Use the circular codon table to fill in the missing DNA triplets, mRNA codons, tRNA anticodons, and amino acids in the table.

Let’s solve this step by step using the codon wheel.

We need to fill in the table with:
- DNA triplet → mRNA codon → tRNA anticodon → Amino Acid

Remember these rules:
1. DNA to mRNA: Replace T with U, and match bases (A→U, T→A, G→C, C→G) — but note: DNA is template strand, so we transcribe it to complementary mRNA.
- Actually, in most school problems like this, they give you the *coding strand* of DNA (same as mRNA except T instead of U). But looking at the first row: DNA = GGC → mRNA = AAG? That doesn’t match if it’s coding strand.

Wait — let’s check the first given row:

> DNA triplet: GGC
> mRNA codon: AAG ← This suggests that the DNA given is the *template strand*, because:
> - Template DNA: GGC → mRNA would be CCG? No, that’s not matching.

Hold on — maybe there’s a mistake in my assumption. Let’s look again.

Actually, in standard transcription:
- If DNA template strand is 3’-GGC-5’, then mRNA is 5’-CCG-3’.
But here, for DNA = GGC, mRNA = AAG? That doesn’t make sense unless...

Wait — perhaps the DNA listed is the *non-template* (coding) strand? Then mRNA should be same as DNA but T→U.

So if DNA coding strand = GGC → mRNA = GGC? But they say mRNA = AAG. That’s not matching.

This is confusing. Let me re-express.

Looking at the first completed row:

DNA triplet: GGC
mRNA codon: AAG
tRNA anticodon: ???
Amino acid: ???

If mRNA is AAG, then tRNA anticodon must be UUC (because tRNA pairs with mRNA: A-U, G-C).

Then amino acid for mRNA codon AAG is Lysine (from codon wheel: A-A-G → Lysine).

Now, how do we get from DNA triplet GGC to mRNA AAG?

That only makes sense if the DNA given is the *template strand*, and we’re transcribing it.

Template DNA: GGC → mRNA should be CCG? Not AAG.

Unless... maybe the DNA is written 5’ to 3’, and we need to reverse complement?

Let’s try:

If DNA template strand is 3’-GGC-5’, then mRNA is 5’-CCG-3’. Still not AAG.

Alternatively, if DNA coding strand is 5’-GGC-3’, then mRNA is 5’-GGC-3’ → which codes for Glycine.

But they have mRNA = AAG → which is Lysine.

There’s inconsistency unless we assume that the DNA triplet given is actually the *complement* of what we think.

Wait — let’s flip it.

Suppose the DNA triplet given is the one that is *complementary* to the mRNA.

For example, if mRNA is AAG, then the DNA template strand would be TTC (since A↔T, G↔C), but they wrote GGC.

No.

Perhaps there's a typo or I'm overcomplicating.

Let me use the codon wheel directly for mRNA codons to find amino acids, and for tRNA anticodons, just take the complement of mRNA.

And for DNA to mRNA: since in many textbooks, when they say "DNA triplet", they mean the coding strand, so mRNA is same as DNA but T→U.

But in the first row, DNA=GGC, mRNA=AAG — that doesn't fit.

Unless... oh! Wait a minute. Look at the second row:

DNA: ???, mRNA: CAG

Third row: DNA: ???, mRNA: ???, tRNA: UUA → so mRNA must be AAU (because tRNA anticodon UUA pairs with mRNA AAU)

Fourth row: DNA: AAA → mRNA should be UUU (if coding strand) → amino acid Phenylalanine

Fifth row: DNA: GTA → mRNA CAU → Histidine

Sixth row: tRNA: CUC → mRNA GAG → Glutamic acid

Seventh row: mRNA: ACA → Threonine

Eighth row: DNA: TAT → mRNA AUA → Isoleucine

Ninth row: mRNA: AGC → Serine

Tenth row: tRNA: AUU → mRNA UAA → Stop

Eleventh row: DNA: CCA → mRNA GGU → Glycine

Twelfth row: tRNA: GGC → mRNA CCG → Proline

But none of this matches the first row where DNA=GGC and mRNA=AAG.

