SME
Grade 9 Science · Problem type
An insertion shifts the reading frame
Use this for “Why does inserting one base change every amino acid after it?”
Diagram
An insertion at base 5, codon 2.
Frameshift: every codon from there on is read in a new frame.
Now Met–Val–Gln–Val.
This is an example. Type your own values to start a new problem.
L
Template bases
example
c
Codons
worked out (not typed)
p
Base the insertion goes before
example
k
Codon holding it
worked out (not typed)
s
Codons read in a shifted frame
worked out (not typed)
Formulas
c
=
L
3
4
=
12
3
k
=
⌈
p
3
⌉
2
=
⌈
5
3
⌉
s
=
c
−
k
+
1
3 = 4 − 2 + 1
Assumptions
•
The template strand is the first L bases of TACCGGTTCATT; an A is inserted before base p.
•
The ribosome reads in threes, so every codon from the one holding the insertion on is read in a shifted frame.
•
Inside the start codon (p = 2 or 3) the insertion breaks AUG, so no protein starts there; a deletion shifts the frame the same way.
Step-by-step
WE KNOW
L = 12, p = 5
FIND
codons (c), codon holding it (k), codons read in a shifted frame (s)
Step 1 · Find codons (c)
Use c = L ÷ 3
Every three bases are one codon.
c
=
12
3
c
=
4
Step 2 · Find codon holding it (k)
Use k = ⌈p ÷ 3⌉
Bases 1 to 3 are codon 1, bases 4 to 6 codon 2, and so on: divide by 3 and round up.
k
=
⌈
5
3
⌉
k
=
2
Step 3 · Find codons read in a shifted frame (s)
Use s = c − k + 1
Codon
k
and
every
codon
after
it,
to
the
last,
are
read
in
the
new
frame.
s
=
4
−
2
+
1
s
=
3
Check
Put the numbers back in. Both sides should match.
4
=
12
3
✓
2
=
⌈
5
3
⌉
✓
3 = 4 − 2 + 1 ✓
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A deletion shifts the reading frame
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