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

Let’s solve each example step by step. Remember:

- DNA → mRNA: Replace T with U, and use complementary bases (A↔U, T↔A, C↔G, G↔C)
- mRNA → tRNA: Use complementary bases again (A↔U, U↔A, C↔G, G↔C)
- mRNA → AA (amino acid): Use the codon chart — read the mRNA codon (3 letters) and find the matching amino acid.

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Example 1:

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

Step 1: Transcribe DNA to mRNA
Remember: In transcription, DNA template strand is used to make mRNA. But here, since it says “give appropriate sequence”, and Example 1 gives DNA as starting point, we assume this is the *template* strand (so mRNA will be complementary).

So for each DNA base:
- T → A
- A → U
- C → G
- G → C

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

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

Step 2: Translate mRNA to amino acids using codon chart.

Look up each mRNA codon:

- AUG → Methionine (start codon)
- CGC → Arginine
- GGA → Glycine
- UCC → Serine
- CCC → Proline
- ACC → Threonine

AA: Methionine Arginine Glycine Serine Proline Threonine

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Example 2:

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

Again, assuming this is the template strand for transcription.

Transcribe DNA → mRNA:

TTC → AAG
GAT → CUA
TAG → AUC
ATG → UAC
CCG → GGC
AAG → UUC

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

Now, translate mRNA → AA:

- AAG → Lysine
- CUA → Leucine
- AUC → Isoleucine
- UAC → Tyrosine
- GGC → Glycine
- UUC → Phenylalanine

AA: Lysine Leucine Isoleucine Tyrosine Glycine Phenylalanine

Now, tRNA: tRNA anticodons are complementary to mRNA codons.

mRNA: AAG → tRNA: UUC
mRNA: CUA → tRNA: GAU
mRNA: AUC → tRNA: UAG
mRNA: UAC → tRNA: AUG
mRNA: GGC → tRNA: CCG
mRNA: UUC → tRNA: AAG

tRNA: U U C G A U U A G A U G C C G A A G

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Example 3:

This one has blanks. We need to fill in missing bases so that all sequences match correctly.

Given:

DNA: C _ _ _ G _ A _ _ _ A _ _ _ C _ T _
mRNA: _ U _ _ _ A _ _ C _ _ A _ G _ _ A _
tRNA: _ A U G _ U _ U G G _ U C C _ G _ A
AA: ? ? ? ? ? ?

We know:

- mRNA is complementary to DNA (with U instead of T)
- tRNA is complementary to mRNA
- So DNA and tRNA should be almost identical, except T in DNA = U in tRNA? Wait — let’s think carefully.

Actually:

If DNA is template strand:

DNA base → mRNA base (complementary, T→U)

Then mRNA base → tRNA base (complementary)

So: DNA → mRNA → tRNA

Therefore, tRNA should be same as DNA, but with U instead of T? Let’s test.

Suppose DNA has “C” → mRNA has “G” → tRNA has “C” → matches DNA.

DNA “T” → mRNA “A” → tRNA “U” → so tRNA has U where DNA has T.

So yes: tRNA sequence should match DNA sequence, except wherever DNA has T, tRNA has U.

Check given tRNA: _ A U G _ U _ U G G _ U C C _ G _ A

Compare to DNA: C _ _ _ G _ A _ _ _ A _ _ _ C _ T _

Let’s align them position by position.

First codon:

DNA: C _ _
tRNA: _ A U → so if tRNA is complementary to mRNA, and mRNA is complementary to DNA, then tRNA should equal DNA with T→U.

So if tRNA first codon is _ A U, then DNA first codon should be _ A T? But DNA starts with C.

Wait — maybe better to go from tRNA to mRNA to DNA.

tRNA: _ A U → mRNA must be complementary: so mRNA = _ U A (because tRNA A pairs with mRNA U, tRNA U pairs with mRNA A)

But given mRNA first codon is _ U _

So if mRNA is _ U _, and tRNA is _ A U, then:

Position 1: tRNA ? , mRNA ? → must be complementary → if tRNA pos1 is X, mRNA pos1 is complement of X.

But we have:

tRNA: _ A U → so positions: 1=?, 2=A, 3=U
mRNA: _ U _ → positions: 1=?, 2=U, 3=?

Since tRNA and mRNA are complementary:

tRNA pos2 = A → mRNA pos2 = U → matches given mRNA pos2 = U → good.

tRNA pos3 = U → mRNA pos3 = A → so mRNA third base is A.

Thus mRNA first codon: _ U A

Now, DNA is complementary to mRNA (with T instead of U).

mRNA: _ U A → DNA: _ A T (because mRNA U → DNA A; mRNA A → DNA T)

But DNA first codon is given as C _ _ → so first base is C.

