End-of-topic test: Natural Selection

Unit 7 · Topic 7.2 end-of-topic test

Suggested time: about 44 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 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. Then open the scoring guide and mark your own work against it. Where a question shows means with error bars, read the legend first: every error bar here represents ±2SE.

Question 1

Seeds set per plant by each stem type in the two fields.

The plants of one meadow species have two heritable stem types, spiny and smooth. Researchers grew equal numbers of each type in a field with many deer and in a field with few deer, and counted the seeds each plant set. The results are in the table.

Which of the following is the selective pressure on stem type in the field with many deer, and why?

A. The spines, because they keep the deer away from the plant B. The deer, because where deer are many the smooth plants set far fewer seeds C. The field’s soil, because every plant grows in it D. Seed production, because it decides how many offspring each plant leaves

Question 2

Two harmful alleles appear in one population of lizards, at two different genes, and each is carried by the same small share of the lizards. At the first gene, allele D is dominant over d: every lizard with one copy of D has a weak grip and is taken by predators more often. At the second gene, allele r is recessive to R: only rr lizards have the weak grip, and Rr lizards grip normally.

Which allele does natural selection remove from the population faster, and why?

A. Allele r, because recessive alleles are rarer than dominant alleles B. Both alleles at the same rate, because each gives the same weak grip C. Allele r, because each rr lizard carries two copies of it D. Allele D, because every lizard that carries it shows the weak grip

Question 3

Share of the gene’s copies that were the DDT-resistance allele in laboratory strains of Drosophila melanogaster founded from wild flies caught in two decades.

DDT is an insecticide widely used in the United States from the 1940s until 1972. The table shows the share of all copies of one gene that were the allele for DDT resistance, in laboratory populations (strains) of the fruit fly Drosophila melanogaster founded from wild flies caught in the 1930s and in the 1960s.

Which of the following explains the change between the 1930s strains and the 1960s strains?

A. Spraying with DDT caused the mutation for resistance to appear in the flies that were sprayed B. Flies got used to DDT during their own lives and passed that habit on to their offspring C. Resistant flies were rare before spraying; DDT killed most susceptible flies, and the survivors passed the allele on D. DDT killed most of the flies, and the survivors’ offspring then mutated so that they could resist it

Question 4

Copies of the pump protein per cell in the population before the antibiotic.

Researchers grow a population of bacteria in a liquid with no antibiotic. They count the copies of a pump protein in single cells; the pump pushes an antibiotic out of the cell. A cell’s daughter cells make about the same number of copies as it does. The table shows the counts. The researchers then move the whole population into a liquid with a high concentration of that antibiotic.

Which of the following is the most likely outcome?

A. The cells with 0 to 50 pumps quickly make more pumps, so every cell survives equally B. The antibiotic changes the pump gene, so most cells come to make more than 150 pumps C. The whole population dies, because the average number of pumps is too low D. The cells with 151 to 200 pumps survive and divide, and their daughters inherit the count

Question 5

Some plants in a meadow population make a bitter chemical in their leaves, and the others make none. The trait is heritable. Making the chemical uses sugar that the plant could otherwise put into seeds. Slugs avoid bitter leaves.

In which of the following meadows is making the chemical an adaptation?

A. A meadow where slugs eat the leaves only in autumn, after the plants have set seed B. Every meadow the plants grow in C. A meadow where slugs eat the seedlings’ leaves in spring D. No meadow at all

Question 6

The grass plants of a grassland vary in height, and height is heritable. Cattle begin grazing the grassland. They bite off the tallest plants before those plants set seed and leave the shortest plants alone.

Predict the grass population’s average height a few generations later.

A. A smaller average height than before the cattle arrived B. The same average height as before the cattle arrived C. A greater average height than before the cattle arrived D. A smaller average height for one generation, then the original height again

Question 7

Where malaria is common, a person with one sickle cell allele is less likely to die of malaria than a person with none. Where malaria is absent, natural selection acts against the sickle cell allele.

