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Practice questions · Topic 6.4

Unit 6 · Practice for the Topic 6.4 end-of-topic test

You’ve gone through everything in this topic. The summary video below recaps it all, so you’re ready for the questions.

Watch first: Translation, summed up

Video coming soon

Where translation happens; codons, the code chart, AUG and the three stops; redundancy and the near-universal code; initiation, elongation, termination; codon and anticodon; genotype to phenotype through the chain; the retrovirus that runs the flow backward.

These are practice questions in the shape of the topic test. Work through them before you take the test; every question tells you what it wanted.
Answer every question. For each multiple-choice question, pick one option and press Check; the feedback gives the reasoning. For the free-response questions, write one short sentence for each step of your reasoning, each on its own line, and make every link clear (so, because, therefore). That is what the exam’s ‘paragraph form’ means for you: linked sentences, not bullet points. The first free-response question walks you through one changed cell one part at a time, and you can open a hint for each part; the second is at the level of the test. When you finish a question, open the scoring guide and mark your own work against it.
Question 1

A cell lining a hen's oviduct builds ovalbumin, a protein it releases into the egg white.

Where is ovalbumin built?

Question 2
A table with three columns, cell, minutes to finish copying the gene into mRNA, minutes until the first ribosome starts translating, and two rowscellminutes to finish copying the geneminutes until translation startsa bacterium30.5a yeast cell37
How long each cell takes to finish copying the gene, and until the first ribosome starts translating.

Suppose a gene switches on in a bacterium and a gene of the same length switches on in a yeast cell. The table shows how long each cell takes to finish copying the gene into mRNA, and how long until the first ribosome starts translating that mRNA.

Which statement explains why translation starts so much sooner in the bacterium?

Question 3
One monospace row: 5′-UAUGGGGUUUAGACCCUAA-3′5′-UAUGGGGUUUAGACCCUAA-3′
The mRNA, both ends written.

Read from the first AUG in threes and stop at the first stop codon; the stop codon adds no amino acid. A ribosome reads the mRNA drawn.

Which codon does the ribosome read third?

Question 4

Three RNA bases make one codon, so there are 64 different codons.

Which statement about the 64 codons is correct?

Question 5
The genetic code chart: a four-by-four grid, first base by row U C A G, second base by column U C A G, four lines in every cell giving the codon and its three-letter amino acid name, AUG marked Met (start) and UAA, UAG and UGA marked stopsecond basefirst basethird baseUCAGUCAGUUUPheUUCPheUUALeuUUGLeuUCUSerUCCSerUCASerUCGSerUAUTyrUACTyrUAAstopUAGstopUGUCysUUGCCysCUGAstopAUGGTrpGCUULeuCUCLeuCUALeuCUGLeuCCUProCCCProCCAProCCGProCAUHisCACHisCAAGlnCAGGlnCGUArgUCGCArgCCGAArgACGGArgGAUUIleAUCIleAUAIleAUGMet (start)ACUThrACCThrACAThrACGThrAAUAsnAACAsnAAALysAAGLysAGUSerUAGCSerCAGAArgAAGGArgGGUUValGUCValGUAValGUGValGCUAlaGCCAlaGCAAlaGCGAlaGAUAspGACAspGAAGluGAGGluGGUGlyUGGCGlyCGGAGlyAGGGGlyG
The genetic code chart.

The genetic code chart is drawn.

How many codons name histidine (His)?

Question 6

A gene from a moss is put into a yeast cell. The yeast cell's ribosomes read the gene's mRNA and build the moss protein, with every amino acid in its place.

What does this result show?

Question 7
One monospace row, 5′-GAUGCCCUGCUUUAGAGGGUAGCA-3′, above The genetic code chart: a four-by-four grid, first base by row U C A G, second base by column U C A G, four lines in every cell giving the codon and its three-letter amino acid name, AUG marked Met (start) and UAA, UAG and UGA marked stop5′-GAUGCCCUGCUUUAGAGGGUAGCA-3′second basefirst basethird baseUCAGUCAGUUUPheUUCPheUUALeuUUGLeuUCUSerUCCSerUCASerUCGSerUAUTyrUACTyrUAAstopUAGstopUGUCysUUGCCysCUGAstopAUGGTrpGCUULeuCUCLeuCUALeuCUGLeuCCUProCCCProCCAProCCGProCAUHisCACHisCAAGlnCAGGlnCGUArgUCGCArgCCGAArgACGGArgGAUUIleAUCIleAUAIleAUGMet (start)ACUThrACCThrACAThrACGThrAAUAsnAACAsnAAALysAAGLysAGUSerUAGCSerCAGAArgAAGGArgGGUUValGUCValGUAValGUGValGCUAlaGCCAlaGCAAlaGCGAlaGAUAspGACAspGAAGluGAGGluGGUGlyUGGCGlyCGGAGlyAGGGGlyG
The mRNA, above the genetic code chart.

