Unit 7 · 13–20% of the exam Beta

Unit 7 practice test: Natural Selection

12 exam-style questions from the unit, free, with an explanation for every option. Sets share one table, graph or experiment, as on the real exam.

Covers: Introduction to Natural Selection; Natural Selection; Artificial Selection; Population Genetics; Hardy-Weinberg Equilibrium; Evidence of Evolution; Common Ancestry; Continuing Evolution; Phylogeny; Speciation; Variations in Populations; Origin of Life on Earth.

Data table

Twelve bacterial populations followed for 40,000 generations

From a single ancestral E. coli cell, researchers started 12 populations in flasks of medium with a small amount of glucose at 37 °C. Each day they moved 1% of every culture into fresh medium, about 6.6 generations a day. Every 500 generations they froze a sample of each population; frozen cells can be revived later. To measure fitness, revived cells from one time point were grown in the same flask as revived ancestral cells (marked so the two can be told apart). Relative fitness is the evolved cells' growth divided by the ancestor's growth in that flask; 1.00 means no difference.

Mean relative fitness of the 12 populations (± 2 SE)
GenerationMean relative fitness± 2 SE
01.000.00
2,0001.320.04
5,0001.440.05
10,0001.500.05
20,0001.600.06
40,0001.680.07

1. Which statement best describes the change in mean relative fitness?

  1. It rose quickly at first, then kept rising more slowly through generation 40,000
  2. It rose at a steady rate, gaining the same amount in each block of 1,000 generations
  3. It rose until generation 10,000 and stayed exactly the same after that
  4. It rose slowly at first and then faster and faster as mutations piled up
Show the answer

Fitness gained 0.32 in the first 2,000 generations but only 0.08 in the last 20,000, and it was still rising at the end.

  • Correct: It rose quickly at first, then kept rising more slowly through generation 40,000: Correct: fast early gains, then slower but continuing gains.
  • It rose at a steady rate, gaining the same amount in each block of 1,000 generations: The gain per 1,000 generations fell from 0.16 early to 0.004 late, so the rate was not steady.
  • It rose until generation 10,000 and stayed exactly the same after that: The mean kept rising after generation 10,000, from 1.50 to 1.60 to 1.68.
  • It rose slowly at first and then faster and faster as mutations piled up: The pattern is the reverse: the fastest gains came first.

2. Why does freezing samples every 500 generations make the fitness measurements more reliable?

  1. Ancestral and evolved cells can be revived and grown together in one flask under the same conditions
  2. Freezing stops new mutations, so the evolved populations have the same genes as the ancestor when tested
  3. Freezing kills the least fit cells, so the cells measured are mostly the ones with beneficial mutations
  4. Frozen samples let researchers skip the daily transfers and still reach 40,000 generations sooner
Show the answer

Because the ancestor is kept alive in the freezer, any later sample can compete against it head to head at the same time, in the same medium and temperature, so differences in growth reflect genetic change rather than different conditions.

  • Correct: Ancestral and evolved cells can be revived and grown together in one flask under the same conditions: Correct: a living ancestor allows a direct, side-by-side comparison.
  • Freezing stops new mutations, so the evolved populations have the same genes as the ancestor when tested: The evolved populations carry new mutations that arose before freezing; freezing does not undo them.
  • Freezing kills the least fit cells, so the cells measured are mostly the ones with beneficial mutations: Freezing is used to store whole samples, not to pick out fitter cells.
  • Frozen samples let researchers skip the daily transfers and still reach 40,000 generations sooner: Generations happen only while cells grow; frozen cells do not reproduce.

3. Using the ± 2 SE intervals, which conclusion is supported?

  1. Fitness at 40,000 generations differs from fitness at 5,000, but 10,000 and 20,000 are not shown to differ
  2. Each mean in the table differs from the mean before it, because each one is higher than the last
  3. The means at generations 20,000 and 40,000 differ from each other, because their two intervals do not overlap
  4. The means are not shown to differ, because each interval overlaps with the one after it
Show the answer

Intervals: 5,000 gives 1.39-1.49 and 40,000 gives 1.61-1.75, which do not overlap. 10,000 gives 1.45-1.55 and 20,000 gives 1.54-1.66, which overlap, so those two means are not shown to differ.

  • Correct: Fitness at 40,000 generations differs from fitness at 5,000, but 10,000 and 20,000 are not shown to differ: Correct: non-overlapping intervals at 5,000 and 40,000; overlapping ones at 10,000 and 20,000.
  • Each mean in the table differs from the mean before it, because each one is higher than the last: A higher mean is not enough; 10,000 (1.45-1.55) and 20,000 (1.54-1.66) overlap.
  • The means at generations 20,000 and 40,000 differ from each other, because their two intervals do not overlap: 20,000 (1.54-1.66) and 40,000 (1.61-1.75) do overlap.
  • The means are not shown to differ, because each interval overlaps with the one after it: 0 and 2,000 (1.00 and 1.28-1.36) do not overlap, and neither do 5,000 and 40,000.

