Unit 6 test

Unit 6 · Unit 6 end-of-unit test (also the test-out)

Suggested time: about 80 minutes. Answer everything, then press Submit the test to see the feedback and scoring guides.

Answer every question, then press Submit the test. Feedback and the scoring guides appear after you submit. For the three 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. After you submit, mark your own free-response work against each scoring guide. Suggested time: 80 minutes. Every count, level, length and sequence in this test is imagined for the question unless the question says otherwise.

Question 1

One strand of a short double-stranded DNA molecule is 25 bases long. It holds 7 adenine, 3 thymine, 9 guanine and 6 cytosine bases.

Which of the following is the percentage of thymine in the whole double-stranded molecule?

A. 12 % B. 20 % C. 28 % D. 40 %

Question 2

The base composition of the virus’s genetic material.

A laboratory measures the base composition of the genetic material of a virus that infects pomegranate trees. The table shows the result.

Which of the following is the virus’s genetic material?

A. RNA B. Double-stranded DNA C. Single-stranded DNA D. A mixture of DNA and RNA

Question 3

Suppose that while copying a gene, DNA polymerase pairs an adenine opposite a cytosine by mistake, and repair enzymes never correct the mismatch. The two molecules made in that copying then each go through one more round of copying, with no further mistakes.

Which of the following describes the four molecules at that site after that round?

A. Two carry the original cytosine–guanine pair and two carry the cytosine–adenine mismatch B. Two carry the original cytosine–guanine pair and two carry a thymine–adenine pair C. All four carry the original cytosine–guanine pair D. Three carry the original cytosine–guanine pair and one carries a thymine–adenine pair

Question 4

Suppose one DNA molecule whose two strands are both old is copied for 5 rounds, so that there are 32 molecules, 64 strands in all, at the end.

How many of the 64 strands are old strands?

A. 0 strands B. 1 strand C. 2 strands D. 32 strands

Question 5

At one replication fork, the lagging strand is built in four pieces.

How many RNA primers have been laid down at this fork in all, on both new strands?

A. 2 primers B. 4 primers C. 5 primers D. 8 primers

Question 6

What is seen in each strain when its DNA is copied.

Four strains of one bacterium each lack one enzyme of the DNA-copying machinery, a different enzyme in each strain. The table shows what is seen when each strain’s cells copy their DNA.

In which strain is the enzyme that lays down the RNA primers missing?

A. Strain J B. Strain Q C. Strain X D. Strain Z

Question 7

The parts of the gene, in order, with their lengths.

A eukaryotic gene of a pukeko, a bird, has four exons and three introns. The table gives the length of each part in order. In one kind of the bird’s cells, the mature mRNA from this gene has a coding length of 1,530 nucleotides.

Which exon do this cell’s splicing enzymes cut out along with the introns?

A. Exon 1 B. Exon 2 C. Exon 3 D. Exon 4

Question 8

Suppose a drug in a eukaryotic cell blocks the enzyme that adds the poly-A tail to a pre-mRNA. The cell goes on transcribing a gene while the drug is present, and each pre-mRNA is still capped at its 5′ end.

Which of the following happens to the mRNAs made from the gene while the drug is present?

A. The mRNAs lack a stop codon B. The mRNAs’ introns are not cut out C. The mRNAs give a longer polypeptide than normal D. The mRNAs are broken down sooner than normal mRNAs

Question 9

Suppose a technician cuts the promoter out of one of a bacterium’s genes and joins it back into the chromosome at a new place, just past the end of that gene, pointing the same way along the DNA as before.

Which of the following does RNA polymerase transcribe after it binds the moved promoter?

A. The DNA that follows the moved promoter, away from the gene B. The gene, as before C. The gene, from its end back toward its start D. Nothing: no RNA is made from the promoter at its new place

Question 10

One gene in a bone cell and a heart cell of the trevally.

A biologist compares one gene in a bone cell and a heart cell of a trevally, a fish. The table gives what she finds.

Which of the following explains why the two cells’ mature mRNAs differ in length?

A. The heart cell’s copy of the gene carries a deletion in one of its exons B. RNA polymerase in the heart cell stops transcribing the gene sooner C. The heart cell adds a shorter poly-A tail to the mRNA D. The two cells keep different sets of exons when they splice the pre-mRNA

Question 11

On an mRNA, the coding sequence runs from the first base of the start codon to the last base of the stop codon. Suppose that stretch is 480 nucleotides long.

