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Practice questions · Topic 6.7

Unit 6 · Practice for the Topic 6.7 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: Mutations and genetic variation, summed up

Video coming soon

Substitution, insertion, deletion; a shifted frame; silent, missense, nonsense; tracing a change through the chain; amount against kind; beneficial, detrimental or neutral in a place; where mutations come from; extra chromosomes and moved pieces; genes that move sideways; variation for selection to act on.

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 change from the base to the animal one link 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
Two monospace rows grouped in threes: 5′-ATG TCC GAT CTG-3′ above 5′-ATG TCC AAA GAT CTG-3′5′-ATG TCC GAT CTG-3′5′-ATG TCC AAA GAT CTG-3′
The original coding strand of the addax’s gene (above) and the changed copy (below), grouped in threes.

A strand is written from its 5′ end to its 3′ end, and both ends are marked. The DNA shown is the coding strand, so its letters match the mRNA’s, with T where the mRNA has U. The original is drawn above the changed copy, both grouped in threes from the start codon. Suppose an addax’s gene for a protein of its nose lining changes. Addaxes are desert antelopes. The original coding strand and the changed copy are drawn below.

Which of the following describes the change?

Question 2
Two monospace rows above the chart: 5′-AUG GCA UUC AAG GAU UGA-3′ grouped in threes, and 5′-AUGGCAUCAAGGAUUGA-3′ written as one unbroken string of letters; below them the code chart5′-AUG GCA UUC AAG GAU UGA-3′5′-AUGGCAUCAAGGAUUGA-3′second basefirst basethird baseUCAGUCAGUUUPheUUCPheUUALeuUUGLeuUCUSerUCCSerUCASerUCGSerUAUTyrUACTyrUAAstopUAGstopUGUCysUUGCCysCUGAstopAUGGTrpGCUULeuCUCLeuCUALeuCUGLeuCCUProCCCProCCAProCCGProCAUHisCACHisCAAGlnCAGGlnCGUArgUCGCArgCCGAArgACGGArgGAUUIleAUCIleAUAIleAUGMet (start)ACUThrACCThrACAThrACGThrAAUAsnAACAsnAAALysAAGLysAGUSerUAGCSerCAGAArgAAGGArgGGUUValGUCValGUAValGUGValGCUAlaGCCAlaGCAAlaGCGAlaGAUAspGACAspGAAGluGAGGluGGUGlyUGGCGlyCGGAGlyAGGGGlyG
The mysid’s mRNA (above) and the same letters with one base removed (below), with the code chart.

A strand is written from its 5′ end to its 3′ end, and both ends are marked. Read from the first AUG in threes and stop at the first stop codon; the stop codon adds no amino acid. An mRNA for a protein of a mysid’s antenna reads 5′-AUG GCA UUC AAG GAU UGA-3′. One base is removed from it, and the remaining letters are written as one unbroken string beneath the original, above the chart.

Which amino acids does the ribosome now join, in order?

Question 3
Two monospace rows above the chart: 5′-CCG-3′ above 5′-CCA-3′; below them the code chart with CCG and CCA ringed5′-CCG-3′5′-CCA-3′second basefirst basethird baseUCAGUCAGUUUPheUUCPheUUALeuUUGLeuUCUSerUCCSerUCASerUCGSerUAUTyrUACTyrUAAstopUAGstopUGUCysUUGCCysCUGAstopAUGGTrpGCUULeuCUCLeuCUALeuCUGLeuCCUProCCCProCCAProCCGProCAUHisCACHisCAAGlnCAGGlnCGUArgUCGCArgCCGAArgACGGArgGAUUIleAUCIleAUAIleAUGMet (start)ACUThrACCThrACAThrACGThrAAUAsnAACAsnAAALysAAGLysAGUSerUAGCSerCAGAArgAAGGArgGGUUValGUCValGUAValGUGValGCUAlaGCCAlaGCAAlaGCGAlaGAUAspGACAspGAAGluGAGGluGGUGlyUGGCGlyCGGAGlyAGGGGlyG
The original codon of the salp’s gene (above) and the changed codon (below), with the code chart.

The DNA shown is the coding strand, so its letters match the mRNA’s, with T where the mRNA has U. The original codon is drawn above the changed codon. Suppose one codon on the coding strand of a salp’s gene for a protein of its outer layer has changed from 5′-CCG-3′ to 5′-CCA-3′, drawn above the chart.

Which kind of mutation is this?

Question 4
A table with three columns, the copy of the gene, mRNA per cell in units and SE in units, and two rows: the usual gene 44 and 3; the changed copy 46 and 3copy of the genemRNA per cell (units)SE (units)the usual gene443the changed copy463means in arbitrary units; SE is the standard error
The mRNA per cell for the usual gene and for the changed copy, each with its SE.

