LONG EXAM QUESTIONS

Question 1

Describe how DNA is replicated. [6 marks]

Answer 1

1~ DNA HELICASE attaches onto DNA molecule & breaks HYDROGEN BONDS between complementary bases.

2~ Each DNA strand acts as a TEMPLATE as 2 polynucleotide strands separate.

3~ FREE nucleotides form H bonds with complementary bases on DNA strand.

4~ DNA POLYMERASE catalyses formation of PHOSPHODIESTER bonds between adjacent nucleotides in condensation reaction.

5~ Forms 2 sets of DAUGHTER DNA helixes that have 1 parent (original) & 1 newly synthesised strand.

Question 2

Relate the structure of a DNA molecule to its function. [6 marks]

Answer 2

Double helix structure w/ many H bonds =
Provides stability to molecule.

Coiling =
Makes DNA compact - stores lots of info in a small space.

Long molecule =
Stores a large amount of info.

Strong covalent sugar-phosphate backbone =
Gives strength - prevents code from being corrupted.

H bonds between bases are weak =
Allows chains to split easily for replication and transcription.

Sequence of bases =
Allows info to be stored.

Question 3

Describe the structure and function of starch

Answer 3

Polymer of α-glucose.
Energy source and storage found in plants.

Insoluble = doesn’t affect water potential (no osmosis)
Large = doesn’t diffuse out of cells.

Made from AMYLOSE and AMYLOPECTIN

Question 4

Describe the features of amylose

Answer 4

Polymer of α-glucose.
Forms ONLY 1-4 glycosidic bonds.

Helix and unbranched structure.
Compact.

Question 5

Describe the features of amylopectin.

Answer 5

Polymer of α-glucose.
Forms BOTH 1,4 & 1,6 glycosidic bonds.

Helix and branched structure.
Compact.

Question 6

Describe the structure and functions of glycogen.

Answer 6

Polymer of α-glucose.
Main energy source / storage in animals.

Forms BOTH 1,4 & 1,6 glycosidic bonds.

Branched structure.= many terminal ends for hydrolysis to glucose
Insoluble = doesn’t affect water potential
Large = doesn’t diffuse out of cells.
Compact = large energy release.

Question 7

Describe the structure and function of cellulose.

Answer 7

Polymer of β-glucose.
Found in plant cell walls.
A structural unit = gives rigidity to plant cell walls (prevents bursting & holds stem up)
Hydrogen bonding between adjacent chains is found.

ONLY 1,4 glycosidic bonds
Straight-chain, unbranched structure
Alternate glucose molecules are rotated 180°

Forms microfibrils! = high tensile strength

Question 8

Describe the Benedict’s test for non-reducing sugars.

Answer 8

Negative result: Benedict’s reagent remains blue.
1~ Add 1cm³ of dilute HCl and boil in water bath for 5mins.
2~ Neutralise mixture with sodium hydrogen carbonate solution.
3~ Proceed with Benedict’s test.

If positive: contains non-reducing sugars.

Question 9

Describe the test for starch.

Answer 9

1~ Place food sample on spotting tile
2~ Add few drops of iodine solution

Positive result: colour change from orange to blue-black.

Question 10

Describe the lipid (emulsion test)

Answer 10

1~ Place food sample in test tube.
Add 3cm³ of ethanol.

2~ Shake lidded tube thoroughly to dissolve any lipid in sample.

3~ Add 3cm³ of water and shake gently.

Positive test: solution turns cloudy (white emulsion forms)

Question 11

How do you prepare a food test?

Answer 11

1~ Grind the food sample with a pestle and mortar while adding distilled water to help dissolve.

2~ Filter the food sample and collect the food sample solution.

3~ Place liquid into a test tube.

Question 12

Relate the structure of triglycerides to their functions.

Answer 12

Non-polar molecule = hydrophobic = insoluble
no affect on water potential (used for waterproofing)

Releases a lot of ENERGY =
High number of C & H atoms - good storage and source of water via respiration.

LARGE = makes it good storage molecule.- can’t leave cell

Slow conductor of heat = (thermal insulation)

Less dense than water = (buoyancy of aquatic animals)

Question 13

Relate the structure of phospholipids to their functions.

