Science You're Expected to Know · Lesson 3

Cells, genetics, photosynthesis and respiration

Two processes explain most gas data in biology passages. Photosynthesis happens in the chloroplasts of plant and algae cells and needs light: it takes in carbon dioxide (CO₂) and water and releases oxygen (O₂). Cellular respiration goes on in nearly every living cell, in light and dark alike (in plant and animal cells, mainly in the mitochondria): it uses O₂ and releases CO₂.

So a plant in the dark only respires, and the O₂ around it falls. In bright light, photosynthesis usually outpaces respiration, and O₂ rises. A setup with no organism is a control: it shows what happens without the living thing.

For genetics, you need a simple cross. Each parent passes one of its two alleles to each offspring. A dominant allele (capital letter, P) shows whenever it is present; a recessive allele (p) shows only when there are two copies (pp). A Punnett square lists one parent’s alleles across the top and the other’s down the side; each of the four boxes is an equally likely offspring.

Cells: know that the nucleus holds the DNA, the cell membrane controls what enters and leaves, and plant cells also have a cell wall and chloroplasts. That level is plenty.

Punnett square for Pp × Pp
Pp
PPP (purple)Pp (purple)
pPp (purple)pp (white)

A Pond Plant in Light and Dark

Students studied Species W, a flowering plant that grows underwater in ponds.

Study 1: The students sealed equal masses of Species W shoots in two water-filled tubes. Tube 1 was kept in bright light and Tube 2 in darkness. Tube 3 held the same water with no shoots and was kept in bright light. All three tubes were held at 20 °C, and the water in each started with 8.0 mg/L of dissolved oxygen. Figure 1 shows the dissolved oxygen over 4 hours.

Study 2: In Species W, flower color is controlled by one gene with two alleles. The allele for purple flowers (P) is dominant over the allele for white flowers (p). The students made three crosses between plants of known genotype and counted the flower colors of the offspring (Table 1).

Figure 1
Figure 1Dissolved oxygen over 4 hours in three tubes. Tube 1 (shoots, light) rises steadily: 8.0, 8.6, 9.2, 9.8 and 10.4 mg/L at 0 to 4 h. Tube 2 (shoots, dark) falls steadily: 8.0, 7.7, 7.4, 7.1 and 6.8 mg/L. Tube 3 (no shoots, light) stays at 8.0 mg/L.6789101101234Time (h)Dissolved oxygen (mg/L)
  • Tube 1 (shoots, light)
  • Tube 2 (shoots, dark)
  • Tube 3 (no shoots, light)
Show the data as a table
Figure 1: data
Time (h)Tube 1 (shoots, light) (mg/L)Tube 2 (shoots, dark) (mg/L)Tube 3 (no shoots, light) (mg/L)
0888
18.67.78
29.27.48
39.87.18
410.46.88

Dissolved oxygen over 4 hours in three tubes. Tube 1 (shoots, light) rises steadily: 8.0, 8.6, 9.2, 9.8 and 10.4 mg/L at 0 to 4 h. Tube 2 (shoots, dark) falls steadily: 8.0, 7.7, 7.4, 7.1 and 6.8 mg/L. Tube 3 (no shoots, light) stays at 8.0 mg/L.

Table 1
CrossParent genotypesPurple-flowered offspringWhite-flowered offspring
Cross 1PP × pp400
Cross 2Pp × Pp6119
Cross 3Pp × pp3842
Offspring of three crosses. Cross 1, PP × pp: 40 purple, 0 white. Cross 2, Pp × Pp: 61 purple, 19 white, close to 3 to 1. Cross 3, Pp × pp: 38 purple, 42 white, close to 1 to 1.

Worked example Easy

Show the whole passage

A Pond Plant in Light and Dark

Students studied Species W, a flowering plant that grows underwater in ponds.

Study 1: The students sealed equal masses of Species W shoots in two water-filled tubes. Tube 1 was kept in bright light and Tube 2 in darkness. Tube 3 held the same water with no shoots and was kept in bright light. All three tubes were held at 20 °C, and the water in each started with 8.0 mg/L of dissolved oxygen. Figure 1 shows the dissolved oxygen over 4 hours.

Study 2: In Species W, flower color is controlled by one gene with two alleles. The allele for purple flowers (P) is dominant over the allele for white flowers (p). The students made three crosses between plants of known genotype and counted the flower colors of the offspring (Table 1).

Figure 1
Figure 1Dissolved oxygen over 4 hours in three tubes. Tube 1 (shoots, light) rises steadily: 8.0, 8.6, 9.2, 9.8 and 10.4 mg/L at 0 to 4 h. Tube 2 (shoots, dark) falls steadily: 8.0, 7.7, 7.4, 7.1 and 6.8 mg/L. Tube 3 (no shoots, light) stays at 8.0 mg/L.6789101101234Time (h)Dissolved oxygen (mg/L)
  • Tube 1 (shoots, light)
  • Tube 2 (shoots, dark)
  • Tube 3 (no shoots, light)
Show the data as a table
Figure 1: data
Time (h)Tube 1 (shoots, light) (mg/L)Tube 2 (shoots, dark) (mg/L)Tube 3 (no shoots, light) (mg/L)
0888
18.67.78
29.27.48
39.87.18
410.46.88

Dissolved oxygen over 4 hours in three tubes. Tube 1 (shoots, light) rises steadily: 8.0, 8.6, 9.2, 9.8 and 10.4 mg/L at 0 to 4 h. Tube 2 (shoots, dark) falls steadily: 8.0, 7.7, 7.4, 7.1 and 6.8 mg/L. Tube 3 (no shoots, light) stays at 8.0 mg/L.

