6.4 Translation

1. AOn ribosomes attached to the cell's rough ER
   A student who thinks prokaryotes have the same organelles as eukaryotes picks this. Prokaryotic cells have no ER; their ribosomes are free in the cytoplasm.
2. BOn the ribosomes that are free in its cytoplasm Correct
   Translation occurs on ribosomes in the cytoplasm of both prokaryotic and eukaryotic cells. Prokaryotic cells have no rough ER, so all of their ribosomes are in the cytoplasm.
3. COn the DNA itself, which is translated without an mRNA
   A student who thinks bacteria translate genes directly from DNA picks this. Ribosomes translate mRNA, even when translation begins while the mRNA is still being transcribed.
4. DWithout ribosomes, as prokaryotic cells lack them
   A student who thinks prokaryotes lack ribosomes because they lack organelles picks this. Ribosomes are not membrane-bound organelles; prokaryotes have them and translate mRNA on them.

CED 6.4.A.1 · Read this in Fix

1. AIt is a prokaryotic cell, because its ribosomes are copying the DNA to make the mRNA.
   A student who thinks ribosomes carry out transcription picks this. The mRNA is made by RNA polymerase; the ribosomes on each strand are translating the mRNA.
2. BIt is a eukaryotic cell, because translation is taking place beside the DNA in its nucleus.
   A student who thinks translation happens in the nucleus picks this. In eukaryotic cells, mRNA is made in the nucleus and translated in the cytoplasm, so ribosomes are not found on mRNA still attached to the DNA.
3. CIt is a prokaryotic cell, because ribosomes are translating mRNA that is still being made. Correct
   Each mRNA is still attached to an RNA polymerase on the DNA, so it is still being transcribed, yet ribosomes are already translating it. This overlap occurs in prokaryotes, which have no nuclear envelope separating the DNA from the ribosomes.
4. DIt is a eukaryotic cell, because prokaryotic cells lack ribosomes to translate any mRNA.
   A student who thinks prokaryotes have no ribosomes picks this. Prokaryotes do have ribosomes, and translation of mRNA that is still being transcribed is a feature of prokaryotic cells.

CED 6.4.A.2 · Read this in Fix

1. AA single codon can code for several different amino acids.
   A student who misreads the redundancy of the code picks this. Each codon specifies one amino acid or a stop signal; the redundancy is that several codons can share an amino acid.
2. BMany amino acids are each specified by more than one codon. Correct
   Many amino acids are encoded by more than one codon (for example, several codons code for glycine), while each codon specifies one amino acid or a stop signal.
3. CEach species uses its own codons for the amino acids.
   A student who thinks each species has its own code picks this. Nearly all organisms use the same genetic code.
4. DThe start codon AUG is a start signal that codes for no amino acid.
   A student who thinks the start codon is only a signal picks this. AUG marks where translation begins and also codes for methionine; only stop codons code for no amino acid.

CED 6.4.A.3.iii · Read this in Fix

1. ANew viral proteins are still made, because ribosomes translate the viral RNA genome directly.
   A student who thinks a retroviral RNA genome is translated directly picks this. New viral proteins are made from the integrated viral DNA, which cannot form if reverse transcriptase is blocked.
2. BThe cells stop transcribing their own genes, because the drug blocks their transcription enzyme.
   A student who thinks reverse transcriptase is the cell's ordinary transcription enzyme picks this. Host genes are transcribed by the cell's own RNA polymerase, which the drug does not target.
3. CThe virus switches to copying its RNA directly into new RNA, since it needs to go on reproducing.
   A student who thinks organisms or viruses change because they need to picks this. A need does not give the virus a new way to copy its genome.
4. DViral RNA enters the cells, but no viral DNA is made, so no new viral progeny are assembled. Correct
   Reverse transcriptase copies the viral RNA genome into DNA, which integrates into the host genome and is transcribed and translated to make new viruses. Blocking it stops this flow of information at its first step.

