Practice questions · Topic 7.2

Unit 7 · Practice for the Topic 7.2 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: Natural selection, summed up

Video coming soon

Selection sees the phenotype, so a hidden recessive allele survives; the selective pressure is the factor that sorts; adaptive here, harmful there; a chemical selects resistance and does not create it; molecules inside cells are phenotype too; explain the data in the exam’s order; directional, stabilizing and disruptive selection.

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 data set one part 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. Where a question shows means with error bars, read the legend first: every error bar here represents ±2SE.

Question 1

A mountain stream floods every spring. Its insect larvae cling to stones, and larvae vary in how strongly they grip. The floods wash away the larvae that grip weakly. Grip strength is heritable.

Which of the following is the selective pressure on grip strength?

A. The larvae’s grip strength B. The stones the larvae cling to C. The spring floods D. The number of eggs each female lays

Question 2

In a herd of wild horses a recessive allele w gives weak leg bones, so most ww foals die before they are old enough to breed. WW and Ww horses have strong legs and breed normally. A student says: “The w allele will be gone from the herd within a few generations.”

Which statement about the student’s claim is correct?

A. The student is right: every horse that carries the w allele dies before it breeds B. The student is wrong: Ww horses have strong legs, breed normally, and pass w on C. The student is right: a recessive allele is always lost faster than a dominant allele D. The student is wrong: a recessive allele can never be removed from a population

Question 3

Wildfires now sweep a shrubland every five years instead of every twenty. The shrubs vary in how thick their bark is, and bark thickness is heritable. Thick bark protects a shrub’s stem from a fire.

Predict the shrub population’s average bark thickness a few generations later.

A. Greater than before the fires became frequent B. The same as before the fires became frequent C. Smaller than before the fires became frequent D. Greater for one generation, then back to the original thickness

Question 4

Survival of large-leaved and small-leaved seedlings in a normal-rainfall year and a drought year.

A meadow plant’s seedlings have either small leaves or large leaves, and leaf size is heritable. Small leaves lose less water; large leaves make more sugar. Researchers tracked the survival of the two types through a year with normal rainfall and through a year with severe drought. The results are in the table.

Which of the following is supported by the data?

A. Small leaves raise fitness in both a drought year and a normal-rainfall year B. Large leaves raise fitness in both a drought year and a normal-rainfall year C. Rainfall has no effect on which leaf size survives more in either year D. Small leaves raise fitness in a drought year and lower it in a normal-rainfall year

Question 5

A vineyard sprays a fungicide every year. After five years, most of the fungus on the vines survives the spray. A student thinks the resistant fungus was already present before the first spray.

Which finding would show that the resistance allele was present before the fungicide was first used?

A. Resistant fungus becomes more common on the vines after each year’s spray of the fungicide B. Fungus samples stored from before the first spray include a few that carry the resistance allele C. Resistant fungus grows only on the vines that have been sprayed with the fungicide D. The fungicide kills 98% of the fungus on the vines in the first year of spraying

Question 6

Copper begins leaking from a mine into the soil of two meadows, and one grass species grows in both. In the north meadow every plant’s root cells make the same small amount of a protein that binds copper. In the south meadow the amount varies from plant to plant, and a plant’s seedlings make the same amount as it does.

Which meadow’s grass is more likely to survive the copper, and why?

A. The north meadow’s, because all of its plants share one level of protection against the copper B. The south meadow’s, because its plants can make more of the protein once the copper reaches their roots C. The south meadow’s, because some plants already make enough of the protein and their seedlings inherit the amount D. The north meadow’s, because identical plants pass their alleles on more reliably than varied plants do

Question 7

Number of plants and mean leaf hair density in three groups on the hillside.

Goats begin grazing a hillside and eat the plants with the fewest leaf hairs. Researchers counted the leaf hairs of the plants before the goats came, of the plants alive two years later, and of the next generation, which they counted before any goat had grazed it. The results are in the table.

