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Question 1

Enzymes required for complete hydrolysis of proteins to amino acids

Answer 1

• Endopeptidases - hydrolyse internal peptide bonds
• Exopeptidases - remove amino acids/hydrolyse bonds at end
• Dipeptidases

Question 2

The pieces of leaf tissue examined by light microscope were very thin.
Explain why this was important (2)

Answer 2

1. Single/few layer(s) of cells;
2. So light can pass through;

Question 3

Suggest two reasons why the rate of water uptake by a plant might not be the same as the rate of transpiration (2)

Answer 3

1. Water used for support/turgidity;
2. Water used in photosynthesis;
3. Water used in hydrolysis;
4. Water produced during respiration;

Question 4

Xerophyte features

Answer 4

1. Hairs so ‘trap’ water vapour and water potential gradient decreased;
2. Stomata in pits/grooves ** so** ‘trap’ water vapour and water potential gradient decreased;
3. Thick (cuticle/waxy) layer so increases diffusion distance;
4. Waxy layer/cuticle so reduces evaporation/transpiration.
5. Rolled/folded/curled leaves so ‘trap’ water vapour and water potential gradient decreased;
6. Spines/needles so reduces surface area to volume ratio;

Question 5

Explain why the removal of hedges decresases the diversity of birds (3)

Answer 5

1. Removes species/types of plant/insect;
2. Fewer food sources;
3. Fewer habitats/niches;

Question 6

Suggest two reasons why an increase in phosphate can be linked to growth of the embryo (2)

Answer 6

1. (Phosphate required) to make RNA;
2. (Phosphate required) to make DNA;
3. (Phosphate required) to make ATP/ADP;
4. (Phosphate required) to make membranes;
5. (Phosphates required) for phosphorylation;

Question 7

The chilling requirement of seeds of certain plant species is considered to be an adaptation for survival in countries with seasonal changes in environmental conditions.
Suggest how this adaptation may enable these plant species to survive and respond to seasonal changes (3)

Answer 7

1. Seeds/embryo remain dormant/inactive in winter/cold
   OR
   Growth/development of seed/embryo during winter/cold;
2. Seeds/plants develop in spring/summer
   OR
   Seeds/plants develop when temperature/light increases;
3. Plant photosynthesise (in spring/when warm);
4. Produce (more) seeds/offspring in spring/growing season;

Question 8

The mean FEV1 value of non-smokers decreases after the age of 30.
Use your knowledge of ventilation to suggest why (1)

FEV1 = forced expiration volume

Answer 8

1. Internal intercostal muscle(s) less effective
   OR
   Less elasticity (of lung tissue);

Question 9

One of the severe disabilities that results from emphysema is that walking upstairs becomes difficult.
Explain how a low FEV1 value could cause this disability (3)

FEV1 = forced expiration volume

Answer 9

1. Less carbon dioxide removed;
2. Less oxygen (uptake/in blood);
3. Less (aerobic) respiration/ATP
   OR
   (More) anaerobic respiration;

Question 10

Explain how mutation can result in the production of non functional proteins

Answer 10

1. Change in DNA base/nucleotide (sequence);
2. Change in amino acid (sequence)/primary structure;
3. Alters (position of) hydrogen/ionic/disulfide bonds;
4. Change in tertiary structure (of receptor);

Question 11

The frequency of the CCR5 mutation is highest in Europe. Scientists have collected data on the history and number of HIV infections in Europe. Using these data, scientists have concluded that the high frequency of the CCR5 mutation is not due to natural selection in response to HIV.
Suggest two reasons why scientists reached this conclusion (2)

Answer 11

1. Low/lower exposure to HIV (in Europe)
   OR
   Low/lower number of HIV/AIDS (infections/cases);
2. (HIV) has only been present for a short time period
   OR
   (HIV relatively) recently evolved;
3. Mutation/CCR5 has been around for many years;
4. Mutation/CCR5 is advantageous (for something else);

Question 12

Some people produce a much higher ventricular blood pressure than normal. This can cause tissue fluid to build up outside the blood capillaries of these people.
Explain why (2)

Answer 12

1. More fluid forced/filtered out of capillary/blood (due to high pressure);
2. Less return of fluid (into capillary/blood) due to pressure
   OR
   Lymph(atic) (system) cannot drain away all excess fluid;

Question 13

suggeset how widening of blood vessels can reduce ventricular blood pressure (2)

Answer 13

• less resistance
• greater blood flow

Question 14

Describe how the structures of starch and cellulose molecules are related to their functions (5)