CED 6.4.A.4 · Read this in Fix

1. APolypeptides are made by ribosomes, and both free and ER-bound ribosomes make them. Correct
   Free ribosomes (9,600 cpm) and rough ER membranes with ribosomes (8,900 cpm) built labeled amino acids into polypeptides. When the ribosomes were removed from the ER membranes, the value fell to 160 cpm, close to the 150 cpm background with no mRNA, so the ribosomes, free or bound, made the polypeptides.
2. BFree ribosomes translate mRNA faster than ER-bound ribosomes, as 9,600 is above 8,900.
   A student who treats any difference between two values as real picks this. There is one value for each mixture and no replicates or error bars, so a difference of 700 cpm cannot be judged against chance variation.
3. CThe ER membrane assembles the polypeptides, and its ribosomes deliver amino acids to it.
   A student who thinks the rough ER membrane itself makes proteins picks this. ER membranes without ribosomes gave only 160 cpm, about the same as background, while free ribosomes with no ER gave 9,600 cpm.
4. DAmino acids bind to the codons of the mRNA and are joined without ribosomes present.
   A student who thinks amino acids pair directly with codons picks this. The mixture with mRNA and amino acids but no ribosomes (ER membranes with ribosomes removed) gave only background radioactivity.

CED 6.4.A.1 · Read this in Fix

1. ABacterial ribosomes read the gene's DNA directly, so no mRNA needs to be made.
   A student who thinks bacteria translate genes straight from DNA picks this. Bacterial ribosomes translate mRNA; translation starts early only because the mRNA is read while it is still being made.
2. BBacterial ribosomes on the rough ER sit near the nucleus, so the mRNA travels only a short way.
   A student who thinks prokaryotes have the same organelles as eukaryotes picks this. Bacteria have no nucleus and no ER; their ribosomes are free in the cytoplasm.
3. CBacterial ribosomes make the mRNA themselves and then translate it straight away.
   A student who thinks ribosomes carry out transcription picks this. RNA polymerase makes the mRNA; ribosomes translate it, starting while it is still being made.
4. DBacterial ribosomes begin translating an mRNA before its transcription is complete. Correct
   Bacteria have no nucleus, so a ribosome can attach to the 5′ end of an mRNA while RNA polymerase is still making the rest of it. In eukaryotic cells, the mRNA is made in the nucleus and must leave it before translation.

CED 6.4.A.2 · Read this in Fix

1. APolypeptides go on being made, as amino acids can pair with the mRNA codons directly.
   A student who thinks amino acids pair directly with codons picks this. Amino acids are brought to codons by tRNAs on ribosomes, so if no new ribosomes can assemble at start codons, no new polypeptides are started.
2. BCells switch to another way of making polypeptides to obtain the proteins they need.
   A student who thinks cells gain a new process because they need it picks this. A cell has no other way to make polypeptides from mRNA; a need does not create one.
3. CPolypeptides already being made are completed, but no new ones are started. Correct
   Translation proceeds in sequential steps. Ribosomes already in elongation are unaffected, so they continue to the stop codon and release their polypeptides, but no new initiation can occur, so no new polypeptides are started.
4. DNo new mRNA is made, since ribosomes can no longer copy the DNA into mRNA.
   A student who thinks ribosomes carry out transcription picks this. RNA polymerase makes mRNA, so transcription is not blocked; the drug acts only on the start of translation.

CED 6.4.A.3 · Read this in Fix

1. AMet–Phe–Gly Correct
   Translation is initiated at the start codon AUG, which codes for methionine. Reading the following triplets 5′→3′ gives UUU (Phe) and GGC (Gly).
2. BPro–Cys–Leu
   A student who thinks translation starts at the first nucleotide at the 5′ end picks this, reading CCA UGU UUG. Translation begins at the start codon AUG, which here begins at the third nucleotide.
3. CAsp–Lys–Arg
   A student who reads the mRNA from its 3′ end picks this, reading GAU AAA CGG. Codons are read 5′→3′, starting at AUG.
4. DTyr–Lys–Pro
   A student who looks up the complementary sequences (the anticodons UAC, AAA, CCG) in the chart picks this. The chart gives the amino acid for each mRNA codon, so AUG, UUU and GGC are looked up.