Unless... perhaps the DNA is the template strand, and we need to write the mRNA as complementary to it.

So for DNA template strand GGC (say 3'-GGC-5'), mRNA is 5'-CCG-3'. But they have mRNA=AAG.

Still not matching.

Another idea: maybe the DNA triplet is written 5' to 3', and it's the template strand, so to get mRNA, we take the complement and reverse it? No, transcription reads template 3' to 5', synthesizes mRNA 5' to 3'.

Let’s calculate properly.

Assume DNA given is template strand, written 5' to 3'. But template strand is read 3' to 5', so if DNA is given as 5'-GGC-3', then the actual template sequence being read is 3'-CGG-5', so mRNA is 5'-GCC-3'. Not AAG.

I think there might be an error in the problem or my understanding.

Let’s look at the codon wheel. The wheel shows mRNA codons. So for any mRNA codon, we can find the amino acid.

Also, tRNA anticodon is complementary to mRNA codon.

For DNA to mRNA: in standard biology, if DNA coding strand is given, mRNA is identical except T->U.

If DNA template strand is given, mRNA is complementary.

In this worksheet, let's test with a known pair.

Take the row where DNA = AAA.

If DNA coding strand = AAA, then mRNA = UUU, amino acid = Phenylalanine.

If DNA template strand = AAA, then mRNA = UUU? No, template AAA would give mRNA UUU only if it's complementary, but A in DNA template pairs with U in mRNA, so yes, template AAA -> mRNA UUU.

Same result.

But in the first row, DNA=GGC, mRNA=AAG.

If DNA is template strand GGC, then mRNA should be CCG (G->C, G->C, C->G).

But they have AAG.

Unless the DNA is written as the non-template strand, but then mRNA should be GGC, not AAG.

Perhaps it's a different convention.

Another thought: maybe "DNA triplet" means the sequence that is complementary to the mRNA, i.e., the template strand, but written in 5' to 3' direction, and we need to reverse it to get the actual template.

For example, if DNA triplet is given as 5'-GGC-3', and it's the template strand, then the actual template sequence for transcription is 3'-CGG-5', so mRNA is 5'-GCC-3'.

Still not AAG.

Let’s calculate what DNA template would give mRNA AAG.

mRNA AAG means the template DNA must be TTC (because A in mRNA comes from T in DNA, A from T, G from C).

So if mRNA is AAG, DNA template is TTC.

But they have DNA = GGC for that row.

GGC vs TTC — not the same.

Unless it's a typo, and it's supposed to be TTC.

Or perhaps the DNA is the coding strand, and for some reason they have a different mapping.

Let’s ignore the first row for now and do the others, then come back.

Second row: mRNA = CAG

From codon wheel: C-A-G → Glutamine

tRNA anticodon = GUC (complement of CAG: C->G, A->U, G->C)

DNA triplet: if mRNA is CAG, and assuming DNA coding strand, then DNA = CAT (since T in DNA corresponds to U in mRNA)

If DNA template strand, then DNA = GTC (complement of CAG)

But typically in such worksheets, DNA given is coding strand.

Let’s assume that for all rows, the DNA triplet given is the coding strand, so mRNA is the same as DNA but with T replaced by U.

Then for first row: DNA = GGC → mRNA = GGC, but they have mRNA = AAG — contradiction.

Unless the first row is misprinted, or I need to use the given mRNA to find the rest.

Perhaps for the first row, they gave DNA and mRNA, and I need to verify or something.

Let’s read the instruction: "Use the circular codon table to complete the DNA triplets, mRNA codons, tRNA anticodons, and amino acids in the table below."

And in the table, some cells are filled, some are blank.

For the first row, DNA and mRNA are given, so I need to find tRNA anticodon and amino acid.

So regardless of how DNA relates to mRNA, for this row, mRNA is AAG, so:

- tRNA anticodon = UUC (complement of AAG: A->U, A->U, G->C)

- Amino acid for AAG: from codon wheel, A-A-G is Lysine.

Similarly, for other rows, I can use the given information to find the missing parts.

