Label: Protein Synthesis - Free Printable

Let’s go step by step through each question. We’ll use basic biology knowledge about DNA, RNA, and protein synthesis.

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5. DNA is located in the (nucleus/cytoplasm)
→ In eukaryotic cells (like human cells), DNA is stored in the nucleus.
✔ Answer: nucleus

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6. (mRNA/rRNA) is used to carry the genetic code from DNA to the ribosomes.
→ mRNA = messenger RNA → carries the code from DNA to ribosome.
rRNA = part of ribosome structure.
✔ Answer: mRNA

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7. (tRNA/rRNA) makes up the ribosome.
→ Ribosomes are made of rRNA + proteins. tRNA brings amino acids.
✔ Answer: rRNA

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8. (DNA/RNA) uses uracil instead of thymine.
→ DNA has A, T, C, G.
→ RNA has A, U, C, G — so RNA uses uracil instead of thymine.
✔ Answer: RNA

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9. (RNA/amino) acids make up a protein.
→ Proteins are chains of amino acids. RNA helps build them but isn’t the building block.
✔ Answer: amino

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10. Fill in the diagram:

We’re given a DNA strand:
5’ - A T G C C T G G T G G T A T G G G - 3’

Step 1: Make mRNA from DNA (transcription).
Remember:
- DNA A → RNA U
- DNA T → RNA A
- DNA C → RNA G
- DNA G → RNA C

Also, mRNA is built 5’ to 3’, complementary to DNA template (which runs 3’ to 5’). But here, the DNA is written 5’ to 3’, so we need to read it backwards for transcription? Wait — actually, in diagrams like this, if DNA is shown 5’ to 3’, and we’re making mRNA below it, we assume the bottom strand is the template? Or top?

Wait — looking at the diagram: DNA is labeled with 5’ on left, 3’ on right. Below it is mRNA, also 5’ on right? No — mRNA is drawn with 3’ on left and 5’ on right? That doesn’t match standard convention.

Actually, let’s look carefully:

The DNA strand is written:
5’ A T G C C T G G T G G T A T G G G 3’

In transcription, RNA polymerase reads the template strand (antisense strand) from 3’ to 5’, and builds mRNA 5’ to 3’.

But in many textbook diagrams, when they show one DNA strand like this and ask for mRNA, they mean that this is the coding strand (same as mRNA except T→U), or the template?

Standard rule: If the DNA strand shown is the template strand, then mRNA is complementary to it.

But often in such problems, if not specified, and DNA is written 5’ to 3’, and you’re to write mRNA below it aligned base-by-base, they usually intend for you to treat the DNA as the template strand, so mRNA will be complementary and antiparallel.

BUT — look at the mRNA line: it has circles connected, and labels 3’ on left and 5’ on right? That would mean mRNA is written 3’ to 5’? That’s unusual.

Wait — no: in the diagram, under “mRNA”, the first circle is labeled 3’ on the left, last is 5’ on the right? That can’t be right because mRNA is synthesized 5’ to 3’. Probably a typo in labeling? Or maybe the direction is flipped.

Actually, looking again:
Under “mRNA”, the left end says “3’” and right end says “5’”. That suggests the mRNA strand is drawn 3’ to 5’ — which is backwards. But perhaps it's just how it's drawn; we should still write the sequence correctly.

To avoid confusion, let’s do it properly.

Assume the DNA strand given is the template strand (the one RNA polymerase reads). So:

DNA template: 3’ - ??? - 5’? But it’s written 5’ to 3’. So if it’s written 5’ ATG...3’, then the actual template strand being read is the reverse complement? This is messy.

Alternative approach: In many high school worksheets, when they give a DNA strand like this and say "write mRNA", they mean: replace T with U, and that’s the mRNA — assuming the given DNA is the coding strand.

But let’s check question 11 etc. — later questions confirm transcription happens in nucleus, translation in cytoplasm, so this is standard central dogma.

Perhaps safest: since the DNA is written 5’ to 3’, and mRNA is to be written below it, aligned, and mRNA is always 5’ to 3’, but in the diagram mRNA has 3’ on left — that must be a mistake in the diagram labeling. Probably, the mRNA should have 5’ on left and 3’ on right, matching DNA orientation.