How many amino acids does the ribosome join when it translates this mRNA?

A. 158 amino acids B. 159 amino acids C. 160 amino acids D. 161 amino acids

Question 12

A retrovirus infects a cell.

Which of the following shows the flow of information from the virus’s genetic material to the viral proteins the cell builds?

A. RNA → DNA → RNA → protein B. DNA → RNA → protein C. RNA → protein D. RNA → DNA → protein

Question 13

The template strand, above the genetic code chart.

A gene’s template strand reads 3′-TAC TTG CGC ATC-5′, as drawn above the chart. RNA polymerase copies it, and a ribosome translates the mRNA.

Which amino acids does the ribosome join, in order?

A. Met, Thr, Ala B. Tyr, Leu, Arg, Ile C. Asp, Ala, Gln, Val D. Met, Asn, Ala

Question 14

The anticodon, above the genetic code chart.

A tRNA’s anticodon reads 3′-GUA-5′, as drawn above the chart.

Which amino acid does this tRNA carry?

A. His B. Val C. Tyr D. Thr

Question 15

The four operons, their jobs and the level of each molecule in the cell now.

Suppose a bacterium has four operons. The table gives each operon’s job and the level of the molecule it deals with in the cell right now.

Which operon’s genes is the cell transcribing right now?

A. The raffinose operon B. The threonine operon C. The asparagine operon D. The lactulose operon

Question 16

In a cell of a kokako, a bird, one gene is silent. The DNA at the gene is tightly wound, its promoter carries many methyl groups, and the histones it is wound on carry few acetyl groups.

Which of the following changes would make the cell transcribe the gene more often?

A. Adding more methyl groups to the promoter B. Adding acetyl groups to the histones at the gene C. Removing the acetyl groups that remain on the histones D. Changing one base in the gene’s coding sequence

Question 17

One gene’s mRNA and protein in two kinds of monkfish cell.

A biologist measures one gene’s mRNA and protein in two kinds of cell from a monkfish. The table gives the results.

Which of the following is being regulated in the liver cells to give the low protein level?

A. How often the gene is transcribed into mRNA B. Whether the cells carry the gene C. How fast the gene’s DNA is copied D. How much of the mRNA the ribosomes translate

Question 18

Suppose a change in an E. coli cell’s DNA alters its lac repressor so that the inducer still binds the repressor, but the repressor’s shape no longer changes when it does. The cell grows in a broth of lactose only.

Which of the following describes the three lac genes in this cell?

A. The genes are transcribed rarely B. The genes are transcribed often C. The genes are transcribed often, but only until the lactose is used up D. The genes are transcribed more and more often as more inducer binds

Question 19

The mRNA of the gene the protein binds beside, in a normal cell and in the cell with the extra copies.

A biologist gives a cell of a shipworm, a wood-boring clam, extra copies of the gene for one of its regulatory proteins, so the cell makes far more of that protein. The protein binds a short stretch of DNA beside one gene. The table gives that gene’s mRNA in a normal cell and in the cell with the extra copies.

Which of the following is the regulatory protein?

A. A repressor B. A protein that speeds the breakdown of the gene’s mRNA C. An activator D. A protein that blocks ribosomes from the gene’s mRNA

Question 20

A biologist supplies an siRNA that pairs with the mRNA of one gene to cells of a starfruit plant. The treated cells hold far less of that gene’s protein than untreated cells do. She wants to show that the fall comes from the siRNA pairing with that gene’s mRNA, rather than from the cells being given any siRNA, whatever its bases.

Which of the following is the control that shows this?

A. Cells given no siRNA at all, but otherwise handled in exactly the same way B. Cells given an siRNA of the same length whose sequence matches no mRNA in the cell C. Cells given twice as much of the siRNA that pairs with that gene’s mRNA D. Cells in which that gene has been deleted from the DNA before any siRNA is supplied

Question 21

A bellbird’s preen-gland cells build an oil-making enzyme; its bone cells build none of it. A student claims that the two kinds of cell carry the same genes and differ only in which genes they express.

Which of the following observations supports the student’s claim?

