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Marlbridge

AP® Biology · Unit 7 · Topic 7.3

Artificial Selection: Practice Questions (Biology 7.3)

Seven original Marlbridge practice questions on artificial selection, selection experiments, lost genetic variation and human-caused selection, with worked solutions and suggested mark points.

These are original Marlbridge practice questions, not past exam questions. The mark points are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. Every data set is fictional. Round to the precision of the data.

Question 1 (multiple choice · foundation)

Which of these is an example of artificial selection?

  • (A) Each year a farmer saves seed only from the tomato plants that ripened earliest, and sows only that seed the next year.
  • (B) After years of antibiotic use in a hospital, most of the bacteria found on its surfaces are resistant to the antibiotic.
  • (C) A scientist inserts a gene for frost tolerance into strawberry cells and grows plants from them.
  • (D) A gardener adds extra fertiliser every year, and the plants in the garden grow taller each year.
Answer and explanation

Answer: (A). The farmer chooses which plants reproduce, based on a heritable trait (ripening time). Repeated each year, early ripening becomes more common.

  • (B) is caused by human activity, but the antibiotic, not a person, decided which bacteria reproduced. It is natural selection in a human-made environment.
  • (C) is genetic engineering: DNA is changed directly instead of choosing parents from existing variation.
  • (D) is a change caused by the environment. Taller growth from fertiliser is not inherited, and no plants were chosen as parents.

Question 2 (multiple choice · core)

A breeder selects sheep for fine wool. For the first 15 generations, mean fibre diameter falls each generation. After that it stays the same, even though the breeder still uses only the finest-wooled sheep as parents. Which is the most likely explanation?

  • (A) Little heritable variation in fibre diameter remains, because alleles for coarse wool have been lost from the flock.
  • (B) The sheep have become as well adapted as they need to be, so they no longer change.
  • (C) Mutations stop happening once a population has been selected for many generations.
  • (D) Fibre diameter is now controlled only by the environment, so selection cannot act on it.
Answer and explanation

Answer: (A). Selection needs heritable variation. After many generations, alleles for finer wool are fixed and others are lost, so the parents chosen are genetically almost the same as the rest. Offspring then resemble the flock average, and the mean stops changing: a selection plateau.

  • (B) is goal-directed thinking. Populations do not “need” or aim for anything; selection only sorts existing variation.
  • (C) is false. Mutations keep arising at a low rate whether or not a population is selected. They are simply too rare to restart a fast response on their own.
  • (D) overstates the case. Genes still affect the trait; what has run out is genetic variation among the sheep.

Question 3 (multiple choice · core)

A poultry line was selected for egg number. The table shows fictional data.

Generation 0 10 20 30
Mean eggs per hen per year 180 232 271 296
Mean number of alleles per gene (20 genes) 4.8 3.9 3.1 2.6

Which statement is best supported by the data?

  • (A) Egg number rose by about 64% over 30 generations while allele number fell by about 46%, and the rise in egg number slowed in later generations.
  • (B) Egg number rose at a constant rate of about 3.9 eggs per generation.
  • (C) The loss of alleles caused the hens to lay more eggs.
  • (D) Egg number rose by about 164% over 30 generations.
Answer and explanation

Answer: (A). Egg number: (296 − 180) ÷ 180 × 100 = 64.4%. Alleles: (2.6 − 4.8) ÷ 4.8 × 100 = −45.8%. The rise per generation was 5.2 eggs (generations 0–10), then 3.9 (10–20), then 2.5 (20–30), so it slowed.

  • (B) 3.9 is the overall average (116 ÷ 30) and the middle interval’s rate. The rate fell from 5.2 to 2.5, so it was not constant.
  • (C) mistakes a correlation for a cause. Both changes are effects of the same selection; losing alleles does not by itself raise egg number.
  • (D) divides final by start (296 ÷ 180 = 1.64) and reads the ratio as a percentage increase. The increase is 64%, not 164%.

Question 4 (data description · core)

Students grew a fast-cycling radish population. In the selected line, each generation they measured 100 plants and bred only the 10 with the longest roots. In the unselected line, they bred 10 plants chosen at random. Both lines were grown in the same room and compost.

Generation 0 1 2 3 4
Selected line, mean root length / mm 42 47 51 56 60
Unselected line, mean root length / mm 42 43 41 42 43

(a) Describe the data for both lines, using values. (b) Calculate the percentage change in the selected line and its mean change per generation. (c) Explain the purpose of the unselected line. (d) From generation 4 the students breed the selected line from 10 random plants. Predict the mean root length over the next few generations, and justify your prediction.

Worked solution and suggested Marlbridge rubric

(a) Selected line: mean root length rises every generation, from 42 mm to 60 mm, by 4–5 mm each time. Unselected line: stays between 41 and 43 mm with no trend. By generation 4 the selected line is 17 mm longer than the unselected line.

(b) (60 − 42) ÷ 42 × 100 = 18 ÷ 42 × 100 = +42.9%. Per generation: 18 ÷ 4 = 4.5 mm.

(c) The unselected line shows what happens in the same conditions without selection. Because it did not change, the rise in the selected line can be attributed to selection (a genetic change), not to the room, the compost or the measuring method.

(d) The mean should stay about 60 mm, perhaps fluctuating a little. Selection changed allele frequencies; random breeding does not undo that, so the line does not “spring back” to 42 mm. (Accept a slight fall if justified, for example by natural selection against long roots in the lab.)

