Genetic Mutations Assignment worksheet - Free Printable

Let’s solve this step by step.

We are given a normal DNA sequence and need to find the mRNA and protein for it. Then we’ll do the same for two mutations (Mutation 4 and Mutation 5) and figure out what type of mutation each is.

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Step 1: Understand how DNA becomes mRNA and then protein

- DNA has bases: A, T, C, G
- mRNA is made from DNA using base pairing rules:
- DNA “A” → mRNA “U”
- DNA “T” → mRNA “A”
- DNA “C” → mRNA “G”
- DNA “G” → mRNA “C”

So, to get mRNA from DNA, just swap each base with its partner (remember: in RNA, U replaces T).

Then, to get the protein, we use the mRNA codons (groups of 3 bases) and look up which amino acid they code for. We’re told that “CCC” codes for “Pro” (Proline), so we can use that as a reference.

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Normal DNA:
GGG ACA CCA TTC ACC CGT

Let’s convert to mRNA:

- GGG → CCC
- ACA → UGU
- CCA → GGU
- TTC → AAG
- ACC → UGG
- CGT → GCA

So Normal mRNA = CCC UGU GGU AAG UGG GCA

Now, let’s write the protein using standard genetic code (we know CCC = Pro):

- CCC → Pro
- UGU → Cys (Cysteine)
- GGU → Gly (Glycine)
- AAG → Lys (Lysine)
- UGG → Trp (Tryptophan)
- GCA → Ala (Alanine)

So Normal Protein = Pro-Cys-Gly-Lys-Trp-Ala

But since the worksheet only shows “Pro-” for the first one, maybe they want us to fill in the rest similarly — but actually, looking at the mutated examples, they seem to be focusing on comparing changes, not necessarily writing full proteins unless needed.

Wait — let’s check the mutations given.

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Mutation 4 (called “Mutation 1” in the table):

Mutated DNA:
GGG-ACA-CCA-TAT-CAC-CCG-T

Note: This has an extra base at the end (“T”), making it 7 triplets? Wait, let’s count:

Original normal DNA had 6 triplets: GGG ACA CCA TTC ACC CGT → 18 bases

This mutated DNA: GGG ACA CCA TAT CAC CCG T → that’s 7 groups? Let’s write without dashes:

GGGACACCATA TCACCCGT → wait, better to group properly:

Actually, let’s break it into triplets:

Position: 1-3: GGG
4-6: ACA
7-9: CCA
10-12: TAT
13-15: CAC
16-18: CCG
19: T ← extra base!

So total 19 bases → not divisible by 3 → frameshift!

But let’s follow the worksheet’s format.

They show Mutated mRNA starts with “CCC-”, same as normal.

Let’s transcribe Mutated DNA to mRNA:

DNA: GGG ACA CCA TAT CAC CCG T
mRNA: CCC UGU GGU AUA GUG GGC A

Wait — last part: DNA “T” alone? That doesn’t make a full codon. But perhaps we ignore incomplete codons? Or maybe it’s a typo?

Looking back: Original normal DNA ends with CGT → mRNA GCA

Here, after CCG, there’s a single T — so maybe it’s meant to be CCGT? But written as CCG-T? Hmm.

Alternatively, perhaps the dash is just separator, and it’s GGG ACA CCA TAT CAC CCG T — meaning 7 codons? But original was 6.

Wait — let’s compare side by side.

Normal DNA:
GGG ACA CCA TTC ACC CGT → 6 codons

Mutated DNA (Mutation 4):
GGG ACA CCA TAT CAC CCG T → if we take first 6 triplets: GGG ACA CCA TAT CAC CCG → that’s 18 bases, plus extra T.

But in the worksheet, under “Mutated mRNA” they have “CCC-” — implying they want us to start filling.

Perhaps the “T” at the end is a mistake? Or maybe it’s intentional to show insertion.

Let me re-read: The header says “Mutation 4” under “Mutation 1” section — probably a labeling error in the worksheet. Anyway, let’s proceed.

Assume Mutated DNA is: GGG ACA CCA TAT CAC CCG T — but for transcription, we’ll go base by base.

DNA: G G G A C A C C A T A T C A C C C G T
mRNA: C C C U G U G G U A U A G U G G G C A

Group into codons:

CCC UGU GGU AUA GUG GGC A → last one is incomplete (only “A”), so maybe stop at GGC?