A. The enzyme’s gene is found in the DNA of both kinds of cell, and its mRNA in the gland cells only B. The gland cells hold more mRNA in total than the bone cells C. The two kinds of cell differ in shape and size D. The enzyme’s gene is found in the DNA of the gland cells and not in the DNA of the bone cells

Question 22

Suppose a deletion removes 5 bases from the coding sequence of a tamarind tree’s gene for a pod enzyme, starting in the seventh codon of about 240 codons.

Which of the following describes the enzyme built from the changed gene?

A. The enzyme is two amino acids shorter than the original, with the rest unchanged B. Only the enzyme’s seventh amino acid differs from the original’s C. The enzyme is the same as the original D. The enzyme’s amino acids from the seventh onward differ from the original’s

Question 23

The original codon of the fenugreek gene (above) and the changed codon (below), with the code chart.

Suppose one codon on the coding strand of a fenugreek plant’s gene for a leaf enzyme changes from 5′-TCA-3′ to 5′-TGA-3′, as drawn above the chart. The coding strand’s letters match the mRNA’s, with T where the mRNA has U. The codon is the 25th of about 300.

Which of the following names the change?

A. A silent mutation B. A missense mutation C. A nonsense mutation D. A frameshift mutation

Question 24

Suppose a plant’s cell copies 4,200,000,000 base pairs each time it copies its DNA, and about 7 uncorrected errors remain after each copying.

About how many base pairs are copied for each uncorrected error?

A. 6,000,000 base pairs B. 60,000,000 base pairs C. 600,000,000 base pairs D. 6,000,000,000 base pairs

Question 25

The three tests and their results.

Two species of bacterium live in the bark of a kauri tree. Cells of species 1 carry a gene for an enzyme that breaks down a toxin in the bark; cells of species 2 do not. A biologist tests the two species in three ways, shown in the table.

Which route carried the gene into species 2?

A. Transformation B. Transduction C. Conjugation D. Transposition

Question 26

A biologist grows a bacterium from a few cells in each of twenty small flasks, and grows one large flask of it too. She adds a bacteriophage to each small flask and to twenty large-flask samples, each holding about as many cells as a small flask, spreads each on a plate and counts the colonies of cells that resist the bacteriophage. Every large-flask plate gives about the same count. The small-flask plates give counts from zero to several hundred.

Which of the following does the difference between the two sets of counts show about the mutations that let a cell resist the bacteriophage?

A. The mutations arose by chance at different times as the small flasks grew, before the bacteriophage was added B. The bacteriophage caused the mutations when it was added to each flask C. The mutations arose in every flask at the moment the bacteriophage arrived, at one fixed rate for every cell D. Cells in a small flask mutate more often than cells in a large flask, so small flasks give more mutants

Question 27

In a silvereye, a small bird, an inversion has reversed a stretch of one chromosome. One of the two breaks fell between one gene and the enhancer that sat near it. Every gene on the stretch keeps its normal order of bases. Yet the bird’s cells make far too little of that gene’s protein.

Which of the following explains the low amount of that protein?

A. The gene’s codons are now read in a different reading frame from the start codon B. The cells now hold only one copy of the gene instead of the usual two C. The gene’s introns are no longer cut out of its pre-mRNA in the reversed stretch D. The gene now lies far from the regulatory sequence that raised its transcription

Question 28

The PCR product from the healthy plant and from the mutant, beside the ladder.

In the drawing below, the small boxes along the top edge are the wells, and each dark bar is a band. The first lane is the ladder. A laboratory copies one stretch of a gene by PCR from a healthy nasturtium plant and from a mutant nasturtium, with the same two primers, and loads the two products beside the ladder.

Which of the following changes in the mutant’s gene fits the gel?

A. A deletion of a few hundred base pairs between the primers B. An insertion of a few hundred base pairs between the primers C. A substitution of one base between the primers D. A deletion of one base between the primers

Question 29

Suppose a tube holds 3 copies of a target at the start, with both primers, free nucleotides and the heat-stable DNA polymerase, and the machine heats and cools the tube through cycles of PCR.

After how many cycles is the number of copies of the target in the tube first above 1,000?

A. 8 cycles B. 9 cycles C. 10 cycles D. 333 cycles

Question 30

Suppose a technician seals a jackfruit tree’s gene for a fruit enzyme into a plasmid that also carries a gene for resistance to an antibiotic, mixes the plasmid with treated bacteria, and spreads them on jelly with the antibiotic. Fourteen colonies grow. A student says the colonies prove that the bacteria build the fruit enzyme.