Answer 14

Starch (max 3)
1. Helical/ spiral shape so compact;
2. Large (molecule)/insoluble so osmotically inactive/does not affect Ψ
3. Branched so glucose is (easily) released for respiration;
4. Large (molecule) so cannot leave cell/cross cell-surface membrane;

Cellulose (max 3)
5. Long, straight/unbranched chains of β glucose;
6. Joined by hydrogen bonding;
7. To form (micro/macro)fibrils;
8. Provides rigidity/strength;

6. Note: references to ‘strong hydrogen bonds’ disqualifies this mark point.

Question 15

Describe the processes involved in the transport of sugars in plant stems (5)

Answer 15

1. (At source) sucrose is actively (transported) into the phloem/sieve element/tube;
2. By companion/transfer cells;
3. Lowers water potential in phloem/sieve element/tube and water enters by osmosis;
4. (Produces) high (hydrostatic) pressure;
5. Mass flow/transport towards sink/roots/storage tissue;
6. At sink/roots sugars are removed/unloaded;

Question 16

Describe the role of a ribosome in the production of a polypeptide.
Do not include transcription in your answer (3)

Answer 16

1. mRNA binds to ribosome;
2. Idea of two codons/binding sites;
3. (Allows) tRNA with anticodons to bind/associate;
4. (Catalyses) formation of peptide bond between amino acids (held by tRNA molecules);
5. Moves along (mRNA to the next codon)/translocation described;

Question 17

Describe the structure of glycogen

Answer 17

1. Polysaccharide of α-glucose;
   OR
   polymer of α-glucose;
2. (Joined by) glycosidic bonds
   OR
   Branched structure;

Question 18

Suggest how glycogen acts as a source of energy

Answer 18

1. Hydrolysed (to glucose);
2. Glucose used in respiration;

Question 19

Suggest and explain two ways the cell-surface membranes of the cells lining the uterus may be adapted to allow rapid transport of nutrients

Answer 19

1. Membrane folded so increased/large surface area;
   OR
   Membrane has increased/large surface area for (fast) diffusion/facilitated diffusion/active transport/co-transport;
2. Large number of protein channels/carriers (in membrane) for facilitated diffusion;
3. Large number of protein carriers (in membrane) for active transport;
4. Large number of protein (channels/carriers in membrane) for co-transport;

Question 20

Sodium ions from salt (sodium chloride) are absorbed by cells lining the gut. Some of these cells have membranes with a carrier protein called NHE3.

NHE3 actively transports one sodium ion into the cell in exchange for one proton (hydrogen ion) out of the cell.

Use your knowledge of transport across cell membranes to suggest how NHE3 does this (3)

Answer 20

1. Co-transport;
2. Uses (hydrolysis of) ATP;
3. Sodium ion and proton bind to the protein;
4. Protein changes shape (to move sodium ion and/or proton across the membrane);

Question 21

High absorption of salt from the diet can result in a higher than normal concentration of salt in the blood plasma entering capillaries. This can lead to a build-up of tissue fluid.

Explain how (2)

Answer 21

1. (Higher salt) results in lower water potential of tissue fluid;
2. (So) less water returns to capillary by osmosis (at venule end);
   OR
3. (Higher salt) results in higher blood pressure/volume;
4. (So) more fluid pushed/forced out (at arteriole end) of capillary;

Question 22

Suggest and explain one advantage to a bacterium of secreting an extracellular protease in its natural environment (2)

Answer 22

1. To digest protein;
2. (So) they can absorb amino acids for growth

Question 23

Describe the action of these membrane-bound dipeptidases and explain their importance (2)

Answer 23

1. Hydrolyse (peptide bonds) to release amino acids;
2. Amino acids can cross (cell) membrane;

Question 24

Describe how phagocytosis of a virus leads to presentation of its antigens (3)

Answer 24

1. Phagosome/vesicle fuses with lysosome;
2. (Virus) destroyed by lysozymes/hydrolytic enzymes;
3. Peptides/antigen (from virus) are displayed on the cell membrane

Question 25

Describe how presentation of a virus antigen leads to the secretion of an antibody against this virus antigen (3)

Answer 25

1. Helper T cell/TH cell binds to the antigen (on the antigen-presenting cell/phagocyte); 2. This helper T/TH cell stimulates a specific B cell; 3. B cell clones **OR** B cell divides by mitosis; 4. (Forms) plasma cells that release antibodies;

Question 26

Explain three ways in which an insect’s tracheal system is adapted for efficient gas exchange