But let’s see what changed compared to normal.

Normal DNA: ...TTC ACC CGT... → positions 10-18: TTC ACC CGT

Mutated: ...TAT CAC CCG T...

So:

Normal: TTC → mRNA AAG → Lys
Mutated: TAT → mRNA AUA → Ile

Normal: ACC → mRNA UGG → Trp
Mutated: CAC → mRNA GUG → Val

Normal: CGT → mRNA GCA → Ala
Mutated: CCG → mRNA GGC → Gly

And then extra T → mRNA A (incomplete)

So the protein would change starting at position 4:

Normal protein: Pro-Cys-Gly-Lys-Trp-Ala
Mutated protein: Pro-Cys-Gly-Ile-Val-Gly-? (and possibly truncated or shifted)

But also, because there’s an extra base at the end, it might cause a frameshift later — but since it’s at the very end, maybe not affecting earlier codons.

Wait — actually, the mutation is in the middle: TTC became TAT, etc., and then an extra T added at the end.

But let’s look at the type of mutation.

Comparing normal and mutated DNA:

Normal: ...TTC ACC CGT
Mutated: ...TAT CAC CCG T

So:

- TTC → TAT : change in first base of codon (T→T? No: TTC vs TAT — second base T→A, third C→T? Wait:

TTC: T-T-C
TAT: T-A-T → so second base changed from T to A, third from C to T → two changes? But that seems unlikely for a single mutation.

Wait — perhaps it’s a point mutation followed by insertion?

Alternatively, maybe the entire sequence after CCA is different.

Another idea: Perhaps “Mutation 4” is mislabeled, and it’s supposed to be compared directly.

Let’s list both sequences aligned:

Normal DNA: GGG ACA CCA TTC ACC CGT
Mutated DNA: GGG ACA CCA TAT CAC CCG T

So up to CCA, same. Then:

Normal: TTC ACC CGT
Mutated: TAT CAC CCG T

So:

- Codon 4: TTC → TAT → change from Phe to Tyr? Wait no — in DNA, TTC codes for Phe? Let’s recall:

In DNA, the coding strand is usually given, and mRNA is complementary.

I think I made a mistake here.

Important: In molecular biology, when we say "DNA sequence" like this, it’s often the *coding strand* (same as mRNA except T instead of U). But sometimes it’s the template strand.

Looking at the normal example:

DNA Normal: GGG ACA CCA TTC ACC CGT
mRNA Normal: CCC ...

If DNA is coding strand, then mRNA should be same as DNA but T→U.

But here, DNA GGG → mRNA CCC — that suggests DNA is the *template strand*, because template GGG would give mRNA CCC (since RNA polymerase reads template and makes complementary RNA).

Yes! That must be it.

In transcription, mRNA is complementary to the template DNA strand.

So if DNA given is template strand, then:

Template DNA: GGG → mRNA: CCC (because G pairs with C)

Similarly, DNA ACA → mRNA UGU (A→U, C→G, A→U? Wait no:

Base pairing for transcription:

DNA template base → RNA base
A → U
T → A
C → G
G → C

So:

DNA template: G G G → RNA: C C C
DNA: A C A → RNA: U G U
DNA: C C A → RNA: G G U
DNA: T T C → RNA: A A G
DNA: A C C → RNA: U G G
DNA: C G T → RNA: G C A

Which matches what I had earlier.

Now for Mutated DNA (Mutation 4): GGG ACA CCA TAT CAC CCG T

As template strand:

mRNA: CCC UGU GGU AUA GUG GGC A

Codons: CCC, UGU, GGU, AUA, GUG, GGC, and A (incomplete)

Protein: Pro, Cys, Gly, Ile, Val, Gly, ?

Compared to normal: Pro, Cys, Gly, Lys, Trp, Ala

So changes start at codon 4: Lys → Ile, Trp → Val, Ala → Gly, and extra amino acid or truncation.

Also, there’s an extra base at the end (the final T in DNA gives A in mRNA, but no complete codon).

But more importantly, let’s see what kind of mutation this is.