Which of the following results would support the student’s claim?

A. PCR on DNA from the colonies gives a band at the size of the fruit-enzyme gene B. The fruit enzyme is detected in cells taken from the colonies C. The colonies grow again when spread on fresh jelly with the antibiotic D. Untreated bacteria spread on jelly with no antibiotic grow into a lawn

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 · Interpreting and Evaluating Experimental Results · 9 points

Suppose a single-celled alga takes up molybdenum from lake water through a transport protein coded by one gene, the uptake gene. Researchers propose the model in Figure 1. A repressor, protein R, can bind a regulatory sequence beside the uptake gene. A kinase, protein M, adds a phosphate group to R only when little molybdenum is in the cell; molybdenum bound to M stops it working. R with a phosphate group on it does not bind the DNA. From a wild-type strain the researchers made two mutant strains, one with no working M and one with no working R. They grew each strain with plenty of molybdenum and with molybdenum scarce, and measured molybdenum uptake and the relative amount of the uptake gene’s mRNA (Table 1). The numbers are imagined.

Figure 1. The proposed model of the uptake gene in the two conditions. Table 1. Molybdenum uptake and the uptake gene’s relative mRNA in the three strains, in the two conditions.

(a)(i) Describe the effect that adding a phosphate group can have on protein R that stops R binding the regulatory sequence. (1 point)

A full-credit answer: The added phosphate group changes the shape of R.
R binds the regulatory sequence only when its binding site fits the DNA.
So the changed shape no longer fits, and R cannot bind.

Check the box for each point your answer earns

the phosphate group changes the shape (structure) of R, so its binding site no longer fits the DNA.

Accept: ‘changes R’s shape’ alone.

Common slip: Saying the phosphate group blocks the DNA. The phosphate group acts on R, not on the DNA.

(a)(ii) Explain how protein R lowers transcription of the uptake gene when R is bound to the regulatory sequence. (1 point)

A full-credit answer: Bound R brings in proteins that wind the DNA at the uptake gene tightly.
RNA polymerase and the transcription factors cannot reach a tightly wound promoter.
So the uptake gene is rarely transcribed.

Check the box for each point your answer earns

Accept one of the following: bound R brings in proteins that wind the DNA tightly (keeps the DNA tightly wound / condensed, as Figure 1 draws), so RNA polymerase (and the transcription factors) cannot reach the promoter; bound R sits in RNA polymerase’s path (blocks it), so the gene is not transcribed.

Scoring note: ‘R switches the gene off’ with no mechanism does not earn the point.

(b)(i) Identify a dependent variable in the researchers’ experiment. (1 point)

A full-credit answer: The molybdenum uptake of the cells.

Check the box for each point your answer earns

Accept one of the following: molybdenum uptake; the relative amount of the uptake gene’s mRNA.

Common slip: Naming the amount of molybdenum in the water. That is what the researchers set: an independent variable.

(b)(ii) Justify the researchers’ making both mutant strains from the wild-type strain. (1 point)

A full-credit answer: Each mutant then differs from the wild type only in the one protein knocked out.
So any difference in uptake or mRNA is due to that missing protein, not to other genetic differences between strains.

Check the box for each point your answer earns

the strains are genetically identical except for the introduced mutation, so a difference in the results is due to that mutation (and not to other genetic differences).

Common slip: Writing ‘the wild type is the control’ and stopping. The point needs what making the mutants from it ensures: no other genetic differences.

(b)(iii) Justify the researchers’ using strains in which only one protein of the model is missing in each. (1 point)

A full-credit answer: Knocking out one protein at a time tests each protein’s effect on its own.
So the results show which protein is responsible for a change in uptake or mRNA.

Check the box for each point your answer earns

the effect of each protein can be tested separately, so the researchers can tell which protein is responsible for an observed change.

(c)(i) Based on Table 1, determine which strain takes up molybdenum quickly whether molybdenum is plentiful or scarce, and state what in the table your decision rests on. (1 point)

A full-credit answer: The r-mutant, the strain with no working repressor R.
Its uptake is 21.0 with plenty of molybdenum and 21.5 with molybdenum scarce, both near the wild type’s uptake with molybdenum scarce.

