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IB Biology – Theme C: Interaction and Interdependence

Enzymes, respiration, photosynthesis, signalling, defence, ecosystems.

12 questions  ·  37 marks  ·  answers included  ·  IB Biology SL & HL (2023 syllabus)
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This worksheet covers:

  • This worksheet covers Theme C
  • Enzymes and cell respiration
  • Photosynthesis and cell signalling
  • Defence against disease and ecosystems

A calculator is allowed for the numerical questions.

Questions

Section A – Enzymes and Cell Respiration

1

(a) Using the induced-fit model, describe how an enzyme’s active site interacts with its substrate.(b) Explain why enzyme activity decreases sharply above the optimum temperature.

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(a) The active site is complementary in shape to the substrate but is flexible; when the substrate binds, the active site changes shape slightly to mould around it, improving the fit and straining bonds in the substrate to help catalyse the reaction.(b) Above the optimum temperature, increased kinetic energy causes the hydrogen and other bonds maintaining the enzyme’s tertiary structure to break; the active site changes shape (denaturation) and can no longer bind the substrate, so the rate of reaction falls rapidly.
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2

(a) Write the overall balanced chemical equation for aerobic cell respiration of glucose.(b) State the location in the cell where the majority of ATP is produced during aerobic respiration.

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(a) $\text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2 \rightarrow 6\text{CO}_2 + 6\text{H}_2\text{O}$ (releasing energy, transferred to ATP)(b) The mitochondria (mainly the inner mitochondrial membrane, via oxidative phosphorylation).
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3

Yeast can respire anaerobically, producing ethanol, while human muscle cells can respire anaerobically, producing lactate.(a) Write a word equation for anaerobic respiration in yeast.(b) State one similarity and one difference between anaerobic respiration in yeast and in human muscle cells.

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(a) glucose $\rightarrow$ ethanol + carbon dioxide (+ energy)(b) Similarity: both regenerate $\text{NAD}^+$ from $\text{NADH}$ so glycolysis can continue, and both produce far less ATP per glucose than aerobic respiration. Difference: yeast produces ethanol and carbon dioxide, whereas human muscle cells produce lactate (no carbon dioxide is released).
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4

Germinating seeds were placed in a respirometer. Over a fixed time, they consumed $17.1\,\text{cm}^3$ of oxygen and produced $12.0\,\text{cm}^3$ of carbon dioxide.(a) Calculate the respiratory quotient (RQ), given $\text{RQ} = \dfrac{\text{volume of CO}_2\text{ produced}}{\text{volume of O}_2\text{ consumed}}$.(b) Suggest, with a reason, the main respiratory substrate being used.

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(a) $\text{RQ} = \dfrac{12.0}{17.1} \approx 0.70$(b) An RQ of about $0.7$ indicates lipids are the main respiratory substrate, since lipids contain proportionally less oxygen than carbohydrates and require more $\text{O}_2$ relative to $\text{CO}_2$ produced during their oxidation.
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Section B – Photosynthesis and Cell Signalling

5

(a) Write the overall balanced equation for photosynthesis.(b) State the region of the chloroplast where the light-dependent reactions occur and the region where the light-independent reactions (Calvin cycle) occur.

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(a) $6\text{CO}_2 + 6\text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2$ (using light energy)(b) Light-dependent reactions occur in the thylakoid membranes; the light-independent reactions occur in the stroma.
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6

In an experiment, the rate of photosynthesis of an aquatic plant was measured (as bubbles of oxygen produced per minute) at increasing light intensity, with $\text{CO}_2$ concentration and temperature kept constant. The rate increased steadily at low light intensities but then levelled off at high light intensity.Explain, in terms of limiting factors, why the rate of photosynthesis levels off at high light intensity even though more light is available.

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At low light intensity, light is the limiting factor, so increasing it increases the rate. At high light intensity, light is no longer limiting; instead another factor — here $\text{CO}_2$ concentration (since temperature is constant) — becomes the limiting factor, so further increases in light intensity no longer increase the rate, and the graph levels off (plateaus).
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7

The hormone insulin binds to a receptor protein on the surface of liver cells, triggering a series of intracellular events that result in glucose uptake and glycogen synthesis.Outline the three general stages of a cell-signalling pathway (reception, transduction, response), using insulin as your example.

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Reception: insulin (the signalling molecule) binds to a specific, complementary receptor protein on the liver cell membrane. Transduction: binding activates a cascade of intracellular relay proteins/enzymes that converts and relays the signal within the cell. Response: the cell’s response is triggered, such as movement of glucose transporter proteins to the membrane and activation of enzymes that convert glucose to glycogen.
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8

Explain how an enzyme cascade during signal transduction can amplify a single signalling event into a much larger cellular response.

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Each activated enzyme in the cascade can activate many molecules of the next enzyme in the sequence, and each of those can activate many more downstream molecules; because this multiplies at each step, binding of a single signalling molecule to one receptor can ultimately activate a very large number of final target molecules, greatly amplifying the response.
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Section C – Defence Against Disease and Ecosystems

9

(a) Distinguish between the innate and adaptive (specific) immune response.(b) Explain why the secondary immune response to a pathogen is faster and stronger than the primary response.

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(a) The innate immune response is non-specific and acts immediately against any pathogen (e.g. phagocytosis, inflammation), while the adaptive response is specific to a particular antigen, involves lymphocytes, and takes longer to develop on first exposure.(b) After the primary response, memory B-cells and memory T-cells specific to that antigen remain in the body; on re-exposure these memory cells recognise the antigen and proliferate rapidly into plasma cells, producing antibodies faster and in greater quantity than during the first exposure.
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10

HIV (human immunodeficiency virus) infects and destroys helper T-cells, eventually leading to AIDS (acquired immunodeficiency syndrome).Explain why the destruction of helper T-cells severely weakens the overall immune response.

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Helper T-cells activate B-cells (to produce antibodies) and cytotoxic T-cells (to destroy infected cells) by releasing signalling molecules (cytokines); without sufficient helper T-cells, both the antibody-mediated and cell-mediated adaptive responses fail to be properly activated, leaving the body unable to fight off infections that it would normally control.
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11

In a grassland ecosystem, producers have a gross energy content of $20\,000\,\text{kJ m}^{-2}\,\text{yr}^{-1}$, and the energy available to the primary consumer trophic level is $1800\,\text{kJ m}^{-2}\,\text{yr}^{-1}$.(a) Calculate the percentage efficiency of energy transfer between these two trophic levels.(b) State two reasons why energy transfer between trophic levels is never 100% efficient.

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(a) $\text{efficiency} = \dfrac{1800}{20\,000}\times100 = 9\%$(b) Any two of: energy is lost as heat through respiration; not all of an organism is eaten (e.g. roots, bones); not all ingested material is digested and absorbed (some is lost in faeces); some energy is used for movement and other life processes rather than stored as biomass.
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12

A student estimates the population size of snails in a field using the mark–release–recapture (Lincoln index) method. She marks and releases $50$ snails. Later, she catches $60$ snails, of which $12$ are already marked.(a) Use the Lincoln index, $N = \dfrac{(\text{number marked}) \times (\text{total caught second time})}{\text{number of marked individuals recaptured}}$, to estimate the total population size.(b) State one assumption of this method.

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(a) $N = \dfrac{50 \times 60}{12} = 250$ snails(b) Any one of: the marked individuals redistribute randomly/evenly throughout the population before recapture; there is no significant immigration, emigration, birth or death between the two sampling times; the marking does not affect survival or behaviour (e.g. does not make them more visible to predators).
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