IB DP Biology: Hormonal and feedback integration in animals, and plant growth regulation (DP Biology) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

These questions cover statements C3.1.11 to C3.1.23 of the IB Biology guide, first assessment 2025, for DP Biology. Questions 1 to 6 and 11 are for SL and HL. Questions 7 to 10 test additional higher level content, so they are HL only. Read the study guide first, and see the revision notes, the IB DP Biology hub and the printable checklist.

Questions

1. The table shows the concentration of melatonin in the blood plasma of one adult over a day.

Time	18:00	20:00	22:00	00:00	02:00	04:00	06:00	08:00	12:00
Melatonin (pg cm⁻³)	5	12	45	78	95	60	15	6	4

(a) State the time at which melatonin concentration was highest. [1] (b) Calculate how many times greater the peak concentration was than the concentration at 12:00. Give your answer to 3 significant figures. [1] (c) Explain how these data relate to the sleep-wake cycle. [2]

2. At rest, a person has a heart rate of 70 beats per minute and a stroke volume of 72 cm³. During vigorous exercise the heart rate is 160 beats per minute and the stroke volume is 120 cm³. At rest, 20% of cardiac output goes to skeletal muscle and 25% to the gut. During exercise these shares are 80% and 3%.

(a) Calculate cardiac output at rest and during exercise. [2] (b) Calculate the blood flow to skeletal muscle during exercise. [1] (c) Calculate the blood flow to the gut at rest and during exercise. [1] (d) Explain how epinephrine contributes to these changes. [2]

3. Outline how the hypothalamus and pituitary gland control the secretion of thyroxine, including the feedback involved. [4]

4. An ECG showed an interval of 0.75 s between successive R waves in a person lying down. Immediately after standing up it was 0.60 s. (a) Calculate the heart rate lying down, the heart rate after standing and the percentage increase. [3] (b) Explain how baroreceptors bring about this change. [2]

5. A volunteer breathed air containing different carbon dioxide concentrations. Minute ventilation was recorded.

Carbon dioxide in inhaled air (%)	0.04	2	4	6
Minute ventilation (dm³ min⁻¹)	6.5	12.0	26.0	58.0

(a) Calculate how many times greater minute ventilation was at 6% than at 0.04%. [1] (b) Describe the trend shown. [1] (c) Explain the control of ventilation rate that produces this trend. [4]

6. Outline how the central nervous system and the enteric nervous system control the movement of food through the digestive system. [3]

7. (HL only) Seedlings were exposed to light from one side. The angle of curvature of six seedlings per treatment was measured.

Treatment	Angles of curvature (degrees)
Control	39, 44, 41, 37, 46, 43
Foil over the base	40, 42, 38, 45, 41, 40
Foil over the tip	2, 0, 3, 1, 2, 4

(a) Calculate the mean angle for the control and for the tip covered with foil. [2] (b) Deduce what the data show about the site where light is detected. [2] (c) Suggest two ways of making the results more reliable. [2]

8. (HL only) Segments of stem, 12.0 mm long, were incubated for six hours. Final lengths were: pH 7 buffer, 12.9 mm; pH 4.5 buffer, 15.0 mm; pH 7 buffer with auxin, 14.7 mm; pH 7 buffer with auxin and a chemical that blocks proton pumps, 12.9 mm. (a) Calculate the percentage increase in length in the pH 7 buffer and in the pH 4.5 buffer. [2] (b) Using all the data, evaluate the hypothesis that auxin promotes growth by acidifying the cell wall. [3]

9. (HL only) (a) When the shoot tip of a young plant is cut off, lateral buds begin to grow. Explain this using auxin and cytokinin. [2] (b) Outline how the exchange of these two hormones helps coordinate root and shoot growth. [2]

10. (HL only) Green bananas kept alone took 7 days to ripen. Bananas kept in a closed bag with one ripe apple took 4 days. (a) Calculate the percentage reduction in ripening time, to the nearest whole number. [1] (b) Explain the difference. [2] (c) Explain the advantage to a plant of controlling ripening by positive feedback. [2]

11. A person runs hard up a hill. Explain how epinephrine and the monitoring of blood chemistry help the body meet the demands of the muscles. [6]

Answers

1. (a) 02:00 [1] (b) 95 / 4 = 23.8 times [1] (c) Melatonin is high during the night, when sleep normally occurs, and low in the day [1]. It begins to rise in the evening and falls by morning, so it signals the cycle of sleeping and waking; light suppresses secretion, which resets the cycle each day [1]. Examiner insight: A “calculate” answer is best given with the working shown and the requested number of significant figures; 23.75 left unrounded would not match the instruction.

2. (a) Rest: 70 × 72 = 5040 cm³ min⁻¹ [1]. Exercise: 160 × 120 = 19 200 cm³ min⁻¹ [1]. (b) 0.80 × 19 200 = 15 360 cm³ min⁻¹ [1] (c) Rest 0.25 × 5040 = 1260; exercise 0.03 × 19 200 = 576; 1260 and 576 cm³ min⁻¹ [1] (d) Epinephrine increases heart rate and stroke volume, so cardiac output rises [1]. It widens blood vessels in skeletal muscle and narrows those to the gut, so a larger share of blood goes to muscle [1]. Examiner insight: Part (d) asks about epinephrine, so naming the hormone’s effect on both the heart and the vessels is what the question rewards; a general statement that “blood flows to muscle” is thin.

