SME
Grade 9 Science · Problem type
Expected genotypes among many offspring
Use this for “Among 100 children of carrier parents (Ff × Ff), how many are expected to be Ff?”
Diagram
F
f
F
FF
Ff
f
Ff
ff
Ff × Ff: 3 of 4 boxes have an F, so 3 in 4 show the trait.
This is an example. Type your own values to start a new problem.
a
F alleles in the first parent
example
c
F alleles in the second parent
example
n
Offspring in all
example
t
Boxes of 4 showing the trait
worked out (not typed)
nFF
Expected FF
worked out (not typed)
nFf
Expected Ff
worked out (not typed)
nff
Expected ff
worked out (not typed)
Pff
Chance of ff
worked out (not typed) · %
Formulas
t
=
4
−
(2
−
a
)
×
(2
−
c
)
3 = 4 − (2 − 1) × (2 − 1)
nFF
=
n
×
a
×
c
4
25
=
100
×
1
×
1
4
nFf
=
n
×
a
× (2 −
c
) +
c
× (2 −
a
)
4
50
=
100
×
1 × (2 − 1) + 1 × (2 − 1)
4
nff
=
n
×
(2
−
a
)
×
2 −
c
4
25
=
100
×
(2
−
1)
×
2 − 1
4
Pff
=
25
×
(2
−
a
)
×
(2
−
c
)
25 = 25 × (2 − 1) × (2 − 1)
Assumptions
•
F is dominant and f recessive; each parent is written by its count of F alleles: 2, 1 or 0.
•
Each of the 4 boxes is a quarter of the offspring, so a genotype in k boxes is k quarters of them.
•
Expected counts are averages: real families vary, as coin flips do.
Step-by-step
WE KNOW
a = 1, c = 1, n = 100
FIND
boxes of 4 showing the trait (t), expected FF (nFF), expected Ff (nFf), expected ff (nff), chance of ff (Pff)
Step 1 · Find boxes of 4 showing the trait (t)
Use t = 4 − (2 − a) × (2 − c)
Only boxes with a recessive allele from both parents lack the dominant trait; the rest of the 4 show it.
t
=
4
−
(2
−
1)
×
(2
−
1)
t
=
4
−
1
t
=
3
Step 2 · Find expected FF (nFF)
Use nFF = n × a × c ÷ 4
An
F
from
each
parent:
a
×
c
of
the
4
boxes
are
FF,
each
a
quarter
of
the
offspring.
nFF
=
100
×
1
×
1
4
nFF = 25
Step 3 · Find expected Ff (nFf)
Use nFf = n × (a × (2 − c) + c × (2 − a)) ÷ 4
F from one parent and f from the other, either way round.
nFf
=
100
×
1 × (2 − 1) + 1 × (2 − 1)
4
nFf
=
100
×
2 − 1 + 1
4
nFf
=
100
×
2
4
nFf
=
200
4
nFf = 50
Step 4 · Find expected ff (nff)
Use nff = n × (2 − a) × (2 − c) ÷ 4
An f from each parent: the f alleles of one times the f alleles of the other.
nff
=
100
×
(2
−
1)
×
2 − 1
4
nff
=
100
4
nff = 25
Step 5 · Find chance of ff (Pff)
Use Pff = 25 × (2 − a) × (2 − c)
Each
box
is
25%
of
the
offspring,
and
(2
−
a
)
×
(2
−
c
)
boxes
are
ff.
Pff = 25 × (2 − 1) × (2 − 1)
Pff = 25%
Check
Put the numbers back in. Both sides should match.
3 = 4 − (2 − 1) × (2 − 1) ✓
25
=
100
×
1
×
1
4
✓
50
=
100
×
1 × (2 − 1) + 1 × (2 − 1)
4
✓
25
=
100
×
(2
−
1)
×
2 − 1
4
✓
25 = 25 × (2 − 1) × (2 − 1) ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
Refresh
Related lessons
Mendelian and non-Mendelian inheritance
Main lesson
Incomplete dominance
Use this for “Two pink snapdragons are crossed. How many of 4 offspring are expected to be red, pink and white?”
Codominance
Use this for “A roan bull and a white cow: how many calves in 4 are roan?”
A sex-linked trait
Use this for “A carrier mother and a father with normal vision: what is the chance a son is color-blind?”
ABO blood types from genotypes
Use this for “Which two genotypes give the same blood type?”
A sex-linked pedigree
Use this for “Can an affected son have a mother who does not show the trait?”