Biology Higher Level

MOCK2010/BIO

22090085

BIOLOGY – HIGHER LEVEL

Thursday 22nd March 2012(pm)

3 hours

Student Name ______

Teacher Name ______

INSTRUCTIONS TO CANDIDATES

Donotopenthisexaminationpaperuntilinstructedtodoso.

Section A: answer all of Section A by placing a X in the correct box on the multiple choice answer sheet.

Section B: answer all of Section B in the spaces provided.

Section C: answer two questions from Section C. Write your answers on the lined paper

provided.Write your name on each sheet, and attach them to this examination paper.

You may use a calculator.

SECTION A

1. C

2. B

3. C

4. B

5. A

6. D

7. D

8. D

9. C

10. B

11. C

12. C

13. A

14. D

15. A

16. C

17. C

18. B

19. A

20. B

21. D

22. C

23. D

24. C

25. B

26. A

27. A

28. B

29. C

30. C

SECTION B

Answer all the questions in the spaces provided.

1. Genetic engineering allows genes for resistance to pest organisms to be inserted into various

crop plants. Bacteria such as Bacillus thuringiensis (Bt) produce proteins that are highly toxic

to specific pests.

Stem borers are insects that cause damage to maize crops. In Kenya, a study was carried out

to see which types of Bt genes and their protein products would be most efficient against three

species of stem borer. The stem borers were allowed to feed on nine types of maize (A–I),

modified with Bt genes. The graph below shows the leaf areas damaged by the stem borers after

feeding on maize leaves for five days.

(This question continues on the following page)
(Question 1 continued)

(a) Calculate the percentage difference in leaf area damaged by Busseola fuscabetween

the control and maize type A. Show your working. [2]

43-23 = 20

20/43 x 100 = 46.5% (47%)

(b) Discuss which species of stem borer was most successfully controlled by the genetic

engineering of the maize plants. [3]

Sesamia (was most successfully controlled);

in control plants Sesamia caused most damage;

all types of Bt/genetically modified maize/A–I show (significant) decrease in damage by Sesamia;

mark for correct numerical comparison;

Sesamia caused no damage to type E/ in one instance;

Busseola not controlled/affected by Bt/genetically modified maize/caused largest amount of damage in types A–I/increased damage in some varieties;

Eldana controlled by some types of maize / B/C/D but not others / Eldana caused least damage in control and not much difference in many maize types;

(This question continues on the following page)
(Question 1 continued)

Before the use of genetically modified maize as a food source, risk assessment must be carried

out. A 90-day study was carried out in which adult male and female rats were fed either:

• seeds from a Bt maize variety

• seeds from the original non-Bt maize variety

• commercially prepared rat food.

All the diets had similar nutritional qualities.

day 28 to 56. [2]

males: ; (Accept answers in range 60–80 g) (460 - 390 )= 70g

females: ; (Accept answers in range 25–35 g) (280 - 250 )= 30g

Units required, no workings required.

(d) Evaluate the use of Bt maize as a food source on the growth of the rats. [2]

(promotes) highest rate of growth at start of study / tapering off later in the study;

Bt maize appears to cause less growth/mass gain than rat food / vice versa;

more pronounced difference in females;

no difference in growth/mass gain between Bt and non-Bt maize;

(e) Comment on the use of Bt maize as a food source compared to the other diets tested. [1]

(Bt) maize may not be as good as the (commercially prepared) rat food;

Bt maize appears to be as good a food source as non-Bt maize;

Bt maize an acceptable/safe food source; [1 max]

Answers require a judgement about Bt maize as a food source rather than a description. (This question continues on the following page)

(Question 1 continued)

Studies have shown that Bt proteins are released by plant roots and remain in the soil.

One study looked at the biomass of microorganisms in soil surrounding the roots of:

• Bt maize

• non-Bt maize

• non-Bt maize with an insecticide (I).

The graph below shows the biomass of microorganisms at two different times in the growth

cycle of the plants (Flower and Harvest). Error bars represent standard error of the mean.

(f) State one role of bacteria in a soil ecosystem. [1]

decomposers / recycle nutrients / cause decay / nitrification/nitrogen fixation / denitrification

(g) Compare the biomass of microbes in the soils surrounding the roots of Bt maize and

non-Bt maize. [2]

(for both groups) overall biomasses were higher during flowering than harvest / vice versa

the microbial biomass for the Bt crop was (slightly) lower than for the non-Bt crops at flower time;

the microbial biomass for the Bt crop was (slightly) higher than for the non-Bt crops at harvest time;

(This question continues on the following page)

(h) The researchers’ original hypothesis stated that microorganisms would be negatively

affected by the Bt protein released by the plant roots. Discuss whether the data supports

the hypothesis. [2]

data does not support the hypothesis as there is little difference between biomass found in the soil (surrounding) roots (of the Bt and non-Bt) at either time;

data does not support the hypothesis as there is a slightly positive effect at harvest;

data supports hypothesis as there is a slightly negative effect at flowering;

Bt proteins act as toxins to insects, primarily by destroying epithelial cells in the insect’s

digestive system. Below is the three-dimensional structure of one such protein.

