Ms Drs Biology 621 Answer Key - Fill Online, Printable, Fillable ... - Free Printable

Let’s go step by step to solve this worksheet on mutations.

We’re looking at how changes in DNA (mutations) affect the protein made from it. The key idea is:
DNA → mRNA → Protein (amino acids).
Each group of 3 letters in mRNA is called a codon, and each codon tells the cell which amino acid to add next in the protein chain.

The original DNA sequence given is:
T G C A T A A A A T G G

First, let’s convert that to mRNA. Remember:
- In RNA, T becomes U.
- So we replace every T with U.

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

Now, break the mRNA into groups of 3 (codons):
ACG - UAU - UUU - ACC

Now use the genetic code (which you’d usually have as a chart — but I’ll give you what you need here):

- ACG → Thr (Threonine)
- UAU → Tyr (Tyrosine)
- UUU → Phe (Phenylalanine)
- ACC → Thr (Threonine)

So the original protein is:
Thr - Tyr - Phe - Thr

Wait — hold on! Looking back at the worksheet, under “Here are two possible outcomes...” it says:

> Original: T G C A T A A A A T G G → Val - Asp - Asn - Trp

That doesn’t match what I just did. Let me check again.

Ah — maybe they’re using the *coding strand* of DNA? Or perhaps there’s a different convention.

Actually, in many worksheets, when they give you a DNA sequence like this, they mean the template strand — the one that gets transcribed into mRNA.

But if the original DNA is written as the template strand, then mRNA would be complementary to it.

Let’s try that.

If DNA template strand is:
T G C A T A A A A T G G

Then mRNA is built complementary to it:

DNA: T G C A T A A A A T G G
mRNA: A C G U A U U U U A C C ← same as before!

Still gives ACG UAU UUU ACC → Thr-Tyr-Phe-Thr

But the worksheet says the original protein is: Val - Asp - Asn - Trp

Hmm. That suggests they might be treating the given DNA as the coding strand (same as mRNA except T instead of U).

Let’s test that.

If DNA coding strand = T G C A T A A A A T G G
Then mRNA = U G C A U A A A A U G G

Break into codons:
UGC - AUA - AAA - UGG

Now translate:

- UGC → Cys (Cysteine) — not Val
Still not matching.

Wait — look at the worksheet again. Under section 1, it says:

> Substitution: One base pair is replaced by another
> Example: Original: T G C A T A A A A T G G
> Mutated: T G C A T A A A A T G C
> mRNA: A C G U A U U U U A C G
> Protein: Thr - Tyr - Phe - Thr

And then it says: “Val - Asp - Asn - Trp” for the original? No — actually, looking closely, under “Here are two possible outcomes...”, it lists:

> 1. Silent mutation: ... no change in amino acid
> Example: CUU and CUC both code for Leu
> 2. Substitution: When one base pair is substituted, the new codon codes for a different amino acid
> Example: If U codes for Serine and CC codes for Proline — wait, that doesn’t make sense. Codons are 3 bases.

I think there’s some formatting issues in the image text. Let me re-read carefully.

Actually, in the “Here are two possible outcomes...” section, it says:

> Here are two possible outcomes of mutations:
> 1. Silent mutation: When a nucleotide substitution results in the same amino acid being produced.
> Example: CUU and CUC both code for Leucine.
> 2. Substitution: When one base pair is substituted, the new codon codes for a different amino acid.
> Example: If UUU codes for Phenylalanine and CCU codes for Proline — still messy.

Then it says:

> Now let’s see what happens in the following examples. Below, all the codons have been translated into their respective amino acids.

Then it shows:

> Original: T G C A T A A A A T G G → Val - Asp - Asn - Trp

Wait — so they’re saying the original DNA sequence translates to Val-Asp-Asn-Trp.

How?

Let’s assume the DNA given is the coding strand (so mRNA is same but T→U).

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

Codons: UGC - AUA - AAA - UGG

Translate:

- UGC → Cys (not Val)
- AUA → Ile (not Asp)
Nope.

