Worksheet illustrating DNA replication, transcription, and translation processes with sequences and genetic code application.
DNA Replication, Transcription & Translation Worksheet showing two examples with DNA strands, mRNA codons, tRNA anticodons, and amino acids.
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Step-by-step solution for: Solved DNA Replication, Transcription & Translation | Chegg.com
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Show Answer Key & Explanations
Step-by-step solution for: Solved DNA Replication, Transcription & Translation | Chegg.com
Let’s solve this step by step.
We are given two DNA sequences (Example #1 and Example #2). For each, we need to:
1. Replicate the DNA → write the complementary strand (remember: A-T, G-C; and direction matters — 5’ to 3’ vs 3’ to 5’)
2. Transcribe mRNA from the template strand → remember: mRNA is made from the 3’→5’ template strand, and uses U instead of T
3. Write tRNA anticodons → these are complementary to mRNA codons (and run 3’→5’ to match mRNA 5’→3’)
4. Translate to amino acids → use genetic code chart (we’ll assume standard one since not provided)
---
- The coding strand is 5’→3’ and matches mRNA (except T→U).
- The template strand is 3’→5’ and is used to make mRNA.
- mRNA is synthesized 5’→3’, reading template 3’→5’.
- tRNA anticodons are written 3’→5’ to pair with mRNA codons (which are 5’→3’).
- Each codon = 3 bases → codes for 1 amino acid.
---
## EXAMPLE #1
Given DNA coding strand (top):
5’ - A T G G T A G C T A A C C T T - 3’
So template strand (bottom) must be complementary and antiparallel:
Complement of A=T, T=A, G=C, C=G
And reverse direction → so if top is 5’→3’, bottom is 3’→5’
Template strand:
3’ - T A C C A T C G A T T G G A A - 5’
✔ Step 1: Replicated DNA (complementary strand) = above template strand.
Now transcribe mRNA from template strand (read template 3’→5’, build mRNA 5’→3’):
Template: 3’ - T A C C A T C G A T T G G A A - 5’
mRNA: 5’ - A U G G U A G C U A A C C U U - 3’
Wait — that’s identical to the coding strand, except T→U. That makes sense!
So mRNA: 5’ - AUG GUA GCU AAC CUU - 3’
Break into codons:
AUG | GUA | GCU | AAC | CUU
Now tRNA anticodons: complementary to mRNA codons, written 3’→5’
mRNA codon: 5’-AUG-3’ → tRNA anticodon: 3’-UAC-5’
mRNA codon: 5’-GUA-3’ → tRNA anticodon: 3’-CAU-5’
mRNA codon: 5’-GCU-3’ → tRNA anticodon: 3’-CGA-5’
mRNA codon: 5’-AAC-3’ → tRNA anticodon: 3’-UUG-5’
mRNA codon: 5’-CUU-3’ → tRNA anticodon: 3’-GAA-5’
Now translate using genetic code:
- AUG → Methionine (Met) [start]
- GUA → Valine (Val)
- GCU → Alanine (Ala)
- AAC → Asparagine (Asn)
- CUU → Leucine (Leu)
✔ So amino acids: Met - Val - Ala - Asn - Leu
---
## EXAMPLE #2
Given DNA coding strand (top):
5’ - C A G G A A T T G C T C G A T - 3’
Template strand (complementary, antiparallel):
3’ - G T C C T T A A C G A G C T A - 5’
mRNA transcribed from template (read 3’→5’, write 5’→3’):
Same as coding strand but T→U →
5’ - CAG GAA UUG CUC GAU - 3’
Wait — let’s do it properly from template:
Template: 3’ - G T C C T T A A C G A G C T A - 5’
mRNA: 5’ - C A G G A A U U G C U C G A U - 3’
Yes.