Working Translation starts at the start codon AUG (nucleotides 3–5 from the 5′ end). Reading in triplets from there: AUG UUU GGC AAA UAG = Met–Phe–Gly–Lys, then stop. Distractors: from the first nucleotide, CCA UGU UUG = Pro–Cys–Leu; from the 3′ end, GAU AAA CGG = Asp–Lys–Arg; complements (anticodons) of AUG UUU GGC, UAC AAA CCG = Tyr–Lys–Pro.

CED 6.4.A.3.i · Read this in Fix

1. AThe AAG mRNA gave significantly more protein than no mRNA, since its mean value is higher.
   A student who thinks any difference between means is significant picks this. The AAG bar (6 to 14) overlaps the no-mRNA bar (2 to 10), so a difference between them has not been shown.
2. BChanging AUG to AAG significantly lowered protein output, since those two bars do not overlap. Correct
   The AUG bar (92 to 108) and the AAG bar (6 to 14) are far apart, so the drop in protein when the start codon is changed is likely to be significant. This supports the role of AUG in initiating translation.
3. CThe AAG mRNA and no mRNA gave exactly equal amounts of protein, as their bars overlap.
   A student who thinks overlapping error bars prove equal means picks this. Overlap means that a difference has not been shown, not that the amounts are equal; their means are 10 and 6.
4. DNo conclusion can be drawn, since error bars show the measurements were done wrongly.
   A student who reads error bars as a record of mistakes picks this. The bars show the uncertainty in each mean from variation among the six tubes and are what allow the treatments to be compared.

Working Intervals (mean ± 2 SE): AUG mRNA 100 ± 8, from 92 to 108; AAG mRNA 10 ± 4, from 6 to 14; no mRNA 6 ± 4, from 2 to 10. The AUG and AAG bars do not overlap, so the difference between them is likely to be statistically significant. The AAG and no-mRNA bars overlap, so a difference between them has not been shown; this does not prove the two means are equal.

CED 6.4.A.3.i · Read this in Fix

1. A903
   A student who thinks each nucleotide specifies one amino acid picks this. The mRNA is read in triplets, so 903 nucleotides make 301 codons.
2. B301
   A student who thinks the stop codon adds a final amino acid picks this. 903 ÷ 3 = 301 codons, but the stop codon codes for no amino acid.
3. C299
   A student who thinks the start codon is only a signal picks this, subtracting both the start and stop codons. AUG codes for methionine, the first amino acid of the polypeptide.
4. D300 Correct
   903 ÷ 3 = 301 codons. The start codon AUG codes for methionine, and the final stop codon codes for no amino acid, so the polypeptide has 301 − 1 = 300 amino acids.

Working Codons are triplets: 903 ÷ 3 = 301 codons. The last codon is the stop codon, which codes for no amino acid, so the polypeptide has 301 − 1 = 300 amino acids (the first is methionine, coded by AUG). Distractors: counting the stop codon as an amino acid gives 301; leaving out the start codon as well as the stop codon gives 299; one amino acid per nucleotide gives 903.

CED 6.4.A.3.ii · Read this in Fix

1. AMet–Gly–Ser–Asp
   A student who thinks a change affects only the codon where it occurs picks this. The extra G shifts the grouping of every later nucleotide, so UCA and GAU are no longer codons in the changed mRNA.
2. BMet–Gly–Ala–Ser
   A student who thinks each nucleotide codes for one amino acid picks this, adding one amino acid for the inserted G. Codons are triplets, so one extra nucleotide regroups all later codons.
3. CMet–Gly–Phe–Arg Correct
   Reading in triplets from AUG, the changed mRNA gives AUG GGC UUC AGA: methionine, glycine, phenylalanine, arginine. The insertion changes the grouping of every codon after it.
4. DTyr–Pro–Lys–Ser
   A student who looks up the complementary sequences (anticodons) in the chart picks this. The chart is used with the mRNA codons themselves: AUG, GGC, UUC, AGA.

Working Original: AUG GCU UCA GAU = Met–Ala–Ser–Asp. Changed: reading in triplets from AUG, AUG GGC UUC AGA = Met–Gly–Phe–Arg; every codon after the insertion is regrouped. Distractors: changing only the codon with the insertion gives Met–Gly–Ser–Asp; adding one amino acid for the extra nucleotide gives Met–Gly–Ala–Ser; looking up the complements of AUG GGC UUC AGA (UAC CCG AAG UCU) gives Tyr–Pro–Lys–Ser.