Alternative: Maybe they’re reading the DNA as the template strand, but translating directly without converting to mRNA? That wouldn’t be right.

Another possibility: Perhaps the sequence is meant to be read as mRNA already? But it has T’s.

Let’s look at the mutated version they show:

Mutated: T G C A T A A A A T G C
Protein: Val - Asp - Asn - Thr

Original protein: Val - Asp - Asn - Trp

So only the last amino acid changed: Trp → Thr

What codon makes Trp? UGG
What codon makes Thr? ACU, ACC, ACA, ACG

In the DNA, last three bases: T G G → if this is coding strand, mRNA would be U G G → Trp

Mutated: T G C → mRNA U G C → Cys? Not Thr.

Unless...

Perhaps the DNA sequence is written 5’ to 3’, and we need to take the complement for mRNA.

Let’s try that properly.

Standard transcription:
Template DNA strand is read 3’ to 5’, mRNA is synthesized 5’ to 3’.

But often in problems, they give you the template strand written 5’ to 3’, and you have to reverse-complement it to get mRNA.

Let’s try that.

Given DNA: T G C A T A A A A T G G (assume this is template strand, written 5’ to 3’)

To get mRNA, we need the complement, and since mRNA is antiparallel, we should reverse it first? No.

Better way:
If template DNA is 5’-TGCA TAAA ATGG-3’, then mRNA is synthesized starting from the 3’ end of template.

So template 3’ end is GGTA... wait, let's write it clearly.

Write DNA template strand 5’ to 3’:
5’ - T G C A T A A A A T G G - 3’

Then the complementary mRNA will be synthesized 5’ to 3’, pairing with the template from 3’ to 5’.

So read template from 3’ to 5’:
3’ - G G T A A A A T A C G T - 5’ [this is the template read backwards]

Now build mRNA 5’ to 3’ complementary to that:

Template 3’-G G T A A A A T A C G T-5’
mRNA 5’-C C A U U U U A U G C A-3’

So mRNA: CCA UUU UAU GCA

Codons: CCA - UUU - UAU - GCA

Translate:

- CCA → Pro
- UUU → Phe
- UAU → Tyr
- GCA → Ala

Not matching Val-Asp-Asn-Trp.

This is frustrating. Let’s look at the answer they imply.

They say original protein is Val - Asp - Asn - Trp

What codons make those?

Val: GUU, GUC, GUA, GUG
Asp: GAU, GAC
Asn: AAU, AAC
Trp: UGG

So mRNA must be something like: GUx - GAx - AAx - UGG

For example: GUA - GAC - AAU - UGG

Which would come from DNA template strand: CAT - CTG - TTA - ACC (complementary)

Or if DNA coding strand: GTA - GAC - AAT - TGG

Look at the given DNA: T G C A T A A A A T G G

If we split it as: TGC ATA AAA TGG

Compare to GTA GAC AAT TGG — not the same.

TGC vs GTA — no.

Another idea: Perhaps the sequence is misaligned or there's a typo in my reading.

Let’s count the bases: T G C A T A A A A T G G — that's 12 bases, so 4 codons.

Suppose the mRNA is: GUA GAC AAU UGG

Then DNA coding strand would be: GTA GAC AAT TGG

But given is TGC ATA AAA TGG — close but not quite.

TGC vs GTA — first base T vs G, second G vs T, third C vs A — completely different.

Unless... perhaps the "original" protein listed is for a different sequence? But the worksheet says "Original: T G C A T A A A A T G G → Val - Asp - Asn - Trp"

Maybe it's a mistake in the worksheet, or in my understanding.

Let’s look at the mutated version they show:

Mutated: T G C A T A A A A T G C
Protein: Val - Asp - Asn - Thr

So only last amino acid changed from Trp to Thr.

Trp is UGG, Thr is ACx.

So if last codon was UGG, now it's UGC or something.

In DNA, if last three were TGG, mutated to TGC.

If DNA is coding strand, mRNA was UGG (Trp), now UGC (Cys) — not Thr.

