SME
Principles of Biology I
Genetics
Use this for “Two AaBb parents are crossed. What share of the offspring show both dominant traits, and how many of the 16 boxes show neither?”
Diagram
AaBb × AaBb: 9:3:3:1 of 16 boxes (A_B_, A_bb, aaB_, aabb).
This is an example. Choose your own values to start a new problem.
A₁
A alleles in parent 1
example
B₁
B alleles in parent 1
example
A₂
A alleles in parent 2
example
B₂
B alleles in parent 2
example
tA
Boxes of 4 showing A
worked out (not typed)
tB
Boxes of 4 showing B
worked out (not typed)
D
Both dominant, boxes of 16
worked out (not typed)
N
Neither dominant, boxes of 16
worked out (not typed)
P
Chance of both dominant
worked out (not typed)
Formulas
tA
=
4
−
(2
−
A₁
)
×
(2
−
A₂
)
3 = 4 − (2 − 1) × (2 − 1)
tB
=
4
−
(2
−
B₁
)
×
(2
−
B₂
)
3 = 4 − (2 − 1) × (2 − 1)
D
=
tA
×
tB
9 = 3 × 3
N
=
(4
−
tA)
×
(4
−
tB)
1 = (4 − 3) × (4 − 3)
P
=
D
16
9
16
=
9
16
Assumptions
•
A is dominant to a, and B to b; each parent is typed by its count of dominant alleles for each gene: 2, 1 or 0.
•
The two genes are on different chromosomes, so they assort independently into the gametes.
•
Each parent’s 4 gametes are equally likely, so each of the 16 boxes is 1/16 of the offspring.
•
A test cross pairs a parent with aabb: type 0 and 0 for the second parent.
Step-by-step
WE KNOW
A₁ = 1, B₁ = 1, A₂ = 1, B₂ = 1
FIND
boxes of 4 showing A (tA), boxes of 4 showing B (tB), both dominant, boxes of 16 (D), neither dominant, boxes of 16 (N), chance of both dominant (P)
Step 1 · Find boxes of 4 showing A (tA)
Use tA = 4 − (2 − A₁) × (2 − A₂)
Of the 4 boxes for gene A, only those with a recessive allele from both parents lack the dominant trait.
tA = 4 − (2 − 1) × (2 − 1)
tA = 4 − 1
tA = 3
Step 2 · Find boxes of 4 showing B (tB)
Use tB = 4 − (2 − B₁) × (2 − B₂)
Of the 4 boxes for gene B, only those with a recessive allele from both parents lack the dominant trait.
tB = 4 − (2 − 1) × (2 − 1)
tB = 4 − 1
tB = 3
Step 3 · Find both dominant, boxes of 16 (D)
Use D = tA × tB
The genes assort independently, so the boxes multiply: A in tA of 4 times B in tB of 4.
D
=
3
×
3
D
=
9
Step 4 · Find neither dominant, boxes of 16 (N)
Use N = (4 − tA) × (4 − tB)
Neither dominant trait: the recessive boxes of one gene times those of the other.
N
=
(4
−
3)
×
(4
−
3)
N
=
1
Step 5 · Find chance of both dominant (P)
Use P = D ÷ 16
Each of the 16 boxes is equally likely.
P
=
9
16
P
=
9
16
Check
Put the numbers back in. Both sides should match.
3 = 4 − (2 − 1) × (2 − 1) ✓
3 = 4 − (2 − 1) × (2 − 1) ✓
9 = 3 × 3 ✓
1 = (4 − 3) × (4 − 3) ✓
9
16
=
9
16
✓
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The product rule for three genes
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