Answer 13

Amphipathic molecule =
Glycerol backbone attached to 2 HYDROPHOBIC fatty acid tails & 1 HYDROPHILIC phosphate head.

Amphipathic = forms phospholipid bilayer in water =
makes up cell membranes

Tails can splay outwards =
waterproofing

Question 14

How do you test for proteins in a food sample?

Answer 14

Biuret test confirms presence of peptide bonds,

1~ Add equal volume of NaOH to sample at room temp.
2~ Add drops of dilute copper (II) sulfate solution.
Swirl to mix.

Positive result: colour change - blue to purple.
Negative result: solution remains blue.

Steps 1 & 2 make Biuret reagent.

Question 15

What is the induced fit model of enzyme action?

Answer 15

1) Enzyme BINDS onto substrate.
2) Tertiary structure of the active site changes and becomes COMPLEMENTARY
3) It MOULDS itself tightly around the substrate - ensures the active site fits PERFECTLY
4) Temporary BONDS are formed - helps catalyse reaction
5) Bonds are STRESSED and products are formed.

Question 16

Compare the lock & key theory with the induced-fit model.

Answer 16

Lock & key theory =
Active site doesn’t change shape when substrate binds.

Induced-fit model =
Tertiary structure of enzyme changes as substrate approaches so active site moulds around substrate.

Question 17

Describe the effect of pH on the rate of reaction.

Answer 17

As the pH moves away from the optimum, the rate decreases.
Bc change in pH changes the CHARGES on the R groups of amino acids. Since H+ ions are donated.
So it changes the IONIC and HYDROGEN bonding between R groups.
Tertiary structure of active site changes, and enzyme has DENATURES.
Active site can no longer become complementary to substrate.
Fewer ES complexes formed.

Question 18

Explain the effect of temperature on the rate of reaction.

Answer 18

Increase in temperature increases kinetic energy of molecules - move around more quickly and collide more frequently - more ES - complexes formed - rate of reaction increases.

At higher temperatures, hydrogen / other bonds in enzyme begin to break - changes tertiary structure to change - active site changes - substrate no longer fits in site - enzyme denatured - doesn’t function again.

OPTIMUM temp usually 40 degrees.

Question 19

Describe the effect of enzyme concentration on the rate of reaction.

Answer 19

Given that substrate is in excess, rate increases proportionally to enzyme concentration.

Higher enzyme concentration means that more enzymes are available for ES complexes to form, more frequent successful collisions, increasing the rate.

Rate levels off when the maximum number of ES complexes form - when substrate is limiting.

Question 20

Describe the effect of substrate concentration on the rate of reaction.

Answer 20

Given that enzyme concentration is fixed, rate increases proportionally to substrate concentration.

Higher substrate concentration means that more substrates are available for ES complexes to form, more frequent successful collisions, increasing the rate.

Rate levels off when the maximum number of ES complexes form - when enzymes are limiting.

Question 21

Describe the structure of proteins [6 marks]

Answer 21

> Polymer of amino acids
Joined by peptide bonds
That are formed by condensation

> Primary structure is the sequence/order of amino acids

> Secondary structure is folding of polypeptide chain due to hydrogen bonding - forms alpha helixes / beta pleated sheets.

> Tertiary is 3D folding due to hydrogen, ionic and disulfide bonds.

> Quaternary structure is 2 or more polypeptide chains.

Question 22

What is the role of ATP?

Answer 22

1~ ATP hydrolysis is exergonic, so there’s a release of free energy.
2~ Enzymes help couple the energy from ATP hydrolysis to endergonic reactions (which absorbs energy)
3~ The molecules in the endergonic are PHOSPHORYLATED and is more reactive than before.
4~ Molecules join together to form product, and Pi is produced (phosphate is removed).
5~ ATP is resynthesised as Pi joins to and ADP to form ATP.

Question 23

What are the features of ATP?

Answer 23

A = amount of energy released is SMALL - can be coupled to energy-requiring reactions (endergonic) which don’t need too much energy - less energy wasted as heat.

T = ta da! Energy is released in a SINGLE reaction - very rapid, less time-consuming.