Suppose a biologist compares the gene for a wool protein in two vicunas. The second vicuna’s copy carries one changed base. Both copies give an mRNA 2,160 nucleotides long. Both proteins have 720 amino acids, and one of them differs between the two. The table gives the mean mRNA per cell for each copy, with its standard error (SE).

Where did the changed base land?

Question 5

Suppose one codon of a barley plant’s gene for a root enzyme changes from CTT to CCT, so each enzyme molecule carries Pro where it carried Leu. The root cells hold the usual amount of the enzyme’s mRNA and the usual amount of the enzyme. Each enzyme molecule makes its product at 19 % of the usual rate.

Which outcome is this for each molecule of the enzyme?

Question 6
A table with three columns, the soil, plants with the usual gene and plants with the changed gene, and two rows: ordinary soil 20 and 13; soil beside an old zinc mine 2 and 19the soilusual genechanged geneordinary soil2013soil beside an old zinc mine219mean mass of one plant after a season, in grams
Mean mass of one plant after a season, on ordinary soil and on the mine soil, for the two kinds of grass.

Suppose a substitution in a grass’s gene changes one amino acid of a protein that pumps zinc ions out of the root cells, and the changed pump works faster. Biologists grow grasses with the usual gene and grasses with the changed gene for a season on ordinary soil and on soil beside an old zinc mine, where zinc is plentiful. The table gives the mean mass of one plant at the end of the season.

Which of the following does the table support?

Question 7

A leaf of a plant carries one pale patch. A biologist finds that every cell in the patch holds three copies of one chromosome and two of every other kind, while every cell outside the patch, on this leaf and on the rest of the plant, holds two copies of every kind.

Which event accounts for the pale patch?

Question 8
Four rows of lettered boxes. Before, row 1: R S T U V W; row 2: E F G H. After, row 1: R S T U W V; row 2: E F G HRSTUVWEFGHRSTUWVEFGHbeforeafter
The two chromosomes before the change (above) and after it (below).

In the drawing, each box is one segment of a chromosome, and each segment carries many genes. The chromosome before the change is drawn above, and after it beneath. A second chromosome, where drawn, sits beneath the first with its own letters. A change in the structure of the chromosomes in a cell is drawn below.

Which kind of change is this?

Question 9

In a jar of pickling brine, a bacterium of one species grows a thin tube to a cell of a second species, and a copy of a plasmid passes along the tube into the second cell. Both cells swim away carrying the plasmid.

Which route is this?

Question 10
A table with three columns, the repair enzymes, ultraviolet light and colonies with the substitution of 60, and four rows: working, none, 1; working, ultraviolet light, 4; blocked, none, 2; blocked, ultraviolet light, 35repair enzymesultraviolet lightcolonies with the substitution (of 60)workingnone1workingfor one minute4blockednone2blockedfor one minute35sixty colonies grown from each line, each colony read for the substitution
Colonies with a substitution in the marker gene, of 60 grown from each line.

A biologist grows four lines of one bacterium. In two lines the repair enzymes work as usual; in the other two a drug blocks them. One line of each pair is lit with ultraviolet light for one minute, which links neighboring bases on a strand in about 80 places per cell in both lit lines. She then grows 60 colonies from each line and reads one marker gene in each colony for a substitution. The table gives the counts.

Which of the following explains the difference between the two lines that were lit?

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 · 5 points
The DNA shown is the coding strand, so its letters match the mRNA’s, with T where the mRNA has U. The original codon is drawn above the changed codon. Suppose one codon on the coding strand of a bongo’s gene for an eyelash protein changes from 5′-AAA-3′ to 5′-ATA-3′, drawn above the chart. Bongos are antelopes of dense woodland. In each eyelash, molecules of this protein grip one another to make the eyelash stiff. The changed amino acid sits where two molecules grip, and the changed molecules do not grip one another at all. Bongos with two copies of the changed gene have eyelashes that bend and break, so dust and flies reach their eyes.
Two monospace rows above the chart: 5′-AAA-3′ above 5′-ATA-3′; below them the code chart5′-AAA-3′5′-ATA-3′second basefirst basethird baseUCAGUCAGUUUPheUUCPheUUALeuUUGLeuUCUSerUCCSerUCASerUCGSerUAUTyrUACTyrUAAstopUAGstopUGUCysUUGCCysCUGAstopAUGGTrpGCUULeuCUCLeuCUALeuCUGLeuCCUProCCCProCCAProCCGProCAUHisCACHisCAAGlnCAGGlnCGUArgUCGCArgCCGAArgACGGArgGAUUIleAUCIleAUAIleAUGMet (start)ACUThrACCThrACAThrACGThrAAUAsnAACAsnAAALysAAGLysAGUSerUAGCSerCAGAArgAAGGArgGGUUValGUCValGUAValGUGValGCUAlaGCCAlaGCAAlaGCGAlaGAUAspGACAspGAAGluGAGGluGGUGlyUGGCGlyCGGAGlyAGGGGlyG
The original codon of the bongo’s gene (above) and the changed codon (below), with the code chart.