3. The hypothalamus releases a releasing hormone that acts on the pituitary [1]. The pituitary secretes thyroid-stimulating hormone, which stimulates the thyroid to secrete thyroxine [1]. Rising thyroxine in the blood inhibits the hypothalamus and pituitary [1], so less stimulation occurs and thyroxine returns towards its set level (negative feedback) [1]. Examiner insight: Answers tend to gain credit when the sequence is complete and the inhibition step is stated; listing the glands alone leaves the feedback unexplained.

4. (a) Lying: 60 / 0.75 = 80 beats per minute [1]. Standing: 60 / 0.60 = 100 beats per minute [1]. Increase: (100 − 80) / 80 × 100 = 25% [1]. (b) Standing lowers blood pressure in the neck arteries, and baroreceptors in the carotid arteries and aortic arch send fewer impulses to the medulla [1]. The medulla sends more impulses along sympathetic nerves to the heart, raising heart rate and stroke volume, which restores pressure [1]. Examiner insight: A heart rate from an interval needs the conversion 60 divided by the interval in seconds; 1 / 0.75 gives beats per second, so state the unit.

5. (a) 58.0 / 6.5 = 8.9 times [1] (b) Minute ventilation increases as the carbon dioxide concentration rises, and the increase becomes steeper (positive correlation) [1] (c) More carbon dioxide in the blood forms carbonic acid, which lowers blood pH [1]. Chemoreceptors in the brainstem detect the pH fall [1]. The medulla sends impulses to the diaphragm and intercostal muscles, so ventilation rate and depth increase [1]. This removes carbon dioxide, pH rises back towards normal and the stimulus declines (negative feedback) [1]. Examiner insight: The chain from carbon dioxide to pH to receptor to medulla to muscles is usually expected in order; naming the receptor type without saying what it detects loses the link.

6. Swallowing is started voluntarily, under control of the CNS [1]. After this, peristalsis in the gut is involuntary and controlled by the ENS in the gut wall, which coordinates waves of contraction so material moves in one direction [1]. Egestion of faeces is again under voluntary control from the CNS [1]. Examiner insight: “Outline” with three marks suggests three distinct points; naming the two systems with their regions of control is clearer than describing the muscles alone.

7. (a) Control: 250 / 6 = 41.7 degrees [1]. Tip covered: 12 / 6 = 2.0 degrees [1]. (b) Covering the tip almost abolished curvature [1], while covering the base made no difference (mean 41.0 against 41.7), so light is detected at the tip, not the lower region where bending occurs [1]. (c) Any two: use more seedlings; use seedlings of the same species and age; keep light intensity and temperature the same; repeat the experiment [1] [1]. Examiner insight: When “deduce” is used, quoting the comparison between treatments in the data is generally needed, not just the conclusion.

8. (a) pH 7: (12.9 − 12.0) / 12.0 × 100 = 7.5% [1]. pH 4.5: (15.0 − 12.0) / 12.0 × 100 = 25.0% [1]. (b) Auxin at pH 7 gave a 22.5% increase, far above the 7.5% of the buffer alone [1], similar to the effect of acid alone (25.0%) [1]. When proton pumps were blocked, auxin gave only 7.5%, so auxin needs proton pumping to act, which supports the hypothesis [1]. Examiner insight: An evaluation is stronger when it uses the inhibitor treatment and the acid treatment together with figures from the table; a single comparison supports less.

9. (a) The shoot tip makes auxin, which inhibits lateral buds [1]. With the tip removed, auxin falls, so cytokinin from the root tips promotes the growth of lateral buds [1]. (b) Root tips make cytokinin, which is transported to the shoots, and shoot tips make auxin, which is transported to the roots [1]. Each organ’s growth is regulated by a signal from the other, so root and shoot growth stay in step [1]. Examiner insight: State where each hormone is made and which way it is transported, as the explanation depends on this.

10. (a) (7 − 4) / 7 × 100 = 43% [1] (b) Ripe apples release ethylene gas, which in the closed bag reaches the bananas and stimulates ripening [1]. The bananas also make more ethylene as they ripen, so ripening speeds up (positive feedback) [1]. (c) Positive feedback makes ripening rapid [1] and synchronised across the fruits, so seeds are available to animal dispersers at the same time [1]. Examiner insight: Answers on advantage gain more by naming rapidity and synchrony than by saying only that fruit “ripens faster”.

11. Exercise increases respiration, so more carbon dioxide enters the blood and blood pH falls [1]. Chemoreceptors detect the pH change and signal the medulla [1]. The medulla increases ventilation through the diaphragm and intercostal muscles [1] and increases heart rate and stroke volume [1], so more oxygen is delivered and carbon dioxide removed. Epinephrine from the adrenal glands adds to the heart effects and widens muscle blood vessels, diverting blood to muscle [1]. It also stimulates the liver to convert glycogen to glucose for respiration [1]. Examiner insight: A six-mark explanation needs several distinct linked points; rewording one effect adds nothing.

Where marks are usually lost

• Giving a rate calculation without a unit or without the working.
• Naming baroreceptors for chemical changes, or chemoreceptors for pressure.
• Locating baroreceptors “in the heart”.
• Missing the pH step in ventilation answers.
• Forgetting to state that the feedback is negative.
• Leaving out the inhibition step in hypothalamus and pituitary answers.
• Saying that auxin “makes cells grow” with no mechanism.
• Mixing up where auxin and cytokinin are made.
• Quoting one result instead of comparing treatments.

Next steps

• Review the revision notes and the study guide.
• Return to the IB DP Biology hub and tick off the printable checklist.
• Try all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

This page follows the IB Biology guide, first assessment 2025, published by the International Baccalaureate Organization.