Solve the following problem…In mice, black hair (B) is dominant over

brown (b) hair. Two heterozygous black mice breed and over time produce many offspring.

A. If the mice have 48 offspring, how many would be expected to be black? brown?

B. What are the odds (in %) of having a black baby mouse? brown baby mouse?

Answers: Black – 36 (75%) Brown – 12 (25%)

WRONG!!Instead, these would be the expected

answers…Black – 27 (56.25%)Brown – 9 (18.75%)White – 12 (25%)

How is this possible?Hair color in mice is an example of EPISTASIS.

What is EPISTASIS?When one gene alters the expression of

another gene.For example – in miceB – black; b – brownC – Pigment will be deposited in the hairc – No pigment will be deposited in the hairDo the following cross BbCc x BbCc1. If the crossing above yield 96 mice, how

many will be white? (no pigment) – What percentage does this represent?

2. What other phenotypes are possible and in what proportions?

Epistasis modifies Mendelian ratios…

Not a 9:3:3:1

What ratio do you see?

9:3:4 (this is a common ratio when heterozygote x heterozygote

Another example of epistasis…

Black – BBE_ or BbE_

Brown – bbE_

Yellow – _ _ee

Yellow – bbee – light brown/pink nose with no black lining around it’s eyes and light eye color

Yellow – B_ee – black nose with black lining around eye with dark eye color

What color are the following dogs?1. BbEe –

Black2. Bbee –

Yellow (with black nose/eyes)3. BBee –

Yellow (with black nose/eyes)4. bbEe –

Brown 5. BBEE –

Black6. BBEe –

Black7. bbee –

yellow (with light nose/eyes)

Describe all possible phenotypes and the proportions for each of the

following crosses:

1. BbEe x Bbee – is brown possible?

2. BBee x bbEe –

3. BBee x BBEE –

4. BBEe x BBEe –

5. Two yellows (Bbee x Bbee)

Parents?