2019 paper 2

Question 1

Succession occurs in natural ecosystems. Describe and explain how succession
occurs.

Answer 1

1. (Colonisation by) pioneer species;
2. Pioneers/species/organisms change the environment/habitat/conditions/factors;
3. (Environment becomes) less hostile for other/new species
   OR
   (Environment becomes) more suitable for other/new species
   OR
   (Environment becomes) less suitable for previous species;
4. Change/increase in diversity/biodiversity;
5. (To) climax community;

Question 2

Use Figure 1 to calculate the percentage of sunlight energy that would be transferred into the faeces and urine of a zebra. Give your answer to 3 significant figures.

Answer 2

0.155;

Question 3

In this ecosystem the net productivity of the vegetation is 24 525 kJ m−2 year−1
Use this information and Figure 1 to calculate the energy stored in new tissues of the
zebra in kJ m−2 year−1

Answer 3

Answer of 180/178/177.5 = 2 marks;;

Question 4

Sickle cell disease (SCD) is a group of inherited disorders. People with SCD have sickle-shaped red blood cells. A single base substitution mutation can cause one type of SCD. This mutation causes a change in the structure of the beta polypeptide chains in haemoglobin.

Answer 4

1. Change in (sequence of) amino acid(s)/primary structure;

2. Change in hydrogen/ionic/disulfide bonds; 3. Alters tertiary/30 structure;

Question 5

Explain how a single base substitution causes a change in the structure of this polypeptide.
Do not include details of transcription and translation in your answer.

Answer 5

1. Produce healthy (red blood) cells
   OR
   Produce (normal) polypeptide/haemoglobin;
2. No sickle/faulty/SCD (red blood) cells (produced)
   OR
   No defective polypeptide/haemoglobin;
3. Stem/marrow cells (continuously) divide/replicate
   OR
   Less chance of rejection (from brother/sister);

Question 6

Use this information to explain how HSCT is an effective long-term treatment for SCD.

Answer 6

(For gene therapy)
1. No destruction of bone marrow
OR
No destruction of stem cells; 2. Donors are not required;
3. Less/no chance of rejection (own stem cells); (Against gene therapy)
4. Sickle/faulty (red blood) cells still produced
5. Immune response against genetically modified cells/virus
OR
Long-term effect not known (as is new treatment)
OR
Virus could cause side effects;

Question 7

Some scientists have concluded that this method of gene therapy will be a more effective long-term treatment for SCD than HSCT. Use all the information provided to
evaluate this conclusion.

Answer 7

1. Tip produces IAA;
2. Affects concentration of IAA
   OR
   Affects (shoot) length/growth/elongation;
3. Mitosis/division occurs in shoot tips;
4. Affects (shoot) length/growth/elongation;

Question 8

Explain why the student removed the shoot tip from each seedling

Answer 8

1. For respiration;

2. Provide ATP/energy (for growth);

Question 9

Explain why the student added glucose solution to each Petri dish.

Answer 9

1. To prevent/reduce evaporation;
2. (Which) alters concentration of (IAA) solution
   OR
   (Which) alters water potential;

Question 10

Explain why the lids were kept on the Petri dishes.

Answer 10

. Increase in IAA concentration the higher/greater the mean (change in) length;

2. (High) IAA stimulates cell elongation;
3. In roots, growth/elongation less/inhibited;

Question 11

The student produced the different concentrations of IAA using a stock
1 g dm–3 solution of IAA (1 g dm–3 = 1 part per thousand) and distilled water.
Complete Table 2 with the volumes of stock IAA solution and distilled water required
to produce 40 cm3 of 10 ppm (parts per million) IAA solution.

Answer 11

0.4 and 39.6;

Question 12

A student looked at the results and concluded that a decrease in pH does cause a decrease in the force of muscle contraction.
Use Figure 3 to evaluate this conclusion.

Answer 12

1. Lower (force of contraction) in mouse/B (than control/100%) below 29 °C
   OR
   Lower (force of contraction) in rabbit/D (than
   control/100%) below 26.5 °C;
2. Higher (force of contraction) in mouse/B (than control/100%) above 29 °C
   OR
   Higher (force of contraction) in rabbit/D (than
   control/100%) above 26.5 °C; 3. Only (used) mouse and rabbit
   OR
   No other organism/species used;
3. Body temperature of mouse/rabbit higher (than temperatures investigated);
4. Only used one/0.5 pH (below typical pH)
   OR
   (Should) use more pH values;
5. (Used) isolated muscle tissue;
6. No stats test to see if (difference is) significant;

Question 13

Another group of scientists suggested that a decrease in the force of muscle contraction is caused by an increase in the concentration of inorganic phosphate, Pi, in muscle tissues.
Their hypothesis is that an increase in the concentration of Pi prevents the release of calcium ions within muscle tissues.
Explain how a decrease in the concentration of calcium ions within muscle tissues
could cause a decrease in the force of muscle contraction.