Check the box for each point your answer earns

the r-mutant (the strain with no working R), because its uptake is large in both conditions (21.0 and 21.5).

Common slip: Naming the wild type. Its molybdenum uptake is large only when molybdenum is scarce.

(c)(ii) Calculate the percent change in molybdenum uptake in wild-type cells with plenty of molybdenum compared with wild-type cells with molybdenum scarce, giving a fall as a negative value. (1 point)

Write down the values in the question:

uptake with plenty of molybdenum (new) = 1.1

uptake with molybdenum scarce (old) = 27.5

Write down the equation:

percent change = \frac{new - old}{old} \times 100

Substitute the values into the equation:

percent change = \frac{1.1 - 27.5}{27.5} \times 100 = −96 %

A full-credit answer: Uptake is 1.1 with plenty of molybdenum and 27.5 with molybdenum scarce.
(1.1 − 27.5) ÷ 27.5 × 100 = −96 %.

(d)(i) In a follow-up experiment, the researchers make a strain whose protein R has lost the site where kinase M adds the phosphate group, while protein M works normally. Based on Figure 1, predict how often the uptake gene is transcribed in this strain when molybdenum is scarce. (1 point)

A full-credit answer: The uptake gene is rarely transcribed, as in wild-type cells with plenty of molybdenum.

Check the box for each point your answer earns

the uptake gene is rarely transcribed (transcription stays low / off), as with plenty of molybdenum.

Common slip: Predicting frequent transcription because M is active. M acts only by adding a phosphate group to R, and this R has no site for one.

(d)(ii) Justify your prediction in part (d)(i). (1 point)

A full-credit answer: With molybdenum scarce, M is active, but this R has no site for the phosphate group, so none is added.
R keeps the shape that fits the regulatory sequence, so R stays bound beside the uptake gene.
Bound R keeps the DNA wound tight, so RNA polymerase cannot reach the promoter and the gene stays rarely transcribed.

Check the box for each point your answer earns

R without a phosphate group keeps its DNA-binding shape and stays bound beside the uptake gene (even though M is active), so transcription stays low.

Scoring note: ‘M cannot switch the gene on’ with no mention of R staying bound does not earn the point.

Free-response score: 0 of 9

Free response 2 · Scientific Investigation · 4 points

Suppose growers raise watercress in six beds of flowing water. Each bed is fed by its own channel from a spring, and the water leaves each bed by its own outflow. A bacterium rots watercress roots, and the roots in three of the beds are rotting (Table 1). Scientists claim that the bacterium is present in the water leaving the infected beds and absent from the water leaving the healthy beds. They take a water sample from each bed’s outflow, extract the DNA from it, add two primers that pair with a stretch of DNA found only in this bacterium, free nucleotides and heat-stable DNA polymerase, heat and cool each tube through 30 cycles of PCR, and load each product on a gel beside a ladder. They also include a tube holding the bacterium’s own DNA in place of a water sample.

Table 1. The state of the watercress roots in each bed.

(a) Describe how the two primers make PCR copy only the bacterium’s stretch of DNA. (1 point)

A full-credit answer: A primer pairs only with a stretch of DNA whose bases match its own.
DNA polymerase starts a new strand only from a paired primer.
So only the stretch between the two primers is copied, and DNA without the matching stretch is not.

Check the box for each point your answer earns

each primer pairs only where its bases match, and DNA polymerase extends only from a paired primer, so only the stretch between the primers (the bacterium’s stretch) is copied.

Accept: ‘DNA that does not match the primers is not copied’ only when both the base pairing and DNA polymerase’s need for a paired primer are stated.

Scoring note: restating the stimulus (‘the primers pair with a stretch found only in this bacterium’) with no mechanism does not earn the point.

Common slip: Saying the primers ‘find’ the bacterium. The mechanism is base pairing: a primer pairs only where its bases match.

(b) Justify the scientists’ including the tube with the bacterium’s own DNA. (1 point)

A full-credit answer: That tube holds the target, so it must give a band if the reaction works.
Its band shows the primers, nucleotides and polymerase worked in this batch of tubes.
So a water lane with no band is a true absence of the target, not a failed reaction.