(i) (i) State the type of structure shown in the region marked A in the diagram above. [1]

alpha helix

(ii) Outline how this structure is held together. [2]

hydrogen bonds;

between the turns of the helix (rather than between R-groups);

bonds between carboxyl and NH groups/C-O---H-N;

(iii) Region A inserts into the membrane. Deduce, with a reason, the nature of the

amino acids that would be expected to be found in this region. [2]

non-polar amino acids/R-groups;

(inner part of phospholipid) bilayer is hydrophobic/non-polar;
2. (a) In some maize plants the seed is enclosed in a green sheath called a tunica.

The allele (T) for this is dominant to the allele (t) for normal, unenclosed seeds.

The endosperm of the seed can be starchy (allele E) or sugary (allele e). The genes

for these two characteristics are linked. The table below shows the outcome of crosses

between a plant heterozygous for both characteristics and one that is homozygous

recessive for both characteristics.

Phenotype / Number
Tunica present, starchy / 326
Unenclosed seeds, starchy / 111
Tunica present, sugary / 118
Unenclosed seeds, sugary / 295

(i) State the genotype of the homozygous parent using the correct notation. [1]

t e

____

t e

(ii) Identify which individuals are recombinants in this cross. [1]

unenclosed seeds, starchy and tunica present, sugary

T e

____

t e

t E

____ both needed

t e

(iii) Explain what has occurred to cause these results. [2]

crossing over;

between non-sister chromatids (in prophase I);

results in exchange of alleles / change in linkage groups;

so some gametes are or ; (linkage notation not expected) T e t E

test cross expect ratio of two phenotypes / correct Punnett Square showing test cross;

but instead get four phenotypes with smaller percentage of recombinants;

Above points can be shown in diagrams.

(b) Maize belongs to the group of plants known as angiospermophyta. Distinguish between

angiospermophytes and filicinophytes. [2]

Angiospermophytes / Filicinophytes
Produce flowers / No flowers
Woody stems / Non-woody stems
Max height 15m / Max height 100m
Leaves present / Pinnate leaves present
Seeds produced / Spores produced

3. (a) Draw a labelled diagram showing two different complementary pairs of nucleotides in

a molecule of DNA. [4]

(b) Outline the structure of nucleosomes. [2]

(eight) histone (proteins);

DNA wrapped around histones/nucleosome;

further histone holding these together; [2 max]

Do not allow histone wrapped around DNA.
SECTION C

Answer two questions. Up to two additional marks are available for the construction of your answers.

Write your answers on the answer sheets provided. Write your session number on each answer sheet, and

attach them to this examination paper and your cover sheet using the tag provided.

4. (a) Draw a labelled diagram showing the tissues present in a dicotyledonous leaf. [4]

Award [1] for each structure clearly drawn and correctly labelled. Accept a plan diagram without individual cells.

upper and lower epidermis;

palisade mesophyll under upper epidermis 1/3 to 1/2 of leaf thickness;

spongy mesophyll/layer in lower half of leaf;

vein showing separate areas of xylem above phloem;

stoma/stomata labelled in (lower) epidermis;

two guard cells; (at least one must be labelled for mark)

(b) Outline the light-dependent reactions of photosynthesis. [6]

(chlorophyll/antenna) in photosystem II absorbs light;

absorbing light/photoactivation produces an excited/high energy/free electron;

electron passed along a series of carriers;

reduction of NADP+ / generates ; NADPH + H+

absorption of light in photosystem II provides electron for photosystem I;

photolysis of water produces H+ /O2;

called non-cyclic photophosphorylation;

in cyclic photophosphorylation electron returns to chlorophyll;

generates ATP by H+ pumped across thylakoid membrane / by chemiosmosis / through ATP synthetase/synthase;

both light and temperature can be limiting factors;

other factors can be limiting;

graph showing increase and plateau with increasing light / description of this;

graph showing increase and decrease with increasing temperature / description of this;

light:

affects the light-dependent stage;

at low intensities insufficient ATP;

and insufficient produced; NADPH H

this stops the Calvin cycle operating (at maximum rate);

temperature:

affects light-independent stage / Calvin cycle;

temperature affects enzyme activity;

less active at low temperatures / maximum rate at high temperatures;

but will then be denatured (as temperature rises further);

Award [5 max] if only one condition is discussed.

