SME
Principles of Biology I · Problem type
The product rule for three genes
Use this for “Two AaBbCc parents: what is the chance of an aabbcc child?”
Diagram
P
=
p₁
×
p₂
×
p₃
=
1
4
×
1
4
×
1
4
P
=
1
64
(1
in
64)
The stages are independent, so each one keeps its share of what the stages before it left: the chances multiply.
This is an example. Choose your own values to start a new problem.
1
64
P
=
p₁
×
p₂
×
p₃
Formulas
P
=
p₁
×
p₂
×
p₃
1
64
=
1
4
×
1
4
×
1
4
Assumptions
•
The three genes are on different chromosomes, so they assort independently.
•
Each gene’s chance comes from its own one-gene cross: Aa × Aa gives AA 1/4, Aa 1/2, aa 1/4, and the dominant trait 3/4.
•
The product rule: the chance of the whole genotype is the three genes’ chances multiplied.
•
A gene the question leaves free counts 1.
Step-by-step
WE KNOW
p₁ = 1/4, p₂ = 1/4, p₃ = 1/4
FIND
chance of all three (P)
Step 1 · Find chance of all three (P)
Use P = p₁ × p₂ × p₃
The genes assort independently, so multiply the chances along the path.
P
=
1
4
×
1
4
×
1
4
P
=
1
16
×
1
4
P
=
1
64
Check
Put the numbers back in. Both sides should match.
1
64
=
1
4
×
1
4
×
1
4
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.