SME
Grade 9 Science · Biology
Evidence for evolution, population genetics and speciation
Use this for “16% of a population shows a recessive trait. What are the allele frequencies, and how many are carriers?” with Hardy–Weinberg.
Diagram
p + q = 0.6 + 0.4 = 1
p
2
+
2pq
+
q
2
=
0.36
+
0.48
+
0.16
p
2
+
2pq
+
q
2
=
1
60 of 100 alleles are A.
This is an example. Type your own values to start a new problem.
p
+
q
=
1
p²
+
2pq
+
q²
=
1
N
People in the population
example
nAA
Expected AA people
worked out (not typed)
nAa
Expected carriers
worked out (not typed)
nₐₐ
Expected aa people
worked out (not typed)
Formulas
q
2
=
q
×
q
0.16 = 0.4 × 0.4
p
=
1
−
q
0.6 = 1 − 0.4
2
pq
=
2
×
p
×
q
0.48 = 2 × 0.6 × 0.4
p
2
=
p
×
p
0.36 = 0.6 × 0.6
nAA
=
N
×
p
2
180 = 500 × 0.36
nAa
=
N
×
(
2
pq
)
240 = 500 × 0.48
nₐₐ
=
N
×
q
2
80 = 500 × 0.16
Assumptions
•
Hardy–Weinberg holds in a large population with random mating, no mutation, no migration and no selection.
•
Only aa shows the recessive trait, so start from its share q²: q is its square root.
•
Carriers (Aa) show nothing but make up 2pq, and p² + 2pq + q² = 1.
Step-by-step
WE KNOW
q² = 0.16, N = 500
FIND
frequency of allele a (q), frequency of allele A (p), share of carriers (Aa) (2pq), share that is AA (p²), expected AA people (nAA), expected carriers (nAa), expected aa people (nₐₐ)
Step 1 · Find frequency of allele a (q)
Use q² = q × q
Only
aa
shows
the
recessive
trait,
so
q
is
the
square
root
of
its
share.
q
=
√
q
2
q
=
√
0.16
q
=
0.4
Step 2 · Find frequency of allele A (p)
Use p = 1 − q
The two alleles’ frequencies add to 1.
p
=
1
−
0.4
p
=
0.6
Step 3 · Find share of carriers (Aa): 2pq
Use 2pq = 2 × p × q
A
from
one
parent
and
a
from
the
other,
or
the
other
way
round:
twice
p
×
q
.
2
pq
=
2
×
0.6
×
0.4
2
pq
=
1.2
×
0.4
2
pq
=
0.48
Step 4 · Find share that is AA (p²)
Use p² = p × p
Two
A
alleles
meet
with
chance
p
×
p
.
p
2
=
0.6
×
0.6
p
2
=
0.36
Step 5 · Find expected AA people (nAA)
Use nAA = N × p²
The
AA
share
of
the
N
people.
nAA = 500 × 0.36
nAA = 180
Step 6 · Find expected carriers (nAa)
Use nAa = N × (2pq)
The
carrier
share
of
the
N
people.
nAa = 500 × 0.48
nAa = 240
Step 7 · Find expected aa people (nₐₐ)
Use nₐₐ = N × q²
The
aa
share
of
the
N
people.
nₐₐ = 500 × 0.16
nₐₐ = 80
Check
Put the numbers back in. Both sides should match.
0.16 = 0.4 × 0.4 ✓
0.6 = 1 − 0.4 ✓
0.48 = 2 × 0.6 × 0.4 ✓
0.36 = 0.6 × 0.6 ✓
180 = 500 × 0.36 ✓
240 = 500 × 0.48 ✓
80 = 500 × 0.16 ✓
Topic check
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