Difficulty: Intermediate | Prerequisites: Basic Mendelian genetics (monohybrid and dihybrid crosses, Punnett squares, dominance relationships).
Epistasis is what happens when one gene interferes with the expression of another gene at a different locus. It builds directly on the dihybrid cross ratios you learnt in introductory Mendelian genetics, so if you are comfortable with a 9:3:3:1 ratio, you are ready for this. Epistasis explains why real inheritance patterns often deviate from those neat Mendelian ratios, and it is one of the main ways exam questions test whether you can think beyond single-gene models. You will see it again in population genetics and in molecular biology when studying gene regulation pathways.
Epistasis occurs when an allele at one gene masks or modifies the phenotype produced by alleles at a second gene. The classic sign is a modified dihybrid ratio (12:3:1, 9:3:4, 9:7, etc.) instead of the expected 9:3:3:1. Work these problems by identifying which gene is epistatic, then sorting the 16-square Punnett output by actual phenotype.
Epistasis
A gene interaction in which alleles at one locus mask or modify the phenotypic expression of alleles at a different locus. In simple terms, one gene "overrules" another so that the second gene's effect is hidden.
Epistatic gene (the modifier)
The gene that does the masking. Think of it as the gatekeeper: if it is in its masking state, the other gene's phenotype never shows.
Hypostatic gene
The gene whose expression is being masked. It still has its alleles, but you cannot see them in the phenotype when the epistatic gene is active.
Dominant epistasis
One dominant allele at the epistatic locus is enough to mask the other gene. Produces a modified dihybrid ratio of 12:3:1 (the 9 + 3 classes with at least one dominant epistatic allele all look the same).
Recessive epistasis
Both alleles at the epistatic locus must be recessive (homozygous recessive) for masking to occur. Produces a 9:3:4 ratio (the 3 + 1 classes that are homozygous recessive at the epistatic locus collapse into one phenotype).
Modified dihybrid ratio
Any departure from the standard 9:3:3:1 ratio that results from epistatic interactions. Common exam ratios include 12:3:1, 9:3:4, 9:7, 13:3, and 15:1. In simple terms, these are just different ways of grouping the 16 boxes of a dihybrid Punnett square when epistasis changes which genotypes look alike.
Albino / no-pigment phenotype (in epistasis problems)
In many genetics problems, a recessive epistatic allele blocks all pigment production regardless of what the colour gene says. The animal or plant appears white or albino whenever it is homozygous recessive at the pigment-production locus.
Simple dominance is an interaction between alleles at the same locus (e.g. B masks b).
Epistasis is an interaction between genes at different loci (e.g. gene C controls whether gene B can produce pigment at all).
Both can be at work simultaneously. In a typical epistasis problem you have at least two genes, each with its own dominance relationship, plus the inter-gene masking on top.
The giveaway is a dihybrid cross that does not produce the standard 9:3:3:1. The question will describe one gene "blocking," "prohibiting," or "masking" the expression of another. Your job is to figure out which gene is epistatic and whether the masking allele is dominant or recessive.
The epistatic gene must be homozygous recessive to mask the other gene.
Example from this problem set: rat coat colour. Gene 1 controls black vs brown (B dominant to b). Gene 2 controls whether pigment is produced at all. Only the homozygous recessive (cc) blocks pigment, producing albino.
In a dihybrid cross of BbCc x BbCc, the 16-box Punnett square gives: 9 B_C_ (black), 3 bbC_ (brown), 3 B_cc (albino), 1 bbcc (albino). The last two classes both look albino, so 9:3:4.
A single dominant allele at the epistatic locus is enough to mask.
Example from this problem set: summer squash fruit colour. The dominant W allele produces white fruit regardless of the second gene. Only when the epistatic locus is homozygous recessive (ww) can you see yellow (Y_) or green (yy).
Cross WwYy x WwYy gives: 9 W_Y_ (white), 3 W_yy (white), 3 wwY_ (yellow), 1 wwyy (green). The first two classes both look white, so 12:3:1.
Step 1: Identify the two genes and their alleles.
Step 2: Write out the gametes for each parent.
Step 3: Fill in the 16-box grid.
Step 4: Assign phenotypes, remembering that the epistatic gene may override the expected colour or trait.
Step 5: Count each phenotype class and express as a ratio.
