2013 june

Question 1

Figure 1 shows one pair of homologous chromosomes

Name X.(1 mark)

Answer 1

Centromere;

Question 2

Describe the role of X in mitosis.(1 mark)

Answer 2

1. Holds chromatids together;
2. Attaches (chromatids) to spindle;
3. (Allows) chromatids to be separated/move to (opposite) poles / (centromere) divides/splits at metaphase/ anaphase;

Question 3

Homologous chromosomes carry the same genes but they are not genetically identical.
Explain why.
(1 mark)

Answer 3

(Homologous chromosomes) carry different alleles;

Question 4

The appearance of each chromosome in Figure 2 is different from those shown in
Figure 1. Explain why.
(1 mark)

Answer 4

(In Figure 2)

1. Chromatids have separated (during anaphase);
2. Chromatids have not replicated;
3. Chromosomes formed from only one chromatid;

Question 5

Other than independent segregation, give one way in which meiosis allows the
production of genetically different cells.(1 mark)

Answer 5

• crossing over
• alleles exchanged between chromatids or chromosomes
• genetic recombination

Question 6

What is a species?

Answer 6

1. Group of similar organisms / organisms with similar features / / organisms with same genes/chromosomes;
2. Reproduce / produce offspring;
3. That are fertile;

Question 7

The forest was cleared to make more land available for agriculture.
After the forest was cleared the species diversity of insects in the area decreased.
Explain why.
(3 marks)

Answer 7

1. Decrease in variety of plants / fewer plant species;
2. Fewer habitats/niches;
3. Decrease in variety of food / fewer food sources;
4. Aspect of clearing forest (killing insects) eg machinery, pesticides;

Question 8

a hierarchy

2 marks

Answer 8

Groups within groups;

2. No overlap (between groups);

Question 9

a phylogenetic group

1 mark

Answer 9

(Grouped according to) evolutionary links/history/relationships / common ancestry;

Question 10

Explain how these results suggest that animal A is the most closely related to humans(2 marks)

Answer 10

1. (Only) one amino acid different / least differences / similar amino acid sequence / similar primary structure;
2. (So) similar DNA sequence/ base sequence;

Question 11

Cytochrome c is more useful than haemoglobin for studying how closely related
different organisms are. Suggest one reason why.(1 mark)

Answer 11

1. All organisms respire/have cytochrome c;

2. (Cytochrome c structure) is more conserved / less varied (between organisms);

Question 12

DNA helicase is important in DNA replication. Explain why.(2 marks)

Answer 12

1. Separates/unwinds/unzips strands/helix / breaks H-bonds;

2. (So) nucleotides can attach/are attracted / strands can act as templates;

Question 13

Use information in the diagram to suggest how cytarabine prevents DNA replication.(2 marks)

Answer 13

1. Similar shape/structure (to cytosine) / added instead of cytosine / binds to guanine;
2. Prevents (complementary) base pairing / prevents H-bonds forming / prevents formation of new strand / prevents strand elongation / inhibits/binds to (DNA) polymerase;

Question 14

Cytarabine has a greater effect on cancer cells than on healthy cells. Explain why.(1 mark)

Answer 14

(Cancer cells/DNA) divide/replicate fast(er)/ uncontrollably;

Question 15

Give one way in which antibiotics can prevent the growth of bacteria.(1 mark)

Answer 15

Prevent cell wall formation / cause (cell) lysis / inhibit ribosomes / inhibit protein synthesis / prevent DNA replication / affect function of cell membrane;

Question 16

Describe how bacteria can become resistant to antibiotics by vertical gene
transmission.
(1 mark)

Answer 16

Plasmid/genes transmitted through) cell division/reproduction/replication/generations

Question 17

The doctors used samples of this bacterium taken from many patients in the hospital.
Explain why this was important.
(1 mark)

Answer 17

Representative/typical/reliable / different types of bacteria;

Question 18

Doctors in a different hospital repeated this investigation. They found that hypochlorite
had little effect on samples of P. aeruginosa they obtained. Suggest how this different
result may have arisen.
(2 marks)

Answer 18

1. Mutation (in bacterium);

2. Gene/allele for resistance;

Question 19

The oxygen dissociation curve of the fetus is to the left of that for its mother. Explain
the advantage of this for the fetus.
(2 marks)

Answer 19

1. Higher affinity / loads more oxygen;
2. At low/same/high partial pressure/pO2;
3. Oxygen moves from mother/to fetus;

Question 20

After birth, fetal haemoglobin is replaced with adult haemoglobin. Use the graph to
suggest the advantage of this to the baby.
(2 marks)

Answer 20

1. Low affinity / oxygen dissociates;

2. (Oxygen) to respiring tissues/muscles/cells;

Question 21

After birth, fetal haemoglobin is replaced with adult haemoglobin. Use the graph to
suggest the advantage of this to the baby.
(2 marks)

Answer 21

Enough adult Hb produced / enough oxygen released / idea that curves/affinities/Hb are similar / more red blood cells produced;

