IB Biology SL Theme A: Unity and diversity questions

Exam-style multiple choice and written questions with full markschemes. Check multiple choice answers straight away, or type, photograph or draw a written answer and get it marked.

Type
Multiple choiceWritten
Difficulty
EasyMediumHard

Question 1

Paper 2Medium16 marks

Figure 1 represents a water molecule. The positions of the electron pairs are shown.

OHH

A class investigated respiration in yeast cells with different water contents. The cultures were supplied with equal glucose and oxygen and kept at 25 °C. Table 1 gives the mean rate of carbon dioxide production.

Water content of yeast cells / %Mean rate of carbon dioxide production / μmol g⁻¹ dry yeast min⁻¹
200.4
351.2
502.4
653.6
804.0
(a)

In Figure 1, identify the type of bond represented by each solid line and state what is shared in this bond.

[2]
(b)

Using Figure 1, explain how water molecules can form hydrogen bonds with one another.

[3]
(c)

Draw two water molecules, showing the partial charges and one hydrogen bond between them. Use solid lines for covalent bonds and a dotted line for a hydrogen bond.

[3]
(d)

At 65% water content, calculate the amount of carbon dioxide produced in 30 min by 0.50 g of dry yeast. Assume that the rate remains constant.

[2]
(e)

Describe the relationship between water content and the rate of carbon dioxide production shown in Table 1.

[2]
(f)

Evaluate the conclusion that these results demonstrate that water is a medium for life and that the first cells originated in water. Use at least one value from Table 1 and your biological knowledge.

[4]

Question 2

Paper 2Medium16 marks

A short region of a gene is transcribed in a cell. The DNA template strand is shown in the direction in which it is read. Spaces separate groups of three bases.

DNA template: 3′–TAC GGA CTT ACG–5′

(a)

Compare DNA and RNA with respect to their number of strands, nitrogenous bases and pentose sugar. Include the structural difference between the sugars.

[5]
(b)

Determine the mRNA sequence produced, writing it in the 5′ to 3′ direction.

[4]
(c)

During translation, the second codon of the mRNA is read at a ribosome. For parts (i) and (ii), use the codon 5′–CCU–3′, even if your answer to part (b) differs.

(i)

Write the tRNA anticodon that pairs with this codon, in the 3′ to 5′ direction.

[1]
(ii)

Explain how this pairing contributes to the formation of a polypeptide.

[2]
(d)

Before a cell divides, its DNA is copied. Explain how complementary base pairing enables this copying.

[4]

Question 3

Paper 2Medium16 marks

Figure 1 presents diploid chromosome numbers for six organisms. Humans and chimpanzees are the named examples in the IB Biology guide. Plant P, Plant Q, Animal R and Animal S are hypothetical species; no prior knowledge of their chromosome numbers is required. For each of these four hypothetical species, the plotted count was obtained from one diploid cell.

HumanChimpanzeePlant PPlant QAnimal RAnimal S46481428123001020304050Chromosomes in one diploid cell
(a)

Calculate the difference between the diploid chromosome numbers of chimpanzees and humans.

[2]
(b)

State the highest and lowest diploid chromosome numbers shown in Figure 1, and calculate the range of the values.

[3]
(c)

Explain why diploid cells have an even number of chromosomes.

[2]
(d)

Plant Q has a diploid chromosome number of 28. Determine the number of chromosomes in a gamete produced by meiosis and in the zygote formed when two such gametes fuse. Explain how the chromosome number changes between these stages.

[3]
(e)

A student proposes that an animal species always has a higher diploid chromosome number than a plant species. Evaluate this proposal using the data for Plant P, Plant Q, Animal R and Animal S in Figure 1.

[4]
(f)

Suggest an improvement to the sampling method that would make the chromosome number recorded for each hypothetical species more reliable.

[2]

Question 4

Multiple choiceMedium1 mark

In one case study, mixed dipterocarp forest in Southeast Asia is cleared for oil-palm plantations. In another, a local wetland is drained for housing as population growth increases demand for land. Which option identifies a direct cause in the forest case and an indirect human driver in the wetland case?

Show answer and explanation

Answer: D. A direct cause is the immediate land-use change; an indirect driver is an underlying human pressure that leads to that change. Clearing the mixed dipterocarp forest for plantations is direct. Population growth raises demand for land, indirectly driving wetland loss through drainage for housing. A: identifies the indirect demand for palm oil rather than the direct forest clearance. B: gives the immediate drainage action for the wetland, not its indirect driver. C: gives an indirect forest pressure and the direct wetland action, reversing both requested categories. Remember: distinguish the land conversion itself from the human pressure that motivates it.

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