I think there’s a labeling error in the diagram. Let me reinterpret:

Typically, in such diagrams:
- Top DNA strand: 5’ to 3’ (given)
- mRNA is synthesized complementary to the *other* DNA strand, but if only one strand is given, and it’s labeled as the template, then mRNA is complementary.

But to resolve this, let’s look at the start codon. The DNA starts with ATG — which codes for Methionine, start codon. In mRNA, that would be AUG.

So if DNA is 5’ ATG ... 3’, and if this is the coding strand, then mRNA is 5’ AUG ... 3’.

If it’s the template strand, then mRNA would be 3’ UAC ... 5’, which is not standard.

Given that ATG is start, and it’s at the beginning, likely the DNA shown is the coding strand, so mRNA is same as DNA but T→U.

Moreover, in the diagram, the mRNA is drawn directly below DNA, suggesting base-pairing, but if DNA is coding strand, mRNA shouldn't pair with it — it pairs with template.

This is confusing. Let me try both ways.

Option 1: Assume DNA given is template strand.

Then mRNA is complementary and antiparallel.

DNA: 5’ A T G C C T G G T G G T A T G G G 3’

Template strand is usually written 3’ to 5’ for transcription, but here it’s 5’ to 3’, so to transcribe, we read it from 3’ end to 5’ end.

So reverse the DNA to get template direction: 3’ G G G T A T G G T G G T C C G T A 5’? No.

Better: to find mRNA sequence, take the complement of the DNA strand, and reverse it? No.

Standard method: if DNA template strand is 3’-TAC-5’, mRNA is 5’-AUG-3’.

Here, DNA is given as 5’-ATG-3’. If this is the template strand, then the actual sequence being read is from 3’ end, so the first base transcribed is the 3’ end of this strand.

So DNA template: 5’ A T G C C T G G T G G T A T G G G 3’

When transcribing, RNA pol binds and reads from 3’ to 5’ along the template, so it starts at the 3’ end of this strand.

So the 3’ end is G (last base), then G, G, T, A, T, G, G, T, G, G, T, C, C, G, T, A — wait, let's list the DNA from 3’ to 5’:

Original DNA 5’ to 3’: positions 1 to 15: A(1),T(2),G(3),C(4),C(5),T(6),G(7),G(8),T(9),G(10),G(11),T(12),A(13),T(14),G(15),G(16)? Wait, count the bases.

DNA: A T G C C T G G T G G T A T G G G — that's 15 bases? Let's count:

1.A 2.T 3.G 4.C 5.C 6.T 7.G 8.G 9.T 10.G 11.G 12.T 13.A 14.T 15.G 16.G 17.G — 17 bases? From the image, it's 17 circles.

In the image: DNA has 17 bases: A,T,G,C,C,T,G,G,T,G,G,T,A,T,G,G,G — yes, 17.

So DNA 5’ to 3’: base1=A,2=T,3=G,4=C,5=C,6=T,7=G,8=G,9=T,10=G,11=G,12=T,13=A,14=T,15=G,16=G,17=G

If this is the template strand, then mRNA is synthesized 5’ to 3’ complementary to it, reading the template from 3’ to 5’.

So template from 3’ to 5’: base17=G,16=G,15=G,14=T,13=A,12=T,11=G,10=G,9=T,8=G,7=G,6=T,5=C,4=C,3=G,2=T,1=A

So template 3’ to 5’: G G G T A T G G T G G T C C G T A

Now mRNA 5’ to 3’ is complementary to this:

For each base in template 3’ to 5’:
G -> C
G -> C
G -> C
T -> A
A -> U
T -> A
G -> C
G -> C
T -> A
G -> C
G -> C
T -> A
C -> G
C -> G
G -> C
T -> A
A -> U

So mRNA 5’ to 3’: C C C A U A C C A C C A G G C A U

But that seems odd, and the start codon is not at the beginning.

Alternatively, if the DNA given is the coding strand (non-template), then mRNA is identical except T->U, so 5’ A U G C C U G G U G G U A U G G G 3’

And that makes sense because it starts with AUG, start codon.