So DNA: C _ _ → must be complementary to mRNA: _ U A

So DNA pos1 = C → mRNA pos1 = G (since C in DNA → G in mRNA)

Thus mRNA first codon: G U A

Then tRNA first codon: complementary to mRNA GUA → C A U

But given tRNA first codon is _ A U → so first base must be C → tRNA: C A U → matches!

So now we can fill:

First codon:

DNA: C _ _ → we have mRNA: G U A → so DNA must be complementary: C A T (because mRNA G ← DNA C, mRNA U ← DNA A, mRNA A ← DNA T)

So DNA: C A T

mRNA: G U A

tRNA: C A U

AA: Look up mRNA codon GUA → Valine (from chart: G U A → Valine)

Second codon:

DNA: _ G _
mRNA: _ _ A
tRNA: G _ U

tRNA: G _ U → complementary to mRNA → so mRNA must be C _ A (because tRNA G → mRNA C, tRNA U → mRNA A)

Given mRNA second codon: _ _ A → so third base is A → matches.

So mRNA: C _ A

Then DNA: complementary to mRNA → G _ T (mRNA C → DNA G, mRNA A → DNA T)

Given DNA: _ G _ → so second base is G → so DNA: G G T? Wait, DNA is _ G _, so first base ?, second G, third ?

From above: DNA should be G _ T → so first base G, third base T → so DNA: G G T

Then mRNA: C C A (complementary to DNA GGT → CCA)

tRNA: G G U (complementary to mRNA CCA → GGU)

Given tRNA: G _ U → so middle base should be G → tRNA: G G U → matches.

AA: mRNA CCA → Proline (chart: C C A → Proline)

Third codon:

DNA: A _ _
mRNA: _ _ C
tRNA: _ U G

tRNA: _ U G → complementary to mRNA → so mRNA: _ A C (tRNA U → mRNA A, tRNA G → mRNA C)

Given mRNA: _ _ C → third base C → matches.

So mRNA: _ A C

DNA: complementary → _ T G (mRNA A → DNA T, mRNA C → DNA G)

Given DNA: A _ _ → first base A → so DNA: A T G

Then mRNA: U A C (complementary to ATG)

tRNA: A U G (complementary to UAC)

Given tRNA: _ U G → first base should be A → tRNA: A U G → matches.

AA: mRNA UAC → Tyrosine

Fourth codon:

DNA: _ A _
mRNA: _ _ A
tRNA: G _ U

tRNA: G _ U → mRNA: C _ A (as before)

Given mRNA: _ _ A → third base A → matches.

So mRNA: C _ A

DNA: G _ T (complementary)

Given DNA: _ A _ → second base A → so DNA: G A T

mRNA: C U A

tRNA: G A U

Given tRNA: G _ U → middle base A → tRNA: G A U → matches.

AA: mRNA CUA → Leucine

Fifth codon:

DNA: _ _ C
mRNA: _ G _
tRNA: C C _

tRNA: C C _ → complementary to mRNA → so mRNA: G G _ (tRNA C → mRNA G)

Given mRNA: _ G _ → second base G → matches.

So mRNA: G G _

DNA: C C _ (complementary to mRNA GG_ → CC_)

Given DNA: _ _ C → third base C → so DNA: C C C

mRNA: G G G

tRNA: C C C

Given tRNA: C C _ → third base should be C → tRNA: C C C → matches.

AA: mRNA GGG → Glycine

Sixth codon:

DNA: _ T _
mRNA: _ A _
tRNA: G _ A

tRNA: G _ A → complementary to mRNA → mRNA: C _ U (tRNA G → mRNA C, tRNA A → mRNA U)

Given mRNA: _ A _ → second base A → so mRNA: C A U

Then DNA: G T A (complementary to CAU → GTA)

Given DNA: _ T _ → second base T → matches.

tRNA: G U A (complementary to CAU)

Given tRNA: G _ A → middle base U → tRNA: G U A → matches.

AA: mRNA CAU → Histidine

Now let’s compile all filled sequences.

Example 3:

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

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

tRNA:
C A U G G U A U G G A U C C C G U A

AA:
Valine Proline Tyrosine Leucine Glycine Histidine

Double-check with codon chart:

- GUA → Valine ✔️
- CCA → Proline ✔️
- UAC → Tyrosine ✔️
- CUA → Leucine ✔️
- GGG → Glycine ✔️
- CAU → Histidine ✔️

All correct.

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Final Answer:

Example 1:
mRNA: A U G C G C G G A U C C C C C A C C
AA: Methionine Arginine Glycine Serine Proline Threonine

Example 2:
mRNA: A A G C U A A U C U A C G G C U U C
tRNA: U U C G A U U A G A U G C C G A A G
AA: Lysine Leucine Isoleucine Tyrosine Glycine Phenylalanine

Example 3:
DNA: C A T G G T A T G G A T C C C G T A
mRNA: G U A C C A U A C C U A G G G C A U
tRNA: C A U G G U A U G G A U C C C G U A
AA: Valine Proline Tyrosine Leucine Glycine Histidine