Codons:
CAG | GAA | UUG | CUC | GAU
tRNA anticodons (complementary, 3’→5’):
mRNA: 5’-CAG-3’ → tRNA: 3’-GUC-5’
mRNA: 5’-GAA-3’ → tRNA: 3’-CUU-5’
mRNA: 5’-UUG-3’ → tRNA: 3’-AAC-5’
mRNA: 5’-CUC-3’ → tRNA: 3’-GAG-5’
mRNA: 5’-GAU-3’ → tRNA: 3’-CUA-5’
Amino acids (using standard genetic code):
- CAG → Glutamine (Gln)
- GAA → Glutamic acid (Glu)
- UUG → Leucine (Leu)
- CUC → Leucine (Leu)
- GAU → Aspartic acid (Asp)
✔ So amino acids: Gln - Glu - Leu - Leu - Asp
---
## Final Answers:
DNA template strand (replicated):
3’ - T A C C A T C G A T T G G A A - 5’
mRNA:
5’ - A U G G U A G C U A A C C U U - 3’
tRNA anticodons (each group is one anticodon, written 3’→5’):
3’ - U A C - 5’
3’ - C A U - 5’
3’ - C G A - 5’
3’ - U U G - 5’
3’ - G A A - 5’
Amino Acids:
Met - Val - Ala - Asn - Leu
---
DNA template strand (replicated):
3’ - G T C C T T A A C G A G C T A - 5’
mRNA:
5’ - C A G G A A U U G C U C G A U - 3’
tRNA anticodons:
3’ - G U C - 5’
3’ - C U U - 5’
3’ - A A C - 5’
3’ - G A G - 5’
3’ - C U A - 5’
Amino Acids:
Gln - Glu - Leu - Leu - Asp
---
Final Answer:
Example #1:
- DNA template: 3’-TACC ATCG ATTG GAA-5’
- mRNA: 5’-AUG GUA GCU AAC CUU-3’
- tRNA anticodons: 3’-UAC-5’, 3’-CAU-5’, 3’-CGA-5’, 3’-UUG-5’, 3’-GAA-5’
- Amino acids: Met - Val - Ala - Asn - Leu
Example #2:
- DNA template: 3’-GTCC TTAA CGAG CTA-5’
- mRNA: 5’-CAG GAA UUG CUC GAU-3’
- tRNA anticodons: 3’-GUC-5’, 3’-CUU-5’, 3’-AAC-5’, 3’-GAG-5’, 3’-CUA-5’
- Amino acids: Gln - Glu - Leu - Leu - Asp
We are given two DNA sequences (Example #1 and Example #2). For each, we need to:
1. Replicate the DNA → write the complementary strand (remember: A-T, G-C; and direction matters — 5’ to 3’ vs 3’ to 5’)
2. Transcribe mRNA from the template strand → remember: mRNA is made from the 3’→5’ template strand, and uses U instead of T
3. Write tRNA anticodons → these are complementary to mRNA codons (and run 3’→5’ to match mRNA 5’→3’)
4. Translate to amino acids → use genetic code chart (we’ll assume standard one since not provided)
---
Important Rules Recap:
- The coding strand is 5’→3’ and matches mRNA (except T→U).
- The template strand is 3’→5’ and is used to make mRNA.
- mRNA is synthesized 5’→3’, reading template 3’→5’.
- tRNA anticodons are written 3’→5’ to pair with mRNA codons (which are 5’→3’).
- Each codon = 3 bases → codes for 1 amino acid.
---
## EXAMPLE #1
Given DNA coding strand (top):
5’ - A T G G T A G C T A A C C T T - 3’
So template strand (bottom) must be complementary and antiparallel:
Complement of A=T, T=A, G=C, C=G
And reverse direction → so if top is 5’→3’, bottom is 3’→5’
Template strand:
3’ - T A C C A T C G A T T G G A A - 5’
✔ Step 1: Replicated DNA (complementary strand) = above template strand.
Now transcribe mRNA from template strand (read template 3’→5’, build mRNA 5’→3’):
Template: 3’ - T A C C A T C G A T T G G A A - 5’
mRNA: 5’ - A U G G U A G C U A A C C U U - 3’
Wait — that’s identical to the coding strand, except T→U. That makes sense!
