A2 Biology 9700: Epistasis in a Dihybrid Cross

In a species of leaf beetle, body colour is controlled by two genes on different chromosomes. Gene E codes for an enzyme needed to make any pigment at all: allele E (functional enzyme, dominant) allows pigment to be produced, while allele e (non-functional enzyme, recessive) means no pigment can be made, regardless of the genotype at the second gene. Gene B then determines which pigment is made, but only in beetles that can produce pigment at all: allele B (black pigment, dominant) is dominant to allele b (brown pigment, recessive).

A beetle that is heterozygous at both genes (genotype EeBb, black) is crossed with another beetle that is also heterozygous at both genes (genotype EeBb).

(a) Define the term epistasis, referring to genes E and B in your answer. [2] (b) Construct a genetic diagram for this cross and use it to determine the expected phenotypic ratio of the offspring, showing how the genotypes are grouped into phenotypes. [4] (c) A beetle has the genotype eeBB. Explain why this beetle is not black, even though it carries two copies of the dominant allele B for black pigment. [2]

Part (a): Defining epistasis

Epistasis is the phenomenon in which the allele(s) present at one gene locus mask or alter the phenotypic expression of the allele(s) at a different gene locus.

In this cross, gene E is epistatic to gene B. Gene E determines whether any pigment can be made at all: a beetle with genotype ee cannot produce the functional enzyme needed for pigment production, so no pigment is made regardless of which alleles are present at gene B. Only in a beetle that has at least one functional E allele (E_) does the genotype at gene B, which pigment colour to produce, have any visible effect.

Part (b): Genetic diagram and phenotypic ratio

Parents: EeBb (black) x EeBb (black)

Gametes from each parent: EB, Eb, eB, eb

	EB	Eb	eB	eb
EB	EEBB	EEBb	EeBB	EeBb
Eb	EEBb	EEbb	EeBb	Eebb
eB	EeBB	EeBb	eeBB	eeBb
eb	EeBb	Eebb	eeBb	eebb

Grouping the 16 genotypes by phenotype:

E_B_ (at least one E and at least one B): 9 out of 16, the enzyme is functional and B directs black pigment, black.
E_bb (at least one E, but bb): 3 out of 16, the enzyme is functional but only brown pigment is directed, brown.
eeB_ and eebb (ee, regardless of the B genotype): 3 + 1 = 4 out of 16 (no functional enzyme, so no pigment is made at all) pale.

So the expected phenotypic ratio is 9 black : 3 brown : 4 pale.

Part (c): Why genotype eeBB is not black

Gene E acts “upstream” of gene B in this pathway: it controls whether the enzyme needed to make any pigment is produced in the first place. A beetle with genotype eeBB has two copies of the dominant B allele, so if it could make pigment at all, that pigment would be directed to be black. However, because its genotype at gene E is homozygous recessive (ee), it cannot produce the functional enzyme required to make pigment in the first place. With no pigment made, the effect of the B allele is never expressed, and the beetle is pale rather than black, a clear example of gene E being epistatic to gene B.

Final answers

(a) Epistasis: the alleles of one gene (E) mask the phenotypic expression of another gene’s alleles (B); here, ee prevents any pigment being made, hiding the effect of the B/b genotype.
(b) The cross EeBb x EeBb gives offspring in the phenotypic ratio 9 black : 3 brown : 4 pale.
(c) Genotype eeBB is pale because the epistatic ee genotype blocks pigment production entirely, so the dominant B allele, which would otherwise direct black pigment, is never expressed.