skin Colour in Humans
Studied by Davenport. Human skin colour is controlled by 3 gene pairs (A, B, C):
Parents: AABBCC (Negro/Black) × aabbcc (White)
F1: AaBbCc (Intermediate/Mullato)
F2 Phenotype Distribution:
| Phenotype | No. of Dominant Alleles | F2 Frequency |
|---|---|---|
| Negro (Black) | 6 | 1 |
| Very Dark Brown | 5 | 6 |
| Dark Brown | 4 | 15 |
| Mullato (Brown) | 3 | 20 |
| Light Brown | 2 | 15 |
| Very Light Brown | 1 | 6 |
| White | 0 | 1 |
F2 Ratio: 1 : 6 : 15 : 20 : 15 : 6 : 1
Number of phenotypes = 2n + 1 = 2(3) + 1 = 7 phenotypes
Cytoplasmic Inheritance (Extra-nuclear Inheritance)
Discovered by Carl Correns in Mirabilis jalapa (4 O'Clock plant). This type of inheritance breaks all of Mendel's rules because the genes involved are not on nuclear chromosomes.
Key Features
- Inheritance of traits controlled by genes located in cytoplasmic organelles (not nuclear chromosomes)
- The gene is located in cytoplasm — in Cytogene or Plasmagene
- Gene located in the nucleus is called Karyogene
- In higher plants, cytoplasmic inheritance is through the mother only (maternal inheritance), because the egg cell is large and contributes most of the cytoplasm
- The male gamete (pollen) contributes the male nucleus only — it has very little cytoplasm, so only karyogene inheritance occurs through males
- Thus, cytoplasmic inheritance is uniparental (maternal)
- If reciprocal crosses give different results, cytoplasmic inheritance is indicated
- Cytoplasmic genes are found in organelles like:
- Plastids (chloroplasts) — Plasmagene
- Mitochondria — Mitochondrial DNA (Cytoplasmin)
IMPORTANT
The key diagnostic test for cytoplasmic inheritance is the reciprocal cross. If swapping the male and female parents gives different F1 results, the trait is cytoplasmically inherited. In nuclear (karyogene) inheritance, reciprocal crosses give the same results.
Examples of Cytoplasmic Inheritance
(a) Leaf Variegation in Mirabilis jalapa (4 O'Clock Plant)
Branch colour inheritance depends entirely on the maternal parent:
| Cross | Female (♀) | Male (♂) | F1 Offspring |
|---|---|---|---|
| (a) | Pale (white) | Green | All Pale |
| (b) | Green | Pale | All Green |
| (c) | Variegated | Any male | Pale, Green, and Variegated |
- The colour of offspring depends entirely on the mother plant — the father's contribution does not matter
- This is because plastids (chloroplasts) are inherited through the egg cytoplasm
- Variegated mothers produce all three types because their eggs contain different proportions of normal and abnormal plastids — some eggs get mostly normal plastids (green), some get mostly abnormal (pale), and some get a mix (variegated)
(b) Male Sterility in Maize (Corn)
- Some maize plants have cytoplasmic male sterility (CMS) — they cannot produce functional pollen
- The gene for male sterility is located in mitochondrial DNA
- If a male sterile plant is crossed with a normal female plant, the offspring are normal (fertile)
- But if a normal plant is crossed with a male sterile plant (as female parent), the offspring are male sterile
- This occurs because the mitochondrial gene for sterility is passed through the egg cytoplasm
- Male sterility gene is inherited maternally
TIP
CMS (Cytoplasmic Male Sterility) is extremely important in agriculture — it is used commercially in hybrid seed production. By using CMS lines as female parents, breeders can produce hybrid seeds without manually emasculating thousands of flowers.