(a) Identify the codon on the mRNA after the change. (1 point)

Hint: The DNA shown is the coding strand: copy its letters, with U in place of T.

A full-credit answer: On the mRNA, the changed codon reads AUA.

Check the box for each point your answer earns

(b) Identify the amino acid the changed codon names, using the chart. (1 point)

Hint: Find the codon’s first base down the side of the chart, its second base across the top, and its third base at the right.

A full-credit answer: The changed codon names Ile, in place of the Lys that AAA named.

Check the box for each point your answer earns

(c) Describe the effect of the change on the amino-acid sequence of the eyelash protein. (1 point)

Hint: One base changed to another: how many bases are there now, and how many codons changed?

A full-credit answer: The eyelash protein has one amino acid different, Ile in place of Lys, and every other amino acid is the same.
The chain keeps its length, because the count of bases is the same and no codon regroups.

Check the box for each point your answer earns

(d) Explain how the change in one amino acid leads to dust and flies reaching the bongo’s eyes. (1 point)

Hint: Follow the links in order: the protein’s shape, its job, the eyelash, the animal.

A full-credit answer: The changed amino acid sits where two molecules grip, so the changed molecules do not grip one another.
Molecules that do not grip make an eyelash that is not stiff.
An eyelash that is not stiff bends and breaks.
So the bongo’s eyelashes no longer keep dust and flies away from its eyes.

Check the box for each point your answer earns

(e) Identify the outcome of the change for each molecule of the eyelash protein, and state the fact about the changed molecules that decides it. (1 point)

Hint: Does each changed molecule do none of its job, some of it, all of it, or its job when it should be off? Then name what the changed molecules do, or fail to do, that told you.

A full-credit answer: For each molecule of the eyelash protein, the outcome is loss of function.
A changed molecule does not grip its neighbors at all, so it does none of its job.

Check the box for each point your answer earns

Free-response score: 0 of 5
Free response 2 · Analyze Data · 4 points
A technician spreads a bacterium over a plate of plain jelly, and seventy-five colonies grow. She presses a velvet stamp onto the colonies, then onto three plates whose jelly holds an extract of garlic that kills most bacteria, and last onto one more plate of plain jelly. In every plate drawing, a dot is one colony, the large plate at the top is the first plate, and the small plates beneath it are the plates the stamp was pressed onto, in a row. The plates are drawn below.
A large round plate at the top scattered with seventy-five small colonies; an arrow down to the line the velvet stamp is pressed onto; a row of four smaller plates: three each carrying two dots at the same two positions, and a fourth carrying seventy-five dotsthe velvet stamp is pressed onto:the first plate: plain jellygarlic extractgarlic extractgarlic extractplain jelly
The first plate, the three garlic plates and the plain plate the stamp was pressed onto.

(a) Identify the number of positions on each garlic plate where a colony grew. (1 point)

A full-credit answer: Two positions on each garlic plate, and the same two positions on all three.

Check the box for each point your answer earns

(b) Describe what the plate of plain jelly shows about the velvet stamp. (1 point)

A full-credit answer: A colony grew at every one of the seventy-five positions on the plain plate.
So the stamp carried living cells from every colony onto every plate it touched.

Check the box for each point your answer earns

(c) The technician claims that the cells of the two resistant colonies were resistant before the garlic extract reached any cell. Support the claim using the drawing. (1 point)

A full-credit answer: The stamp puts each colony’s cells at the same position on every plate.
The resistant colonies sit at the same two positions on all three garlic plates, so they grew from the same two colonies on the first plate.
Those two colonies grew on plain jelly, with no garlic extract near them.
So their cells were resistant before the extract reached any of them.

Check the box for each point your answer earns

Common slip: Saying the garlic extract made two cells resistant on each plate. Cells changed on each plate separately would sit at different positions on each plate.

(d) Predict what the technician would see on a fourth garlic plate stamped from the same first plate. (1 point)

A full-credit answer: Colonies at the same two positions as on the other three garlic plates, and nowhere else.

Check the box for each point your answer earns

Free-response score: 0 of 4
Multiple choice checked: 0 of 10 correct.
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