Which of the following explains why the next generation’s mean is close to the survivors’ mean rather than to the mean before the goats came?

A. The goats also grazed the next generation and left only hairy plants B. The next generation grew where the goats had trampled the soil C. The surviving plants grew more hairs after being grazed and passed the new hairs on D. Leaf hair density is inherited, so the survivors’ offspring resemble the survivors

Question 8

The field’s grasshoppers by hind-leg length: dashed, before; solid, many generations later.

The grasshoppers of a field vary in hind-leg length. The curves show the population by hind-leg length 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 hind-leg length

Question 9

A meadow plant’s petals vary in length. Researchers follow petal length in the population over many generations.

Which of the following observations shows disruptive selection?

A. The whole population shifted toward longer petals B. Middle-length petals became rare, and short and long petals common C. The range of petal lengths narrowed around the middle length D. Petal lengths did not change

Question 10

Share of the pond’s frogs in each jump-length class in generation 1 and generation 5.

A pond’s frogs vary in how far they can jump, and jump length is heritable. A heron that hunts frogs arrives at the pond between generation 1 and generation 2. The table shows the share of frogs in each jump-length class in generation 1 and in generation 5.

Which of the following explains the change by generation 5?

A. Each frog lengthened its own jump during its life to escape the heron and passed the longer jump on B. Short-jumping frogs mutated into long-jumping frogs in order to survive the heron’s hunting C. The heron ate the short-jumping frogs, which made the allele for long jumps dominant in the pond D. Long-jumping frogs escaped the heron more often and reproduced more, passing long jumps to their offspring

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 Data · 5 points

A desert shrub lives for many years and sets seed every year. Its seeds vary in the thickness of their coats. A seed-eating rodent arrives in the desert. It cracks thin-coated seeds and eats them, and it cannot crack thick-coated seeds. Researchers measured the coat thickness of the seeds set before the rodents arrived, of the seeds still lying uneaten on the ground two years later, and of the seeds set by the new shrubs that grew from those uneaten seeds. The table gives each group’s mean and its ±2SE.

Mean seed-coat thickness of three groups of seeds, with ±2SE for each mean.

(a) Identify the variation that was present in the shrubs’ seeds before the rodents arrived. (1 point)

Hint: Which measured feature differs from seed to seed in the stimulus?

A full-credit answer: Before the rodents arrived, the seeds varied in the thickness of their coats.

Check the box for each point your answer earns

variation in seed-coat thickness among the seeds.

Common slip: Naming the rodents. The rodents are the pressure; the variation is a feature of the seeds.

(b) Identify the selective pressure on seed-coat thickness. (1 point)

Hint: Which factor in the environment decides which seeds survive?

A full-credit answer: The selective pressure is the seed-eating rodent, because it eats the thin-coated seeds and leaves the thick-coated seeds.

Check the box for each point your answer earns

the rodent (its seed eating) as the selective pressure.

Common slip: Naming the seed coat. The coat is the trait being sorted; the rodent does the sorting.

(c) Explain why the seeds still uneaten two years later have a greater mean coat thickness than the seeds set before the rodents arrived. (1 point)

Hint: Think about which seeds the rodents could crack and which they could not.

A full-credit answer: The rodents ate the thin-coated seeds, so the seeds left uneaten are mostly thick-coated.
So the mean coat thickness of the uneaten seeds is greater than the mean before the rodents arrived.

Check the box for each point your answer earns

the rodents ate the thin-coated seeds, so the uneaten seeds are mostly thick-coated, which raises the mean.

Common slip: Saying the seeds grew thicker coats to protect themselves. No seed changed; the rodents removed the thin-coated seeds.

(d) Describe the evidence in the table that seed-coat thickness is heritable. (1 point)

Hint: Compare the new shrubs’ seeds with the uneaten seeds they grew from and with the seeds before the rodents.