In most educational contexts, when they show a DNA strand like this and ask for mRNA without specifying, and it starts with ATG, they intend for you to replace T with U to get mRNA.

Moreover, in the diagram, the mRNA is drawn directly below DNA, suggesting alignment, which would only make sense if it's the coding strand comparison.

Also, for tRNA, it will be complementary to mRNA.

So I think it's safe to assume that the DNA shown is the coding strand, so mRNA is the same sequence with T replaced by U.

Thus:

DNA: 5’ A T G C C T G G T G G T A T G G G 3’

mRNA: 5’ A U G C C U G G U G G U A U G G G 3’

But in the diagram, mRNA is labeled with 3’ on left and 5’ on right — that must be a mistake. Probably, it should be 5’ on left, 3’ on right. I'll proceed with mRNA 5’ to 3’ as above.

For the diagram filling:

mRNA row: should be A,U,G,C,C,U,G,G,U,G,G,U,A,U,G,G,G — but since the diagram has mRNA with 3’ on left, perhaps they want it written in reverse? To match the drawing.

Look at the diagram: under "mRNA", the first circle (left) is labeled 3’, last (right) is 5’. So they are drawing mRNA from 3’ to 5’. That is unconventional, but we have to follow the diagram.

So if mRNA is normally 5’ AUGCCUGGUGGUAUGGG 3’, then when drawn 3’ to 5’, it would be the reverse: 3’ GGGUAUGGUGGUCCGUA 5’? Let's calculate.

Normal mRNA 5’ to 3’: A U G C C U G G U G G U A U G G G

So from 5’ to 3’: pos1=A,2=U,3=G,4=C,5=C,6=U,7=G,8=G,9=U,10=G,11=G,12=U,13=A,14=U,15=G,16=G,17=G

If drawn 3’ to 5’, then left end is 3’ end, which is G (pos17), then G(16), G(15), U(14), A(13), U(12), G(11), G(10), U(9), G(8), G(7), U(6), C(5), C(4), G(3), U(2), A(1)

So mRNA in diagram (3’ to 5’): G, G, G, U, A, U, G, G, U, G, G, U, C, C, G, U, A

Similarly, tRNA anticodons are complementary to mRNA codons.

First, group mRNA into codons. Since mRNA is drawn 3’ to 5’, but codons are read 5’ to 3’ by ribosome, this is messy.

Perhaps for the diagram, we should fill based on pairing.

In the diagram, DNA is 5’ to 3’ left to right.

mRNA is below it, with 3’ on left, 5’ on right — so it's antiparallel to DNA? But DNA and mRNA are not directly paired; mRNA is complementary to template DNA.

I think there's a fundamental issue with the diagram labeling.

To simplify for a student, and given that this is likely a basic worksheet, I'll assume that the DNA shown is the template strand, and mRNA is complementary and antiparallel, and the diagram intends for us to write mRNA bases complementary to DNA, with mRNA running opposite direction.

So DNA: 5’ A T G C C T G G T G G T A T G G G 3’

Since mRNA is antiparallel, its 5’ end pairs with DNA 3’ end.

So for each DNA base, the complementary RNA base is:

DNA A -> RNA U
DNA T -> RNA A
DNA G -> RNA C
DNA C -> RNA G

And since mRNA is antiparallel, when DNA is 5’ to 3’ left to right, mRNA should be 3’ to 5’ left to right, which matches the diagram labeling (mRNA has 3’ on left, 5’ on right).

Perfect! So in the diagram, mRNA is drawn 3’ to 5’ left to right, complementary to DNA 5’ to 3’ left to right.

So for each position:

DNA base (5’to3’) : A T G C C T G G T G G T A T G G G

Complementary RNA base: U A C G G A C C A C C A U A C C C

And since mRNA is written 3’ to 5’ in the diagram, this sequence U A C G G A C C A C C A U A C C C is from 3’ to 5’, so the actual mRNA sequence 5’ to 3’ would be the reverse: C C C A U A C C A C C A G G C A U, but for the diagram, we fill the circles with U,A,C,G,G,A,C,C,A,C,C,A,U,A,C,C,C from left to right (since left is 3’ end).