Check the box for each point your answer earns

the tube shows the reaction works when the target is present, so a lane with no band means the target was absent, not that the reaction failed.

Scoring note: ‘it is a positive control’ with no purpose stated does not earn the point.

Common slip: Writing ‘it is the control’ and stopping. The point needs what the tube shows: the reaction works.

(c) If the scientists’ claim is supported, predict which beds’ outflow samples give a band. (1 point)

A full-credit answer: The samples from beds 2, 4 and 5, and no others.

Check the box for each point your answer earns

beds 2, 4 and 5 (the beds with rotting roots) only.

Common slip: Predicting a band in every bed. The claim puts the bacterium only in water leaving infected beds.

(d) Justify your prediction in part (c). (1 point)

A full-credit answer: A band appears only when the bacterium’s stretch of DNA was in the sample, because only then do the primers pair and the polymerase copy.
The claim puts the bacterium in the water leaving the infected beds, so those samples carry the stretch and give a band.
The healthy beds’ water lacks the bacterium, so those samples give no band.

Check the box for each point your answer earns

a band forms only when the target stretch is in the sample; under the claim the bacterium (and its DNA) is in the water leaving infected beds and not the healthy ones, so only those lanes show a band.

Scoring note: a justification consistent with a wrong prediction in part (c) earns the point if the reasoning is sound.

Free-response score: 0 of 4

Free response 3 · Analyze Model or Visual Representation · 4 points

The model shows a ribosome part-way through translating an mRNA. The mRNA is written from its 5′ end to its 3′ end, and boxes mark the codons the ribosome reads. One tRNA stands on the second codon, carrying the growing chain; a second tRNA is arriving at the third codon.

The ribosome, the mRNA with its codons boxed, the tRNA on the second codon and the tRNA arriving at the third.

(a) Describe a characteristic of a tRNA that lets it bring the right amino acid to a codon. (1 point)

A full-credit answer: A tRNA’s anticodon pairs, base by base, only with the codon whose bases match it.
An enzyme loads each kind of tRNA with the one amino acid its codon means.
So the tRNA that pairs with a codon carries that codon’s amino acid.

Check the box for each point your answer earns

Accept one of the following: its anticodon pairs by base pairing only with the matching codon; it is loaded with (carries) only the one amino acid that its codon means.

(b) Based on the model, explain why the ribosome began translating at the third base of the mRNA rather than at its first base. (1 point)

A full-credit answer: The small subunit binds the mRNA near its 5′ end and moves along it to the first AUG.
AUG is the start codon, and here it is bases 3 to 5.
The Met tRNA pairs with that AUG, so the reading frame starts there and the first two bases are not read.

Check the box for each point your answer earns

the ribosome (small subunit) moves from the 5′ end to the first AUG, the start codon, which begins at base 3, and reads in threes from there.

Scoring note: ‘it starts at AUG’ with no link to the first AUG being at base 3 does not earn the point.

Common slip: Saying the ribosome reads from the very first base. The small subunit moves along to the first AUG before reading begins.

(c) Using the code chart, identify the amino acid that the tRNA arriving at the third codon carries. (1 point)

The genetic code chart.

A full-credit answer: The third codon is 5′-GCU-3′.
GCU reads Ala on the chart, so the arriving tRNA carries alanine (Ala).

Check the box for each point your answer earns

alanine (Ala).

Common slip: Reading the second codon, GAC, instead. The arriving tRNA stands over the third codon.

(d) Based on the model, explain how a substitution that changes the fourth codon from 5′-UGG-3′ to 5′-UGA-3′ would affect the polypeptide the ribosome releases. (1 point)

The genetic code chart.

A full-credit answer: UGA is a stop codon, so no tRNA pairs with it.
The ribosome releases the chain when it reaches the fourth codon, after three amino acids: Met, Asp and Ala.
The polypeptide is shorter than the original, which had five amino acids, so it is unlikely to fold into a working protein.

Check the box for each point your answer earns

UGA is a stop codon (no tRNA pairs with it), so the chain is released after the third amino acid and the polypeptide is shorter (three amino acids instead of five) and likely nonfunctional.

Scoring note: ‘the protein changes’ with no link to the stop codon does not earn the point.

Common slip: Saying tryptophan is replaced by another amino acid. UGA names no amino acid; it ends the message.

Free-response score: 0 of 4

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