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End-of-topic test: Translation

Unit 6 · Topic 6.4 end-of-topic test

Suggested time: about 47 minutes. Answer everything, then press Submit the test to see the feedback and scoring guides.

Answer every question. For each multiple-choice question, pick one option. When you have answered every question, press Submit the test; the feedback then gives the reasoning for each. 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. Then open the scoring guide and mark your own work against it.
Question 1
A table with two columns, protein and where it ends up, and four rows lettered W, X, Y, Zproteinwhere it ends upWcarries oxygen; stays inside the blood cell that made itXspeeds up one step of fermentation in a yeast cell's cytoplasmYforms part of a muscle cell's contracting fibers, inside the cellZa mucus protein released onto the surface of the airway
Four proteins and where each ends up.

The table describes four proteins made by eukaryotic cells and where each protein ends up.

Which protein is built on a ribosome attached to the rough ER?

Question 2
A table with four columns, time after the gene switches on, the 5′ end of the mRNA, the first amino acids of the polypeptide, the 3′ end of the mRNA, and three rowstime after switch-on5′ end of the mRNAfirst amino acids3′ end of the mRNA15 spresentabsentabsent40 spresentpresentabsent90 spresentpresentpresent
Which of the three things are present, at three times after the gene switches on.

A gene in a soil bacterium switches on. Scientists can tell when three things first exist in the cell: the 5′ end of the gene's mRNA, the first amino acids of its polypeptide, and the 3′ end of the mRNA. The table shows which of the three are present at three times after the gene switches on.

Which conclusion do the observations support?

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

A ribosome reads the mRNA drawn.

Which codon does the ribosome read second?

Question 4
A table with two columns, drug and what is seen in treated cells, and four rows lettered W, X, Y, Zdrugseen in treated cellsWsmall subunits bind mRNAs; no complete ribosome formsXcomplete ribosomes form; no chain grows past a few amino acidsYribosomes sit at the stop codons, full-length chains still attachedZfull-length chains are released but do not fold into their working shapes
What is seen in cells treated with each drug.

A scientist treats eukaryotic cells with each of four drugs. The table shows what is seen in the treated cells.

Which drug blocks termination?

Question 5
One monospace row, 5′-UAU-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′-UAU-3′second basefirst basethird baseUCAGUCAGUUUPheUUCPheUUALeuUUGLeuUCUSerUCCSerUCASerUCGSerUAUTyrUACTyrUAAstopUAGstopUGUCysUUGCCysCUGAstopAUGGTrpGCUULeuCUCLeuCUALeuCUGLeuCCUProCCCProCCAProCCGProCAUHisCACHisCAAGlnCAGGlnCGUArgUCGCArgCCGAArgACGGArgGAUUIleAUCIleAUAIleAUGMet (start)ACUThrACCThrACAThrACGThrAAUAsnAACAsnAAALysAAGLysAGUSerUAGCSerCAGAArgAAGGArgGGUUValGUCValGUAValGUGValGCUAlaGCCAlaGCAAlaGCGAlaGAUAspGACAspGAAGluGAGGluGGUGlyUGGCGlyCGGAGlyAGGGGlyG
The codon, above the genetic code chart.

The codon 5′-UAU-3′ is drawn above the genetic code chart.

Which amino acid does this codon name?

Question 6
A table with two columns, mRNA written 5′ to 3′ and polypeptide made, and three rowsmRNA (5′ to 3′)polypeptideAUG CUU CCC UAAMet–Leu–ProAUG CUC CCC UAAMet–Leu–ProAUG CUU UUU UAAMet–Leu–Phe
Three mRNAs and the polypeptide each gave.

Three short mRNAs were translated in a cell. The table shows each mRNA and the polypeptide it gave. A fourth mRNA reads 5′-AUG CUC UUU UAA-3′.

Which polypeptide does the fourth mRNA give?

Question 7

A soil bacterium, an octopus and an owl each read the codon 5′-UUU-3′ as phenylalanine (Phe), and so does every other organism tested.

Which of the following best explains why the shared code is evidence of common ancestry?

Question 8

The small subunit of a ribosome has bound a eukaryotic mRNA near its 5′ end and moved along it to the first AUG.

Which of the following happens next?