Let's list:

Position: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
DNA: A T G C C T G G T G G T A T G G G
mRNA comp: U A C G G A C C A C C A U A C C C (this is complementary, and since mRNA is antiparallel, this is correct for the diagram where mRNA is 3’ to 5’ left to right)

So mRNA row: fill with U, A, C, G, G, A, C, C, A, C, C, A, U, A, C, C, C

Now tRNA: tRNA has anticodons that are complementary to mRNA codons.

First, mRNA is 3’ U A C G G A C C A C C A U A C C C 5’ (as filled)

But codons are read 5’ to 3’ by ribosome, so we need to consider the mRNA sequence in 5’ to 3’ direction for codon grouping.

Actual mRNA sequence 5’ to 3’ is the reverse of what's in the diagram: so from right to left of the diagram: C C C A U A C C A C C A G G C A U

So mRNA 5’ to 3’: C C C A U A C C A C C A G G C A U

Group into codons (triplets): CCC AUACCAC CAG GCA U — but 17 bases, not divisible by 3. 17 div 3 is 5 codons with 2 extra? That can't be.

Count the DNA bases: in the image, it's 17 bases, but typically genes have multiples of 3. Perhaps I miscounted.

Looking back at the user's image description: "DNA" row has circles: A,T,G,C,C,T,G,G,T,G,G,T,A,T,G,G,G — that's 17 characters.

But 17 is not divisible by 3, so for translation, it might be incomplete, or perhaps it's 18? Let me double-check.

In the text: "A T G C C T G G T G G T A T G G G" — let's count: 1A,2T,3G,4C,5C,6T,7G,8G,9T,10G,11G,12T,13A,14T,15G,16G,17G — yes 17.

Perhaps the last G is not part of it, or maybe it's a typo. For the sake of the problem, we'll proceed with 17 bases, and for tRNA, we'll pair with mRNA as per diagram.

In the diagram, tRNA is below mRNA, and it's shown with lines connecting to mRNA, and tRNA has 5’ on left, 3’ on right? Let's see: "tRNA" row has "5’" on left and "3’" on right, while mRNA has "3’" on left and "5’" on right, so tRNA is parallel to DNA? This is complicated.

Perhaps for tRNA, the anticodon is complementary to the mRNA codon, and in the diagram, since mRNA is 3’ to 5’ left to right, and tRNA is drawn with 5’ to 3’ left to right, then for each mRNA base, the tRNA anticodon base is complementary, and since they are antiparallel, it should match.

Standard: mRNA codon 5’-AUG-3’ pairs with tRNA anticodon 3’-UAC-5’.

In the diagram, if mRNA is written 3’ to 5’ left to right, then for a codon, say the first three bases of mRNA in diagram are 3’ U A C 5’ (left to right), which corresponds to mRNA 5’ C A U 3’ for that codon? I'm getting confused.

To make it simple for the student, and since this is a common type of problem, I'll assume that for the diagram, we fill mRNA as complementary to DNA, with the directions as labeled, and for tRNA, complementary to mRNA.

So:

DNA: 5’ A T G C C T G G T G G T A T G G G 3’

mRNA (complementary, and since antiparallel, in diagram 3’ to 5’): so for each DNA base, RNA complement:

A->U, T->A, G->C, C->G, etc.

So mRNA: 3’ U A C G G A C C A C C A U A C C C 5’ (written left to right as 3’ to 5’)

So in the mRNA row, fill: U, A, C, G, G, A, C, C, A, C, C, A, U, A, C, C, C

Now tRNA: tRNA anticodons are complementary to mRNA codons. But in the diagram, tRNA is drawn with 5’ on left, 3’ on right, and it's below mRNA, with lines connecting, so likely, for each mRNA base, the tRNA base is complementary, and since tRNA is 5’ to 3’ left to right, and mRNA is 3’ to 5’ left to right, they are antiparallel, so complementary bases will pair.

For example, mRNA first base (left) is U (3’ end), so tRNA first base (5’ end) should be A (since A pairs with U).

General rule: A-U, G-C.