4. All 12 populations gained fitness, though each gained it through mostly different mutations. What does this best suggest?

  1. The shared conditions selected for faster growth, while which mutations arose differed by chance
  2. The ancestor already carried the beneficial mutations, so the 12 flasks needed no new mutations
  3. Genetic drift by itself raised fitness, since drift tends to move separate populations in one direction
  4. The populations exchanged genes during the daily transfers, so the 12 ended with matching mutations
Show the answer

Twelve independent populations all improving in the same conditions points to natural selection by those conditions. Which beneficial mutations arose first was a matter of chance, so the genetic routes differed.

  • Correct: The shared conditions selected for faster growth, while which mutations arose differed by chance: Correct: the same pressure, different chance mutations, the same outcome.
  • The ancestor already carried the beneficial mutations, so the 12 flasks needed no new mutations: All 12 began from one cell, but they gained fitness through different new mutations that arose later.
  • Genetic drift by itself raised fitness, since drift tends to move separate populations in one direction: Drift is random in direction; it would not raise fitness in all 12 populations.
  • The populations exchanged genes during the daily transfers, so the 12 ended with matching mutations: Each population is transferred into its own flask, and the stem says the mutations mostly differed.

Experimental setup

Rock layers in a canyon wall

Geologists described six undisturbed layers exposed in a canyon wall, numbered from the top. Three layers are volcanic ash, whose crystals were dated by the decay of a radioactive isotope. The other layers are sedimentary rock containing fossils of horse-like mammals.

Layers from the top of the canyon wall (layer 1) to the bottom (layer 6)
LayerRock typeFossils or date
1 (top)SandstoneHorse-like mammal with one large toe on each foot, about 1.5 m tall
2Volcanic ashDated to 5.3 million years ago
3ShaleHorse-like mammal with three toes on each foot, the middle one largest, about 1 m tall
4Volcanic ashDated to 23 million years ago
5MudstoneDog-sized mammal with four toes on each front foot and three on each hind foot
6 (bottom)Volcanic ashDated to 56 million years ago

5. What is the best estimate of the age of the three-toed fossils in layer 3?

  1. Between 5.3 and 23 million years old
  2. Exactly 14 million years, halfway between the ash layers
  3. Older than 23 million years
  4. Younger than 5.3 million years
Show the answer

Layer 3 lies above the ash dated to 23 million years (layer 4) and below the ash dated to 5.3 million years (layer 2), so it formed between those times.

  • Correct: Between 5.3 and 23 million years old: Correct: the ash layers bracket the fossil.
  • Exactly 14 million years, halfway between the ash layers: Sediment does not build up at a steady rate, so the midpoint is not a reliable estimate.
  • Older than 23 million years: Layer 3 lies above the 23-million-year ash, so it is younger.
  • Younger than 5.3 million years: Layer 3 lies below the 5.3-million-year ash, so it is older.

6. Why did the geologists date the ash layers rather than the fossil-bearing sandstone, shale and mudstone?

  1. Ash crystals form as molten rock cools, starting the clock; sedimentary grains are older than the layer itself.
  2. Sedimentary rocks contain no radioactive isotopes, so no radiometric method works on them.
  3. Volcanic ash decays faster than sedimentary rock, so it gives a more precise age in fewer years.
  4. Fossils are destroyed when sedimentary rock is heated, so the rock around them is not tested.
Show the answer

The radiometric clock measures time since a mineral crystallized. In ash, that is the eruption, so the date is the age of the layer. Sandstone grains were worn from older rocks, so their isotopes record the age of those older rocks.

  • Correct: Ash crystals form as molten rock cools, starting the clock; sedimentary grains are older than the layer itself.: Correct: the clock must start when the layer forms.
  • Sedimentary rocks contain no radioactive isotopes, so no radiometric method works on them.: Sedimentary grains do contain radioactive isotopes; they give the age of the grains' source rocks, not of the layer.
  • Volcanic ash decays faster than sedimentary rock, so it gives a more precise age in fewer years.: Decay rates depend on the isotope, not on the rock type.
  • Fossils are destroyed when sedimentary rock is heated, so the rock around them is not tested.: Dating does not require heating the rock; the problem is when its minerals formed.

7. Which claim about the three fossils is best supported by these layers?

  1. Horse-like mammals changed over time toward larger bodies and fewer toes: descent with modification.
  2. Each fossil is the direct parent species of the fossil in the layer above it.
  3. The one-toed mammal lived at the same time as the dog-sized mammal, but its fossils were buried higher.
  4. The animals lost toes during their lives as they ran on harder ground, and passed the change on.
Show the answer

Ordered by age, the fossils show bodies getting larger and side toes becoming smaller and fewer: change over time in one group, as descent with modification predicts. The record cannot show which fossil, if any, was a direct ancestor of another.