Golden Key Points
| Point | Detail |
|---|---|
| Incomplete dominance F2 ratio | 1:2:1 (phenotypic = genotypic) |
| Codominance | Both alleles expressed equally |
| Lethal gene modified ratio | 2:1 (instead of 3:1) |
| Complementary genes | 9:7 |
| Dominant epistasis | 12:3:1 |
| Recessive epistasis | 9:3:4 |
| Duplicate genes | 15:1 |
| Inhibitory genes | 13:3 |
| Polygenic inheritance phenotypes | (2n + 1) where n = number of gene pairs |
| Multiple alleles genotypes formula | n(n+1)/2 |
| Cytoplasmic inheritance | Maternal (uniparental), discovered by Correns |
| Pleiotropic gene | One gene → multiple phenotypic effects |
Practice Questions (Beginner's Box)
-
In codominance, the F2 phenotypic ratio for a monohybrid cross will be:
- (1) 3:1 (2) 1:2:1 (3) 1:1:1:1 (4) 2:2
- Answer: (2) 1:2:1
-
In polygenic inheritance, if one trait is controlled by two gene pairs, the two parents that are heterozygous for both, the phenotypic ratio from their cross will be:
- (1) 1:2:1 (2) 9:3:3:1 (3) 1:4:6:4:1 (4) 1:6:15:20:15:6:1
- Answer: (3) 1:4:6:4:1
Explanation
With 2 gene pairs (n=2), the number of phenotypic classes = 2n + 1 = 5. The F2 ratio follows the binomial distribution: 1:4:6:4:1. This represents the frequency of individuals with 0, 1, 2, 3, and 4 dominant alleles respectively.-
For Lathyrus odoratus flowers, to produce purple colour, dominant C and P genes are both needed. The cross CcPp x ccPp would produce white-flowered offspring in the ratio:
- (1) 3:5 (2) 9:7 (3) 2:6 (4) 4:4
- Answer: (2) — Modified based on complementary genes
-
If the mother's blood group is AB and father's blood group is A, children's blood group will be:
- (1) A (2) B (3) AB (4) O
- Answer: (3) AB (possible groups are A, B, AB)
Explanation
Mother (AB) genotype = I^A I^B. Father (A) genotype = I^A I^A or I^A i. If father is I^A I^A: children can be I^A I^A (type A) or I^A I^B (type AB). If father is I^A i: children can be I^A I^A (A), I^A I^B (AB), I^A i (A), or I^B i (B). Blood group O is not possible since mother always contributes either I^A or I^B.Summary Cheat Sheet
| Concept / Topic | Key Details / Explanation |
|---|---|
| Non-allelic gene interactions | Interactions between genes at different loci; modify the 9:3:3:1 dihybrid ratio |
| Epistasis | One non-allelic gene suppresses another; epistatic gene suppresses, hypostatic gene is suppressed |
| All modified ratios total | 16 (since 9+3+3+1 = 16) |
| Complementary genes | Ratio 9:7; both genes needed in dominant form for expression Example: Flower colour in sweet pea (Lathyrus) — C_ and P_ both needed |
| Dominant epistasis | Ratio 12:3:1; one dominant gene masks the other Example: Fruit colour in summer squash — W_ = white (masks Y), wwY_ = yellow, wwyy = green |
| Recessive epistasis | Ratio 9:3:4; homozygous recessive at one locus masks the other Example: Coat colour in mice — cc = albino regardless of A gene; C_A_ = agouti, C_aa = black |
| Duplicate genes | Ratio 15:1; either gene alone produces the trait; only aabb shows alternative Example: Kernel colour in wheat |
| Inhibitory genes | Ratio 13:3; one dominant gene inhibits expression of another (inhibitor has no own phenotype) Example: Feather colour in poultry — I_ inhibits colour gene C |
| Polygenic inheritance | Many genes control one trait; each has small additive effect; produces continuous variation and bell-shaped curve |
| Polygenic phenotype classes | (2n + 1) where n = number of gene pairs |
| Wheat kernel colour | Studied by Nilsson-Ehle; 2 gene pairs → F2 ratio 1:4:6:4:1 (5 phenotypes) |
| Human skin colour | Studied by Davenport; 3 gene pairs → F2 ratio 1:6:15:20:15:6:1 (7 phenotypes) |
| Cytoplasmic inheritance | Genes in cytoplasmic organelles (plastids, mitochondria); maternal (uniparental) inheritance |
| Discovered by | Carl Correns in Mirabilis jalapa |
| Cytoplasmic vs nuclear | Reciprocal crosses give different results → cytoplasmic; same results → nuclear (karyogene) |
| Plasmagene vs Karyogene | Plasmagene = gene in cytoplasm; Karyogene = gene in nucleus |
| Mirabilis variegation | Offspring colour depends entirely on mother; plastids inherited through egg cytoplasm |
| CMS (Cytoplasmic Male Sterility) | Gene in mitochondrial DNA; inherited maternally; used in hybrid seed production in agriculture |
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