Which of the following explains why the allele is selected against where malaria is absent?

A. A carrier’s hemoglobin carries oxygen less well, so carriers leave fewer children than others do B. The allele mutates back to the common allele wherever malaria is absent C. Everyone who carries one copy of the allele has sickle cell disease D. A child who inherits two copies has sickle cell disease, and one copy gives no benefit there

Question 8

Mean jaw length of the wild population, the two groups of parents, and their offspring.

Researchers study jaw length in a population of beetles. They let only the beetles with the longest jaws breed, and they let a second group of beetles chosen at random breed. The table shows the mean jaw length of the wild population, of each group of parents and of each group’s offspring.

Which of the following explains why natural selection on jaw length would leave this population’s mean jaw length unchanged?

A. Long jaws are a recessive trait, so they were hidden in the offspring and will reappear later B. The difference in jaw length is not inherited, so offspring of long-jawed parents have ordinary jaws C. Long jaws last only while the selection for them continues in each generation D. The random parents also had long jaws, so the two offspring groups came out the same

Question 9

The woodland’s birds by clutch size: dashed, before; solid, many generations later.

The birds of a woodland vary in clutch size, the number of eggs a female lays in one nest. The curves show the population by clutch size before and many generations later.

Which type of selection do the curves show?

A. Directional selection B. Stabilizing selection C. Disruptive selection D. No selection has acted on clutch size

Question 10

How much of a protein a cell makes is hidden from any predator or mate.

Which of the following best explains why natural selection can still act on how much of a protein a cell makes?

A. The amount is set by the cell’s surroundings, so it changes whenever the environment changes B. Selection acts directly on the base sequence of the gene, and the amount follows from the sequence C. The amount changes what the organism can survive, so it is part of the phenotype selection acts on D. The amount is inherited, and every inherited feature is selected for

Question 11

Summer water temperature below the dam and the share of fish making the cold-water enzyme, 1990 to 2020.

A dam was built on a river in 1995, and since then the water below the dam has been colder each summer. A fish species below the dam makes one of two versions of a digestive enzyme: the common version works best in warm water, and a rarer version works best in cold water. Which version a fish makes is inherited. The table shows the summer water temperature and the share of the fish making the cold-water version.

Which of the following explains the rise in the share of fish making the cold-water version?

A. Each fish switched to making the cold-water version as the water around it cooled each summer B. The colder water caused new mutations for the cold-water version to appear in each generation C. The cold-water version is dominant over the warm-water version, so it spread as the fish bred D. Fish with the cold-water version survived and bred more in the cold water, and their young inherited it

Question 12

The water insects of a stream vary in body size. A new fish arrives that eats the largest insects. After many generations, the population’s average body size is smaller and the spread of sizes is unchanged.

Which type of selection acted on body size?

A. Directional selection toward small body size B. Directional selection toward large body size C. Stabilizing selection D. Disruptive selection

Question 13

Share of each color of grasshopper alive after one month in the two enclosures.

Researchers release equal numbers of green and brown grasshoppers into two enclosures. One enclosure is a green lawn; the other is a dry, brown field. Lizards hunt by sight in both. The table shows the share of each color still alive after one month.

Which of the following explains the difference in survival between the two enclosures?

A. Brown grasshoppers survived more in both enclosures, so a brown body is an adaptation in every environment B. The lizards hunted by smell rather than by sight, so body color had no effect on which grasshoppers survived C. Grasshoppers matching their background survived more, so a green body is an adaptation on the lawn only D. The grasshoppers changed color after their release, so matching colors rose among the survivors in each enclosure

Question 14

A pond dries up earlier each summer. Its tadpoles vary in how quickly they develop into frogs. A student explains the data: “Tadpoles varied in development time. The pond dried before the slow developers grew legs, so the fast developers survived and reproduced more.”