For example, second row: mRNA = CAG, so amino acid = Glutamine (C-A-G), tRNA anticodon = GUC, and DNA triplet: if we assume DNA coding strand, then DNA = CAT (since mRNA CAG, DNA coding is same as mRNA but U->T, so CAG -> CAT)

If DNA template strand, then DNA = GTC (complement of CAG).

But in most educational contexts, when they say "DNA triplet" without specification, and provide mRNA, they often mean the coding strand.

Moreover, in the fourth row, DNA = AAA, which would give mRNA = UUU if coding strand, and UUU is Phenylalanine.

In fifth row, DNA = GTA, mRNA = CAU (if coding strand), which is Histidine.

In eighth row, DNA = TAT, mRNA = AUA, Isoleucine.

In eleventh row, DNA = CCA, mRNA = GGU, Glycine.

All consistent if DNA is coding strand.

But for first row, DNA = GGC, if coding strand, mRNA should be GGC, but they have AAG — so either it's a mistake, or for that row, DNA is template strand.

Perhaps the DNA given is always the template strand.

Let’s test that.

If DNA is template strand, then mRNA is complementary to it.

For first row: DNA = GGC (template), so mRNA = CCG (G->C, G->C, C->G)

But they have mRNA = AAG — not matching.

Unless the DNA is written 3' to 5', but usually it's 5' to 3'.

I think there might be an error in the problem, or I need to proceed with the given mRNA for each row to find the amino acid and tRNA, and for DNA, if not given, infer from mRNA.

For rows where DNA is given, and mRNA is not, I can find mRNA from DNA.

But for first row, both are given, so perhaps it's correct as is, and I should use mRNA=AAG for that row.

So let's do that.

Row 1:
- DNA: GGC (given)
- mRNA: AAG (given)
- tRNA anticodon: complement of mRNA AAG → UUC
- Amino acid: from codon wheel, AAG is Lysine

Row 2:
- DNA: ?
- mRNA: CAG (given)
- tRNA anticodon: complement of CAG → GUC
- Amino acid: CAG is Glutamine
- DNA: if we assume DNA coding strand, then DNA = CAT (since mRNA CAG, DNA coding is C A T)
If DNA template strand, then DNA = GTC
But to be consistent, let's see later rows.

Row 3:
- DNA: ?
- mRNA: ?
- tRNA: UUA (given)
- So mRNA = complement of tRNA UUA → AAU (U->A, U->A, A->U)
- Amino acid: AAU is Asparagine
- DNA: if coding strand, DNA = AAT; if template, DNA = TTA

Row 4:
- DNA: AAA (given)
- mRNA: ? → if DNA coding, mRNA = UUU
- tRNA: complement of UUU → AAA
- Amino acid: UUU is Phenylalanine

Row 5:
- DNA: GTA (given)
- mRNA: ? → if DNA coding, mRNA = CAU
- tRNA: complement of CAU → GUA
- Amino acid: CAU is Histidine

Row 6:
- DNA: ?
- mRNA: ?
- tRNA: CUC (given)
- mRNA = complement of CUC → GAG
- Amino acid: GAG is Glutamic acid
- DNA: if coding, DNA = GAG; if template, DNA = CTC

Row 7:
- DNA: ?
- mRNA: ACA (given)
- tRNA: complement of ACA → UGU
- Amino acid: ACA is Threonine
- DNA: if coding, DNA = ACT; if template, DNA = TGT

Row 8:
- DNA: TAT (given)
- mRNA: ? → if DNA coding, mRNA = AUA
- tRNA: complement of AUA → UAU
- Amino acid: AUA is Isoleucine

Row 9:
- DNA: ?
- mRNA: AGC (given)
- tRNA: complement of AGC → UCG
- Amino acid: AGC is Serine
- DNA: if coding, DNA = AGC; if template, DNA = TCG

Row 10:
- DNA: ?
- mRNA: ?
- tRNA: AUU (given)
- mRNA = complement of AUU → UAA
- Amino acid: UAA is Stop
- DNA: if coding, DNA = TAA; if template, DNA = ATT

Row 11:
- DNA: CCA (given)
- mRNA: ? → if DNA coding, mRNA = GGU
- tRNA: complement of GGU → CCA
- Amino acid: GGU is Glycine

Row 12:
- DNA: ?
- mRNA: ?
- tRNA: GGC (given)
- mRNA = complement of GGC → CCG
- Amino acid: CCG is Proline
- DNA: if coding, DNA = CCG; if template, DNA = GGC

Now, to resolve the DNA issue, let's look at row 1: DNA=GGC, mRNA=AAG

If we assume DNA is coding strand, then mRNA should be GGC, but it's AAG, so not matching.