Table 1
CrossParent genotypesPurple-flowered offspringWhite-flowered offspring
Cross 1PP × pp400
Cross 2Pp × Pp6119
Cross 3Pp × pp3842
Offspring of three crosses. Cross 1, PP × pp: 40 purple, 0 white. Cross 2, Pp × Pp: 61 purple, 19 white, close to 3 to 1. Cross 3, Pp × pp: 38 purple, 42 white, close to 1 to 1.

Which process best explains the change in dissolved oxygen in Tube 1?

  1. Respiration, which releases oxygen into the water
  2. Photosynthesis, which uses up oxygen in the water
  3. Oxygen from the air slowly leaking into the tube
  4. Photosynthesis, which releases oxygen into the waterAnswer

How to solve it

  1. From the passage: Tube 1 rises 8.0 → 10.4 mg/L; the no-shoot control, Tube 3, stays at 8.0 mg/L, so the shoots caused the rise.
  2. From background: photosynthesis needs light and releases O₂; respiration uses O₂.
  3. Shoots plus light plus rising O₂ means photosynthesis.

Why each choice is right or wrong

  • A. Incorrect. Respiration uses oxygen; it doesn’t release it. That is why the dark tube, Tube 2, lost oxygen.
  • B. Incorrect. Photosynthesis releases oxygen, and Tube 1’s oxygen went up, from 8.0 to 10.4 mg/L, not down.
  • C. Incorrect. Tube 3 was also in light but had no shoots, and its oxygen stayed at 8.0 mg/L. Nothing was leaking in.
  • D. Correct. Figure 1 shows Tube 1 (shoots, light) rising from 8.0 to 10.4 mg/L, while Tube 3 (no shoots) stayed at 8.0 mg/L (passage). From background, photosynthesis in light releases oxygen.

Worked example Medium

Show the whole passage

A Pond Plant in Light and Dark

Students studied Species W, a flowering plant that grows underwater in ponds.

Study 1: The students sealed equal masses of Species W shoots in two water-filled tubes. Tube 1 was kept in bright light and Tube 2 in darkness. Tube 3 held the same water with no shoots and was kept in bright light. All three tubes were held at 20 °C, and the water in each started with 8.0 mg/L of dissolved oxygen. Figure 1 shows the dissolved oxygen over 4 hours.

Study 2: In Species W, flower color is controlled by one gene with two alleles. The allele for purple flowers (P) is dominant over the allele for white flowers (p). The students made three crosses between plants of known genotype and counted the flower colors of the offspring (Table 1).

Figure 1
Figure 1Dissolved oxygen over 4 hours in three tubes. Tube 1 (shoots, light) rises steadily: 8.0, 8.6, 9.2, 9.8 and 10.4 mg/L at 0 to 4 h. Tube 2 (shoots, dark) falls steadily: 8.0, 7.7, 7.4, 7.1 and 6.8 mg/L. Tube 3 (no shoots, light) stays at 8.0 mg/L.6789101101234Time (h)Dissolved oxygen (mg/L)
  • Tube 1 (shoots, light)
  • Tube 2 (shoots, dark)
  • Tube 3 (no shoots, light)
Show the data as a table
Figure 1: data
Time (h)Tube 1 (shoots, light) (mg/L)Tube 2 (shoots, dark) (mg/L)Tube 3 (no shoots, light) (mg/L)
0888
18.67.78
29.27.48
39.87.18
410.46.88

Dissolved oxygen over 4 hours in three tubes. Tube 1 (shoots, light) rises steadily: 8.0, 8.6, 9.2, 9.8 and 10.4 mg/L at 0 to 4 h. Tube 2 (shoots, dark) falls steadily: 8.0, 7.7, 7.4, 7.1 and 6.8 mg/L. Tube 3 (no shoots, light) stays at 8.0 mg/L.

Table 1
CrossParent genotypesPurple-flowered offspringWhite-flowered offspring
Cross 1PP × pp400
Cross 2Pp × Pp6119
Cross 3Pp × pp3842
Offspring of three crosses. Cross 1, PP × pp: 40 purple, 0 white. Cross 2, Pp × Pp: 61 purple, 19 white, close to 3 to 1. Cross 3, Pp × pp: 38 purple, 42 white, close to 1 to 1.

Using a Punnett square for Cross 2, what fraction of the offspring would be expected to have white flowers?

  1. 0
  2. 1/4Answer
  3. 1/2
  4. 3/4

How to solve it

  1. From the passage: Cross 2’s parents are Pp × Pp, and P (purple) is dominant.
  2. Punnett square: P and p across the top, P and p down the side. The boxes are PP, Pp, Pp, pp.
  3. Only pp is white: 1/4. Check with the data: 19 ÷ (61 + 19) = 19/80, close to 1/4.

Why each choice is right or wrong

  • A. Incorrect. 0 assumes the dominant allele always wins. Two Pp parents can each pass on p, giving pp white flowers, and Table 1 shows 19 white offspring.
  • B. Correct. Table 1 gives Cross 2’s parents as Pp × Pp (passage). The Punnett square (background) has PP, Pp, Pp and pp, and only pp is white: 1 of 4. Table 1’s 19 white of 80 total is close to 1/4.
  • C. Incorrect. 1/2 is the expected white fraction for Pp × pp, Cross 3, not Cross 2.
  • D. Incorrect. 3/4 is the expected fraction with purple flowers (PP, Pp, Pp), not white.

Trap Reaching past the passage

Use only the basics: photosynthesis needs light and releases O₂, respiration goes on all the time and uses O₂. You don’t need chemical equations, and you shouldn’t add facts the passage never gives, such as how fast Species W grows in warm water. A choice built on an outside detail the data can’t check is almost always wrong, even when it sounds scientific.