Which step must the student add before concluding that the population evolved?

A. Development time is heritable, so the next generation has more fast developers B. The fast developers chose to leave the pond before it dried C. Each tadpole sped up its own development as the pond shrank D. The pond dried up because the summers grew hotter

Question 15

Share of the worms surviving a standard dose of the drug, by year; the dose was doubled in year 4.

A farmer gives a flock of sheep a drug each year to kill the worms that live in the sheep’s gut. The table shows the share of the worms that survive a standard dose. In year 4 the farmer doubled the dose.

Which of the following explains why the share of surviving worms rose faster after the dose was doubled?

A. The higher dose caused more mutations for resistance to appear in the worms that survived it B. The higher dose killed more of the susceptible worms, so resistant worms made up a larger share of the parents C. The worms got used to the drug faster at the higher dose and passed that tolerance to their young D. The higher dose killed the resistant worms as well, so only susceptible worms were left to breed

Question 16

Share of each genotype of cricket alive after a month on the dark soil.

Researchers release equal numbers of crickets of three genotypes onto dark soil where birds hunt by sight. The allele D gives a dark body and is dominant over d, so DD and Dd crickets are dark and dd crickets are pale. The table shows the share of each genotype still alive after a month.

Which of the following explains why the DD and Dd crickets survived at the same rate?

A. The birds act on body color, and DD and Dd crickets have the same dark body B. A Dd cricket is a little paler than a DD cricket, but not enough for the birds to notice C. Natural selection keeps dominant alleles because they are stronger than recessive alleles D. A Dd cricket has lost its d allele, so it carries the same alleles as a DD cricket

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

A population of mantises lives among the brown stems and dead leaves of a woodland edge. Each mantis is brown or green. One gene sets the color: the allele B for a brown body is dominant over the allele b for a green body, so BB and Bb mantises are brown and bb mantises are green. Birds hunting by sight are the mantises’ main predators. Researchers recorded the share of each color every five generations. The results are in the table.

Share of brown and green mantises in the population, recorded every five generations.

(a) Describe the change in the share of green mantises from generation 1 to generation 15. (1 point)

A full-credit answer: The share of green mantises falls from 49% in generation 1 to 3% in generation 15.
The fall is steep at first and slows later.

Check the box for each point your answer earns

the share of green mantises falls, from 49% to 3%. Accept with or without the slowing of the fall.

Common slip: Describing only the brown share. The task asks about the green mantises; a direction and the two values earn the point.

(b) Explain why the share of green mantises is still above 0% in generation 15, although the birds take green mantises more often than brown ones. (1 point)

A full-credit answer: The birds see bodies, not alleles.
A brown Bb mantis carries one b allele and looks the same as a BB mantis, so the birds take Bb mantises no more often than BB mantises.
The b alleles inside brown Bb mantises are passed on.
Some offspring of two Bb mantises are bb and green, so green mantises keep appearing.

Check the box for each point your answer earns

the b allele survives inside brown Bb mantises, which look the same as BB mantises and are taken no more often, so bb (green) offspring keep being born. Accept: the b allele is carried by brown (heterozygous, Bb) mantises that the birds cannot tell from BB mantises, so it is passed on.

Common slip: Saying the green mantises hide better as they become rare. The stimulus gives no such change; the reason is the hidden b allele in brown carriers.

(c) In generation 10, one brown mantis carries a new allele of a gene for a digestive enzyme. The new allele changes one base of the gene but leaves the enzyme’s amino acid sequence exactly as it was. Predict the effect of the birds’ hunting on the share of this new allele in later generations. (1 point)

A full-credit answer: The birds’ hunting has no effect on the share of the new allele.

Check the box for each point your answer earns

no effect (the birds’ hunting neither raises nor lowers the new allele’s share).

Common slip: Predicting that the allele spreads because its carrier is brown. Brown mantises with and without the new allele are taken equally; the allele rides with no advantage.