If DNA is template strand, then for DNA=GGC (template), mRNA should be CCG, not AAG.

But if DNA template is TTC, then mRNA is AAG.

So perhaps for row 1, the DNA is miswritten, or it's a different convention.

Maybe "DNA triplet" means the sequence that is complementary to the mRNA, i.e., the template strand, but written in the same order as mRNA.

For example, for mRNA AAG, the template DNA is TTC, but they wrote GGC — still not.

Another idea: perhaps the DNA is given as the non-template strand, but for some rows, it's template.

I think for the sake of completing the worksheet, I'll assume that for rows where DNA is given and mRNA is not, I'll derive mRNA from DNA by replacing T with U (assuming DNA is coding strand).

For rows where mRNA is given and DNA is not, I'll derive DNA by replacing U with T (coding strand).

For row 1, since both are given, I'll use the given mRNA to find tRNA and amino acid, and accept that DNA=GGC might be a mistake or special case.

Perhaps in this context, the DNA triplet is the one that is used to determine the mRNA via complementarity, but let's calculate what DNA would give mRNA AAG.

As above, if mRNA is AAG, then the template DNA is TTC, so if they give DNA=GGC, it might be the coding strand for a different gene, but that doesn't help.

Let's look at the codon wheel. The wheel has mRNA codons, so for any mRNA, we can find amino acid.

For tRNA, it's always complementary to mRNA.

For DNA, if not specified, I'll use the convention that DNA coding strand is the same as mRNA with T instead of U.

So for row 1: mRNA=AAG, so amino acid=Lysine, tRNA=UUC

DNA=GGC is given, but if it's coding strand, it should be AAG for mRNA, but it's GGC, so perhaps it's template, but then mRNA should be CCG.

I think there's a possibility that in this worksheet, the DNA given is the template strand, and we need to write the mRNA as complementary.

Let me try that for all rows.

Assume DNA given is template strand.

Then mRNA is complementary to DNA.

For row 1: DNA=GGC (template), so mRNA=CCG (G->C, G->C, C->G)

But they have mRNA=AAG — not matching.

Unless the DNA is written 3' to 5', but usually it's 5' to 3'.

Perhaps the DNA is given as the sequence, and we need to take the complement to get mRNA.

For DNA GGC, complement is CCG, so mRNA=CCG.

But they have AAG.

I give up. Let's use the given mRNA for each row to find the amino acid and tRNA, and for DNA, if not given, infer from mRNA by replacing U with T for coding strand.

For row 1, DNA is given as GGC, but mRNA is AAG, so perhaps it's an exception, or I should leave DNA as is.

Perhaps "DNA triplet" means the sequence that is identical to the mRNA except T/U, but for row 1, it's not.

Another thought: in some contexts, the DNA triplet refers to the codon in the DNA that corresponds to the amino acid, but it's ambiguous.

Let's look at the answer for row 4: DNA=AAA, which is likely coding strand, mRNA=UUU, amino acid=Phe.

Row 5: DNA=GTA, mRNA=CAU, His.

Row 8: DNA=TAT, mRNA=AUA, Ile.

Row 11: DNA=CCA, mRNA=GGU, Gly.

All good if DNA is coding strand.

For row 1, if DNA=GGC, and it's coding strand, mRNA should be GGC, which is Glycine, but they have mRNA=AAG, which is Lysine.

So perhaps for row 1, the DNA is not GGC, or it's a typo, and it should be TTC or something.

Maybe the DNA is the template, and for row 1, DNA=GGC means 3'-GGC-5', so mRNA=5'-CCG-3', but they have AAG.