Question 9
One monospace row, 5′-UGC-3′, with the note the next codon in the ribosome, 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′-UGC-3′the next codon in the ribosomesecond basefirst basethird baseUCAGUCAGUUUPheUUCPheUUALeuUUGLeuUCUSerUCCSerUCASerUCGSerUAUTyrUACTyrUAAstopUAGstopUGUCysUUGCCysCUGAstopAUGGTrpGCUULeuCUCLeuCUALeuCUGLeuCCUProCCCProCCAProCCGProCAUHisCACHisCAAGlnCAGGlnCGUArgUCGCArgCCGAArgACGGArgGAUUIleAUCIleAUAIleAUGMet (start)ACUThrACCThrACAThrACGThrAAUAsnAACAsnAAALysAAGLysAGUSerUAGCSerCAGAArgAAGGArgGGUUValGUCValGUAValGUGValGCUAlaGCCAlaGCAAlaGCGAlaGAUAspGACAspGAAGluGAGGluGGUGlyUGGCGlyCGGAGlyAGGGGlyG
The next codon, above the genetic code chart.

A ribosome has joined the first two amino acids of a polypeptide. The next codon in the ribosome is 5′-UGC-3′, drawn above the genetic code chart.

Which of the following arrives next?

Question 10
One monospace row, 5′-UCAUGUUUCCCAGAUAUCUCGGGUAGGCC-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′-UCAUGUUUCCCAGAUAUCUCGGGUAGGCC-3′second basefirst basethird baseUCAGUCAGUUUPheUUCPheUUALeuUUGLeuUCUSerUCCSerUCASerUCGSerUAUTyrUACTyrUAAstopUAGstopUGUCysUUGCCysCUGAstopAUGGTrpGCUULeuCUCLeuCUALeuCUGLeuCCUProCCCProCCAProCCGProCAUHisCACHisCAAGlnCAGGlnCGUArgUCGCArgCCGAArgACGGArgGAUUIleAUCIleAUAIleAUGMet (start)ACUThrACCThrACAThrACGThrAAUAsnAACAsnAAALysAAGLysAGUSerUAGCSerCAGAArgAAGGArgGGUUValGUCValGUAValGUGValGCUAlaGCCAlaGCAAlaGCGAlaGAUAspGACAspGAAGluGAGGluGGUGlyUGGCGlyCGGAGlyAGGGGlyG
The mRNA, above the genetic code chart.

The mRNA drawn above the genetic code chart is translated.

How many amino acids does the polypeptide contain?

Question 11
One monospace row, 5′-AUGCCAGAUUGA-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′-AUGCCAGAUUGA-3′second basefirst basethird baseUCAGUCAGUUUPheUUCPheUUALeuUUGLeuUCUSerUCCSerUCASerUCGSerUAUTyrUACTyrUAAstopUAGstopUGUCysUUGCCysCUGAstopAUGGTrpGCUULeuCUCLeuCUALeuCUGLeuCCUProCCCProCCAProCCGProCAUHisCACHisCAAGlnCAGGlnCGUArgUCGCArgCCGAArgACGGArgGAUUIleAUCIleAUAIleAUGMet (start)ACUThrACCThrACAThrACGThrAAUAsnAACAsnAAALysAAGLysAGUSerUAGCSerCAGAArgAAGGArgGGUUValGUCValGUAValGUGValGCUAlaGCCAlaGCAAlaGCGAlaGAUAspGACAspGAAGluGAGGluGGUGlyUGGCGlyCGGAGlyAGGGGlyG
The mRNA, above the genetic code chart.

The mRNA drawn above the genetic code chart is translated.

Which amino acids does the polypeptide contain, in order?

Question 12
One monospace row, 3′-TAC GAG TAT ATC-5′, with the note the template strand, 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 stop3′-TACGAGTATATC-5′the template strandsecond basefirst basethird baseUCAGUCAGUUUPheUUCPheUUALeuUUGLeuUCUSerUCCSerUCASerUCGSerUAUTyrUACTyrUAAstopUAGstopUGUCysUUGCCysCUGAstopAUGGTrpGCUULeuCUCLeuCUALeuCUGLeuCCUProCCCProCCAProCCGProCAUHisCACHisCAAGlnCAGGlnCGUArgUCGCArgCCGAArgACGGArgGAUUIleAUCIleAUAIleAUGMet (start)ACUThrACCThrACAThrACGThrAAUAsnAACAsnAAALysAAGLysAGUSerUAGCSerCAGAArgAAGGArgGGUUValGUCValGUAValGUGValGCUAlaGCCAlaGCAAlaGCGAlaGAUAspGACAspGAAGluGAGGluGGUGlyUGGCGlyCGGAGlyAGGGGlyG
The template strand, above the genetic code chart.

A gene's template strand reads 3′-TAC GAG TAT ATC-5′, drawn above the genetic code chart. RNA polymerase copies it, and a ribosome translates the mRNA.

Which amino acids does the polypeptide contain, in order?

Question 13

A student describes the flow of information in a eukaryotic cell with four labels: ‘in the nucleus’, ‘transcription by RNA polymerase’, ‘read 5′ to 3′’ and ‘begins with Met’. The student now describes the same flow in a bacterium.