If DNA is template strand, mRNA was ACC (Thr)? Let's calculate.

Assume DNA given is template strand.

Original DNA template: T G C A T A A A A T G G

mRNA complementary: A C G U A U U U U A C C (as I had initially)

Codons: ACG UAU UUU ACC → Thr Tyr Phe Thr

But they say protein is Val Asp Asn Trp — not matching.

Perhaps they mean the DNA sequence is the non-template strand, and we need to find the template first.

I recall that in some curricula, when they give a DNA sequence and say "this codes for", they mean the coding strand, and you just change T to U for mRNA.

So let's force it.

Suppose DNA coding strand: T G C A T A A A A T G G
mRNA: U G C A U A A A A U G G

Codons: UGC AUA AAA UGG

Translate:

- UGC = Cys
- AUA = Ile
- AAA = Lys
- UGG = Trp

So Cys-Ile-Lys-Trp — not Val-Asp-Asn-Trp.

Closest is Trp at the end.

For Val-Asp-Asn-Trp, mRNA should be GUN GAY AAY UGG, where N is any, Y is U or C.

For example: GUA GAC AAU UGG

DNA coding strand: GTA GAC AAT TGG

Given DNA: TGC ATA AAA TGG

Compare position by position:

Position 1-3: TGC vs GTA — different
4-6: ATA vs GAC — different
7-9: AAA vs AAT — similar, AAA vs AAT
10-12: TGG vs TGG — same

Only last codon matches.

Perhaps there's a shift or something.

Another thought: Maybe the sequence is written, but we need to start from a different point? But it's 12 bases, divisible by 3.

Let’s look at the other parts of the worksheet.

In section 4, it says:

> Deletion: One base is deleted. A whole new amino acid is added or one is missing from the mutated protein.

Example:
Original: T G C A T A A A A T G G → Val - Asp - Asn - Trp
Mutated: T G C A T A A A A T G → Val - Asp - Asn - ?

With deletion of last G, so DNA: T G C A T A A A A T G (11 bases)

mRNA would be incomplete, but they might expect us to see that the last codon is cut off.

But still, we need to know what the original mapping is.

Perhaps for the sake of this problem, we should accept that the original DNA sequence T G C A T A A A A T G G corresponds to protein Val - Asp - Asn - Trp, and proceed with the mutations based on that, even if the biochemistry isn't perfect. Maybe it's simplified.

In many high school worksheets, they provide the translation directly, so we don't have to derive it.

Looking back at the user's image description, under "Here are two possible outcomes...", it lists:

> Original: T G C A T A A A A T G G → Val - Asp - Asn - Trp

And then for substitution example:

> Mutated: T G C A T A A A A T G C → Val - Asp - Asn - Thr

So they're telling us that changing the last base from G to C changes Trp to Thr.

Similarly, for insertion:

> Inserted an extra A after position 6: T G C A T A A A A A T G G → Val - Asp - Asn - Lys - Gly

Original protein has 4 amino acids, inserted has 5, so frameshift.

For deletion:

> Deleted the last G: T G C A T A A A A T G → Val - Asp - Asn - (incomplete)

Frameshift mutation.

For nonsense mutation:

> Changed a base to create a stop codon: e.g., T G C A T A A A A T G A → Val - Asp - Asn - Stop

So for the purpose of this worksheet, we can take the given translations as fact, and focus on identifying the type of mutation and its effect.

The questions are likely to identify the type of mutation in each case.

But the user didn't specify which question to answer; they just uploaded the worksheet.

Perhaps the task is to understand the types of mutations described.

Since the user said "solve the problem accurately", and the worksheet has several parts, but no specific question asked, I think the intent is to explain the mutations shown.

Maybe the "problem" is to fill in or understand the examples.

Another idea: perhaps the student is supposed to complete the worksheet, but since it's already filled in the image, maybe we need to verify or explain.

Let's assume that the main task is to recognize the types of mutations in the examples provided.