So for each position, tRNA base is complementary to mRNA base at that position.

mRNA: pos1=U, so tRNA pos1=A (complement)
mRNA pos2=A, tRNA pos2=U
mRNA pos3=C, tRNA pos3=G
mRNA pos4=G, tRNA pos4=C
mRNA pos5=G, tRNA pos5=C
mRNA pos6=A, tRNA pos6=U
mRNA pos7=C, tRNA pos7=G
mRNA pos8=C, tRNA pos8=G
mRNA pos9=A, tRNA pos9=U
mRNA pos10=C, tRNA pos10=G
mRNA pos11=C, tRNA pos11=G
mRNA pos12=A, tRNA pos12=U
mRNA pos13=U, tRNA pos13=A
mRNA pos14=A, tRNA pos14=U
mRNA pos15=C, tRNA pos15=G
mRNA pos16=C, tRNA pos16=G
mRNA pos17=C, tRNA pos17=G

So tRNA row: A, U, G, C, C, U, G, G, U, G, G, U, A, U, G, G, G

Now for amino acids: each tRNA carries an amino acid corresponding to its anticodon.

But to find the amino acid, we need the codon on mRNA.

Since mRNA is 3’ U A C G G A C C A C C A U A C C C 5’, the actual coding sequence 5’ to 3’ is the reverse: 5’ C C C A U A C C A C C A G G C A U 3’

Group into codons from 5’ end: CCC, AUA, CCA, CCA, GGC, AU — but 17 bases, so last codon incomplete. Perhaps ignore the last two or something.

Codons:
1. CCC - Proline
2. AUA - Isoleucine
3. CCA - Proline
4. CCA - Proline
5. GGC - Glycine
6. AU - incomplete, perhaps not used.

But in the diagram, there are 5 ovals for amino acids, so probably they expect 5 amino acids.

From the mRNA 5’ to 3’: let's write it as 5’ CCC AUA CCA CCA GGC AU 3’ — so first five codons: CCC, AUA, CCA, CCA, GGC

Amino acids:
- CCC: Proline (Pro)
- AUA: Isoleucine (Ile)
- CCA: Proline (Pro)
- CCA: Proline (Pro)
- GGC: Glycine (Gly)

So the amino acid chain: Pro, Ile, Pro, Pro, Gly

In the diagram, the amino acids are connected to tRNA, and there are 5 groups, each group of three tRNA bases connected to one amino acid oval.

Looking at the diagram: "Amino Acids" row has 5 ovals, and above them, the tRNA is grouped in sets of three, each set connected to one oval.

So for tRNA, the first three bases (left) correspond to first amino acid, next three to second, etc.

tRNA sequence: A U G | C C U | G G U | G G U | A U G | G G — but 17 bases, so perhaps the last two are not used, or grouped differently.

From earlier, tRNA: positions 1-3: A,U,G
4-6: C,C,U
7-9: G,G,U
10-12: G,G,U
13-15: A,U,G
16-17: G,G — incomplete

But there are 5 ovals, so likely they consider 5 codons, so perhaps the DNA is meant to be 15 bases? Or maybe the last two are ignored.

Perhaps in the diagram, the grouping is done every three, and for 17 bases, it's not perfect, but for the sake of the problem, we'll take the first 15 bases for 5 codons.

So tRNA first 15 bases: A U G, C C U, G G U, G G U, A U G

Each of these is the anticodon.

Anticodon is complementary to mRNA codon.

For example, tRNA anticodon 3’-AUG-5’ would pair with mRNA 5’-UAC-3’, but in our case, the tRNA is written 5’ to 3’ in the diagram, so the sequence given is 5’ to 3’ for tRNA.

Standard: tRNA anticodon is written 3’ to 5’ when pairing, but in sequences, it's often written 5’ to 3’.

To avoid confusion, let's use the mRNA codon to find amino acid.

From mRNA 5’ to 3’: as above, 5’ CCC AUA CCA CCA GGC ... 3’

So codons:
1. CCC -> Proline
2. AUA -> Isoleucine
3. CCA -> Proline
4. CCA -> Proline
5. GGC -> Glycine

So amino acids: Pro, Ile, Pro, Pro, Gly

In the diagram, the first oval (left) corresponds to the first codon, which is CCC, Proline, and so on.