P = Pi (inorganic phosphate) released during ATP hydrolysis can be used to phosphorylate substances to make them MORE REACTIVE.

= Soluble =
Small & soluble so it can be easily transported around the cell.

= Cannot diffuse =
Can’t diffuse out of cell, meaning the cell will not run out of energy.

Question 24

What are the advantages of hydrolysing ATP instead of glucose directly?

Answer 24

+ Energy is released very rapidly - only single bond broken, whereas in glucose a number of bonds are broken (which takes more time) - energetically more favourable.

+ Provides energy in small and usable amounts - glucose releases too much energy.

Question 25

What are the properties of water?

Answer 25

Strong cohesion =
H₂O has strong cohesion between water molecules due to H bonding - produces surface tension when air meets water.
Strong cohesion also supports columns of water in the transport cells of plants.

Metabolite =
H₂O is a metabolite in many metabolic reactions, including condensation & hydrolysis.

Solvent =
So metabolic reactions occur AND allows transport of substances.

High specific heat capacity =
So good at buffering changes in temperature.

High latent heat of vaporisation =
Provides a cooling effect through evaporation.

Question 26

Outline how you would prepare a temporary mount of tissue for an optical microscope [4 marks]

E.G.
How would you prepare an onion cell slide?

Answer 26

1) Pipette a drop of WATER onto a slide.

2) Use tweezers/FORCEPS to separate epidermis tissue from onion, and place the tissue onto the slide.

3) Pipette a drop of a STAIN (E.G. iodine dissolved in potassium iodide)

4) Lower cover slip gently on top using a MOUNTED NEEDLE.

5) Place slip onto stage of microscope and clip it.

Question 27

What are the properties of the solution used during cell fractionation? Explain why these properties are necessary. [4 marks]

Answer 27

> ICE-COLD =
stops enzyme activity - prevents breakdown of organelles.

> BUFFERED =
maintains pH so enzymes/proteins aren’t denatured.

> ISOTONIC =
concentration is kept the same inside & outside the cell to prevent osmosis.
Prevents osmosis so organelles don’t burst or shrink from gain/loss of water.

Question 28

Describe how a sample of chloroplasts could be isolated from leaves. [4 marks]

Answer 28

1) Homogenise plant cell walls/membrane with blender/pestle & mortar and add a cold, buffered, isotonic solution.
2) Filter sample to remove debris & collect organelle-rich solution.
3) Centrifuge sample at low speed to separate the dense organelle first (nucleus).
4) Ultracentrifuge by repeating spinning at higher speeds. Chloroplast forms a pellet.
5) Syringe supernatant liquid out of tube to separate from pellet.

Question 29

State the order of density of organelles from most dense to least dense. [2 marks]

Answer 29

MOST DENSE

> nucleus
chloroplasts
mitochondria
lysosomes
membranes / RER / SER
ribosomes

LEAST DENSE

Question 30

Describe how binary fission is carried out [4 marks]

Answer 30

1# Plasmids and circular DNA makes copies of itself.

2# The 2 copies of DNA move to polar sides of cell and the cell ELONGATES.

3# The plasmids divide RANDOMLY between the 2 sides, and the organelles grow.

4# Cytoplasm SPLITS and membrane folds in.

5# 2 new cells produced with different number of plasmids.

Question 31

Describe how viral replication is carried out [4 marks]

Answer 31

1# Virus ATTACHES to the protein markers on a host cell using attachment proteins.

2# Virus INJECTS its genetic material into the HOST cell.

3# Host cell transcribes / translates the viral genes.

4# Protein forms NEW virus particles.

5# Virus particles BURST out of host cells and the host cell is destroyed.

Question 32

Describe the structure and function of a virus. [4 marks]

Answer 32

A virus consists of…

Genetic material =
ALL viruses contain DNA / RNA.

Capsid =
Protein structure which contains the genetic material.

Attachment proteins =
Allow virus particles to attach and enter host cell.

Lipid envelope =
SOME viruses are surrounded by a lipid envelope formed from the host cell membrane.

Question 33

Describe the process of how phagocytes destroy pathogens?