Answer 13

04.2
1. (Less/No) tropomyosin moved from binding site
OR
Shape of tropomyosin not changed so binding site not exposed/available;
2. (Fewer/No) actinomyosin bridges formed;
3. Myosin head does not move
OR
Myosin does not pull actin (filaments)
OR
(Less/No) ATP (hydrol)ase (activation);

Question 14

In muscles, pyruvate is converted to lactate during prolonged exercise.
Explain why converting pyruvate to lactate allows the continued production of ATP by
anaerobic respiration.

Answer 14

1. Regenerates/produces NAD
   OR
   oxidises reduced NAD;
2. (So) glycolysis continues;

Question 15

Describe the role of glucagon in gluconeogenesis.

Do not include in your answer details on the second messenger model of glucagon action.

Answer 15

1. (Attaches to receptors on target cells and) activates/stimulates enzymes;
2. Glycerol/amino acids/fatty acids into glucose;. Correct answer of 3.24 = 2 marks;;

Question 16

The gene that codes for glucagon is 9.531 kilobases in length. The DNA helix makes one complete turn every 10 base pairs. Every complete turn is 3.4 nm in length.
Use this information to calculate the length in micrometres (μm) of the gene for
glucagon. Give your answer to 3 significant figures.

Answer 16

Correct answer of 3.24 = 2 marks

Question 17

Explain how increasing a cell’s sensitivity to insulin will lower the blood glucose
concentration.

Answer 17

1. (More) insulin binds to receptors;
2. (Stimulates) uptake of glucose by channel/transport proteins
   OR
   Activates enzymes which convert glucose to glycogen;

Question 18

Explain how inhibiting adenylate cyclase may help to lower the blood glucose
concentration.

Answer 18

1. Less/no ATP is converted to cyclic AMP/cAMP; 2. Less/no kinase is activated;
2. Less/no glycogen is converted to glucose
   OR
   Less/no glycogenolysis;

Question 19

Give the full genotype of the fly numbered 6 in Figure 4.

Answer 19

GgXRXr ;

Question 20

Give one piece of evidence from Figure 4 to show that the allele for grey body colour
is dominant.

Answer 20

If it were recessive all flies of 3 and 4 would be grey
OR
3 and 4 produce 9/black (fly)
OR
Grey parents produce black (fly);

Question 21

Explain one piece of evidence from Figure 4 to show that the gene for body colour is
not on the X chromosome.

Answer 21

1. (Fly) 3 (and 4) produce 9/black (fly)
OR
(Fly) 9 would not be black
OR
(Fly) 9 would be grey
OR
Grey parents/male produce black female (fly);
2. (Fly) 3 would pass dominant allele to 9; 3.(Fly) 2 (and 1) produce 5/grey (fly)
OR Black female produces grey male OR
(Fly) 5 could not be grey
OR
(Fly) 5 would be black;
4.(Fly) 5 would receive recessive allele from 2;

Question 22

A heterozygous grey-bodied, white-eyed female fly was crossed with a black-bodied, red-eyed male fly.
Complete the genetic diagram below to show all the possible genotypes and the ratio
of phenotypes expected in the offspring from this cross.

Answer 22

1.GgXrXr and ggXRY;
2. GgXRXr, ggXRXr, GgXrY and ggXrY;
3. Grey-bodied red-eyed female, black-bodied
red-eyed female, grey-bodied white-eyed male, black-bodied white-eyed male and ratio 1 : 1 : 1 : 1;

Question 23

A population of fruit flies contained 64% grey-bodied flies. Use the Hardy–Weinberg
equation to calculate the percentage of flies heterozygous for gene G.

Answer 23

1. Correct answer of 48% = 2 marks;;

Question 24

In photosynthesis, which chemicals are needed for the light-dependent reaction?
Tick () one box.

Answer 24

NADP, ADP, Pi and water;

Question 25

Describe what happens during photoionisation in the light-dependent reaction.

Answer 25

1. Chlorophyll absorbs light
   OR
   Light excites/moves electrons in chlorophyll;
2. Electron/s are lost
   OR
   (Chlorophyll) becomes positively charged;

Question 26

Explain why the student marked the origin using a pencil rather than using ink.

Answer 26

Ink and (leaf) pigments would mix
OR
(With ink) origin/line in different position
OR
(With pencil) origin/line in same position
OR
(With pencil) origin/line still visible;

Question 27

Describe the method the student used to separate the pigments after the solution of pigments had been applied to the origin.

Answer 27

1. Level of solvent below origin/line;

2. Remove/stop before (solvent) reaches top/end;

Question 28

Use Figure 5 to calculate the Rf value of pigment C.

Answer 28

Accept any answer in range of 0.58 to 0.62;

Question 29

The pigments in leaves are different colours. Suggest and explain the advantage of
Use Figure 5 to calculate the Rf value of pigment C.
[1 mark]
having different coloured pigments in leaves.