From the worksheet:

1. Substitution: one base replaced. Example: TGG -> TGC in DNA, protein Trp -> Thr. This is a missense mutation (different amino acid).

2. Insertion: extra base added. Example: added A after position 6, so sequence becomes longer, and protein has extra amino acids: Val-Asp-Asn-Lys-Gly instead of Val-Asp-Asn-Trp. This is a frameshift mutation because adding one base shifts the reading frame.

3. Deletion: one base removed. Example: removed last G, so DNA shorter, protein truncated or frameshifted. They show Val-Asp-Asn-? , implying the last codon is incomplete or changed.

4. Nonsense mutation: substitution that creates a stop codon. Example: TGG -> TGA in DNA, which would be UGG -> UGA in mRNA, and UGA is stop codon, so protein stops early: Val-Asp-Asn-Stop.

5. Frameshift mutation: when insertion or deletion causes the reading frame to shift, altering all subsequent amino acids. In the insertion example, it's frameshift.

In the worksheet, for insertion, they show the protein as Val-Asp-Asn-Lys-Gly, which suggests that after the insertion, the codons are regrouped.

Let's simulate that.

Original DNA: T G C A T A A A A T G G
Assume this is coding strand, mRNA: U G C A U A A A A U G G → codons UGC AUA AAA UGG → but they say protein is Val-Asp-Asn-Trp, so perhaps it's not.

To match Val-Asp-Asn-Trp, let's assume the mRNA is GUA GAC AAU UGG

So DNA coding strand: GTA GAC AAT TGG

But given is TGC ATA AAA TGG — so perhaps there's a typo, and it's supposed to be GTA GAC AAT TGG.

Maybe "T G C" is "G T A" misread, but in the image it's clear.

Another possibility: the sequence is written, but we need to consider it as the template, and the protein is given for reference.

For the sake of progressing, I'll use the given protein sequences as provided in the worksheet, and explain the mutations based on that.

So, for each example:

- Substitution (TGG to TGC): changes last amino acid from Trp to Thr — this is a point mutation, specifically missense.

- Insertion (added A after position 6): original DNA positions 1-12, insert A after 6th base, so new sequence: T G C A T A A A A A T G G (13 bases)

Original codons: say positions 1-3,4-6,7-9,10-12

After inserting A after position 6, the sequence is: bases 1-6 same, then A, then 7-12.

So new grouping: 1-3, 4-6, 7-9 (but 7 is now the inserted A, etc.)

Let's index:

Original: 1:T 2:G 3:C 4:A 5:T 6:A 7:A 8:A 9:A 10:T 11:G 12:G

Insert A after position 6, so new sequence: 1:T 2:G 3:C 4:A 5:T 6:A 7:A (inserted) 8:A 9:A 10:A 11:T 12:G 13:G

Now codons: 1-3: TGC, 4-6: ATA, 7-9: AAA, 10-12: ATG, 13:G (incomplete)

But they say protein is Val-Asp-Asn-Lys-Gly

So perhaps with the insertion, the reading frame shifts.

If original mRNA was GUA GAC AAU UGG for Val-Asp-Asn-Trp

After inserting a base, say after the 6th base of DNA, which corresponds to after the 2nd codon.

Original DNA coding: let's say it's GTA GAC AAT TGG for simplicity, even though given is different.

To match the worksheet's protein, I'll assume that the DNA sequence is intended to be such that it codes for Val-Asp-Asn-Trp, and the mutations are as described.

For the insertion example: they say mutated DNA: T G C A T A A A A A T G G (with extra A)

Protein: Val - Asp - Asn - Lys - Gly

So compared to original Val-Asp-Asn-Trp, now it's Val-Asp-Asn-Lys-Gly, so the last two amino acids are different, and there's an extra one.

This indicates a frameshift mutation due to insertion.

Similarly, for deletion: removed last G, so DNA: T G C A T A A A A T G (11 bases)

Protein: Val - Asp - Asn - ? , and they might imply it's incomplete or frameshifted.

In the worksheet, for deletion, it says: "Val - Asp - Asn - " and then blank, or perhaps "Val - Asp - Asn - " meaning the last is missing or changed.