Read from the first AUG in threes and stop at the first stop codon; the stop codon adds no amino acid. The mRNA drawn above the genetic code chart is translated.

How many amino acids does the polypeptide contain?

Question 8
One monospace row: 3′-GAA-5′3′-GAA-5′
The anticodon, written from its 3′ end.

A strand is written from its 5′ end to its 3′ end, and both ends are marked. Codons are written as they read on the mRNA, 5′ to 3′; an anticodon is written 3′ to 5′ against its codon. A tRNA's anticodon reads 3′-GAA-5′, as drawn.

Which codon on an mRNA does this tRNA pair with?

Question 9
A bar chart: x-axis codon, four bars labeled GCU GCC GCA GCG; y-axis time per codon in relative units, 0 to 4 with a gridline every 0.5 numbered every 1.0; no values printed on the bars01.02.03.04.0time per codon (relative units)codonGCUGCCGCAGCG
Time the ribosome spends at each of the four alanine codons, relative to the fastest.

A scientist measures how long a ribosome spends at each codon of an mRNA. The bar chart shows the time spent at the four codons that name alanine (Ala), relative to the fastest of them.

At which codon does the ribosome spend the longest time?

Question 10
A bar chart: x-axis codon, four bars labeled GCU GCC GCA GCG; y-axis time per codon in relative units, 0 to 4 with a gridline every 0.5 numbered every 1.0; no values printed on the bars01.02.03.04.0time per codon (relative units)codonGCUGCCGCAGCG
Time the ribosome spends at each of the four alanine codons, relative to the fastest.

The bar chart shows the time a ribosome spends at each of the four codons that name alanine (Ala), relative to the fastest. A cell makes two mRNAs for the same polypeptide: one uses 5′-GCA-3′ at every alanine, the other uses 5′-GCC-3′ at every alanine.

Which mRNA is translated faster, and why?

How to tackle the free-response questions. Read the verb first: describe asks what you see or know; explain asks why or how, so name the mechanism; predict asks what will happen and why; justify asks for the evidence that supports a claim. Each point is earned by one idea, stated in a sentence that names the thing and the mechanism. Extra words earn nothing; a wrong extra can lose the point. If there is a figure or table, use what it shows. When you finish, check the box for each point your answer earns and compare your sentences with the full-credit answer.
Free response 1 · Conceptual Analysis · 4 points
Codons are written as they read on the mRNA, 5′ to 3′; an anticodon is written 3′ to 5′ against its codon. Read from the first AUG in threes and stop at the first stop codon; the stop codon adds no amino acid. Suppose that in a yeast cell the enzyme that loads cysteine (Cys) onto its tRNAs stops working. The cell's tRNAs, ribosomes and mRNAs are unchanged, but no tRNA carries Cys. A ribosome moves on from a codon only after a loaded tRNA has paired with it and its amino acid is joined, so a ribosome never skips a codon. A ribosome begins reading the mRNA drawn above the genetic code chart.
One monospace row, 5′-AUG CUU UGC CCC GGG UAA-3′, above The genetic code chart: a four-by-four grid, first base by row U C A G, second base by column U C A G, four lines in every cell giving the codon and its three-letter amino acid name, AUG marked Met (start) and UAA, UAG and UGA marked stop5′-AUGCUUUGCCCCGGGUAA-3′second basefirst basethird baseUCAGUCAGUUUPheUUCPheUUALeuUUGLeuUCUSerUCCSerUCASerUCGSerUAUTyrUACTyrUAAstopUAGstopUGUCysUUGCCysCUGAstopAUGGTrpGCUULeuCUCLeuCUALeuCUGLeuCCUProCCCProCCAProCCGProCAUHisCACHisCAAGlnCAGGlnCGUArgUCGCArgCCGAArgACGGArgGAUUIleAUCIleAUAIleAUGMet (start)ACUThrACCThrACAThrACGThrAAUAsnAACAsnAAALysAAGLysAGUSerUAGCSerCAGAArgAAGGArgGGUUValGUCValGUAValGUGValGCUAlaGCCAlaGCAAlaGCGAlaGAUAspGACAspGAAGluGAGGluGGUGlyUGGCGlyCGGAGlyAGGGGlyG
The mRNA, above the genetic code chart.