Answer 33

1) Pathogen releases chemicals called chemoattractants which attract phagocytes which moves along the concentration gradient
2) The pathogen is ENGULFED in a vesicle called a PHAGOSOME
3) The phagosome membrane FUSES with the membrane of the lysosomes in the phagocyte
4) Lysosome contains hydrolytic enzymes, lysozymes which are released into phagosome
5) Lysozymes hydrolyse the molecules the pathogen is made of
6) Soluble products of breakdown are absorbed into cytoplasm (e.g. glucose, amino acids, antigens)
7) Non-soluble products expelled out of the cell

Question 34

Outline the process of the cell-mediated response [6 marks]

Answer 34

1) Phagocyte engulfs pathogen
2) Phagocyte presents antigens on its cell surface membrane = APC
3) Specific Helper T-cells with RECEPTORS complementary to the specific antigen BIND to the APC. The TH cells become ACTIVATED
4) The activated TH cells divide by mitosis (clonal selection)
> They also secrete chemicals which activate other T-cells and B-cells which have the same receptors.

5)
> Killer T cells - `Cytotoxic T-Cells (TC) are stimulated –> Produces chemicals which make the APC’s membrane permeable which can kill the cell. The holes also allow toxins to get in, which kills the cell and everything in it (e.g. viruses that invaded)

> B-cells are stimulated

> Stimulate phagocytosis by phagocytes

> Form T-memory cells

Question 35

What can occur after Helper-T cells divide by mitosis in the cell-mediated response?

Answer 35

1) Cytotoxic T-Cells are stimulated –> Produces chemicals which make the APC’s membrane permeable which can kill the cell. The holes also allow toxins to get in, which kills the cell and everything in it (e.g. viruses that invaded)

2) B-cells are stimulated

3) Stimulate phagocytosis by phagocytes

4) Form T-memory cells

Question 36

What kind of cells or molecules stimulate an immune response?

Answer 36

1) Pathogens

2) Abnormal cells - (cancer)

3) Foreign cells from other organisms of the same species - transplants

4) Toxins (released by pathogens)

Question 37

Describe the structure of an antibody

Answer 37

1) Antibodies are proteins with a quaternary structure

2) Made of 4 polypeptide chains…
> 2 heavy chains (large polypeptide chain)
> 2 light chains (small polypeptide chain)

3) The heavy and light polypeptide chains are held by DISULPHIDE bonds

4) Contains VARIABLE regions which form antigen binding sites

5) The rest of the antibody is CONSTANT.

http://www.biology.arizona.edu/immunology/tutorials/antibody/structure.html

Question 38

Describe how B-lymphocytes respond when they are stimulated by antigens [4 marks] MS

Answer 38

1) Divide by mitosis / forms clones

2) To produce plasma cells

3) The plasma cells make antibodies

4) The plasma cells produce memory cells

Question 39

Describe how antibodies are produced in the body following a viral infection [6 marks]
(humoral response)

Answer 39

1) Virus contains antigen

2) Virus is engulfed by phagocyte and presents the antigens on its surface

3) T cells with a specific complementary receptor bind to the antigen and are activated.

4) T cells undergo mitosis/differentiation (stimulate macrophages, become cytotoxic T cells/form memory T cells (Store memory of that antigen)

5) Remain T helper cells can also activate B cells

6) B cells divide by mitosis and undergo clonal selection

7) …(and) form plasma cells which produce antibodies (monoclonal)

Question 40

Describe the Meselson and Stahl experiment

https://www.youtube.com/watch?v=O-P_NJwa-h4

Answer 40

Bacteria (with 14N) are grown in a broth containing the heavy (15N) isotope

A sample of DNA from the 15N culture of bacteria was extracted and spun in a centrifuge

The bacteria containing only 15N DNA was then taken out of the 15N broth and added to a broth containing only the lighter 14N nitrogen. The bacteria were left for enough time for one round of DNA replication to occur before their DNA was extracted and spun in a centrifuge

The DNA molecules would now contain both the heavy 15N and light 14N nitrogen and would therefore settle in the middle of the tube (one strand of each DNA molecule would be from the original DNA containing the heavier nitrogen and the other (new) strand would be made using only the lighter nitrogen)