Which label must change?

Question 14
A bar chart: x-axis codon, four bars labeled ACU ACC ACA ACG; 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)codonACUACCACAACG
Time the ribosome spends at each of the four threonine 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 threonine (Thr), relative to the fastest of them. The scientist can rewrite the mRNA, keeping the same polypeptide.

Which one change would make the mRNA's translation faster?

Question 15
Two monospace rows, 5′-AUA-3′ noted allele 1 and 5′-AGA-3′ noted allele 2, 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′-AUA-3′allele 15′-AGA-3′allele 2second basefirst basethird baseUCAGUCAGUUUPheUUCPheUUALeuUUGLeuUCUSerUCCSerUCASerUCGSerUAUTyrUACTyrUAAstopUAGstopUGUCysUUGCCysCUGAstopAUGGTrpGCUULeuCUCLeuCUALeuCUGLeuCCUProCCCProCCAProCCGProCAUHisCACHisCAAGlnCAGGlnCGUArgUCGCArgCCGAArgACGGArgGAUUIleAUCIleAUAIleAUGMet (start)ACUThrACCThrACAThrACGThrAAUAsnAACAsnAAALysAAGLysAGUSerUAGCSerCAGAArgAAGGArgGGUUValGUCValGUAValGUGValGCUAlaGCCAlaGCAAlaGCGAlaGAUAspGACAspGAAGluGAGGluGGUGlyUGGCGlyCGGAGlyAGGGGlyG
The differing codon of each allele's mRNA, above the genetic code chart.

An animal's two alleles of one gene for an enzyme differ at one base of DNA. One codon of the mRNA differs as a result: 5′-AUA-3′ in allele 1 is 5′-AGA-3′ in allele 2. The two codons are drawn above the genetic code chart.

How does the enzyme built from allele 2 differ from the enzyme built from allele 1?

Question 16
A table with five columns, virus, genome, size, host and enzyme carried inside the coat, and four rows lettered W, X, Y, Zvirusgenomesizeinfectsenzyme carried inside the coatWDNAlarger than mostplantsnoneXRNAsmaller than mostinsectsone that copies RNA into RNAYRNAabout averagebirdsone that copies RNA into DNAZDNAsmaller than mostbacteriaone that copies DNA into RNA
Four viruses, described.

The table describes four viruses.

Which virus is a retrovirus?

Question 17

A retrovirus's DNA copy has joined one of the host cell's chromosomes.

Which enzyme reads that DNA copy to make the virus's RNA?

Question 18

A drug that blocks reverse transcriptase reaches two cells of the same kind. A retrovirus enters cell W one hour after the drug arrives. A retrovirus entered cell X a week before the drug arrived, and one of cell X's chromosomes already carries the DNA copy.

Which cell builds new virus particles while the drug is present?

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 · Analyze Model or Visual Representation · 4 points
The model drawn shows the flow of genetic information in a eukaryotic cell. Four of its labels are missing, and the blanks are lettered Q, R, S and T.
Three boxes in a row joined by two arrows pointing right. The first two boxes read DNA and mRNA; the third holds a circled Q. Above the first two boxes their places are written; above the third box a circled R. Under the first arrow the step and what acts are written; under the second arrow a circled S. Beneath the second and third boxes a direction is written; beneath the first box a circled TDNAmRNAQin the nucleuscapped, tailed and spliced in the nucleus,then leaves through a poreRtranscriptionby RNA polymeraseSTread 5′ to 3′begins with Met
A model of the flow of genetic information, with four blanks lettered Q, R, S and T.

(a) Represent the flow of information by stating what belongs at each blank: the molecule at Q, the place at R, the step at S and, at T, the direction in which the template strand is read. (2 points)

A full-credit answer: Q is the polypeptide.
R is at a ribosome in the cytoplasm.
S is translation.
T is the template strand read 3′ to 5′.
The completed model is drawn.

Three boxes in a row joined by two arrows pointing right, reading DNA, mRNA and polypeptide, every label on: places above the boxes, the step and what acts under each arrow, a direction beneath each boxDNAmRNApolypeptidein the nucleuscapped, tailed and spliced in the nucleus,then leaves through a poreat a ribosomein the cytoplasmtranscriptionby RNA polymerasetranslationby a ribosome, with tRNAstemplate strand read 3′ to 5′read 5′ to 3′begins with Met
The completed model: every label on.

Check the box for each point your answer earns

Accept at T 'the mRNA is built 5′ to 3′' only where the student names it as the mRNA's direction; a bare '5′ to 3′' at T earns nothing.

