A2 Level

AS Level

Unit 3

Unit 4

Unit 5

A2 Human Biology Unit 6

Text questions

P115/Q1

P116/Q2

P118/Q3

The increase in the atmospheric CO2 concentration will be even greater due to a huge increase in the combustion of fossil fuels (caused by ever increasing number of cars).

P119/Q4

Higher concentration of CO₂:

Increase in weed growth
Change in the proportion of organic C : organic N, hence lower proportion of protein in the plant tissues.

P120/Q5

Nitrification	Denitrification
1) Ammonia converted 2) Aerobic conditions	1) Nitrates converted (oxidised) to nitrites (reduced) to nitrogen and/or nitrates 2) Anaerobic conditions

P121/Q6

ATP supplies energy necessary for the process of nitrogen fixation (reduction of N to NH₃).

Assignment

P126/Q1

Total length
of hedges
in 1946/m

Total length
of hedges
in 1965/m

Total length
of hedges lost
b. 1946-65/m

% loss of hedge
between
1946-1965

% loss of hedge between 1946-1965. It is a proportion/ratio used to express the changes in the
lengths of the hedges over time. Proportion/ratio can be used to carry out fair comparisons between
different fields in different areas.

P126/Q2

More surface area for growing crops and easier use of heavy machinery.

P126/Q3

Field boundary	Diversity of breeding birds
D (overgrown)	(1) Highest number of species (2) Generally, highest number of individuals representing each species
A (no hedge)	(1) Smallest number of species present
B (remnant)	(1) Smallest number of species present (2) Smallest number of individuals representing each species

P128/Q4

Change in the mean crop yield = -10%
10% of 5.3 = 0.53t
Hence, 5.3 - 0.53 = 4.77t

Change in the mean crop yield = +10%
10% of 5.3 = 0.53t
Hence, 0.53 + 5.3 = 5.83t

P128/Q5

Higher number of species and diversity of breeding birds in the hedges may affect the size of the crop. The crop may be reduced in the strip of land (even 2 metres wide) adjacent to the hedge, because animals feed on the plants growing there.

P129/Q6

(1) Loss of surface area taken up by the hedge;
(2) Loss of crops in the strip of land 2 m wide, adjacent to the hedge;

(1) Increased crop yields in the strip of land, which is between 2 and 15 metres from the hedge; the increase in the crop yield more than offsets losses described in (a) point (2) above.

Examinations

P129/Q1

(i) N2 is nitrogen gas found in the atmosphere (79%)

(ii) Chemical compounds containing N;

Your drawing.

Both the host (leguminous plant) and the bacterium mutually benefit from the feeding relationship, i.e.:

(1) Plant receives most of nitrogenous compounds fixed by the bacterium. N is needed for the synthesis of aminoacids (→proteins);
(2) Bacterium receives the source of energy (glucose) for the process of N-fixation;

(i) Nitrogen gas (N₂) from atmosphere

(ii) Diffusion

Leghaemoglobin is present in the membranes surrounding each bacteroid group. Leghaemoglobin may be responsible for the transfer of protons (H⁺) and electrons (e^-) required for the reaction to occur:

15ATP + 6H⁺ + 6e- + N₂ → 2NH₃ + 15ADP + 15P_i

(i) ATP for the reaction is supplied from the respiration of glucose;

(ii) 15 ATP molecules needed for the fixation of 1 molecule of N₂/dinitrogen;

N in leguminous plants → death → decomposers → NH₃ compounds → nitrite → nitrate → N in (other) green plants;

Inter-specific competition (for light, space, water):

(1) On a soil with low nitrate (NO₃)^- content leguminous plants (LP) have the advantage resulting from their symbiosis with N-fixing bacteria → LP will meet their demand for combined N. Hence, they are able to compete successfully with non-leguminous plants which may not be able to meet their demand for combined N [due to low (NO₃)^- content in the soil].

(2) On a soil with high (NO₃)^- content, non-leguminous plants will meet their demand for combined N, hence they will compete successfully with leguminous plants. LP will use a large proportion of glucose (from photosynthesis) to produce ATP needed in the symbiosis with N-fixing bacteria, rather than "invest" this ATP in synthesis of new tissues/growth.

P131/Q2

(i)

Rates of breakdown	(% per week)
Lignin: 98 - 95 = 3% / 8 weeks	= 0.37%
Cellulose: 83 - 33 = 50% / 8	= 6.25%
pectin: 80 - 22 = 58% / 8	= 7.25%

(ii) Rate of breakdown of lignin is low, hence parts with high lignin content (stems) take longer to break down.

C-compounds → decomposers → respiration → CO₂ → photosynthesis → glucose

P131/Q3

Soil N in kg ha-1 year-1

Gain	Loss
B 14	A 58
C 332	D 71
F 31	E 195
Total 377	324
Balance + 53 (gain)

(i) F

(ii) B

(i) Nitrates (NO₃)^- are reduced to N₂ gas: 2(NO₃)^- → N₂ + 3O₂ + 2e^-

(ii) Denitrification is a process by which denitrifying bacteria obtain oxygen for respiration.

Waterlogged soils are poorly aerated → have low O₂ content, hence are an ideal environment for the growth of anaerobic bacteria, including denitrifying bacteria. Therefore, increased process of denitrification.

On sandy soils → better aeration → less denitrifying bacteria → lower denitrification.