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Grade 9 Science · Problem type

A deletion shifts the reading frame

Use this for “One base is deleted from a gene. Which codons change, and is there still a stop codon?”

Diagram

3′5′template strandcomplementary strandTAATCGCGGCTATACGATTATAG deletedmRNA 5′ (from the template)AUGGCAAGUAAAUGGCAAGUBeforeMetAlaLysStopAfterMetAlaSer
A deletion at base 5, codon 2.
Frameshift: every codon from there on is read in a new frame.
Now Met–Ala–Ser.
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 deleted
example
k
Codon holding it
worked out (not typed)
s
Codons read in a shifted frame
worked out (not typed)
B
Bases left
worked out (not typed)
w
Whole codons left
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
B
=
L
−
1
11 = 12 − 1
w
=
⌊
B
3
⌋
3
=
⌊
11
3
⌋

Assumptions

•
The template strand is the first L bases of TACCGGTTCATT; base p is deleted.
•
The ribosome reads in threes, so from the codon that held the lost base on, every codon is read in a shifted frame.
•
With one base fewer the last codon is cut short; in the full 12-base gene the stop codon UAA at the end is lost. Deleting three bases together would keep the frame.

Step-by-step

WE KNOW
L = 12, p = 5
FIND
codons (c), codon holding it (k), codons read in a shifted frame (s), bases left (B), whole codons left (w)
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
Step 4 · Find bases left (B)
Use B = L − 1
One base is gone.
B
=
12
−
1
B
=
11
Step 5 · Find whole codons left (w)
Use w = ⌊B ÷ 3⌋
Read in threes: the whole codons, with 2 bases left over that make no codon.
w
=
⌊
11
3
⌋
w
=
3
Check
Put the numbers back in. Both sides should match.
4
=
12
3
✓
2
=
⌈
5
3
⌉
✓
3 = 4 − 2 + 1 ✓
11 = 12 − 1 ✓
3
=
⌊
11
3
⌋
✓
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