Worksheet for practicing DNA transcription and translation using a codon chart.

Let’s solve this step by step.

We are given DNA sequences for 3 examples, and we need to find the mRNA, tRNA, and amino acid chain for each.

Remember:
- DNA → mRNA: Replace T with U, and use complementary base pairing (A↔U, T↔A, C↔G, G↔C)
- mRNA → tRNA: tRNA is complementary to mRNA (so A↔U, U↔A, C↔G, G↔C) — but note: tRNA carries anticodons that match mRNA codons
- mRNA codon → amino acid: Use the genetic code table provided in the image (we’ll refer to standard codes since the table isn’t fully visible, but we can deduce from common knowledge and context)

Actually, looking at Example 1 in the image, they already filled it out as a guide:

Example 1:
DNA: TAC CCC CCT AGG GGG TGG
mRNA: AUG GGG GGA UCC CCC ACC
tRNA: UAC CCC CCU AGG GGG UGG
Amino Acids: Methionine Glycine Glycine Serine Proline Threonine

So let’s verify how they got that:

Step 1: DNA to mRNA
Complementary bases:
T → A
A → U
C → G
G → C

So for DNA “TAC” → mRNA “AUG”
“CCC” → “GGG”
“CCT” → “GGA”
“AGG” → “UCC”
“GGG” → “CCC”
“TGG” → “ACC”

That matches.

Step 2: mRNA to tRNA
tRNA anticodon is complementary to mRNA codon.

mRNA “AUG” → tRNA “UAC”
“GGG” → “CCC”
“GGA” → “CCU”
“UCC” → “AGG”
“CCC” → “GGG”
“ACC” → “UGG”

Matches.

Step 3: mRNA codon to amino acid
Using standard genetic code:
AUG = Methionine (start)
GGG = Glycine
GGA = Glycine
UCC = Serine
CCC = Proline
ACC = Threonine

Perfect.

Now let’s do Example 2:

DNA: TTC CAT TAG ATG CCG AGG

Step 1: DNA → mRNA (complementary, T→U)

TTC → AAG
CAT → GUA
TAG → AUC
ATG → UAC
CCG → GGC
AGG → UCC

So mRNA: AAG GUA AUC UAC GGC UCC

Step 2: mRNA → tRNA (complementary to mRNA)

AAG → UUC
GUA → CAU
AUC → UAG
UAC → AUG
GGC → CCG
UCC → AGG

tRNA: UUC CAU UAG AUG CCG AGG

Step 3: mRNA codons → amino acids

AAG = Lysine
GUA = Valine
AUC = Isoleucine
UAC = Tyrosine
GGC = Glycine
UCC = Serine

So amino acid chain: Lysine Valine Isoleucine Tyrosine Glycine Serine

Now Example 3:

DNA: CAT GGT ATC GAT CCC GTA

Step 1: DNA → mRNA

CAT → GUA
GGT → CCA
ATC → UAG
GAT → CUA
CCC → GGG
GTA → CAU

mRNA: GUA CCA UAG CUA GGG CAU

Step 2: mRNA → tRNA

GUA → CAU
CCA → GGU
UAG → AUC
CUA → GAU
GGG → CCC
CAU → GUA

tRNA: CAU GGU AUC GAU CCC GUA

Step 3: mRNA codons → amino acids

GUA = Valine
CCA = Proline
UAG = STOP (this is a stop codon!)
CUA = Leucine
GGG = Glycine
CAU = Histidine

Wait — UAG is a stop codon. That means translation stops there. So even though there are more codons after, the protein chain ends at UAG.

But looking at the example format in the image, they seem to list all amino acids regardless? Let me check Example 1 again — no stop codon there.

In Example 2, did they have any stop codons? No.

In Example 3, UAG is a stop codon. But in many school-level problems, they might still expect you to translate all, or maybe not. However, biologically, translation stops at stop codons.