For nonsense: changed to TGA, protein Val-Asp-Asn-Stop.

So to summarize the types:

- Substitution that changes amino acid: missense mutation

- Insertion of one base: frameshift mutation

- Deletion of one base: frameshift mutation

- Substitution that creates stop codon: nonsense mutation

Now, for the student, the task might be to identify the type of mutation in each case.

Since the user didn't specify a particular question, I'll provide a general explanation of the mutations shown.

But the instruction is to "solve the problem accurately", so perhaps the problem is to understand and classify the mutations.

Maybe the worksheet has blanks to fill, but in the image, it's already filled.

Another idea: perhaps the "problem" is the entire worksheet, and we need to ensure the answers are correct.

Let's verify the insertion example.

Suppose original DNA coding strand: let's define it as D1 D2 D3 D4 D5 D6 D7 D8 D9 D10 D11 D12 = T G C A T A A A A T G G

But to get protein Val-Asp-Asn-Trp, let's assume the mRNA is M1M2M3 M4M5M6 M7M8M9 M10M11M12 = GUA GAC AAU UGG

So DNA coding strand should be GTA GAC AAT TGG

So perhaps there's a typo in the worksheet, and it's supposed to be GTA GAC AAT TGG.

In many fonts, G and T might look similar, but in the image, it's "T G C" which is clear.

Perhaps "T G C" is "G T A" if misread, but unlikely.

Let's calculate what DNA would give Val-Asp-Asn-Trp.

Val: codons GTN, so DNA coding GTA, GTC, GTG, GTT

Asp: GAY, so GAT, GAC

Asn: AAY, so AAT, AAC

Trp: TGG

So possible DNA coding strand: GTA GAC AAT TGG

Given is TGC ATA AAA TGG

So if we compare:

Given: T G C A T A A A A T G G

Intended: G T A G A C A A T T G G

So positions: 1:T vs G, 2:G vs T, 3:C vs A, 4:A vs G, 5:T vs A, 6:A vs C, 7:A vs A, 8:A vs A, 9:A vs T, 10:T vs T, 11:G vs G, 12:G vs G

Only positions 7,8,10,11,12 match; others don't.

This is problematic.

Perhaps the sequence is for the template strand, and we need to find the mRNA.

Let me try once more.

Assume given DNA is template strand: 5'-TGCATAAAATGG-3'

Then mRNA is 5'-CCA UUU UAU GCA-3' as I had earlier, which is Pro-Phe-Tyr-Ala, not matching.

If we take the complement and reverse for mRNA.

Template DNA 5'-TGCATAAAATGG-3'

Complement is 3'-ACGTATTTTACC-5'

mRNA is 5'-CCAUUUUAUGCA-3' same as above.

Not working.

Another possibility: perhaps the "original" protein is for a different sequence, but the worksheet links it.

Let's look at the very beginning of the worksheet.

It says: "There are 4 ways that DNA can be altered when mutations occur in DNA sequences."

Then lists:

1. Substitution: one base pair replaced by another

Example: Original: T G C A T A A A A T G G

Mutated: T G C A T A A A A T G C

mRNA: A C G U A U U U U A C G

Protein: Thr - Tyr - Phe - Thr

Then it says: "Here are two possible outcomes of mutations:"

1. Silent mutation: no change in amino acid

2. Substitution: different amino acid

Then: "Now let's see what happens in the following examples. Below, all the codons have been translated into their respective amino acids."

Then: "Original: T G C A T A A A A T G G → Val - Asp - Asn - Trp"

This is inconsistent because in the first example, with the same DNA, they got Thr-Tyr-Phe-Thr, but here they say Val-Asp-Asn-Trp.

Unless the "Original" in the second part is a different sequence, but it's written the same.

Perhaps it's a different instance.

In the first substitution example, they have mRNA ACG UAU UUU ACG for the mutated, but for original, if DNA is TGCATAAAATGG, mRNA should be ACGUAUUUUACC for template, giving Thr-Tyr-Phe-Thr, and they have that.