CED 6.4.A.3.ii · Read this in Fix

1. AA different amino acid replaces the glycine at that position.
   A student who thinks each amino acid has only one codon picks this. The chart shows that GGC, like GGU, codes for glycine.
2. BNo polypeptide is made, because the changed codon is unreadable.
   A student who thinks any change makes an mRNA unreadable picks this. GGC is an ordinary codon, and the ribosome reads it as glycine.
3. CIt keeps the same length but no longer works normally.
   A student who thinks every change in a sequence damages the protein picks this. The amino acid sequence is unchanged, so there is no basis for predicting a change in function.
4. DIts amino acid sequence is the same as before the change. Correct
   GGU and GGC both code for glycine, as the chart shows. Many amino acids are encoded by more than one codon, so this change does not alter the amino acid sequence.

CED 6.4.A.3.iii · Read this in Fix

1. ABoth organisms read these codons with the same genetic code, evidence that they share a common ancestor. Correct
   The bacteria translated the jellyfish codons into the same amino acids as the jellyfish does, showing that both use the same code. Nearly all organisms share this code, which is evidence for the common ancestry of all living organisms.
2. BThe result shows that bacteria evolved from jellyfish, so the bacteria inherited the jellyfish code.
   A student who thinks a shared feature means one living species descended from another picks this. Bacteria and jellyfish both descended from a common ancestor; neither is the ancestor of the other.
3. COrganisms that need to make similar proteins evolve the same genetic code independently of each other.
   A student who thinks shared features come from similar needs picks this. Organisms with very different proteins and ways of life share the code, which is best explained by inheritance from a common ancestor.
4. DThe bacteria changed their genetic code to read the jellyfish gene, since they needed the protein.
   A student who thinks organisms change because they need to picks this. The bacteria used their existing code, which already matched the jellyfish code.

CED 6.4.A.3.iv · Read this in Fix

1. AA second tube of the bacterial extract with the same mammalian mRNA added
   A student who thinks repeating the treatment provides a control picks this. A replicate shows whether the result is reproducible but cannot show whether the protein came from translating the added mRNA.
2. BA tube that contains neither the bacterial extract nor mammalian mRNA
   A student who thinks a control is a setup with nothing in it picks this. Leaving out both the extract and the mRNA changes two things at once and cannot show whether the extract alone gives a positive test.
3. CThe bacterial extract with the mammalian mRNA, incubated at a higher temperature
   A student who thinks a control changes some other condition picks this. Changing the temperature adds a second variable and does not show what happens without the mRNA.
4. DThe same bacterial extract, treated identically, with no mammalian mRNA added Correct
   This tube differs from the experimental tube only in the added mRNA. If it gives no mammalian protein, the protein in the experimental tube must have come from translating the added mRNA, not from something already in the extract.

CED 6.4.A.3.iv · Read this in Fix

1. AEach amino acid binds directly to its own codon on the mRNA, with no carrier needed.
   A student who thinks amino acids recognize codons directly picks this. An amino acid cannot pair with a codon; it is carried by a tRNA whose anticodon pairs with the codon.
2. BA tRNA whose anticodon pairs with the codon carries the amino acid the codon specifies. Correct
   Each tRNA carries a specific amino acid and has an anticodon complementary to a codon. Base pairing between codon and anticodon brings the correct amino acid to the place specified by the codon.
3. CA tRNA whose anticodon has the same sequence as the codon brings its amino acid there.
   A student who thinks the anticodon is identical to the codon picks this. The anticodon is complementary to the codon; for example, UUC pairs with AAG.
4. DThe ribosome converts the three nucleotides of the codon into the amino acid it codes for.
   A student who thinks mRNA nucleotides are turned into amino acids picks this. The amino acids come from the cell on tRNAs; the mRNA is not converted and can be translated again.

