SME
Grade 9 Science · Problem type
Allele frequencies from genotype counts
Use this for “A sample has 49 AA, 42 Aa and 9 aa. What are p and q?”
Diagram
p + q = 0.7 + 0.3 = 1
p
2
+
2pq
+
q
2
=
0.49
+
0.42
+
0.09
p
2
+
2pq
+
q
2
=
1
70 of 100 alleles are A.
This is an example. Type your own values to start a new problem.
p
=
2
×
+
2
×
nₐₐ
Individuals aa
example
A
Count of A alleles
worked out (not typed)
a
Count of a alleles
worked out (not typed)
q
Frequency of allele a
worked out (not typed)
Formulas
N
=
nAA
+
nAa
+
nₐₐ
100 = 49 + 42 + 9
A
=
2
×
nAA
+
nAa
140 = 2 × 49 + 42
a
=
2
×
nₐₐ
+
nAa
60 = 2 × 9 + 42
p
=
A
2 ×
N
0.7
=
140
2 × 100
q
=
a
2 ×
N
0.3
=
60
2 × 100
Assumptions
•
Each individual carries two alleles: AA has two A, Aa one of each, aa two a.
•
So N individuals carry 2N alleles, and p is the share of them that are A.
•
Counting alleles works for any population, whether or not it is in Hardy–Weinberg equilibrium.
Step-by-step
WE KNOW
nAA = 49, nAa = 42, nₐₐ = 9
FIND
individuals in all (N), count of A alleles (A), count of a alleles (a), frequency of allele A (p), frequency of allele a (q)
Step 1 · Find individuals in all (N)
Use N = nAA + nAa + nₐₐ
Add the three genotype counts.
N
=
49
+
42
+
9
N
=
100
Step 2 · Find count of A alleles (A)
Use A = 2 × nAA + nAa
Two A in each AA individual and one in each Aa.
A
=
2
×
49
+
42
A
=
98
+
42
A
=
140
Step 3 · Find count of a alleles (a)
Use a = 2 × nₐₐ + nAa
Two a in each aa individual and one in each Aa.
a
=
2
×
9
+
42
a
=
18
+
42
a
=
60
Step 4 · Find frequency of allele A (p)
Use p = A ÷ (2 × N)
The
A
alleles
out
of
all
2
N
alleles.
p
=
140
2 × 100
p
=
140
200
p
=
0.7
Step 5 · Find frequency of allele a (q)
Use q = a ÷ (2 × N)
The
a
alleles
out
of
all
2
N
alleles.
q
=
60
2 × 100
q
=
60
200
q
=
0.3
Check
Put the numbers back in. Both sides should match.
100 = 49 + 42 + 9 ✓
140 = 2 × 49 + 42 ✓
60 = 2 × 9 + 42 ✓
0.7
=
140
2 × 100
✓
0.3
=
60
2 × 100
✓
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