SME
Genetics · Problem type
X-linked traits in men and women
Use this for “8% of men are color-blind. What share of women are, and what share carry it?”
Diagram
Men have one X, so affected men (XᵃY) are q = 0.08; affected women (XᵃXᵃ) need two: q² = 0.0064.
Affected
men
outnumber
affected
women
1
q
=
12.5
to
1;
carriers
(XᴬXᵃ)
are
2pq
=
0.1472.
This is an example. Type your own values to start a new problem.
qM
Share of men affected
example
q
Frequency of Xᵃ
calculated
p
Frequency of Xᴬ
worked out (not typed)
q²
Share of women affected
worked out (not typed)
2pq
Share of women who carry it
worked out (not typed)
Formulas
qM
=
q
0.08 = 0.08
p
=
1
−
q
0.92 = 1 − 0.08
q
2
=
q
×
q
0.0064 = 0.08 × 0.08
2
pq
=
2
×
p
×
q
0.1472 = 2 × 0.92 × 0.08
Assumptions
•
The allele is recessive and on the X chromosome; Hardy–Weinberg holds, with q the same in men and women.
•
A man has one X, so he is affected whenever it carries the allele: the share of men affected is q.
•
A woman needs the allele on both X to be affected (q²); with one copy she is a carrier (2pq).
Step-by-step
WE KNOW
qM = 0.08
FIND
frequency of Xᵃ (q), frequency of Xᴬ (p), share of women affected (q²), share of women who carry it (2pq)
Step 1 · Find frequency of Xᵃ (q)
Use qM = q
A
man’s
one
X
carries
the
allele
with
chance
q
.
q
=
qM
q
=
0.08
Step 2 · Find frequency of Xᴬ (p)
Use p = 1 − q
The two alleles’ frequencies add to 1.
p
=
1
−
0.08
p
=
0.92
Step 3 · Find share of women affected (q²)
Use q² = q × q
A woman’s two X both carry the allele.
q
2
=
0.08
×
0.08
q
2
=
0.0064
Step 4 · Find share of women who carry it (2pq)
Use 2pq = 2 × p × q
The
allele
on
her
mother’s
X
or
her
father’s,
not
both:
twice
p
×
q
.
2
pq
=
2
×
0.92
×
0.08
2
pq
=
1.84
×
0.08
2
pq
=
0.1472
Check
Put the numbers back in. Both sides should match.
0.08 = 0.08 ✓
0.92 = 1 − 0.08 ✓
0.0064 = 0.08 × 0.08 ✓
0.1472 = 2 × 0.92 × 0.08 ✓
Topic check
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