Answer 26

1. Tracheoles have thin walls **so** short diffusion distance to cells; 2. Highly branched/large number of tracheoles **so** short diffusion distance to cells; 3. Highly branched/large number of tracheoles **so** large surface area (for gas exchange); 4. Tracheae provide tubes full of air **so** fast diffusion (into insect tissues); 5. Fluid in the end of the tracheoles that moves out (into tissues) during exercise **so** faster diffusion through the air to the gas exchange surface; **OR** Fluid in the end of the tracheoles that moves out (into tissues) during exercise **so** larger surface area (for gas exchange); 6. Body can be moved (by muscles) to move air **so** maintains diffusion/concentration gradient for oxygen/carbon dioxide;

Question 27

Contrast how an optical microscope and a transmission electron microscope work and contrast the limitations of their use when studying cells (6)

Answer 27

1. TEM use electrons **and** optical use light; 2. TEM allows a greater resolution; 3. (So with TEM) smaller organelles/named cell structure can be observed **OR** greater detail in organelles/named cell structure can be observed; 4. TEM view only dead/dehydrated specimens **and** optical (can) view live specimens; 5. TEM does not show colour **and** optical (can); 6. TEM requires thinner specimens; 7. TEM requires a more complex/time consuming preparation; 8. TEM focuses using magnets and optical uses (glass) lenses;

Question 28

What is a homologous pair of chromosomes?

Answer 28

Two chromosomes that carry the same genes

Question 29

Species Richness

Answer 29

A measure of the number of different species in a community

Question 30

How does the formation of an enzyme-substrate complex increase the rate of reaction.

Answer 30

1. Reduces activation energy; 2. Due to bending bonds

Question 31

Why use passive immunity over active immunity when bitten by a venomous snake

Answer 31

1. Antivenom antibodies bind to the antigen and causes its destruction 2. Active immunity would be too slow/slower;

Question 32

Why is the second vaccine dosage larger than the first (humoral response)

Answer 32

1. B cells specific to the venom reproduce by mitosis; 2. B cells produce plasma cells and memory cells; 3. The second dose produces antibodies (in secondary immune response) in higher concentration and quickly OR The first dose must be small so the animal is not killed;

Question 33

Name and Role of enzymes in semi conservative replication

Answer 33

DNA Helicase - catalyses breaking of hydrogen bonds between complementary base pairs DNA Polymerase - joins DNA nucleotides by forming phosphodiester bonds

Question 34

Describe the gross structure of the human gas exchange system and how we breathe in and out

Answer 34

Order: Trachea, Bronchi, Bronchioles, Alveoli Breathing in: - External intercostal muscles actively contract whilst internal intercostal muscles relax - Diaphragm contracts and flattens - Ribcage moves upwards and outwards - volume increase and pressure decrease in thoracic cavity - Air forced into the lungs from the atmosphere down pressure gradient Breathing out: - Internal intercostal muscles actively contract whilst external intercostal muscles relax - Diaphragm relaxes and returns to dome shape - Ribcage moves downwards and inwards - volume decrease and pressure increase in thoracic cavity - Air forced out of the lungs into the atmosphere down pressure gradient

Question 35

Compare and contrast the structure and properties of triglycerides and phospholipids

Answer 35

1. Both contain ester bonds (between glycerol and fatty acid); 2. Both contain glycerol; 3. Fatty acids on both may be saturated or unsaturated; 4. Both are insoluble in water; 5. Both contain C, H and O but phospholipids also contain P; 6. Triglyceride has three fatty acids and phospholipid has two fatty acids plus phosphate group; 7. Triglycerides are hydrophobic/non-polar and phospholipids have hydrophilic and hydrophobic region; 8. Phospholipids form monolayer (on surface)/micelle/bilayer (in water) but triglycerides don’t;

Question 36

Describe how lactose is formed and where in the cell it would be attached to a polypeptide to form a glycoprotein

Answer 36

- Glucose + Galactose - Joined by condensation reaction - Joined by glycosidic bond - Added to polypeptide in Golgi Apparatus

Question 37

Describe the role of micelles in the absorption of fats into the cells lining the ileum (3)

Answer 37

1. Micelles include bile salts and fatty acids; 2. Make the fatty acids (more) soluble in water; 3. Bring/release/carry fatty acids to cell/lining (of the ileum); 4. Maintain high(er) concentration of fatty acids to cell/lining (of the ileum); 5. Fatty acids (absorbed) by diffusion;

Question 38

why is there a small increase in pressure and rate of blood flow in the aorta even after it surpasses the ventricles in pressure

Answer 38

1. elastic recoil 2. maintains blood pressure

Question 39

Describe the role of DNA polymerase in the semi-conservative replication of DNA (2)