Answer 29

(Absorb) different/more wavelengths (of light) for photosynthesis;

Question 30

What is a DNA probe?

Answer 30

1. (Short) single strand of DNA;

2. Bases complementary (with DNA/allele/gene)

Question 31

Describe how the DNA is broken down into smaller fragments.

Answer 31

1. Restriction endonuclease/enzyme;
2. (Cuts DNA at specific) base sequence
   OR
   (Breaks) phosphodiester bonds
   OR
   (Cuts DNA) at recognition/restriction site;

Question 32

The DNA on the nylon membrane is treated to form single strands. Explain why.

Answer 32

(So DNA) probe binds/attaches/anneals;

Question 33

Lane 1 of Figure 7 enabled the size of the different viral fragments to be determined.
Suggest and explain how.

Answer 33

1. (Lane 1 has DNA fragments) of known sizes/lengths;

2. Compare (position of viral fragment/s);

Question 34

Which volunteers had at least one of the viral DNA fragments with 250 base pairs or
535 base pairs?

Answer 34

3, 4, 5 with these numbers in any sequence;

Question 35

Describe how you could estimate the size of a population of sundews in a small
marsh.

Answer 35

1. Use a grid
   OR
   Divide area into squares/sections;
2. Method of obtaining random coordinates/numbers e.g. calculator/computer/random numbers table/generator;
3. Count number/frequency in a quadrat/section;
4. Large sample and calculate mean/average number (per quadrat/section);
5. Valid method of calculating total number of sundews, e.g. mean number of plants per quadrat/section/m2 multiplied by number of quadrats/sections/m2 in marsh;

Question 36

Suggest and explain how digesting insects helps the sundew to grow in soil with very
low concentrations of some nutrients.

Answer 36

1. Digestion/breakdown of proteins; 2. Provides amino acids
   OR
   (Sundew can) produce a named (organic) nitrogen-containing compound e.g. proteins, amino acids, DNA, ATP;
2. Digestion/breakdown of named (organic) phosphate-containing compound e.g. DNA, RNA;
3. Provides named (organic) phosphate-containing product e.g. nucleotides

Question 37

Damage to the myelin sheath of neurones can cause muscular paralysis (lines 2–4).
Explain how.

Answer 37

1. (Refers to) saltatory conduction
   OR
   (Nerve) impulses/depolarisation/ions pass to other neurones
   OR
   Depolarisation occurs along whole length (of axon);
2. (Nerve) impulses slowed/stopped; 3. (Refers to) neuromuscular junction
   OR
   (Refers to) sarcolemma;

Question 38

Sometimes Guillain–Barré syndrome causes heart rate irregularities (lines 4–5).
Suggest and explain why.

Answer 38

1. Slower/fewer impulse(s) along sympathetic/parasympathetic (pathway/neurones);
2. (Impulses) from cardiac centre
   OR
   (Impulses) from medulla; 3.To SAN

Question 39

The first successful drug trial to reduce concentrations of huntingtin in the brain used single-stranded DNA molecules (lines 13–14).
Suggest and explain how this drug could cause a reduction in the concentration of the
protein huntingtin.

Answer 39

1. It/DNA is complementary to (m)RNA;

2. Binds to mRNA (for huntingtin); 3. Prevents translation;

Question 40

Scientists from the first successful drug trial to reduce concentrations of
huntingtin (lines 9–11) reported that the drug is not a cure for Huntington’s disease.
Suggest two reasons why the drug should not be considered a cure. Do not include repeats of the drug trial in your answer

Answer 40

1. Small sample size
OR
Only 46;
2. Only four-months
OR
short period (of trial);
3. Huntingtin/protein reduced
OR
Huntingtin/protein still produced
OR
Huntingtin/protein not removed;
4. Allele/gene/mutation/mRNA (for Huntington’s) still present
OR
(May be) temporary
OR
Drug treatment repeated;
5. Brain already damaged
OR
Brain damage may continue;

Question 41

Suggest two reasons why people had the drug injected into the cerebrospinal fluid (lines 12–13) rather than taking a pill containing the drug.

Answer 41

1. (Drug/DNA) will directly/quickly reach brain
OR
(Cerebrospinal) fluid bathes the brain;
2. (Drug/DNA) not destroyed by acid
OR
(Drug/DNA) not digested (by enzymes);

Question 42

Suggest and explain one way epigenetics may affect the age when symptoms of
Huntington’s disease start.

Answer 42

1. (Increased) methylation of DNA/gene/allele;
2. Inhibits/prevents transcription;
3. Decreased methylation of DNA/gene/allele;
4. Stimulates/allows transcription;
   OR
5. Decreased acetylation of histone(s);
6. Inhibits transcription;
   OR
7. Increased acetylation of histone(s); 8. Stimulates/allows transcription;