Transcription and Translation 18 a Label the | StudyX - Free Printable

Explanation:
Let’s solve part (a) first — labeling the diagram using the word list.

We have two main processes shown:
- Top half: Transcription (DNA → mRNA)
- Bottom half: Translation (mRNA → protein)

Step-by-step labeling:

1. Label 1: The newly made RNA strand exiting the transcription complex. It has U (uracil), so it’s RNA, and it’s being synthesized from DNA → this is mRNA (transcription)

2. Label 2: The enzyme that builds mRNA by reading DNA → RNA polymerase

3. Label 3: The DNA strand that is being read (template). In transcription, RNA is made complementary to the *template strand*, which runs 3’→5’. So this is DNA – template strand

4. Label 4: The other DNA strand — not used as template; same sequence as mRNA (except T instead of U) → DNA – coding strand

5. Label 5: The region where transcription happens — inside the nucleus (in eukaryotes). The double helix is unwinding in the nucleus → nucleus

Now the boxes:

A. This box points to the whole top process (DNA → mRNA) → transcription

B. Arrow going down from transcription to translation → represents the movement of mRNA from nucleus to cytoplasm → the location where translation occurs is the cytoplasm

C. This box points to the bottom process (mRNA + tRNA + ribosome making protein) → translation

D. Arrow going down from translation → indicates the product: a chain of amino acids → peptide

Now part (b): Label 5’ and 3’ ends of all DNA and RNA strands.

Recall:
- Nucleic acids are directional: 5’ end has a phosphate group, 3’ end has a hydroxyl group.
- During transcription, RNA is synthesized 5’ → 3’, antiparallel to the DNA template strand (which runs 3’ → 5’).
- mRNA is read 5’ → 3’ during translation.
- tRNA binds mRNA with its anticodon loop; the anticodon is oriented 3’→5’ to pair with mRNA 5’→3’.

Look at the strands:

- Strand 1 (mRNA): Labeled at left with “5’” — correct. So its right end is 3’ (but not labeled — we just need to know direction).
- DNA template strand (label 3): Runs antiparallel to mRNA. Since mRNA is 5’→3’ left to right, DNA template must be 3’→5’, so left end is 3’, right end is 5’.
- DNA coding strand (label 4): Same direction as mRNA (5’→3’ left to right), so left = 5’, right = 3’
- mRNA in translation (label 7): Shown with A-U-G at left → start codon AUG is at 5’ end, so left = 5’, right = 3’
- tRNAs (labels 8 and 13): Each tRNA has an anticodon loop. Anticodons bind mRNA 3’→5’, so the anticodon sequence is written 3’→5’. But the tRNA molecule itself has a 5’ end (acceptor stem where amino acid attaches) and 3’ end (CCA tail). However, in standard diagrams like this:
- The acceptor stem (where amino acid attaches) is the 3’ end of tRNA (ends in CCA).
- The anticodon loop is in the middle.
- So for labeling 5’/3’ ends of tRNA strands, we look at the backbone: the end with the amino acid (Met, Ala, Glu) attached is the 3’ end of tRNA.
- Therefore, for tRNA labeled 8 (carrying Met), the end attached to Met is 3’, so the opposite end (top of tRNA) is 5’
- Same for tRNA 13 (Glu): amino acid attached end = 3’, other end = 5’

But the question says: *“Label the 5’ and 3’ ends of all DNA and RNA strands”* — meaning we need to assign directions to each numbered strand that is DNA or RNA.

From the diagram, the strands that need 5’/3’ labels are:
- 1 (mRNA) → already shows 5’ on left → so left = 5’, right = 3’
- 3 (DNA template strand) → since mRNA is 5’→3’ left to right, template is antiparallel → left = 3’, right = 5’
- 4 (DNA coding strand) → same orientation as mRNA → left = 5’, right = 3’
- 7 (mRNA in translation) → starts with AUG on left → 5’ left, 3’ right
- 8 (tRNA) → RNA strand; amino acid attached at 3’ end → so the end bound to Met is 3’, the top loop end is 5’
- 13 (tRNA) → same: amino acid (Glu) attached at 3’ end, other end = 5’
- Also, the tRNA labeled 12 (the one in the ribosome with anticodon CGU) — its amino acid (Ala) is attached at 3’ end → so that end is 3’, top is 5’

However, the problem likely expects only the major strands labeled with numbers 1, 3, 4, 7, and possibly 8 and 13 — but since the question says “all DNA and RNA strands”, and the diagram uses numbers 1–13, we focus on those that are nucleic acid strands.

Let’s map each number:

1. mRNA (transcription product) → 5’ on left (given), so: 5’ → 3’ left to right
3. DNA template strand → antiparallel to mRNA → 3’ → 5’ (left to right)
4. DNA coding strand → same as mRNA direction → 5’ → 3’
7. mRNA (translation template) → starts with AUG on left → 5’ → 3’
8. tRNA (Met) → RNA strand; amino acid attached to 3’ end → the end with Met is 3’, so the opposite (top) is 5’
12. tRNA in ribosome (Ala) → same: amino acid end = 3’, other = 5’
13. tRNA (Glu) → amino acid end = 3’, other = 5’

But the worksheet likely expects just the directional labels for the strands shown with arrows/lines — and in many such worksheets, they want you to write “5’” or “3’” next to the ends.

Since the user only asks to *solve* the problem (not fill in the image), and the instruction is to provide the final answer, we give the correct labels for parts (a) and (b) as a list.

For (a), matching numbers/boxes to terms:

1. mRNA (transcription)
2. RNA polymerase
3. DNA – template strand
4. DNA – coding strand
5. nucleus
A. transcription
B. cytoplasm
C. translation
D. peptide

For (b), 5’/3’ labeling:

- Strand 1 (mRNA): left = 5’, right = 3’
- Strand 3 (DNA template): left = 3’, right = 5’
- Strand 4 (DNA coding): left = 5’, right = 3’
- Strand 7 (mRNA in ribosome): left = 5’, right = 3’
- Strand 8 (tRNA): end attached to Met = 3’, other end = 5’
- Strand 12 (tRNA in ribosome): end attached to Ala = 3’, other end = 5’
- Strand 13 (tRNA): end attached to Glu = 3’, other end = 5’

But since the question says “Label the 5’ and 3’ ends of all DNA and RNA strands”, and in standard test format, they expect you to indicate for each numbered strand the orientation — however, the final answer should be concise.

Given the format of the worksheet, the expected answer is likely a list matching each label (1–5, A–D) and for (b) stating directions.

But the user instruction says: *“Solve the problem accurately”* and *“Provide only the final correct answer”* after reasoning.

So we compile the correct labeling:

Final Answer:
(a)
1. mRNA (transcription)
2. RNA polymerase
3. DNA – template strand
4. DNA – coding strand
5. nucleus
A. transcription
B. cytoplasm
C. translation
D. peptide

(b)
Strand 1: 5’ (left) → 3’ (right)
Strand 3: 3’ (left) → 5’ (right)
Strand 4: 5’ (left) → 3’ (right)
Strand 7: 5’ (left) → 3’ (right)
Strand 8: 5’ (top) → 3’ (Met end)
Strand 12: 5’ (top) → 3’ (Ala end)
Strand 13: 5’ (top) → 3’ (Glu end)