Answer 39

1. Joins (adjacent DNA) nucleotides; 2. (Catalyses) condensation (reactions); 3. (Catalyses formation of) phosphodiester bonds (between adjacent nucleotides);

Question 40

Describe how an enzyme can be phosphorylated

Answer 40

1. Attachment/association of (inorganic) phosphate (to the enzyme); 2. (Released from) hydrolysis of ATP OR (Released from) ATP to ADP + Pi;

Question 41

Suggest how one antibody can be specific to tick protein and to alpha gel

Answer 41

1. similar tertiary structure/shape 2. antibody is complementary to both

Question 42

Non coding base sequences

Answer 42

1. DNA that does not code for protein / polypeptides OR DNA that does not code for (sequences of) amino acids OR DNA that does not code for tRNA/rRNA; 2. (Positioned) between genes;

Question 43

Describe how mRNA is formed by transcription in eukaryotes (5)

Answer 43

1. Hydrogen bonds (between DNA bases) break; 2. (Only) one DNA strand acts as a template; 3. (Free) RNA nucleotides align by complementary base pairing; 4. (In RNA) Uracil base pairs with adenine (on DNA) OR (In RNA) Uracil is used in place of thymine; 5. RNA polymerase joins (adjacent RNA) nucleotides; 6. (By) phosphodiester bonds (between adjacent nucleotides); 7. Pre-mRNA is spliced (to form mRNA) OR Introns are removed (to form mRNA);

Question 44

Describe how a polypeptide is formed by translation of mRNA (6)

Answer 44

1. (mRNA attaches) to ribosomes OR (mRNA attaches) to rough endoplasmic reticulum; 2. (tRNA) anticodons (bind to) complementary (mRNA) codons; 3. tRNA brings a specific amino acid; 4. Amino acids join by peptide bonds; 5. (Amino acids join together) with the use of ATP; 6. tRNA released (after amino acid joined to polypeptide); 7. The ribosome moves along the mRNA to form the polypeptide;

Question 45

Define gene mutation and explain how a gene mutation can have: - no effect on an individual - a positive effect on an individual (4)

Answer 45

(Definition of gene mutation) 1. Change in the base/nucleotide (sequence of chromosomes/DNA); 2. Results in the formation of new allele; (Has no effect because) 3. Genetic code is degenerate (so amino acid sequence may not change); OR Mutation is in an intron (so amino acid sequence may not change); 4. Does change amino acid but no effect on tertiary structure; 5. (New allele) is recessive so does not influence phenotype; (Has positive effect because) 6. Results in change in polypeptide that positively changes the properties (of the protein) OR Results in change in polypeptide that positively changes a named protein; 7. May result in increased reproductive success OR May result in increased survival (chances)

Question 46

Explain the advantage for larger animals of having a specialised system that facilitates oxygen uptake

Answer 46

1. Large organisms have a small surface area:volume 2. Overcomes long diffusion pathway OR Faster diffusion;

Question 47

Explain how the counter-current principle allows efficient oxygen uptake in the fish gas exchange system.

Answer 47

1. Blood and water flow in opposite directions; 2. Diffusion/concentration gradient (maintained) along (length of) lamella/filament;

Question 48

Describe precautions you should take when clearing away after dissections

Answer 48

1. Carry/wash sharp instruments by holding handle 2. Disinfect instruments/surfaces; 3. Disinfect hands OR Wash hands with soap (and water); 4. Put organ/gloves/paper towels in a (separate) bag/bin/tray to dispose; Disinfect = use antiseptic

Question 49

Describe how a sample of chloroplasts could be isolated from leaves (4)

Answer 49

1. Break open cells/tissue and filter 2. In cold, same water potential, pH controlled solution; 3. Centrifuge and remove nuclei/cell debris; (pellet and supernatant) 4. Centrifuge at higher speed, chloroplasts settle out;

Question 50

Describe the structure of DNA (5)

Answer 50

1. Polymer of nucleotides; 2. Each nucleotide formed from a deoxyribose sugar, a phosphate (group) and a nitrogenous base; 3. Phosphodiester bonds (between nucleotides); 4. Double helix held by hydrogen bonds; 5. (Hydrogen bonds/pairing) between adenine, thymine and cytosine, guanine;

Question 51

Name and describe all 5 ways substances can move across the cell-surface membrane into a cell (5)

Answer 51

1. (Simple) diffusion of small/non-polar molecules down a concentration gradient; 2. Facilitated diffusion down a concentration gradient via protein carrier/channel; 3. Osmosis of water down a water potential gradient; 4. Active transport against a concentration gradient via protein carrier using ATP; 5. Co-transport of 2 different substances using a carrier protein;