So mRNA: 5’ - AUG GUA GCU AAC CUU - 3’
Break into codons:
AUG | GUA | GCU | AAC | CUU
Now tRNA anticodons: complementary to mRNA codons, written 3’→5’
mRNA codon: 5’-AUG-3’ → tRNA anticodon: 3’-UAC-5’
mRNA codon: 5’-GUA-3’ → tRNA anticodon: 3’-CAU-5’
mRNA codon: 5’-GCU-3’ → tRNA anticodon: 3’-CGA-5’
mRNA codon: 5’-AAC-3’ → tRNA anticodon: 3’-UUG-5’
mRNA codon: 5’-CUU-3’ → tRNA anticodon: 3’-GAA-5’
Now translate using genetic code:
- AUG → Methionine (Met) [start]
- GUA → Valine (Val)
- GCU → Alanine (Ala)
- AAC → Asparagine (Asn)
- CUU → Leucine (Leu)
✔ So amino acids: Met - Val - Ala - Asn - Leu
---
## EXAMPLE #2
Given DNA coding strand (top):
5’ - C A G G A A T T G C T C G A T - 3’
Template strand (complementary, antiparallel):
3’ - G T C C T T A A C G A G C T A - 5’
mRNA transcribed from template (read 3’→5’, write 5’→3’):
Same as coding strand but T→U →
5’ - CAG GAA UUG CUC GAU - 3’
Wait — let’s do it properly from template:
Template: 3’ - G T C C T T A A C G A G C T A - 5’
mRNA: 5’ - C A G G A A U U G C U C G A U - 3’
Yes.
Codons:
CAG | GAA | UUG | CUC | GAU
tRNA anticodons (complementary, 3’→5’):
mRNA: 5’-CAG-3’ → tRNA: 3’-GUC-5’
mRNA: 5’-GAA-3’ → tRNA: 3’-CUU-5’
mRNA: 5’-UUG-3’ → tRNA: 3’-AAC-5’
mRNA: 5’-CUC-3’ → tRNA: 3’-GAG-5’
mRNA: 5’-GAU-3’ → tRNA: 3’-CUA-5’
Amino acids (using standard genetic code):
- CAG → Glutamine (Gln)
- GAA → Glutamic acid (Glu)
- UUG → Leucine (Leu)
- CUC → Leucine (Leu)
- GAU → Aspartic acid (Asp)
✔ So amino acids: Gln - Glu - Leu - Leu - Asp
---
## Final Answers:
Example #1:
DNA template strand (replicated):
3’ - T A C C A T C G A T T G G A A - 5’
mRNA:
5’ - A U G G U A G C U A A C C U U - 3’
tRNA anticodons (each group is one anticodon, written 3’→5’):
3’ - U A C - 5’
3’ - C A U - 5’
3’ - C G A - 5’
3’ - U U G - 5’
3’ - G A A - 5’
Amino Acids:
Met - Val - Ala - Asn - Leu
---
Example #2:
DNA template strand (replicated):
3’ - G T C C T T A A C G A G C T A - 5’
mRNA:
5’ - C A G G A A U U G C U C G A U - 3’
tRNA anticodons:
3’ - G U C - 5’
3’ - C U U - 5’
3’ - A A C - 5’
3’ - G A G - 5’
3’ - C U A - 5’
Amino Acids:
Gln - Glu - Leu - Leu - Asp
---
Final Answer:
Example #1:
- DNA template: 3’-TACC ATCG ATTG GAA-5’
- mRNA: 5’-AUG GUA GCU AAC CUU-3’
- tRNA anticodons: 3’-UAC-5’, 3’-CAU-5’, 3’-CGA-5’, 3’-UUG-5’, 3’-GAA-5’
- Amino acids: Met - Val - Ala - Asn - Leu
Example #2:
- DNA template: 3’-GTCC TTAA CGAG CTA-5’
- mRNA: 5’-CAG GAA UUG CUC GAU-3’
- tRNA anticodons: 3’-GUC-5’, 3’-CUU-5’, 3’-AAC-5’, 3’-GAG-5’, 3’-CUA-5’
- Amino acids: Gln - Glu - Leu - Leu - Asp
Parent Tip: Review the logic above to help your child master the concept of dna transcription and translation worksheet.