Compare DNA sequences:

Normal: ...TTC ACC CGT
Mutated: ...TAT CAC CCG T

So:

- Position 10: T → T (same)
- Position 11: T → A (change)
- Position 12: C → T (change)
- Position 13: A → C (change)
- Position 14: C → A (change)
- Position 15: C → C (same)
- Position 16: C → C (same)
- Position 17: G → G (same)
- Position 18: T → ? Wait, normal has CGT, so position 16:C,17:G,18:T

Mutated has CCG T — so position 16:C,17:C,18:G,19:T

Ah! So it’s not just substitutions; there’s an insertion.

Specifically, after position 15, instead of CGT, it’s CCG and then T — so perhaps a base inserted between 15 and 16? Let's index:

Normal DNA bases 1-18:
1G 2G 3G 4A 5C 6A 7C 8C 9A 10T 11T 12C 13A 14C 15C 16C 17G 18T

Mutated DNA:
1G 2G 3G 4A 5C 6A 7C 8C 9A 10T 11A 12T 13C 14A 15C 16C 17C 18G 19T

So comparing:

Bases 1-9: same
Base 10: T=T
Base 11: T→A
Base 12: C→T
Base 13: A→C
Base 14: C→A
Base 15: C=C
Base 16: C=C
Base 17: G→C
Base 18: T→G
Plus base 19: T (extra)

This looks messy. Perhaps it's a frameshift due to insertion.

Notice that from base 10 onwards, everything is shifted.

Normal codon 4: bases 10-12: TTC
Mutated codon 4: bases 10-12: TAT — but then codon 5: bases 13-15: CAC (normal was ACC)
Codon 6: bases 16-18: CCG (normal was CGT)
Then extra base 19: T

So the reading frame is altered starting at codon 4, and there's an extra base at the end.

This is likely a frameshift mutation caused by insertion of a base.

But where is the insertion? If we assume that the first three codons are unchanged, then at position 10, instead of TTC, it's TAT, which could be a substitution, but then the following bases don't match.

Perhaps it's easier to see that the sequence after CCA is completely different and longer.

Another way: count the number of bases.

Normal: 18 bases
Mutated: 19 bases — so one extra base inserted somewhere.

Where? If we align:

Normal: GGG ACA CCA TTC ACC CGT
Mutated: GGG ACA CCA TAT CAC CCG T

If we insert a base after position 9 (after CCA), then:

After insertion, the sequence shifts.

Suppose we insert 'A' after position 9:

Then mutated would be: GGG ACA CCA A TAT CAC CCG T — but that's not matching.

Perhaps the mutation is that "TTC" became "TAT", and "ACC" became "CAC", and "CGT" became "CCG", and then "+T" — so multiple point mutations plus insertion.

But that seems complicated for a practice problem.

Looking back at the worksheet, for Mutation 4, they have "Mutated mRNA: CCC-" and "Mutated Protein: Pro-", same as normal start, and ask for "Type of Mutation".

Also, in the mutated DNA, it's written as "GGG-ACA-CCA-TAT-CAC-CCG-T" — with dashes, suggesting 7 items, while normal has 6.

So likely, an extra base is inserted, causing a frameshift.

Moreover, the protein will be different from that point on.

So type of mutation: Frameshift mutation (due to insertion of a nucleotide).

But let's confirm with the other mutation.

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Mutation 5 (under "Mutation 2"):

Mutated DNA: GGT ACA CCA TTC ACC CGT

Compare to normal: GGG ACA CCA TTC ACC CGT

Only difference: first codon GGG → GGT

So DNA change: GGG to GGT — third base G to T.

Now, transcribe to mRNA:

Normal DNA GGG → mRNA CCC
Mutated DNA GGT → mRNA CCA (because G→C, G→C, T→A)

So mRNA: CCA instead of CCC

Protein: CCC = Pro, CCA = Pro (both code for Proline)

So no change in protein!

This is a silent mutation — a point mutation that doesn't change the amino acid.

Because both CCC and CCA code for Proline.

So for Mutation 5:

Mutated mRNA: CCA UGU GGU AAG UGG GCA
Mutated Protein: Pro-Cys-Gly-Lys-Trp-Ala (same as normal)
Type of Mutation: Silent mutation

Now back to Mutation 4.

For Mutation 4, Mutated DNA: GGG ACA CCA TAT CAC CCG T

As template strand, mRNA: CCC UGU GGU AUA GUG GGC A

Protein: Pro-Cys-Gly-Ile-Val-Gly-? (and since last codon incomplete, might stop or be wrong)

Compared to normal: Pro-Cys-Gly-Lys-Trp-Ala

So different amino acids from position 4 onward.