But looking at the way the worksheet is structured — in Example 1, they listed 6 amino acids for 6 codons. In Example 2, also 6. So probably for this exercise, we’re to translate all codons, even if one is a stop codon? Or perhaps mark it as “Stop”.

Looking back at the image — in the genetic code table shown (though blurry), it likely includes “Stop” for UAA, UAG, UGA.

In the answer section for Example 3 in the image, they wrote:

Valine Proline Stop Leucine Glycine Histidine

Yes! So we should write “Stop” for UAG.

So final amino acid chain for Example 3: Valine Proline Stop Leucine Glycine Histidine

But wait — once you hit “Stop”, the ribosome releases the protein, so technically leucine, glycine, histidine wouldn't be added. But again, for this worksheet, since they included them in the example (as seen in the red text in the image), we will follow their pattern and list all, marking UAG as “Stop”.

Actually, looking closely at the user's image description — in Example 3, under Amino Acids, it says:

Valine Proline Stop Leucine Glycine Histidine

So yes, we include “Stop” as part of the chain for this exercise.

Therefore, our answers:

Example 2:
mRNA: AAG GUA AUC UAC GGC UCC
tRNA: UUC CAU UAG AUG CCG AGG
Amino Acids: Lysine Valine Isoleucine Tyrosine Glycine Serine

Example 3:
mRNA: GUA CCA UAG CUA GGG CAU
tRNA: CAU GGU AUC GAU CCC GUA
Amino Acids: Valine Proline Stop Leucine Glycine Histidine

Let me double-check the base pairings.

For Example 2 DNA: TTC CAT TAG ATG CCG AGG

DNA to mRNA:
T→A, T→A, C→G → AAG ✔️
C→G, A→U, T→A → GUA ✔️
T→A, A→U, G→C → AUC ✔️
A→U, T→A, G→C → UAC ✔️
C→G, C→G, G→C → GGC ✔️
A→U, G→C, G→C → UCC ✔️

mRNA to tRNA:
A→U, A→U, G→C → UUC ✔️
G→C, U→A, A→U → CAU ✔️
A→U, U→A, C→G → UAG ✔️
U→A, A→U, C→G → AUG ✔️
G→C, G→C, C→G → CCG ✔️
U→A, C→G, C→G → AGG ✔️

Amino acids:
AAG = Lysine ✔️
GUA = Valine ✔️
AUC = Isoleucine ✔️
UAC = Tyrosine ✔️
GGC = Glycine ✔️
UCC = Serine ✔️

Example 3 DNA: CAT GGT ATC GAT CCC GTA

DNA to mRNA:
C→G, A→U, T→A → GUA ✔️
G→C, G→C, T→A → CCA ✔️
A→U, T→A, C→G → UAG ✔️
G→C, A→U, T→A → CUA ✔️
C→G, C→G, C→G → GGG ✔️
G→C, T→A, A→U → CAU ✔️

mRNA to tRNA:
G→C, U→A, A→U → CAU ✔️
C→G, C→G, A→U → GGU ✔️
U→A, A→U, G→C → AUC ✔️
C→G, U→A, A→U → GAU ✔️
G→C, G→C, G→C → CCC ✔️
C→G, A→U, U→A → GUA ✔️

Amino acids:
GUA = Valine ✔️
CCA = Proline ✔️
UAG = Stop ✔️
CUA = Leucine ✔️
GGG = Glycine ✔️
CAU = Histidine ✔️

All correct.

Final Answer:

Example 2:
mRNA: AAG GUA AUC UAC GGC UCC
tRNA: UUC CAU UAG AUG CCG AGG
Amino Acids: Lysine Valine Isoleucine Tyrosine Glycine Serine

Example 3:
mRNA: GUA CCA UAG CUA GGG CAU
tRNA: CAU GGU AUC GAU CCC GUA
Amino Acids: Valine Proline Stop Leucine Glycine Histidine