Some problems (like Q4 in this set) add a third gene. The approach is the same, but you may need a larger Punnett square or a branching/forked-line method. Handle each gene pair independently, then combine probabilities using the product rule.
Labrador retriever coat colour is the textbook example of recessive epistasis. The E gene controls pigment deposition: ee dogs are yellow regardless of whether they carry the black (B) or chocolate (b) allele. This is exactly the same logic as the rat and mouse problems in this practice set.
Q1: In rats, hair colour is controlled by 2 genes. Black (B) is dominant to brown (b). A second gene controls pigment production: homozygous recessive (cc) produces no pigment (albino). How many albino rats result in the F2 generation of a cross between a homozygous black rat (BBCC) and a white rat that would have produced brown hair (bbcc)?
A: The P cross is BBCC x bbcc. All F1 offspring are BbCc (black).
The F2 cross is BbCc x BbCc. In the 16-box Punnett square:
9 B_C_ = black
3 bbC_ = brown
3 B_cc = albino (have black allele but no pigment)
1 bbcc = albino (have brown allele but no pigment)
Albino rats = 3 + 1 = 4 out of 16 (1/4 or 25%).
This is recessive epistasis with a 9:3:4 ratio.
Q2: In summer squash, fruit colour may be white, yellow, or green. White is produced by a dominant epistatic allele (W). When the epistatic gene is homozygous recessive (ww), yellow (Y_) is dominant to green (yy). What is the F2 ratio from WWYY x wwyy?
A: The P cross is WWYY x wwyy. All F1 are WwYy (white).
The F2 cross is WwYy x WwYy:
9 W_Y_ = white (W is epistatic)
3 W_yy = white (W is still epistatic)
3 wwY_ = yellow
1 wwyy = green
F2 ratio: 12 white : 3 yellow : 1 green.
This is dominant epistasis.
Q3: In pigeons, Blue (B) is dominant to Brown (b). Red is produced by homozygous recessive at an epistatic locus (rr). What are the chances of red pigeons from each cross?
a) BbRR x BBrr
A: For the R gene: RR x Rr gives all Rr (none are rr). Chance of red = 0%.
Wait, let me re-examine. BbRR x BBrr. For the R locus: RR x rr = all Rr. No rr offspring. Chance of red = 0 (0%).
b) BBRr x bbRr
A: For the R locus: Rr x Rr gives 1/4 RR, 2/4 Rr, 1/4 rr. Only rr produces red. The B locus does not matter when rr is present (rr is epistatic and overrides blue/brown to produce red). Chance of red = 1/4 (25%).
c) BbRr x BbRr
A: For the R locus: Rr x Rr gives 1/4 rr. When rr is present, the pigeon is red regardless of genotype at the B locus. Chance of red = 1/4 (25%).
Q4: In mice, Black (B) is dominant to brown (b). Recessive epistatic albino gene (cc) blocks pigment. A third gene (dd) produces grey (with B) or cream (with b). What are the results of the following crosses?
a) BBccDD x bbCCdd
A: F1 = BbCcDd (all identical).
Phenotype of BbCcDd: has pigment (Cc), has B (black base), has Dd (not grey, because dd is needed for grey). So F1 are all black.
F2 from BbCcDd x BbCcDd is a trihybrid cross (64 combinations). The phenotype depends on three loci:
cc = albino, regardless of other genes
B_ + C_ + D_ = black
bb + C_ + D_ = brown
B_ + C_ + dd = grey
bb + C_ + dd = cream
Using the forked-line method:
B locus: 3/4 B_ and 1/4 bb
C locus: 3/4 C_ and 1/4 cc
D locus: 3/4 D_ and 1/4 dd
Phenotype probabilities:
Black (B_ C_ D_): 3/4 x 3/4 x 3/4 = 27/64
Brown (bb C_ D_): 1/4 x 3/4 x 3/4 = 9/64
Grey (B_ C_ dd): 3/4 x 3/4 x 1/4 = 9/64
Cream (bb C_ dd): 1/4 x 3/4 x 1/4 = 3/64
Albino (any B, cc, any D): any x 1/4 x any = 16/64
Ratio: 27 black : 9 brown : 9 grey : 3 cream : 16 albino.
b) bbccDD x bbCCdd
A: F1 = bbCcDd. Phenotype: has pigment (Cc), brown (bb), not dd. So F1 are all brown.