Also, the DNA has 19 bases vs 18, so insertion of one base.

Where is the insertion? If we assume the first 9 bases are same, then at position 10, normal is T, mutated is T — same.

Position 11: normal T, mutated A — change.

But if there was an insertion before position 10, it would shift everything.

Suppose an extra base is inserted after position 9.

Normal: positions 1-9: GGGACACCA
Then 10-12: TTC

If we insert, say, 'A' after position 9, then new sequence: GGGACACCA A TTCACC CGT — but that's not matching mutated.

Mutated is GGGACACCA TATCACCCGT

So after CCA, it's TAT instead of TTC.

Perhaps it's a substitution of TTC to TAT, and then ACC to CAC, and CGT to CCG, and +T — but that's multiple changes.

Another possibility: the "T" at the end is not extra, but part of the last codon.

Normal last codon: CGT
Mutated last: CCG T — if we consider CCG as codon, and T is separate, but in DNA, it should be grouped.

Perhaps the mutated DNA is meant to be GGG ACA CCA TAT CAC CCG, and the "T" is a typo or something.

But in the text, it's "GGG-ACA-CCA-TAT-CAC-CCG-T" — seven groups.

Let's calculate the length: each group is 3 bases except last is 1, so 3*6 +1 = 19 bases.

Normal is 6*3=18.

So definitely an extra base.

The most straightforward explanation is that a base is inserted, causing a frameshift.

For example, if a base is inserted after position 9, then the reading frame shifts.

Normal codons: 1:GGG, 2:ACA, 3:CCA, 4:TTC, 5:ACC, 6:CGT

If we insert 'A' after position 9 (after CCA), then new sequence: GGG ACA CCA A TTA CCA CCG T — but that's not matching.

Mutated is GGG ACA CCA TAT CAC CCG T

So after CCA, it's TAT, which is three bases, then CAC, etc.

Perhaps the insertion is within the fourth codon.

Normal fourth codon: TTC

If we insert a base after the first T, it becomes T X TC, but here it's TAT, which is different.

I think for the purpose of this problem, since the protein changes significantly and there's an extra base, it's a frameshift mutation.

Moreover, in many textbooks, adding or deleting a base causes frameshift.

So for Mutation 4:

- Mutated mRNA: CCC UGU GGU AUA GUG GGC A (but since last is incomplete, perhaps write up to GGC)
- Mutated Protein: Pro-Cys-Gly-Ile-Val-Gly (assuming we stop at complete codons)
- Type of Mutation: Frameshift mutation (insertion)

But to be precise, let's see the exact change.

From normal to mutated DNA:

Normal: ...CCA TTC ACC CGT
Mutated: ...CCA TAT CAC CCG T

So the "TTC" became "TAT", "ACC" became "CAC", "CGT" became "CCG", and "+T"

This could be interpreted as a series of point mutations plus an insertion, but that's unlikely.

Notice that if we delete the first 'T' of "TTC", then "TC" remains, but not matching.

Another idea: perhaps "TAT" is a result of changing "TTC" to "TAT" by two substitutions, but then "ACC" to "CAC" is another, etc.

But that would be multiple point mutations, not typical for a single mutation event.

Given that the worksheet has "Type of Mutation" and for this one, it's likely frameshift due to the extra base.

Perhaps the "T" at the end is meant to be part of the last codon, but then it would be CCGT, which is four bases.

I think the best answer is that it's a frameshift mutation caused by insertion of a nucleotide.

For the protein, since the mRNA is CCC UGU GGU AUA GUG GGC A, and assuming we translate until stop or end, the protein is Pro-Cys-Gly-Ile-Val-Gly, and the last A is not translated.

Compared to normal Pro-Cys-Gly-Lys-Trp-Ala, so different.

So for Mutation 4:

- Mutated mRNA: CCC UGU GGU AUA GUG GGC
- Mutated Protein: Pro-Cys-Gly-Ile-Val-Gly
- Type of Mutation: Frameshift mutation

But to match the worksheet's format, they have "CCC-" for mRNA and "Pro-" for protein, so perhaps they want the full sequence.

