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Biology A-level > AS level content > Flashcards

AS level content Flashcards

1
Q

What is a monomer?

A

The smaller units from which larger molecules are made [e.g. monosaccharides, amino acids & nucleotides].

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2
Q

What elements are carbohydrates made out of?

A

Carbon, hydrogen and oxygen

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3
Q

Name three monosaccharides

A

Glucose, fructose and galactose

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4
Q

What is maltose made up of?

A

Glucose and glucose

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5
Q

What is sucrose made up of?

A

Glucose and fructose

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6
Q

What is lactose made up of?

A

Glucose and galactose

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7
Q

Amylose

A

AMYLOSE = long, UNBRANCHED chain of α-glucose joined by 1-4 glycosidic
bonds, which give it a coiled structure → compact, so good for storage.

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8
Q

Amylopectin

A

AMYLOPECTIN = long, BRANCHED chain of α-glucose joined by 1-4 glycosidic bonds with branches formed by 1-6 glycosidic bonds [branched structure rather than helical] → side branches allow enzymes to access and hydrolyse glycosidic bonds easily, so glucose can be released quickly.

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9
Q

Glycogen

A

Highly branched, very compact so good for storage and insoluble so does not affect water potential

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10
Q

Cellulose

A

Long, straight, unbranched chains of β-glucose. Every other cellulose monomer is INVERTED.Cellulose chains are linked together by hydrogen bonds to form microfibrils → strong fibres to enable cellulose to provide structural support for plant cells.

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11
Q

Non-reducing sugar test

A
  1. Add dilute hydrochloric acid to a new sample of test solution and heat (this hydrolyses any glycosidic
    bonds i.e. hydrolyses non-reducing sugars into reducing sugars).
  2. Neutralise with sodium hydrogen carbonate.
  3. Add Benedict’s reagent and heat again.
  4. If test is positive, it will form a coloured precipitate. If test is negative, solution will stay blue, meaning the
    sample does not contain any sugar (reducing or non-reducing).
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12
Q

What elements are lipids made out of?

A

Carbon, hydrogen and oxygen

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13
Q

What is the ‘R’ group of a fatty acid?

A

The variable hydrocarbon chain

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14
Q

What is an unsaturated fatty acid chain?

A

Is a chain containing at least one carbon-carbon double bond forming a kink in the molecule.

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15
Q

What elements are proteins made up of?

A

Carbon, oxygen, hydrogen and nitrogen

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16
Q

What is activation energy?

A

The minimum energy required for a reaction to take place

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17
Q

What is the induced-fit model?

A
  1. Complimentary substrate binds to the active site of the enzyme
  2. Forms an enzyme-substrate complex
  3. As substrate binds, active site changes shape slightly, which provides a better fit
  4. Substrate is broken down/joined together to form the product(s)
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18
Q

Competitive inhibitors

A
  • Competitive inhibitor = a substance with a SIMILAR
    shape to the substrate and a complementary shape
    to the enzyme’s active site
  • Competes for and binds to the active site.
  • Fewer enzyme-substrate complexes form.
  • Can be overcome by increasing the substrate
    concentration.
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19
Q

Non-competitive inhibitors

A
  • Non-competitive inhibitor = a substance that binds
    to a site on the enzyme other than the active site (allosteric site).
  • Causes the active site to change shape.
  • Substrate can no longer bind to the active site.
  • Fewer enzyme-substrate complexes form.
  • Decreases the rate of reaction.
  • CANNOT be overcome by increasing substrate concentration.
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20
Q

Why is DNA double stranded?

A

Both strands can act as templates for semi-conservative replication

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21
Q

Why is there complementary base pairing in DNA?

A

Allows accurate replication

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22
Q

Why is there a sugar-phosphate backbone in DNA?

A

Protects bases and hydrogen bonds from damage

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23
Q

Why is DNA a long molecule?

A

Allows lots of genetic information to be stored

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24
Q

Describe the process of semi-conservative DNA replication

A
  1. DNA helicase breaks the hydrogen bonds between the two polynucleotide strands.
  2. DNA helix unzips to form two single strands.
  3. Both strands are used as a template.
  4. Free DNA nucleotides are attracted to the exposed bases on each template strand by complementary
    base pairing, A with T and C with G.
  5. DNA polymerase joins adjacent nucleotides via condensation reactions to form phosphodiester bonds
    (forming the sugar-phosphate backbone).
  6. Hydrogen bonds form between the bases on the original and new strand.
  7. The strand twists to form a double helix.
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25
Q

Which scientists carried out experiments to confirm that DNA replicated by semi-conservative replication?

A

Meselson and Stahl

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26
Q

What properties of ATP that make it a suitable source of energy in biological processes?

A

Energy is released in small, manageable amounts → energy is not wasted. Soluble → easily transported around the cytoplasm of a cell. Involves a single reaction → rapid.

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27
Q

Why is ATP not a good long-term energy?

A

Phosphate bonds are very unstable and are very easily broken. ATP cannot be stored has to be continuously made.

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28
Q

Water: High specific heat capacity

A

The hydrogen bonds between the water
molecules can absorb a lot of energy, so it takes a lot of energy to change the temperature of water. This means that water doesn’t experience rapid temperature changes.

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29
Q

Importance of water having a high specific heat capacity

A

Good habitat for aquatic organisms
as the temperature under water is
more stable than on land. Helps organisms to maintain a constant body temperature
(important for enzyme activity).

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30
Q

Water: High latent heat of vapourisation

A

Large amounts of energy are required to
change water from a liquid to a gas because it
takes a lot of energy to break the hydrogen
bonds between water molecules.

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31
Q

Importance of water having a high latent heat of vapourisation

A

Enables land-based organisms to
use evaporation as a cooling
mechanism e.g. sweating and
panting.