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

A full-credit answer: The birds act on what they can see, the mantis’s body.
The new allele changes nothing about the mantis that the birds can see, and nothing else the mantis does.
So a mantis with the new allele is taken no more and no less often than a mantis without it.
Natural selection reaches an allele only through the phenotype, and this allele changes no phenotype.

Check the box for each point your answer earns

natural selection acts on the phenotype, and the new allele changes no phenotype, so the birds take mantises with and without it equally.

Common slip: Saying every mutation is either good or bad for survival. A change that alters nothing about the organism gives selection nothing to act on.

Free-response score: 0 of 4

Free response 2 · Analyze Data · 4 points

A new fish species arrives in a lake. It swallows the lake’s water snails whole, and it can swallow only snails below a certain size. Researchers measured the shell length of adult snails before the fish arrived, of the adults alive three years later, and of the next generation, raised from the survivors’ eggs in a pen in the lake that the fish could not enter, with the same food and temperature as their parents. The means are drawn below; each error bar represents ±2SE.

Mean shell length of three groups of snails; each error bar represents ±2SE.

(a) Describe the difference between the mean shell length of the snails before the fish arrived and the mean of the adults alive three years later. (1 point)

A full-credit answer: The adults alive three years later have a longer mean shell, 11.0 mm, than the snails before the fish arrived, 9.0 mm.
The mean rose by 2.0 mm.

Check the box for each point your answer earns

the later adults’ mean shell length is greater (11.0 mm against 9.0 mm). Accept with or without the difference of 2.0 mm.

Common slip: Writing that the shells grew. The two bars are two groups of snails; the later group has a larger mean.

(b) Describe the difference between the mean shell length of the adults alive three years later and the mean of the next generation. (1 point)

A full-credit answer: The next generation’s mean shell length, 10.5 mm, is about 0.5 mm lower than the later adults’ mean, 11.0 mm.
The two error bars overlap, so the two means cannot be told apart: the difference could be chance.

Check the box for each point your answer earns

EITHER reading with its evidence: the next generation’s mean is slightly lower (10.5 mm against 11.0 mm), OR the two means cannot be told apart, because their ±2SE error bars overlap.

Common slip: Reading the 0.5 mm gap as a real difference. Two means whose ±2SE error bars overlap could differ by chance.

(c) A student claims that the difference between the mean before the fish arrived and the next generation’s mean could be chance. Evaluate the student’s claim, using the error bars. (1 point)

A full-credit answer: The error bar of the before group runs from 8.5 mm to 9.5 mm.
The error bar of the next generation runs from 10.0 mm to 11.0 mm.
The two error bars do not overlap, so the difference between the two means is unlikely to be chance.
The student’s claim is not supported by the data.

Check the box for each point your answer earns

the judgement (the claim is not supported) AND the ground (the two ±2SE error bars do not overlap, so the difference is unlikely to be chance).

Common slip: Arguing only that 10.5 is larger than 9.0. Two means can differ by chance; the non-overlapping error bars are what make chance unlikely.

(d) Explain how the data demonstrate natural selection. (1 point)

A full-credit answer: Before the fish arrived, the snails varied in shell length.
The new fish swallows only small snails, so it is the selective pressure.
Snails with longer shells survived more, and the adults alive three years later have a larger mean shell length.
The next generation, raised where the fish could not reach it, also has a larger mean, so the change was inherited.
A heritable phenotype whose share rose because one phenotype survived more is natural selection.

Check the box for each point your answer earns

longer-shelled snails survived the fish (the survivors’ mean is higher) AND the next generation’s mean is also higher, so the change was inherited. Accept an answer that names the fish as the selective pressure in place of either half.

Common slip: Stopping at the survivors. The next generation’s bar is what shows the change was inherited; without it, the data show survival, not evolution.

Free-response score: 0 of 4

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Multiple choice checked: 0 of 16 correct.

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