Since the normal has six codons, and mutated has six complete codons if we ignore the last T, but mutated DNA has 19 bases, so mRNA has 19 bases, which is 6 full codons and one extra base.

So mRNA: CCC UGU GGU AUA GUG GGC A — so six full codons: CCC, UGU, GGU, AUA, GUG, GGC, and then A.

So protein: Pro, Cys, Gly, Ile, Val, Gly

Yes.

Now for Mutation 5, as above, silent mutation.

So let's summarize.

First, for the normal part, although not asked to fill, but for completeness:

Normal mRNA: CCC UGU GGU AAG UGG GCA
Normal Protein: Pro-Cys-Gly-Lys-Trp-Ala

But the worksheet only shows "CCC-" and "Pro-", so perhaps they expect us to fill the rest similarly for mutations.

For Mutation 4:

Mutated DNA: GGG ACA CCA TAT CAC CCG T
Mutated mRNA: CCC UGU GGU AUA GUG GGC A → but since last is incomplete, perhaps write as CCC UGU GGU AUA GUG GGC (ignoring the last A for protein synthesis)

In practice, ribosomes may not translate incomplete codons, so protein is based on complete codons.

So Mutated mRNA: CCC UGU GGU AUA GUG GGC
Mutated Protein: Pro-Cys-Gly-Ile-Val-Gly
Type of Mutation: Frameshift mutation (due to insertion of a nucleotide)

For Mutation 5:

Mutated DNA: GGT ACA CCA TTC ACC CGT
Mutated mRNA: CCA UGU GGU AAG UGG GCA (since GGT → CCA)
Mutated Protein: Pro-Cys-Gly-Lys-Trp-Ala (same as normal, since CCA also codes for Pro)
Type of Mutation: Silent mutation

Now, to write the answers as per worksheet.

In the worksheet, for Mutation 4, they have:

Mutated mRNA: CCC- → so we need to fill the rest: UGU GGU AUA GUG GGC

Similarly, Mutated Protein: Pro- → Cys Gly Ile Val Gly

Type: Frameshift

For Mutation 5:

Mutated mRNA: CCA UGU GGU AAG UGG GCA

Mutated Protein: Pro Cys Gly Lys Trp Ala

Type: Silent

But let's double-check Mutation 5.

DNA mutated: GGT ACA CCA TTC ACC CGT

As template strand:

GGT → mRNA: CCA (G→C, G→C, T→A)
ACA → UGU
CCA → GGU
TTC → AAG
ACC → UGG
CGT → GCA

So mRNA: CCA UGU GGU AAG UGG GCA

Protein: CCA=Pro, UGU=Cys, GGU=Gly, AAG=Lys, UGG=Trp, GCA=Ala — same as normal.

Yes.

For Mutation 4, DNA: GGG ACA CCA TAT CAC CCG T

mRNA:
GGG → CCC
ACA → UGU
CCA → GGU
TAT → AUA (T→A, A→U, T→A? Wait: DNA TAT as template: T→A, A→U, T→A, so mRNA AUA)
CAC → GUG (C→G, A→U, C→G)
CCG → GGC (C→G, C→G, G→C)
T → A (but incomplete)

So mRNA: CCC UGU GGU AUA GUG GGC A

Protein: CCC=Pro, UGU=Cys, GGU=Gly, AUA=Ile, GUG=Val, GGC=Gly, and A not translated.

So protein: Pro-Cys-Gly-Ile-Val-Gly

Type: Since there is an extra base (19 vs 18), and the reading frame is shifted, it's a frameshift mutation. Specifically, insertion of a nucleotide.

Now, to present the answer.

Final Answer:

For Mutation 4:
- Mutated mRNA: CCC UGU GGU AUA GUG GGC
- Mutated Protein: Pro-Cys-Gly-Ile-Val-Gly
- Type of Mutation: Frameshift mutation

For Mutation 5:
- Mutated mRNA: CCA UGU GGU AAG UGG GCA
- Mutated Protein: Pro-Cys-Gly-Lys-Trp-Ala
- Type of Mutation: Silent mutation

But in the worksheet, for Mutation 4, they have "Mutated mRNA: CCC-" so perhaps they want it written with dashes or spaces.

Similarly, protein with dashes.

Also, for Mutation 5, no red text, so we fill all.

So let's write it clearly.