Worksheet for practicing DNA transcription and translation with a codon chart and fill-in-the-blank exercises.

Let’s solve each example step by step.

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Example 1:

We are given DNA sequence:
TAC GCG CCT AGG GGG TGG

Step 1: Transcribe DNA to mRNA.
Remember: In transcription, DNA → mRNA.
Base pairing rules for DNA → mRNA:
- A → U
- T → A
- C → G
- G → C

So:

DNA: TAC → mRNA: AUG
DNA: GCG → mRNA: CGC
DNA: CCT → mRNA: GGA
DNA: AGG → mRNA: UCC
DNA: GGG → mRNA: CCC
DNA: TGG → mRNA: ACC

mRNA: AUG CGC GGA UCC CCC ACC

Step 2: Translate mRNA to amino acids (AA) using the codon chart.

Look up each mRNA codon in the chart:

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

AA: Methionine Arginine Glycine Serine Proline Threonine

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Example 2:

DNA: TTC GAT TAG ATG CCG AAG

Step 1: DNA → mRNA

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

mRNA: AAG CUA AUC UAC GGC UUC

Step 2: mRNA → tRNA (complementary to mRNA)

tRNA pairs with mRNA:
A ↔ U, U ↔ A, C ↔ G, G ↔ C

So:

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: UUC GAU UAG AUG CCG AAG

Step 3: mRNA → AA (using codon chart)

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

AA: Lysine Leucine Isoleucine Tyrosine Glycine Phenylalanine

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Example 3:

This one is mixed — we have partial sequences and need to fill in blanks.

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’ll work backwards from tRNA or forward from DNA/mRNA — let’s use tRNA as anchor since it’s complementary to mRNA.

Recall: tRNA anticodon pairs with mRNA codon.

So if tRNA is “_ A U”, then mRNA must be “_ U A” (because A↔U, U↔A). But wait — look at mRNA row: first codon is “_ U _”. That matches!

Let’s go codon by codon.

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Codon 1:

tRNA: _ A U → so mRNA must be _ U A (since tRNA anticodon binds to mRNA codon)

But mRNA is written as _ U _ → so middle base is U, last base must be A → mRNA: _ U A

Now, what DNA would make that? DNA → mRNA: A→U, T→A, C→G, G→C

So if mRNA is _ U A, then DNA must be _ A T (because mRNA U comes from DNA A; mRNA A comes from DNA T)

DNA is given as C _ _ → so first base is C → DNA: C A T

Then mRNA: G U A? Wait — no! Let’s double-check.

Wait — if DNA is CAT, then mRNA is GUA (C→G, A→U, T→A)

But we just said mRNA should be _ U A → which would be GUA? Yes! G U A → but written as _ U _ → yes, middle is U, last is A → so mRNA: GUA

But tRNA is _ A U → which pairs with mRNA GUA? Let’s check:

mRNA: G U A
tRNA: C A U (because G↔C, U↔A, A↔U) → but given tRNA is _ A U → so first base is blank, but should be C → so tRNA: C A U → matches _ A U if first is C.

Perfect.

So Codon 1:

DNA: C A T
mRNA: G U A
tRNA: C A U
AA: Look up mRNA GUA → Valine

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Codon 2:

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

mRNA is given as _ _ A → so last base is A → matches.

So mRNA: C _ A

What DNA makes that? DNA → mRNA: C→G, _→_, A→U → so DNA: G _ T

DNA is given as _ G _ → so middle is G → so DNA: G G T

Then mRNA: C C A

Check tRNA: G _ U → should be G G U? Because mRNA CCA → tRNA GGU → yes, G _ U → middle is G → so tRNA: G G U

AA: mRNA CCA → Proline

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Codon 3:

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

mRNA is given as _ _ C → last base C → matches.

So mRNA: _ A C

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

DNA is given as A _ _ → so first base A → DNA: A T G

Then mRNA: U A C

tRNA: A U G (because U↔A, A↔U, C↔G) → but given tRNA is _ U G → so first base should be A → tRNA: A U G → matches _ U G

AA: mRNA UAC → Tyrosine

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Codon 4:

tRNA: G _ U → so mRNA: C _ A

mRNA given: _ _ A → last base A → matches.

DNA → mRNA: C←G, _←_, A←T → DNA: G _ T

DNA given: _ A _ → middle is A → so DNA: G A T

mRNA: C U A

tRNA: G A U → but given tRNA is G _ U → so middle should be A → tRNA: G A U → matches G _ U

AA: mRNA CUA → Leucine

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Codon 5:

tRNA: C C _ → so mRNA: G G _

mRNA given: _ G _ → middle G → matches.

DNA → mRNA: G←C, G←C, _←_ → DNA: C C _

DNA given: _ _ C → last base C → so DNA: C C C

mRNA: G G G

tRNA: C C C → but given tRNA is C C _ → so last base C → matches.

AA: mRNA GGG → Glycine

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Codon 6:

tRNA: G _ A → so mRNA: C _ U

mRNA given: _ A _ → middle A → so mRNA: C A U

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

DNA given: _ T _ → middle T → so DNA: G T A

tRNA: G _ A → mRNA CAU → tRNA GUA → so middle U → tRNA: G U A → matches G _ A

AA: mRNA CAU → Histidine

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Finalizing Example 3:

DNA: CAT GGT ATG GAT CCC GTA
mRNA: GUA CCA UAC CUA GGG CAU
tRNA: CAU GGU AUG GAU CCC GUA
AA: Valine Proline Tyrosine Leucine Glycine Histidine

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Now, compiling all answers:

Example 1:

mRNA: AUG CGC GGA UCC CCC ACC
AA: Methionine Arginine Glycine Serine Proline Threonine

Example 2:

mRNA: AAG CUA AUC UAC GGC UUC
tRNA: UUC GAU UAG AUG CCG AAG
AA: Lysine Leucine Isoleucine Tyrosine Glycine Phenylalanine

Example 3:

DNA: CAT GGT ATG GAT CCC GTA
mRNA: GUA CCA UAC CUA GGG CAU
tRNA: CAU GGU AUG GAU CCC GUA
AA: Valine Proline Tyrosine Leucine Glycine Histidine

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Final Answer:

Example 1:
mRNA: AUG CGC GGA UCC CCC ACC
AA: Methionine Arginine Glycine Serine Proline Threonine

Example 2:
mRNA: AAG CUA AUC UAC GGC UUC
tRNA: UUC GAU UAG AUG CCG AAG
AA: Lysine Leucine Isoleucine Tyrosine Glycine Phenylalanine

Example 3:
DNA: CAT GGT ATG GAT CCC GTA
mRNA: GUA CCA UAC CUA GGG CAU
tRNA: CAU GGU AUG GAU CCC GUA
AA: Valine Proline Tyrosine Leucine Glycine Histidine