A full-credit answer: The new shrubs’ seeds have a mean of 0.50 mm, close to the 0.51 mm of the uneaten seeds they grew from and well above the 0.42 mm before the rodents, which shows that the new shrubs inherited thick seed coats from the seeds that survived.

Check the box for each point your answer earns

the new shrubs’ seeds (0.50 mm) match the uneaten seeds they grew from (0.51 mm) and exceed the before mean (0.42 mm), so the trait was inherited.

Common slip: Pointing to the uneaten seeds alone. The uneaten seeds show which seeds survived; the new shrubs’ seeds show what was inherited.

(e) Determine whether the ±2SE error bars of the seeds set before any rodent arrived and of the new shrubs’ seeds overlap, and state what that means for the difference between the two means. (1 point)

Hint: Add and subtract each ±2SE from its mean, then compare the top of the before range with the bottom of the next generation’s range.

A full-credit answer: The before error bar runs from 0.39 mm to 0.45 mm and the new shrubs’ from 0.47 mm to 0.53 mm, so the two error bars do not overlap.
The difference between the two means is unlikely to be chance.

Check the box for each point your answer earns

the ranges 0.39–0.45 mm and 0.47–0.53 mm do not overlap, so the difference is unlikely to be chance.

Common slip: Comparing the two means alone. The error bars are what show whether a difference of 0.08 mm could be chance.

Free-response score: 0 of 5

Free response 2 · Conceptual Analysis · 4 points

A drug treats one kind of tumor. The drug binds to a receptor protein on the surface of the tumor cells. The more of the drug a cell’s receptors bind, the more likely the cell is to die. Before treatment, doctors count the receptor molecules on single cells taken from the tumor: the count ranges from about 1,000 to more than 50,000 receptors per cell. When a tumor cell divides, its daughter cells make about the same number of receptors as it did.

(a) Describe the variation present in the tumor before treatment. (1 point)

A full-credit answer: The tumor cells differ in the number of receptor molecules each cell makes, from about 1,000 to more than 50,000 per cell.
The number is passed to a cell’s daughter cells.

Check the box for each point your answer earns

the cells vary in the number of receptors per cell (about 1,000 to more than 50,000). Accept with or without the inheritance of the number.

Common slip: Describing the drug. The variation is a feature of the cells, the receptor count.

(b) Explain why a cell with about 1,000 receptors is more likely to survive the treatment than a cell with about 50,000 receptors. (1 point)

A full-credit answer: The drug kills a cell by binding to its receptors.
A cell with about 1,000 receptors binds far less of the drug than a cell with about 50,000.
So the drug is less likely to trigger the death of the low-receptor cell, and it is more likely to survive.

Check the box for each point your answer earns

fewer receptors bind less drug, so the drug is less likely to kill the cell.

Common slip: Saying the low-receptor cell resists the drug because it is a mutant. The stimulus gives the count, not a mutation; the reason is how much drug the cell binds.

(c) Predict how the mean number of receptors per cell in the tumor after several rounds of treatment compares with the mean before treatment. (1 point)

A full-credit answer: After several rounds of treatment, the mean number of receptors per cell is lower than before treatment.

Check the box for each point your answer earns

the mean receptor count per cell falls.

Common slip: Predicting that the mean rises because the surviving cells adapt by making more receptors. More receptors would bind more drug; the survivors are the cells with few.

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

A full-credit answer: Cells with few receptors were already in the tumor before treatment.
Each round of the drug kills mostly the cells with many receptors, and the cells with few receptors survive.
The surviving cells divide, and their daughter cells make the same low number of receptors.
So the share of low-receptor cells rises, and the mean falls.

Check the box for each point your answer earns

the drug kills high-receptor cells, low-receptor cells survive and divide, and their daughters inherit the low count, so the mean falls.

Common slip: Saying the drug caused the cells to make fewer receptors. The low-receptor cells were counted before any treatment; the drug selected them, it did not create them.

Free-response score: 0 of 4

Multiple choice checked: 0 of 10 correct.

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