Then later, when they say "Original: T G C A T A A A A T G G → Val - Asp - Asn - Trp", it must be a mistake, or perhaps it's for a different context.

Maybe "Val - Asp - Asn - Trp" is for a standard sequence, and the DNA is illustrative.

To resolve this, I think for the purpose of this response, I'll use the first example where they consistently use the DNA to mRNA to protein conversion.

In the first part:

DNA original: T G C A T A A A A T G G (template strand)

mRNA: A C G U A U U U U A C C

Protein: Thr - Tyr - Phe - Thr

Then for substitution: DNA mutated: T G C A T A A A A T G C

mRNA: A C G U A U U U U A C G

Protein: Thr - Tyr - Phe - Thr (since ACG and ACC both code for Thr? ACG is Thr, ACC is also Thr, yes! In genetic code, ACG and ACC both code for Threonine.

Is that true?

Recall: Thr codons: ACU, ACC, ACA, ACG — yes, all four code for Threonine.

So in the substitution example, they changed the last base from G to C in DNA template.

DNA original template: ...T G G (last three)

mRNA: ...A C C (since complementary)

DNA mutated template: ...T G C

mRNA: ...A C G

Both ACC and ACG code for Thr, so protein is the same: Thr-Tyr-Phe-Thr

But in the worksheet, for the substitution example, they have protein as Thr-Tyr-Phe-Thr for both, so it's a silent mutation.

Then later, when they say "Substitution: When one base pair is substituted, the new codon codes for a different amino acid", and give an example with UUU for Phe and CCU for Pro, but that's not related to the sequence.

Then in the "now let's see" section, they have a different scenario where substitution changes the amino acid.

For example, in the nonsense mutation, they change to TGA, which would be stop.

So for the sequence T G C A T A A A A T G G, if we take it as template, mRNA ACG UAU UUU ACC, protein Thr-Tyr-Phe-Thr.

If we substitute the last G to A in DNA template: T G C A T A A A A T G A

mRNA: A C G U A U U U U A C U

Codons: ACG UAU UUU ACU — all still Thr, Tyr, Phe, Thr — same protein.

To get a different amino acid, we need to change a base that changes the codon to a different amino acid.

For example, change the 4th base from A to G in DNA template.

DNA original: T G C A T A A A A T G G

Change 4th base A to G: T G C G T A A A A T G G

mRNA: A C G C A U U U U A C C

Codons: ACG CAU UUU ACC → Thr-His-Phe-Thr — so Tyr to His, different amino acid.

But in the worksheet, for the "substitution" that changes amino acid, they might have a different change.

In the nonsense example, they change to TGA.

DNA: T G C A T A A A A T G A

mRNA: A C G U A U U U U A C U — still Thr, as above.

To get a stop codon, we need mRNA to have UAA, UAG, or UGA.

For example, if we change the 7th base from A to T in DNA template.

DNA: T G C A T A T A A T G G

mRNA: A C G U A U A U U A C C

Codons: ACG UAU AUU ACC → Thr-Tyr-Ile-Thr — not stop.

Change the 10th base from T to A: DNA T G C A T A A A A A G G

mRNA: A C G U A U U U U U C C

Codons: ACG UAU UUU UCC → Thr-Tyr-Phe-Ser — not stop.

To get UGA in mRNA, which is stop, we need DNA template to have ACT (since mRNA UGA, template ACT).

So if we change the last three bases from TGG to ACT in DNA template.

DNA: T G C A T A A A A A C T

mRNA: A C G U A U U U U U G A

Codons: ACG UAU UUU UGA → Thr-Tyr-Phe-Stop

Yes! So if we change the last three bases from TGG to ACT in DNA template, mRNA has UGA, stop codon.

In the worksheet, for nonsense mutation, they have: "Changed a base to create a stop codon: e.g., T G C A T A A A A T G A → Val - Asp - Asn - Stop"

But with TGA at the end, if DNA template is ...TGA, mRNA is ...ACU, which is Thr, not stop.

