Disruptions in Ecosystems: Practice Questions (Biology 8.7)

Answer: (B). A few beetles already survived before any spraying, so the resistance variation existed first. The insecticide acted as a selective pressure. Survivors reproduced, and the allele frequency rose over the generations.

• (A) states the misconception that the environment produces the mutations that are needed. Mutations are random with respect to need.
• (C) describes inheritance of acquired characteristics. Changes during an individual’s life are not passed on through its genes.
• (D) Without selection, rare new alleles could not reach 80% in six generations; and resistance was present before spraying.

Answer: (B). In X the heterozygote (1.0) has a higher relative fitness than both homozygotes (0.7 and 0.8). That is the definition of heterozygote advantage. Both alleles will be kept in the population.

• (A) In W, Aa equals AA. That fits a dominant A allele with no extra benefit for the heterozygote.
• (C) In Y, fitness falls steadily from AA to aa. Selection favours A, and the heterozygote is intermediate.
• (D) In Z the heterozygote has the lowest fitness, the opposite of heterozygote advantage.

Answer: (B). Two pieces of the stem matter: the species share food, so they compete, and the newcomer has escaped its parasites and predators. Fewer deaths let its population grow fast, and the native fish lost the competition for larvae.

• (A) contradicts the stem: both species eat the same larvae, so the niches overlap.
• (C) The stem gives no evidence of predation on native fish. Competition explains the data without it.
• (D) Every population is eventually limited by density-dependent factors such as food. The newcomer’s growth slows once it reaches the lake’s carrying capacity for it.

Answer: (C). There are three steps from zooplankton to seal. Concentration = 0.05 × 8 × 8 × 8 = 0.05 × 512 = 25.6 ppm.

• (A) adds the factors instead of multiplying: 0.05 × 8 × 3 = 1.2.
• (B) uses only two steps (0.05 × 8² = 3.2): this is the large fish.
• (D) multiplies by 8 four times (0.05 × 8⁴ = 204.8). It counts the four organisms as four steps, but there are only three steps between them.

(a) (55 − 4) ÷ 4 × 100 = 51 ÷ 4 × 100 = 1275% (the algae increased almost 14-fold).

(b) Phosphate had been limiting algal growth. The extra phosphate caused a bloom that peaked at week 4. As the algae died, they sank and were broken down by aerobic decomposers (bacteria). The decomposers multiplied, and their respiration used up the dissolved oxygen. So oxygen fell to 1.6 mg L⁻¹ at week 6, after the peak of the bloom, not at it.

(c) Fish need dissolved oxygen for aerobic respiration. At 1.6 mg L⁻¹ many fish suffocated or left the deeper water, so the catch fell from 22 to 3.

(d) Any one with a mechanism, for example: plant a buffer strip of vegetation between fields and the lake, so plant roots take up nitrate and phosphate before they reach the water; apply less fertiliser, or not before heavy rain; treat sewage to remove phosphate. Each reduces the nutrient input that triggers the bloom.

Do not award the third point for “algae use up the oxygen”: living algae produce oxygen by photosynthesis in the light.

(a) “The vine patches are distributed in proportion to the area of each habitat” (habitat type has no effect on where the vine grows).

(b) Expected: edge 200 × 0.40 = 80; grassland 200 × 0.35 = 70; interior 200 × 0.25 = 50.

Degrees of freedom = 3 − 1 = 2; critical value 5.99. Since 32.79 > 5.99, reject the null hypothesis. The vine is more common at forest edges and less common in the interior than area alone predicts.

(c) A climbing vine needs light and something to climb. Forest edges give both: trees for support and light from the open side. The canopy shades the interior, so the vine grows poorly there. (Accept other reasonable explanations, such as seeds spread by people along edge paths.)

(d) The claim is not supported. 22 patches were found in the forest interior, so the vine can grow there, though less often than expected. The data show only where the vine is now; it could spread further if the canopy is opened, for example by logging or storms.

Accept χ² carried forward from an error in the expected values.

(a) Mean = (62 + 68 + 55 + 64) ÷ 4 = 249 ÷ 4 = 62.25 chicks per 100 nests. Decrease = (62.25 − 9) ÷ 62.25 × 100 = 85.5% (about 86%).

(b) In an El Niño year, the surface water warms and upwelling of cold, nutrient-rich water weakens. With fewer nutrients, there is less phytoplankton growth, so less food for zooplankton and for the small fish that eat them. Parent birds cannot find enough fish, so fewer chicks are fed well enough to fledge. The disruption passes up the food chain.

(c) More frequent warm years would mean more years of very low breeding success. The population would shrink, because too few young replace adults that die. A population with low genetic diversity is less likely to contain individuals with variations that help them cope, for example by foraging on different prey or nesting at a different time. So selection has less to work with, and the risk of local extinction is higher.

(d) In year 4 the anomaly was small (+0.4 °C), so upwelling and food supply returned to near normal. But year 3’s missing chicks are a lost age group: fewer birds will reach breeding age later.