So for the amino acid row, fill: Proline, Isoleucine, Proline, Proline, Glycine

Or abbreviations: Pro, Ile, Pro, Pro, Gly

Now for the remaining questions.

11. Transcription takes place in the (nucleus/cytoplasm).
→ In eukaryotes, transcription (DNA to RNA) happens in the nucleus.
✔ Answer: nucleus

12. tRNA is used in (translation/transcription).
→ tRNA brings amino acids during translation (protein synthesis).
✔ Answer: translation

13. tRNA uses (anticodons/codons) to match to the mRNA.
→ tRNA has anticodons that bind to mRNA codons.
✔ Answer: anticodons

14. Proteins are made at the (nucleus/ribosome).
→ Ribosomes are the sites of protein synthesis.
✔ Answer: ribosome

15. (rRNA/mRNA) attaches the amino acids into a chain.
→ rRNA is part of the ribosome, which catalyzes peptide bond formation. mRNA carries the code, but rRNA does the attaching.
Actually, the ribosome (made of rRNA and proteins) facilitates the bonding, but technically, it's the ribosome that does it, and rRNA is the catalytic component.
Some might say tRNA brings them, but the attachment is done by the ribosome.
The question says "attaches the amino acids into a chain" — that's the function of the ribosome, which contains rRNA.
mRNA is the template.
So rRNA is correct.
✔ Answer: rRNA

16. tRNA is found in the (nucleus/cytoplasm).
→ tRNA functions in the cytoplasm at the ribosome. It is transcribed in nucleus but works in cytoplasm.
✔ Answer: cytoplasm

17. (Translation/Transcription) converts mRNA into a protein.
→ Translation is the process of making protein from mRNA.
✔ Answer: Translation

18. Translation takes place in the (cytoplasm/nucleus).
→ In eukaryotes, translation occurs in the cytoplasm (on ribosomes).
✔ Answer: cytoplasm

19. (DNA/RNA) can leave the nucleus.
→ DNA stays in nucleus. RNA (mRNA, tRNA, etc.) can leave the nucleus to go to cytoplasm.
✔ Answer: RNA

Now for question 10, to summarize the diagram filling:

- mRNA row (from left to right, 3’ to 5’): U, A, C, G, G, A, C, C, A, C, C, A, U, A, C, C, C

- tRNA row (from left to right, 5’ to 3’): A, U, G, C, C, U, G, G, U, G, G, U, A, U, G, G, G

- Amino acids (5 ovals, from left to right): Proline, Isoleucine, Proline, Proline, Glycine

Since the diagram may expect specific formatting, but for the answer, we can state it.

For the final answer, since the user asked to solve the problem, and question 10 is a diagram, perhaps we need to provide the sequences.

But in the response, we can list the answers for all.

Also, for question 10, the amino acids might be expected as names or abbreviations.

Given that, let's compile all answers.

Final Answers:

5. nucleus
6. mRNA
7. rRNA
8. RNA
9. amino
10. mRNA: U,A,C,G,G,A,C,C,A,C,C,A,U,A,C,C,C (left to right)
tRNA: A,U,G,C,C,U,G,G,U,G,G,U,A,U,G,G,G (left to right)
Amino Acids: Proline, Isoleucine, Proline, Proline, Glycine
11. nucleus
12. translation
13. anticodons
14. ribosome
15. rRNA
16. cytoplasm
17. Translation
18. cytoplasm
19. RNA

For question 10, since it's a diagram, and the user might need to fill it, but in text, we describe.

Perhaps for the boxed answer, we can list all.

Since the instruction is to provide final answer after reasoning, and for multiple parts, we can list them.

I think for clarity, I'll write the answers in order.

Note: For question 10, the amino acid sequence assumes the first 15 bases are used for 5 codons, ignoring the last two bases, as there are 5 ovals.

If the DNA is 17 bases, and mRNA 17, then 5 full codons require 15 bases, so last two are not translated, which is fine.

So proceeding.