(a) Identify the amino acid that the codon 5′-UGC-3′ names. (1 point)

Hint: Find the row for the first base, U, and the column for the second base, G. Then read the line for the third base, C.

A full-credit answer: 5′-UGC-3′ names Cys.

Check the box for each point your answer earns

(b) Describe what the loaded tRNA that pairs with 5′-UGC-3′ does at that codon in a cell where the loading enzyme works. (1 point)

Hint: Say what its anticodon does and what it brings into the ribosome.

A full-credit answer: In a working cell, the tRNA pairs its anticodon with the codon UGC and brings Cys into the ribosome.
The ribosome joins that Cys to the growing chain.

Check the box for each point your answer earns

(c) Predict how the polypeptide built in the changed yeast cell differs from the polypeptide a working cell builds from this mRNA. (1 point)

Hint: Ask what arrives at the ribosome when it reaches the third codon.

A full-credit answer: In the changed cell, the ribosome joins Met–Leu and then stops at the third codon.
The polypeptide is left unfinished, two amino acids long, instead of Met–Leu–Cys–Pro–Gly.

Check the box for each point your answer earns

A prediction that the ribosome skips UGC and builds Met–Leu–Pro–Gly (the chain with Cys left out) earns nothing.

(d) Justify your prediction. (1 point)

Hint: What must pair with the codon and deliver an amino acid before the ribosome can move on?

A full-credit answer: The ribosome stops there because only a tRNA carrying Cys can bring Cys to the codon UGC.
With the loading enzyme stopped, no tRNA carries Cys.
The ribosome joins the next amino acid and moves on only after a loaded tRNA has paired.
So it waits at UGC, and the chain grows no further.

Check the box for each point your answer earns

Free-response score: 0 of 4
Free response 2 · Conceptual Analysis · 6 points
A retrovirus's DNA copy sits in one of a cell's chromosomes. Suppose a drug binds the promoter of that DNA copy and nothing else, so that the host cell's RNA polymerase cannot bind there. The drug reaches the cell before the host cell's RNA polymerase has read the copy even once.

(a) Identify the molecule that the host cell's RNA polymerase builds from the DNA copy in an untreated cell. (1 point)

A full-credit answer: The host's RNA polymerase builds RNA from the DNA copy: new copies of the virus's RNA genome and mRNAs for the virus's proteins.

Check the box for each point your answer earns

(b) Describe what the host cell's ribosomes do with the molecule you identified in part (a), in an untreated cell. (1 point)

A full-credit answer: The host's ribosomes read the virus's mRNA codon by codon.
They build the virus's proteins: its coat proteins and its reverse transcriptase.

Check the box for each point your answer earns

(c) Make a claim about whether the treated cell builds new virus particles while the drug is present, and a second claim about whether the DNA copy is passed on when the treated cell divides. (2 points)

A full-credit answer: The treated cell builds no new virus particles while the drug is present.
The DNA copy is passed to both daughter cells when the cell divides.

Check the box for each point your answer earns

Accept for the first point: no new particles once any viral mRNA already present is used up.

Common slip: Claiming that the drug removes the copy. The drug blocks reading of the copy; the copy itself stays in the chromosome.

(d) Support both claims. (2 points)

A full-credit answer: With the promoter blocked, the host's RNA polymerase makes no RNA from the copy.
So the ribosomes get no viral mRNA and build no coat proteins or reverse transcriptase, and no new genome RNA is made.
So no particles can be assembled.
The copy is part of a chromosome, so the cell copies it in S phase with the rest and each daughter cell receives it.

Check the box for each point your answer earns

Free-response score: 0 of 6
Multiple choice checked: 0 of 10 correct.