  • Correct: Horse-like mammals changed over time toward larger bodies and fewer toes: descent with modification.: Correct: a trend over time, without overclaiming direct parentage.
  • Each fossil is the direct parent species of the fossil in the layer above it.: Fossils show related forms in order, but not that one was the direct ancestor of the next; they may be side branches.
  • The one-toed mammal lived at the same time as the dog-sized mammal, but its fossils were buried higher.: Superposition places layer 1 above the 5.3-million-year ash and layer 5 below the 23-million-year ash, so they did not live together.
  • The animals lost toes during their lives as they ran on harder ground, and passed the change on.: Changes during life are not inherited; the trend comes from selection on heritable variation.

8. A female cod can release millions of eggs a year, yet a cod population can stay roughly the same size for decades. Which conclusion follows?

  1. Almost all offspring die before reproducing, as young fish compete for limited resources or are eaten.
  2. Cod release more eggs in years when the population is small and fewer when it is large.
  3. Most eggs go unfertilized, so the number of young that hatch matches the number of adults that die each year.
  4. The high egg number shows that cod have higher fitness than fish that release fewer eggs.
Show the answer

If each pair's millions of eggs gave even a few dozen breeding adults, the population would explode. A steady size means that on average about two offspring per pair survive to breed: overproduction plus heavy losses, the conditions for selection.

  • Correct: Almost all offspring die before reproducing, as young fish compete for limited resources or are eaten.: Correct: overproduction with a steady population means most offspring die before reproducing.
  • Cod release more eggs in years when the population is small and fewer when it is large.: Nothing in the observation suggests egg number tracks how many cod there are; the point is that most young die.
  • Most eggs go unfertilized, so the number of young that hatch matches the number of adults that die each year.: Even if many eggs go unfertilized, the number that hatch far exceeds the number that live to breed.
  • The high egg number shows that cod have higher fitness than fish that release fewer eggs.: Fitness counts surviving offspring that reproduce, not eggs; comparing egg numbers alone says nothing about fitness.

9. A study of more than 13,000 births in a London hospital between 1935 and 1946 found that babies of middle birth weight had the highest survival, and that survival fell for both lighter and heavier babies. If birth weight is partly heritable, which change in the distribution of birth weights would this selection tend to produce?

  1. Little change in the mean, with fewer babies at both extremes
  2. A shift of the mean toward heavier babies, as heavy babies survive
  3. Two separate peaks, one of light and one of heavy babies
  4. A wider spread, with more babies at both extremes
Show the answer

This is stabilizing selection: the middle is favored and both tails are selected against, so the curve narrows around about the same mean.

  • Correct: Little change in the mean, with fewer babies at both extremes: Correct: stabilizing selection keeps the mean and trims both tails.
  • A shift of the mean toward heavier babies, as heavy babies survive: Heavy babies had lower survival than middle-weight babies, so the mean would not shift upward.
  • Two separate peaks, one of light and one of heavy babies: Two peaks result when both extremes are favored (disruptive), the opposite of this pattern.
  • A wider spread, with more babies at both extremes: Widening is the effect of disruptive selection; stabilizing selection narrows the spread.

10. Kale, cabbage, broccoli and kohlrabi look very different, but they belong to one species and can be crossed. Which explanation best fits?

  1. Farmers bred each crop from one wild mustard by selecting a different enlarged part.
  2. Each crop arose from a different wild plant, and the crops later became one species by crossing.
  3. The crops grow in different soils, which make the same plant grow into different shapes.
  4. Each crop's shape came from one large mutation, which appeared when farmers first planted it.
Show the answer

Choosing leaves (kale), the end bud (cabbage), flower buds (broccoli) or the stem (kohlrabi) generation after generation turned one ancestral plant into several crops that are still one species.

  • Correct: Farmers bred each crop from one wild mustard by selecting a different enlarged part.: Correct: different selection on one ancestor.
  • Each crop arose from a different wild plant, and the crops later became one species by crossing.: They share one wild ancestor; the crops did not come from different species.
  • The crops grow in different soils, which make the same plant grow into different shapes.: Grown side by side in one field, the crops keep their shapes, so the differences are inherited, not caused by soil.
  • Each crop's shape came from one large mutation, which appeared when farmers first planted it.: The differences built up over many generations of selection on many genes, not from one mutation each.

11. A population of 200 snails contains 120 BB, 60 Bb and 20 bb individuals. Calculate the frequency of allele b. Give your answer to two decimal places.

Type a number.

Show the answer

Copies of the gene: 2 × 200 = 400. Copies of b: (2 × 20) + 60 = 100. Frequency of b = 100 ÷ 400 = 0.25.

  • Answer: 0.25

12. A population of 800 frogs is in Hardy-Weinberg equilibrium for a gene with p = 0.7. How many frogs are expected to be heterozygous? Give a whole number.

Type a number in frogs.

Show the answer

q = 1 − 0.7 = 0.3. Heterozygotes = 2pq × 800 = 2 × 0.7 × 0.3 × 800 = 0.42 × 800 = 336 frogs.

  • Answer: 336 frogs

Keep going

Practice has every question in the unit, with feedback after each one; the notes for every topic are free. Write a full answer with the justification trainer. Other units: Unit 1 · Unit 2 · Unit 3 · Unit 4 · Unit 5 · Unit 6 · Unit 7 · Unit 8.