Question 22

Hereditary persistence of fetal haemoglobin (HPFH) is a condition in which production
of fetal haemoglobin continues into adulthood. Adult haemoglobin is also produced.
People with HPFH do not usually show symptoms. Suggest why.
(1 mark)

Answer 22

Enough adult Hb produced / enough oxygen released / idea that curves/affinities/Hb are similar / more red blood cells produced;

Question 23

Suggest how a mutation can lead to the production of a protein that has one amino
acid missing.
(2 marks)

Answer 23

Loss of 3 bases/triplet = 2 marks;;

Loss of base(s) = 1 mark;

Question 24

Suggest how the production of a protein with one amino acid missing may lead to a
genetic disorder such as Ellis-van Creveld syndrome.
2 marks)

Answer 24

1. Change in tertiary structure/ active site;

2. (So) faulty/non-functional protein /enzyme;

Question 25

Explain how water enters xylem from the endodermis in the root and is then transported to the leaves. (6 marks)

Answer 25

(In the root) 1. Casparian strip blocks apoplast pathway / only allows symplast pathway; 2. Active transport by endodermis; 3. (Of) ions/salts into xylem; 4. Lower water potential in xylem / water enters xylem by osmosis /down a water potential gradient; (Xylem to leaf) 5. Evaporation / transpiration (from leaves); 6. (Creates) cohesion / tension / H-bonding between water molecules / negative pressure; 7. Adhesion / water molecules bind to xylem; 8. (Creates continuous) water column;

Question 26

Abdominal pumping increases the efficiency of gas exchange between the tracheoles and muscle tissue of the insect. Explain why. (2 marks)

Answer 26

1. More air/oxygen enters / air/oxygen enters quickly/quicker; 2. (So) maintains/greater diffusion or concentration gradient;

Question 27

Abdominal pumping is an adaptation not found in many small insects. These small insects obtain sufficient oxygen by diffusion. Explain how their small size enables gas exchange to be efficient without the need for abdominal pumping.(1 mark)

Answer 27

Large(r) SA:VOL / short(er) diffusion distance (to tissues);

Question 28

The insect opens its spiracles at a lower frequency in very dry conditions. Suggest one advantage of this.(1 mark)

Answer 28

Less/no water lost / (more) water retained;

Question 29

The ends of tracheoles connect directly with the insect’s muscle tissue and are filled with water. When flying, water is absorbed into the muscle tissue. Removal of water from the tracheoles increases the rate of diffusion of oxygen between the tracheoles and muscle tissue. Suggest one reason why.(1 mark)

Answer 29

1. Greater surface area exposed to air; 2. Gases move/diffuse faster in air than through water; 3. Increases volume/amount of air;

Question 30

Apart from the size of the female, suggest two factors that should have been kept constant in order to obtain reliable results. (2 marks)

Answer 30

Any two suitable suggestions eg 1. Volume/concentration of skin lipid; 2. Age/sexual maturity; 3. Species of snake; 4. Size of male; 5. Colour; 6. Temperature; 7. Light; 8. Time of day/year/breeding season; 9. Duration/length of time observing; 10. Diet; 11. Filter paper; 12. Size of cage;

Question 31

The male snakes used for each trial were selected at random. Explain why this was important.(1 mark)

Answer 31

To avoid bias;

Question 32

The scientists used different male snakes in each trial. Suggest why.(1 mark)

Answer 32

``` 1. To avoid change in (courtship) behaviour (due to past experience); 2. To observe a typical/general/representative (response); ```

Question 33

What should the scientists have placed in the cage as a control, to show that males were responding to lipids from females?(1 mark)

Answer 33

Filter paper without (skin) lipids / untreated filter paper / filter paper with water / (female) snakes with lipids removed;

Question 34

The scientists concluded that male snakes showed more courtship behaviour towards long female snakes and to lipids from the skins of long females. Explain one other conclusion that can be made from the data.(2 marks)

Answer 34

1. Similar response to lipids and (whole) snakes; 2. (So males are) responding to lipids; 3. (So males are) not responding to (whole) snakes/visual clues;

Question 35

Suggest two advantages of male snakes courting with longer females. (2 marks)

Answer 35

(Parent/offspring) 1. Produce more/larger offspring/eggs; 2. Better predators / fitter / more successful at gaining food / less likely to be eaten / more able to protect offspring/eggs; 3. (More) sexually mature / fertile; 4. Have more food stores for offspring/eggs;

Question 36

Use Figure 3 and Figure 4 to suggest why male garter snakes show greater courtship behaviour to longer female snakes.(2 marks)

Answer 36

1. (Males) respond to/sense (unsaturated) fatty acids; | 2. (Long females) produce/have more fatty acids / positive correlation;

Question 37

The females of other species of snake secrete lipids on their skin. These lipids also contain unsaturated fatty acids. Male garter snakes do not show courtship behaviour towards these females. Suggest why.(1 mark)

Answer 37

(Other females/species) produce different fatty acids;