AS Human Biology Unit 5

Text questions

P77/Q1		Exchange surfaces are very thin (usually 1-cell thick), made of membrane bound cells, which are very fragile → these structures need to be internal in order to protect them against damage.
P77/Q2		To allow for an efficient gaseous exchange, because both O₂ and CO₂ dissolve in water well. Once dissolved, they will diffuse across the cell membranes and cytoplasm easily.
P78/Q3		The concentration gradient would drop.
P80/Q4		The breathing centre which co-ordinates/controls breathing is autonomic of our higher centres of the brain.
P80/Q5		Chemoreceptors (1 aortic & 2 carotid bodies) detect low CO₂ / high O₂ concentrations (also higher than usual pH) → signal to the breathing centre → decreased breathing rate & depth.
P82/Q6		15/min x 0.5l = 7.5 l/min

Assignment

P86/Q1

There are parts of the body from which skin cannot be removed, e.g. head or limbs → surface area of these parts will not be included in the calculations, hence error.

Cylinders and cones are regular 3-D shapes. Parts of the body of an animal are never regular 3-D shapes, hence a source of error, if assuming this.

P86/Q2

Length of 1 side	Total surface area	Volume	Ratio (surface area:volume)
1	6	1	6:1
2	24	8	3:1
3	24	27	2:1
4	96	64	3:2
5	150	125	6:5
6	216	216	1:1

A True
B False
C True
D True

P87/Q3

Thanks to the log scale, the plotted graph will be "compacted" to a smaller area. A possible trend can then be analysed more easily, because more data can be presented using the graph.

(i) 1000cm�

(ii) 10⁶ = 1000kg

P87/Q4

The large ears are used for thermoregulation.
Extra heat can be radiated out of the elephant's body to the atmosphere.

P87/Q5

Reptiles are poikilothermic ("cold-blooded"). They use mostly the external sources of heat (e.g. from the Sun) to raise their body temperature. If the body has a large surface area and a small volume, it will be easier for the organism to trap/absorb radiant heat from the Sun (example: lizards). However, if the reptile has a small surface area in relation to the volume of its body, then more time will be needed to use the available radiant heat to raise the body temperature. This becomes even more important if one considers that at night the temperature of the reptile's body will drop.

P88/Q6

real size [0.003] = (size of image [18mm]) / magnification [?]

Hence, magnification = 18 / 0.003 = 6000

P89/Q7

Sample of surface without microvilli: 50 mm.
This sample is covered by 20 microvilli, each adding 38 mm to the sample.

Hence, 20 x 38mm = 760mm
760 / 50 = 15x

Examinations

P89/Q1

Three features:

Large surface area
Good blood supply/many capillaries
Short diffusion pathway/1 cell thick alveoli & 1 cell thick capillary (2 cell thick in total)

P89/Q2

(i)	Small alveoli/air sacs, hence large surface area: volume ratio Huge number of alveoli (750million)
(i)	Ventilation of the lungs/inspiration and expiration Circulation of blood

There is 15% of oxygen in the exhaled air. The critical concentration of O₂ in the inhaled air which is
detected by the chemoreceptors (carotid & aortid bodies) as a minimum, (no need for increase inbreathing rate) is 12%. 15% is more than 12%, so this concentration of O₂ will be sufficient for resuscitation ventilation.

P90/Q3

Muscles stimulated by nerve A:

Diaphragm
Intercostal muscles

As lungs expand, stretch receptors in the walls of the bronchial tree are stimulated. Impulses are sent via the vagus nerve (B) to the expiratory centre, which automatically cuts off inspiratory activity.

Higher centres of the brain can modify the tidal volume and breathing rate (No of breaths/min). Important for some activities (e.g. diving).

(i) Frequent and shallow breaths ("panting").

(ii) Large proportion (75%) of the tidal volume is a "dead space" (it is all the air present in the trachea and bronchi). It is not involved in gas exchange simply because the air does not reach alveoli. Increased breathing rate alone doesn't change this proportion. Hence the total volume of air involved in the gaseous exchange in the lungs is not much larger.