Common slip: Writing 'read 5′ to 3′' at T: that is the direction the mRNA is read, written beneath the middle box. RNA polymerase reads the template strand the other way.

(b) Describe what happens at the step lettered S. (1 point)

A full-credit answer: A ribosome reads the mRNA's codons one after another, from the 5′ end toward the 3′ end.
For each codon a tRNA brings the amino acid the codon names.
The ribosome joins the amino acids by peptide bonds into the polypeptide.

Check the box for each point your answer earns

(c) A student claims that, in this cell, transcription and the step lettered S must happen in two different places. Using the model, support the claim. (1 point)

A full-credit answer: The DNA stays on its chromosome inside the nucleus, so RNA polymerase copies it there.
The ribosomes work in the cytoplasm, outside the nuclear envelope.
So the mRNA is finished in the nucleus and leaves through a pore before a ribosome can read it.

Check the box for each point your answer earns

The label 'then leaves through a pore' copied on its own earns nothing: the point wants where the DNA stays AND where the ribosomes work.

Common slip: Saying the ribosomes enter the nucleus to read the mRNA there. Ribosomes work in the cytoplasm; the mRNA travels to them.

Free-response score: 0 of 4
Free response 2 · Conceptual Analysis · 4 points
A plant carries two alleles of one gene for an enzyme that builds the hard material of the plant's seed coats. The two alleles differ at one base of DNA, so one codon of the mRNA differs: 5′-GGG-3′ in the first allele's mRNA is 5′-GAG-3′ in the second allele's mRNA. The two codons are drawn above the genetic code chart. A plant with two copies of the first allele makes seeds with hard coats. A plant with two copies of the second allele makes seeds with soft coats.
Two monospace rows, 5′-GGG-3′ noted the first allele's mRNA and 5′-GAG-3′ noted the second allele's mRNA, 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′-GGG-3′the first allele's mRNA5′-GAG-3′the second allele's mRNAsecond basefirst basethird baseUCAGUCAGUUUPheUUCPheUUALeuUUGLeuUCUSerUCCSerUCASerUCGSerUAUTyrUACTyrUAAstopUAGstopUGUCysUUGCCysCUGAstopAUGGTrpGCUULeuCUCLeuCUALeuCUGLeuCCUProCCCProCCAProCCGProCAUHisCACHisCAAGlnCAGGlnCGUArgUCGCArgCCGAArgACGGArgGAUUIleAUCIleAUAIleAUGMet (start)ACUThrACCThrACAThrACGThrAAUAsnAACAsnAAALysAAGLysAGUSerUAGCSerCAGAArgAAGGArgGGUUValGUCValGUAValGUGValGCUAlaGCCAlaGCAAlaGCGAlaGAUAspGACAspGAAGluGAGGluGGUGlyUGGCGlyCGGAGlyAGGGGlyG
The differing codon of each allele's mRNA, above the genetic code chart.

(a) Describe how the difference between the two alleles changes the enzyme's order of amino acids. (1 point)

A full-credit answer: On the chart GGG names Gly and GAG names Glu.
So the second allele's enzyme carries Glu at one position where the first allele's enzyme carries Gly.
Every other amino acid is the same.

Check the box for each point your answer earns

Common slip: Saying the second allele's enzyme is shorter or is not made. GAG names an amino acid, so the codon is read and the chain goes on.

(b) Explain how that difference leads to seeds with soft coats. (1 point)

A full-credit answer: Glu's R group carries a charge, and Gly's does not.
The order of amino acids sets the fold, so the enzyme folds into a different shape around that spot.
The shape lets the enzyme do its job, and the changed shape no longer fits the coat's building block at its active site.
So the hard material is not built, and the seed coat stays soft.

Check the box for each point your answer earns

Common slip: Stopping at 'the enzyme is different'. The point is the chain: a different amino acid → a different fold → the job is lost → the trait.

(c) A third allele differs from the first allele at the third base of the same codon: its mRNA carries 5′-GGA-3′ there. Predict the seed coats of a plant with two copies of the third allele. (1 point)

A full-credit answer: Hard seed coats, like those of a plant with two copies of the first allele.

Check the box for each point your answer earns

(d) Justify your prediction. (1 point)

A full-credit answer: GGA and GGG both name Gly: the code is redundant.
So the third allele's enzyme has the same order of amino acids as the first allele's.
The same order gives the same fold and the same job.
So the enzyme builds the hard material, and the seed coats are hard.

Check the box for each point your answer earns

Common slip: Claiming soft coats because a base changed. A base change that keeps the amino acid changes nothing downstream.

Free-response score: 0 of 4
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Multiple choice checked: 0 of 18 correct.
Next: Not every gene, not all the time