CED 6.4.A.3.v · Read this in Fix

1. AtRNA 2 is joined into the chain along with the Ala that it carries.
   A student who thinks tRNA becomes part of the polypeptide picks this. Only the amino acid is added to the chain; the tRNA is later released and can carry another amino acid.
2. BAla stays on tRNA 2 until tRNAs have filled every codon of the mRNA.
   A student who thinks amino acids are joined only after all of them are lined up picks this. Amino acids are added to the chain one at a time as each codon is read.
3. CThe Met–Phe chain on tRNA 1 becomes joined to the Ala on tRNA 2. Correct
   The model shows a growing chain (Met–Phe) on tRNA 1 and the next amino acid (Ala) on tRNA 2, whose anticodon is paired with the next codon. The chain is transferred to Ala, so the polypeptide grows by one amino acid.
4. DThe ribosome moves toward the 5′ end of the mRNA, back to the AUG codon.
   A student who thinks an mRNA can be read toward its 5′ end picks this. The ribosome moves 5′→3′, toward codon ACU, after the chain is transferred.

CED 6.4.A.3.vi · Read this in Fix

1. ATranslation stops at UAG, releasing a polypeptide of just two amino acids, Met–Pro. Correct
   The chart shows that UAG is a stop codon. Translation continues along the mRNA until a stop codon is reached, so it ends after Met and Pro and the short polypeptide is released.
2. BTranslation adds one more amino acid for UAG and then releases a chain of three.
   A student who thinks a stop codon codes for a final amino acid picks this. UAG codes for no amino acid; the chain is released with Met–Pro only.
3. CTranslation runs to the end of the mRNA, giving a polypeptide of full length.
   A student who thinks the whole mRNA is translated, end to end, picks this. Translation ends at the first stop codon in the reading frame, here codon 3.
4. DNo translation takes place, because the changed mRNA can no longer be read.
   A student who thinks any change makes an mRNA unreadable picks this. The start codon is unchanged, so translation begins normally and reads codons until it reaches UAG.

Working Original: AUG CCA UAC GGU AAA … = Met–Pro–Tyr–Gly–Lys …, 299 amino acids (300 codons minus the stop codon). Changed: codon 3 is UAG, a stop codon, so translation stops after Met–Pro and a polypeptide of 2 amino acids is released.

CED 6.4.A.3.vii · Read this in Fix

1. AA tRNA pairs with the stop codon and adds a last amino acid that seals the chain.
   A student who thinks a stop codon codes for an amino acid picks this. A stop codon codes for no amino acid; when it is reached, the finished chain is released.
2. BThe ribosome reaches the last nucleotide of the mRNA and lets go of the chain.
   A student who thinks the whole mRNA is translated picks this. Translation ends at a stop codon, which is usually some distance from the 3′ end of the mRNA.
3. CWhen a stop codon is reached, the completed polypeptide is released. Correct
   Translation continues codon by codon until a stop codon is reached. No amino acid is added for the stop codon, and the newly synthesized polypeptide is released from the ribosome.
4. DThe ribosome releases the amino acids, which then join to form the polypeptide.
   A student who thinks amino acids are joined only after translation picks this. The amino acids are joined one at a time on the ribosome, so a complete chain is released.

CED 6.4.A.3.viii · Read this in Fix

1. AIt rose most rapidly between 12 and 24 h and then leveled off near 92. Correct
   The largest increases, 27 and 33 units per 6 h, were between 12 and 24 h. After about 36 h the amount stayed near 92, so the curve leveled off.
2. BIt rose fastest at 48 h, when the amount of viral DNA was at its greatest.
   A student who reads the height of a curve as its slope picks this. Between 42 and 48 h the amount did not change; the steepest part of the curve is between 12 and 24 h.
3. CIt rose at a constant rate over the whole 48 h after the cells were infected.
   A student who assumes a rising quantity rises at a constant rate picks this. The increases per 6 h range from 0 to 33, so the rate was far from constant.
4. DIt started to fall after 30 h, when the curve became much less steep.
   A student who reads a less steep rise as a fall picks this. After 30 h the amount rose from 84 to 92 and then stayed level; it never fell.

Working No-drug cells, change per 6 h: 0–6 h, +2; 6–12 h, +6; 12–18 h, +27; 18–24 h, +33; 24–30 h, +16; 30–36 h, +6; 36–42 h, +2; 42–48 h, 0. The largest increases are between 12 and 24 h; after 30 h the amount still rises slightly and then stays at about 92.

CED 6.4.A.4 · Read this in Fix