I think I need to proceed with the given information as is for the rows where mRNA is given, and for DNA, when not given, use the coding strand convention.

So for row 1:
- DNA: GGC (given)
- mRNA: AAG (given)
- tRNA anticodon: UUC (complement of AAG)
- Amino acid: Lysine (from codon wheel for AAG)

For row 2:
- DNA: ?
- mRNA: CAG (given)
- tRNA: GUC (complement)
- Amino acid: Glutamine
- DNA: since mRNA=CAG, and assuming DNA coding strand, DNA = CAT (replace U with T, but mRNA has no U, so CAG -> CAT)

For row 3:
- tRNA: UUA (given)
- mRNA: AAU (complement of UUA)
- Amino acid: Asparagine (AAU)
- DNA: AAT (coding strand)

For row 4:
- DNA: AAA (given)
- mRNA: UUU (replace T with U)
- tRNA: AAA (complement of UUU)
- Amino acid: Phenylalanine

For row 5:
- DNA: GTA (given)
- mRNA: CAU (replace T with U)
- tRNA: GUA (complement of CAU)
- Amino acid: Histidine

For row 6:
- tRNA: CUC (given)
- mRNA: GAG (complement)
- Amino acid: Glutamic acid
- DNA: GAG (coding strand)

For row 7:
- mRNA: ACA (given)
- tRNA: UGU (complement)
- Amino acid: Threonine
- DNA: ACT (coding strand)

For row 8:
- DNA: TAT (given)
- mRNA: AUA (replace T with U)
- tRNA: UAU (complement of AUA)
- Amino acid: Isoleucine

For row 9:
- mRNA: AGC (given)
- tRNA: UCG (complement)
- Amino acid: Serine
- DNA: AGC (coding strand)

For row 10:
- tRNA: AUU (given)
- mRNA: UAA (complement)
- Amino acid: Stop
- DNA: TAA (coding strand)

For row 11:
- DNA: CCA (given)
- mRNA: GGU (replace T with U)
- tRNA: CCA (complement of GGU)
- Amino acid: Glycine

For row 12:
- tRNA: GGC (given)
- mRNA: CCG (complement)
- Amino acid: Proline
- DNA: CCG (coding strand)

Now for row 1, even though DNA=GGC and mRNA=AAG don't match under this convention, since both are given, I'll keep them as is, and fill tRNA and amino acid based on mRNA.

So final table:

| DNA triplet | mRNA codon | tRNA anticodon | Amino Acid |
|-------------|------------|----------------|----------------|
| GGC | AAG | UUC | Lysine |
| CAT | CAG | GUC | Glutamine |
| AAT | AAU | UUA | Asparagine |
| AAA | UUU | AAA | Phenylalanine |
| GTA | CAU | GUA | Histidine |
| GAG | GAG | CUC | Glutamic acid |
| ACT | ACA | UGU | Threonine |
| TAT | AUA | UAU | Isoleucine |
| AGC | AGC | UCG | Serine |
| TAA | UAA | AUU | Stop |
| CCA | GGU | CCA | Glycine |
| CCG | CCG | GGC | Proline |

In row 6, tRNA is given as CUC, so mRNA is GAG, DNA coding is GAG.

In row 12, tRNA is GGC, mRNA is CCG, DNA coding is CCG.

For row 3, tRNA UUA, mRNA AAU, DNA AAT.

All seem consistent except row 1, but since it's given, we'll go with it.

So the completed table is as above.

Final Answer:

| DNA triplet | mRNA codon | tRNA anticodon | Amino Acid |
|-------------|------------|----------------|----------------|
| GGC | AAG | UUC | Lysine |
| CAT | CAG | GUC | Glutamine |
| AAT | AAU | UUA | Asparagine |
| AAA | UUU | AAA | Phenylalanine |
| GTA | CAU | GUA | Histidine |
| GAG | GAG | CUC | Glutamic acid |
| ACT | ACA | UGU | Threonine |
| TAT | AUA | UAU | Isoleucine |
| AGC | AGC | UCG | Serine |
| TAA | UAA | AUU | Stop |
| CCA | GGU | CCA | Glycine |
| CCG | CCG | GGC | Proline |