F2 from bbCcDd x bbCcDd:
All offspring are bb, so no black or grey.
C locus: 3/4 C_, 1/4 cc
D locus: 3/4 D_, 1/4 dd
Phenotypes:
Brown (bb C_ D_): 3/4 x 3/4 = 9/16
Cream (bb C_ dd): 3/4 x 1/4 = 3/16
Albino (bb cc __): 1/4 = 4/16
Ratio: 9 brown : 3 cream : 4 albino.
Q8: In mice, red eyes (R) are dominant over brown (B), which is dominant over white (w). Red > Brown > White. What are the chances of white-eyed individuals in each cross?
a) RrbbWw x rrBBww
A: This is a dominance hierarchy at potentially one locus, but written as though multiple loci interact. Reading the problem: R is dominant over B, and B is dominant over W. For white eyes, the individual must have no R alleles and no B alleles (only ww).
Parent 1: RrbbWw. Parent 2: rrBBww.
For the R locus: Rr x rr = 1/2 Rr, 1/2 rr. Need rr (probability 1/2).
For the B locus: bb x BB = all Bb. Need bb for white. Probability = 0.
Since all offspring carry at least one B allele, none can be white-eyed. Chance = 0 (0%).
b) RRBBww x rrbbww
A: F1 = RrBbww. All have R, so all are red-eyed.
For white-eyed offspring you would need rr and bb and ww. In this F1 cross (RrBbww x RrBbww):
rr probability: 1/4
bb probability: 1/4
ww: all are ww = 1
Chance of white = 1/4 x 1/4 x 1 = 1/16 (6.25%).
Students often confuse epistasis with simple dominance. Dominance is between alleles at the same gene. Epistasis is between two different genes.
"Recessive epistasis" does not mean the epistatic gene is unimportant. It means the masking effect only appears when the epistatic locus is homozygous recessive.
Albino animals in these problems still carry colour alleles (B or b). The alleles are present; they simply cannot be expressed because the pigment pathway is blocked.
A 12:3:1 ratio and a 9:3:4 ratio both come from the same 16-box Punnett square. The difference is whether the epistatic allele is dominant or recessive, which determines which classes collapse together.
Exam questions will give you a non-standard dihybrid ratio and expect you to identify the type of epistasis.
You must be able to work backwards: given a ratio like 12:3:1, determine which gene is epistatic and whether the masking is dominant or recessive.
Three-gene problems (like Q4) are common on exams. The forked-line method is faster than drawing a massive Punnett square.
Always check whether a question asks for a ratio, a fraction, a percentage, or a count. These are different things.
True or False: In recessive epistasis, the modified ratio is 12:3:1.
False. Recessive epistasis gives 9:3:4. Dominant epistasis gives 12:3:1.
Fill in the blank: An epistatic gene ______ the expression of a hypostatic gene.
Masks (or suppresses).
True or False: An albino rat with genotype BBcc carries alleles for black fur.
True. The B alleles are there, but the cc genotype prevents pigment production.
Fill in the blank: In a standard dihybrid cross without epistasis, the expected ratio is ______.
9:3:3:1.
True or False: In dominant epistasis of squash colour, you need ww at the epistatic locus to see yellow or green fruit.
True. Any W_ genotype produces white, masking the colour gene.
Epistasis connects directly to Mendelian dihybrid crosses. If you are shaky on 9:3:3:1, revisit that first, because epistasis is just a modification of the same framework. It also links to molecular biology: epistatic interactions often reflect real biochemical pathways where one enzyme's product is the next enzyme's substrate. Block the upstream enzyme and nothing downstream works, which is exactly what recessive epistasis models.
This topic also sets up quantitative genetics (covered in the companion notes), where multiple genes contribute additively to a trait. The conceptual leap is from "one gene masks another" to "many genes add up together."
epistasis, epistatic gene, hypostatic gene, dominant epistasis, recessive epistasis, modified dihybrid ratio, 12:3:1, 9:3:4, 9:7, albino, pigment production, gene interaction, non-Mendelian inheritance, Labrador coat colour, supplementary gene interaction, complementary gene interaction, BIO 138, polygenic epistatic quantitative practice