Unless they mean the DNA coding strand.

Assume DNA coding strand for nonsense: if they change to TGA, and if DNA coding is TGA, mRNA is UGA, stop.

So perhaps in the "now let's see" section, they are using the DNA as coding strand.

Let's assume that for the "Original: T G C A T A A A A T G G → Val - Asp - Asn - Trp" , they mean DNA coding strand, and mRNA is U G C A U A A A A U G G, but as we saw, that's Cys-Ile-Lys-Trp, not Val-Asp-Asn-Trp.

Unless "T G C" is "G T A" , but it's written as TGC.

Perhaps it's "GTA" but misprinted as "TGC".

In many fonts, G and T can be confused, but in the image, it's likely TGC.

To move forward, I'll use the consistent part from the beginning.

In the first substitution example, with DNA template TGCATAAAATGG, mRNA ACGUAUUUUACC, protein Thr-Tyr-Phe-Thr, and when mutated to TGCA TAAA ATGC, mRNA ACGUAUUUUACG, protein still Thr-Tyr-Phe-Thr, so silent mutation.

Then for the other examples, they may have different sequences, but for the student, the key is to understand the types.

Perhaps the "problem" is to identify the type of mutation in each case described.

So for the worksheet:

- In the first substitution example: silent mutation (no change in amino acid)

- In the insertion example: frameshift mutation (changes the reading frame, alters subsequent amino acids)

- In the deletion example: frameshift mutation

- In the nonsense example: nonsense mutation (creates stop codon)

- In the substitution that changes amino acid: missense mutation

So for the final answer, since the user might want the classification, I'll provide that.

But to be precise, let's box the types.

Perhaps the task is to complete the worksheet, but since it's already done, maybe explain.

I think for this response, I'll provide a summary of the mutation types based on the examples.

So, step by step:

1. For the substitution where DNA changes from TGG to TGC in the last three bases (template strand), mRNA changes from ACC to ACG, both code for Thr, so no change in protein — this is a silent mutation.

2. For the insertion of an extra A after position 6: this adds one base, shifting the reading frame for all subsequent codons, so the amino acid sequence changes after the insertion point — this is a frameshift mutation.

3. For the deletion of the last G: removes one base, also shifting the reading frame, so the protein is altered — frameshift mutation.

4. For the substitution that creates a stop codon (e.g., changing to TGA in coding strand or appropriate change in template to make UGA in mRNA): this causes premature termination of protein synthesis — nonsense mutation.

5. For a substitution that changes one amino acid to another (e.g., if a change caused Tyr to become His): this is a missense mutation.

In the worksheet, for the "substitution" that changes amino acid, they might have a specific example, but in the text, it's general.

So for the final answer, I'll state the types.

Since the user may expect the answer for the entire worksheet, but there's no specific question, I'll provide the classifications.

Perhaps the "problem" is the first substitution example, and we need to say what type it is.

In that case, for the first example: substitution of G to C in last base of DNA template, protein unchanged, so silent mutation.

And they have it as such.

So I think that's it.

Final Answer: The first substitution example is a silent mutation because the amino acid sequence does not change. Other examples include frameshift mutations (for insertion and deletion), nonsense mutation (when a stop codon is created), and missense mutation (when a different amino acid is coded).

To put it neatly.

Since the worksheet has multiple parts, but the initial example is clear, I'll box that.

Or perhaps the student needs to know for each.

I think for this context, I'll provide the type for the first mutation described.

So:

The mutation in the first example (changing the last base from G to C in the DNA template strand) is a silent mutation because both the original and mutated codons code for the same amino acid (Threonine).

For completeness, let's confirm the codons.

DNA template original last three: T G G → mRNA: A C C → Thr

DNA template mutated last three: T G C → mRNA: A C G → Thr

Yes, both Thr.

So silent mutation.

In the worksheet, they have it as such.

So I think that's the answer for the first part.

Perhaps the "problem" is to identify that.

So I'll go with that.

Final Answer: silent mutation