Point What earns it
1 Both trends described with at least two quoted values each
1 +42.9% (or 43%) and 4.5 mm per generation
1 Control explained: same conditions without selection, so the change is due to selection
1 Prediction (mean stays near 60 mm) justified by changed allele frequencies

Question 5 (constructed response · core)

A fictional fruit crop is grown almost entirely as one variety. Growers propagate it from cuttings, not seeds, because cuttings keep the fruit’s taste exactly the same. A new strain of a fungus appears that kills plants of this variety.

(a) Explain why the variety has almost no genetic variation. (b) Explain why the fungus could destroy nearly all plantations, rather than only some plants. (c) Describe two ways breeders could increase genetic variation in the crop, and give one disadvantage of doing so.

Model answer and suggested Marlbridge rubric

(a) Cuttings are produced by mitosis, so every plant is a clone of the original selected plant, with the same genotype. Earlier selection had already kept only plants with the wanted fruit, removing other alleles.

(b) Because all plants share one genotype, they share the same susceptibility. If the fungus can infect one plant, it can infect all, and there are no resistant individuals to survive and slow the spread. In a variable population some plants would carry alleles for resistance.

(c) Any two: cross the variety with wild relatives or other varieties and select resistant offspring that still taste good; use seeds from a gene bank of old varieties; grow several varieties instead of one. Disadvantage (one): offspring of crosses vary, so fruit is less uniform; restoring the wanted traits takes many generations.

Point What earns it
1 Clonal propagation gives identical genotypes (and earlier selection removed alleles)
1 Identical susceptibility, no resistant individuals
1 Two valid ways to add variation
1 A sensible disadvantage linked to uniformity or time

Question 6 (constructed response · stretch)

A fictional fish stock was heavily fished with nets that catch mainly large fish. Scientists recorded the mean age and length at which females first spawn.

Year 1980 1990 2000 2010 2020
Mean age at maturity / years 6.1 5.6 5.0 4.6 4.3
Mean length at maturity / cm 62 58 54 51 49

A report claims: “Fishing has caused this population to evolve to mature earlier.”

(a) Describe the trends, including the percentage change in age at maturity. (b) Explain how fishing could act as a selective force on age at maturity. (c) Evaluate the claim. Give one other explanation for the data and describe how it could be tested.

Model answer and suggested Marlbridge rubric

(a) Mean age at maturity fell every decade, from 6.1 to 4.3 years, a change of −1.8 years or (4.3 − 6.1) ÷ 6.1 × 100 = −29.5%. Mean length at maturity also fell, from 62 to 49 cm (−21.0%). The fall in age was largest from 1990 to 2000 (0.6 years) and smallest from 2010 to 2020 (0.3 years).

(b) Nets remove large fish, which are usually older. Fish that inherit a tendency to mature young and small can spawn before they reach catchable size, so they leave more offspring. Alleles for early maturity therefore become more common in each generation.

(c) The data fit the claim but do not prove a genetic change. Alternative: with fewer fish there is more food per fish, or the water has warmed, so fish grow faster and mature earlier without any change in their genes (a non-heritable response to the environment). Test: rear offspring from this stock and from a lightly fished stock in identical tanks with the same food and temperature. If the heavily fished stock’s offspring still mature earlier, the difference is genetic.

Note: no one chose the winners on purpose, so strictly this is natural selection in a human-made environment. Accept either label if the mechanism is explained.

Point What earns it
1 Both trends described with values and −29.5% (or about −30%)
1 Mechanism: large fish removed, early maturers reproduce more, heritable
1 States data are correlational / do not prove a genetic change, with a valid alternative
1 A test that holds environment constant (common conditions) and says what result would support the claim

Question 7 (data analysis · stretch)

A plant line was selected for seed oil content for 12 generations (fictional data).

Generation 0 2 4 6 8 10 12
Mean oil content / % of seed mass 4.0 4.9 5.7 6.3 6.6 6.7 6.7

(a) Describe the pattern, comparing the gain in the first six generations with the gain in the last six. (b) Explain the pattern in terms of genetic variation. (c) Breeders cross the line with an unrelated wild population and continue selecting. Four generations later, mean oil content is 7.4%. Explain this result.

Worked solution and suggested Marlbridge rubric

(a) Oil content rose from 4.0% to 6.7%, an overall rise of 67.5%. The gain was 2.3 percentage points in generations 0–6 (4.0 to 6.3) but only 0.4 in generations 6–12 (6.3 to 6.7). The gain per two generations fell each time: 0.9, 0.8, 0.6, 0.3, 0.1, then 0. Oil content reached a plateau of 6.7% from generation 10.

(b) At first the population held many alleles affecting oil content, so the chosen parents differed genetically from the average and the response was fast. Each round fixed favourable alleles and removed others, so heritable variation shrank. When almost none remained, choosing parents no longer changed the offspring, and the mean stopped rising.

(c) The cross brought in new alleles from the wild population (gene flow into the line), including some that raise oil content and were not present before. This restored heritable variation, so selection could act again: the mean rose 0.7 percentage points, from 6.7% to 7.4%.

Point What earns it
1 Description: rise slowing to a plateau, with 2.3 vs 0.4 (or equivalent values)
1 Plateau explained by loss of heritable variation (fixation and loss of alleles)
1 Cross explained as adding new alleles / variation
1 Links restored variation to renewed response, quoting the 0.7 rise

How did you do?

  • Q1 wrong: re-read “Humans as the selecting force” and “Selection without meaning to” in the study guide.
  • Q2 or Q7 wrong: re-read “What selection does to a population” and Figure 2.
  • Q3 or Q4 descriptions or percentages wrong: rework Worked example 1; quote values and compare with the control.
  • Q5 incomplete: rework Worked example 2 on crop diversity.
  • Q6 incomplete: check you separated a correlation from a genetic change and proposed a fair test.

Then tick off the topic checklist.

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