5. (a) Outline a possible cause of Down syndrome. [4]

non-disjunction (can cause Down syndrome);

occurs when pair of homologous chromosomes fails to separate during meiosis;

one gamete/daughter cell receives two chromosomes / diagram showing this;

occurs in anaphase I/II of meiosis;

fertilization involving this gamete leads to change in chromosome number/47 chromosomes;

most common form of Down is trisomy 21/extra chromosome 21;

increase risk of Down syndrome with increased age of mother;

(b) Outline the processes involved in oogenesis within the human ovary. [8]

oogenesis is process by which female gametes/eggs are produced;

begins during fetal development; oogonia/large number of cells formed by mitosis;

oogonia/cells enlarge/undergo cell growth/become primary oocytes;

begin first meiotic division but stop in Prophase I;

until puberty;

(at puberty) some follicles develop each month in response to FSH;

(primary oocyte) completes first meiotic division;

forms two cells of different/unequal sizes / unequal distribution of cytoplasm;

(creating a) polar body;

polar body eventually degenerates;

larger cell/secondary oocyte proceeds to meiosis II;

stops at prophase II;

meiosis II completed if cell is fertilized;

ovum and second polar body formed;

pros/positive considerations: [3 max]

chance for infertile couples to have children;

decision to have children is clearly a conscious one due to difficulty of becoming pregnant;

genetic screening of embryos could decrease suffering from genetic diseases;

spare embryos can safely be stored for future pregnancies/used for stem cell research;

cons/negative considerations: [3 max]

IVF is expensive and might not be equally accessible;

success rate is low therefore it is stressful for the couple;

it is not natural/cultural/religious objections;

could lead to eugenics/gender choice;

could lead to (unwanted) multiple pregnancies with associated risks;

production and storage of unused embryos / associated legal issues / extra embryos may be used for (stem cell) research;

inherited forms of infertility might be passed on to children;

Accept any other reasonable answers.

6. (a) Outline the various means of transfer of different types of molecules through the plasma

membrane. [4]

diffusion is the movement of particles down a concentration gradient / higher to

lower concentration / is passive;

osmosis is passive transport / diffusion of water;

osmosis is movement from lower solute concentration to higher / higher to lower

water potential;

facilitated diffusion involves channels (in membranes);

active transport requires protein pumps/ATP/energy;

active transport is movement against concentration gradient;

correct reference to endocytosis/exocytosis/pinocytosis;

(b) Describe the transport of water through an angiosperm root system. [6]

root hairs/epidermal cells take up water by osmosis;

correct reference to root pressure;

symplastic pathway through cytoplasm (of cells);

by diffusion / down concentration gradient;

apoplastic pathway through (cortex) cell walls;

by capillary action;

Casparian strip blocks apoplastic pathway;

water passes into xylem;

transpiration causes the pull of water/transpiration stream (through plant);

maintenance within narrow limits / constant level;

by negative feedback (mechanisms);

occurs in nephron (of kidney);

collecting duct is main site of osmoregulation;

water levels (in blood) detected / monitored;

by hypothalamus;

if (blood) water level is low (more) ADH is secreted;

by pituitary gland;

collecting duct (and distal convoluted tubule) more permeable;

correct reference to aquaporins;

more water reabsorbed;

urine concentrated;

if water level is high ADH not secreted / less ADH secreted;

less water reabsorbed;

urine dilute;

7. (a) Outline the effects of factors that increase or decrease the size of a population. [4]

natality/immigration causes increased population size;

mortality/emigration causes decreased population size;

predation / disease / any other limiting factor decreases population size;

population change (natality immigration) −(mortality emigration);

natural disasters / density independent factors;

(b) Describe the production of antibodies as the response of the immune system to the entry of pathogens into the body. [6]

pathogen is a disease causing organism/virus;

antigen is foreign protein/molecule in body;

macrophages engulf antigen/pathogen;

display antigen on membrane;

activate helper T-cells;

which activate B-cells;

to divide and clone;

to produce plasma cells;

which produce antibodies;

which inactivate/bind to the antigens;

memory cells for long-term protection;

a food chain includes a producer and consumers;

represents the direction of energy flow;

energy loss occurs between trophic levels;

due to material not consumed/assimilated;

and from heat loss due to cell respiration;

energy passed on from one level to next is 10−20%;

which limits length of food chain;

photosynthesis / producers convert solar energy to chemical energy (in organic

molecules);

consumers obtain necessary energy from eating organisms of previous trophic level;

an energy pyramid shows the flow of energy from one tropic level to the next

(in a community);

units are energy per unit area per unit time / Jm−2 yr−1 ;

Pyramid of energy – properly drawn, each level no more than one fifth the width of

the level below it, with three correctly labelled trophic levels

e.g. producer, primary consumer;