SME
Genetics · Problem type
Chi-square test of a Mendelian ratio
Use this for “An F₂ gives 95, 28, 27 and 10. Does it fit 9:3:3:1?”
Chart
Chi-square curve with df = 3.
Critical value at α = 0.05: χ² = 7.815.
χ² = 0.7111: p = P(χ² ≥ 0.7111) = 0.8706 ≥ α = 0.05: fail to reject H₀.
This is an example. Type your own values to start a new problem.
O₁
Offspring in class 1
example
O₂
Offspring in class 2
example
O₃
Offspring in class 3
example
O₄
Offspring in class 4
example
ratio
Expected ratio
example
N
Offspring in all
worked out (not typed)
df
Degrees of freedom
worked out (not typed)
χ²
Chi-square statistic
worked out (not typed)
α
Significance level
example
χ²_c
Critical value
worked out (not typed)
Formulas
N
=
O₁
+
O₂
+
O₃
+
O₄
160 = 95 + 28 + 27 + 10
df = classes in ratio − 1
3 = classes in 9:3:3:1 − 1
χ
2
=
(
O₁
− E₁)
2
÷
E₁
+
(
O₂
− E₂)
2
÷
E₂
+
(
O₃
− E₃)
2
÷
E₃
+
(
O₄
− E₄)
2
÷
E₄,
Eᵢ
=
N
×
its
share
of
ratio
0.7111
=
(95 − E₁)
2
÷
E₁
+
(28 − E₂)
2
÷
E₂
+
(27 − E₃)
2
÷
E₃
+
(10 − E₄)
2
÷
E₄,
Eᵢ
=
160
×
its
share
of
9:3:3:1
χ
2
_c
=
the
χ
2
with
right
tail
α
at
df
df
7.8147
=
the
χ
2
with
right
tail
0.05
at
3
df
Assumptions
•
Each class’s expected count is the total times its share of the ratio: 9/16, 3/16, …
•
Every expected count should be at least 5; the counts are offspring, not percents.
•
df = classes − 1; the counts fit the ratio when χ² is below the critical value at α.
•
A 9:3:4, 9:7 or 3:1 ratio has fewer classes: the classes past it hold 0.
Step-by-step
WE KNOW
O₁ = 95, O₂ = 28, O₃ = 27, O₄ = 10, ratio: 9:3:3:1, α = 0.05
FIND
offspring in all (N), degrees of freedom (df), chi-square statistic (χ²), critical value (χ²_c)
Step 1 · Find offspring in all (N)
Use N = O₁ + O₂ + O₃ + O₄
Add the classes for the offspring in all.
N
=
95
+
28
+
27
+
10
N
=
160
Step 2 · Find degrees of freedom (df)
Use df = classes in ratio − 1
The total fixes the last class’s count, so one class fewer is free.
df = 4 − 1
df = 3
Step 3 · Find chi-square statistic (χ²)
Use χ² = (O₁ − E₁)² ÷ E₁ + (O₂ − E₂)² ÷ E₂ + (O₃ − E₃)² ÷ E₃ + (O₄ − E₄)² ÷ E₄, Eᵢ = N × its share of ratio
Each
expected
count
is
E
=
N
×
the
class’s
share
of
the
ratio.
Add
(O − E)
2
÷
E
over
the
classes.
χ
2
=
(
O₁
− 90)
2
÷
90
+
(
O₂
− 30)
2
÷
30
+
(
O₃
− 30)
2
÷
30
+
(
O₄
− 10)
2
÷
10
χ
2
=
(95 − 90)
2
÷
90
+
(28 − 30)
2
÷
30
+
(27 − 30)
2
÷
30
+
(10 − 10)
2
÷
10
E₁
=
160
×
9
16
=
90;
E₂
=
160
×
3
16
=
30;
E₃
=
160
×
3
16
=
30;
E₄
=
160
×
1
16
=
10
χ
2
=
0.2778
+
0.1333
+
0.3
+
0
=
0.7111
χ
2
=
0.7111
Step 4 · Find critical value (χ²_c)
Use χ²_c = the χ² with right tail α at df df
The
critical
value
cuts
off
the
right
tail
of
area
α
under
the
chi-square
curve.
χ
2
_c
=
the
χ
2
with
right
tail
0.05
at
3
df
χ
2
_c
=
7.8147
→
0.7111
<
7.8147:
the
counts
fit
9:3:3:1
Check
Put the numbers back in. Both sides should match.
160 = 95 + 28 + 27 + 10 ✓
3 = 4 − 1 ✓
0.7111
=
(95 − 90)
2
÷
90
+
(28 − 30)
2
÷
30
+
(27 − 30)
2
÷
30
+
(10 − 10)
2
÷
10
✓
7.8147
=
the
χ
2
with
right
tail
0.05
at
3
df
✓
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Pedigrees and Mendelian genetics
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Expected counts for a modified dihybrid ratio
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