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32
Q

Water: Cohesive

A

Water molecules are cohesive because they’re polar. This enables water to flow in columns of water.

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33
Q

Importance of water being cohesive

A

Transport of water in xylem relies on
the cohesive properties of water.

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34
Q

Water: Good solvent

A

Can dissolve ionic substances (such as NaCl)
a positive end of water molecule is attracted to negative ion (e.g. Cl-) and negative end of water is attracted to a positive ion (e.g. Na+). Important for metabolic reactions they take place in solution.

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35
Q

Importance of water being a good solvent

A

Metabolic reactions take place in
solution e.g. respiratory and
photosynthetic gases dissolve before
entering cells. Able to act as a transport medium e.g. transport of mineral ions in
xylem.

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36
Q

Water: Less dense when solid

A

When water freezes, the water molecules are
held further apart because each water molecule forms 4 hydrogen bonds to other water molecules → produces a lattice shape. This makes ice less dense than liquid water, so
it floats.

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37
Q

Importance of water being less dense when solid

A

Ice forms an insulating layer on top
of water, so the water below doesn’t
freeze. Enabling aquatic organisms to live in
the water below the ice.

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38
Q

What is an inorganic ion?

A

An ion that doesn’t contain carbon

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39
Q

Function of hydrogen ions

A

Affects pH [pH is calculated based on the concentration of H+ → the more H+ present, the lower the pH]. Enhances pepsin activity in the stomach.

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40
Q

Function of iron ions

A

Iron ions are a key component of haemoglobin. The iron ion in haemoglobin binds to oxygen.Haemoglobin transports oxygen around the body in red blood cells.

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41
Q

Function of sodium ions

A

Needed for the co-transport of glucose and amino acids into cells [sodium is actively transported out of the cell via the sodium-potassium pump, which creates a sodium ion concentration gradient]. Affects osmosis. Diffusion of sodium ions creates an action potential.

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42
Q

Nucleus structure

A

Nucleus is surrounded by double membrane
(nuclear envelope) containing pores (allow
molecules to enter and leave the nucleus).
Nucleus contains chromatin (DNA and histone
proteins) and a nucleolus (site of ribosome
production).

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43
Q

Nucleus function

A

Controls the cell’s activities through the
production of mRNA, which carries
genetic information from the nucleus to
the site of protein synthesis. Manufactures rRNA and ribosomes.

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44
Q

RER structure

A

Series of membrane-bound sacs enclosed by a
membrane. With ribosomes on the surface.

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45
Q

RER function

A

Folds and processes proteins made
on the ribosomes.

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46
Q

SER structure

A

Series of membrane-bound sacs (NO ribosomes
attached).

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47
Q

SER function

A

Produces and processes lipids.

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48
Q

Golgi apparatus structure

A

Stack of membrane-bound sacs (called
cisternae). Vesicles at the edges.

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49
Q

Golgi apparatus function

A
  • modifies (e.g. addition of a
    carbohydrate chain) and packages
    proteins.
  • transports/modifies/stores lipids.
  • makes lysosomes.
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50
Q

Ribosomes structure

A

Made of proteins and rRNA. Not membrane-bound.

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51
Q

Ribosomes function

A

Site of protein synthesis.

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52
Q

Mitochondria structure

A

Bound by a double membrane (inner membrane is folded to form cristae). Inside is the matrix (contains enzymes for
aerobic respiration).

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53
Q

Mitochondria function

A

Site of aerobic respiration which produces ATP.

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54
Q

Lysosomes structure

A

Vesicles (bound by a single membrane)
containing digestive enzymes.

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55
Q

Lysosomes function

A

Digest invading cells or break down
worn-out cell components.

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56
Q

Plasma membrane structure

A

Membrane found on surface of animal cells &
inside the cell wall of plant cells and prokaryotic
cells. Composed of phospholipid bilayer & proteins.

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57
Q

Plasma membrane function

A

Regulates the movement of substances
into and out of the cell. Has receptors enabling cell to respond
to hormones etc.

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58
Q

Chloroplasts structure

A

Surrounded by double membrane with
membranes inside called thylakoid membranes. Stroma is the fluid inside.

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59
Q

Chloroplasts function

A

Site of photosynthesis

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60
Q

Cell wall structure

A

Made mainly of cellulose (in plant cells).
[main cell wall component in fungi = chitin, in
prokaryotic cells = murein]

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61
Q

Cell wall function

A

Provides strength and support to
prevent plant cells bursting from the
entry of water by osmosis.

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62
Q

Vacuole structure

A

Membrane-bound organelle (surrounding
membrane called the tonoplast). Contains cell sap.

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63
Q

Vacuole function

A

Helps to maintain pressure inside the
cell. Involved in isolating unwanted
chemicals inside the cell.

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64
Q

Function of a cell wall in prokaryotes

A

Made of a peptidoglycan called murein.
* supports cell and prevents it from changing shape.

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65
Q

Function of capsule

A

Helps protect bacteria from attack by the immune system.

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66
Q

Function of flagellum

A

A tail-structure which rotates to move the cell.

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67
Q

Structure of viruses

A
  • Acellular and non-living.
  • They are nucleic acids surrounded by a protein coat called a
    capsid.
  • Attachment proteins allow the virus to attach to specific
    host cells.
  • Viruses have no plasma membrane or cytoplasm or
    ribosomes.
  • Viruses invade and reproduce inside the cells of other organisms (host cells).
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68
Q

Light microscope

A
  • Maximum resolution of 0.2μm because light has a relatively long wavelength → so only able to view whole
    cells & tissues.
  • Maximum magnification of x1500.
  • Can organelle that can be seen is the nucleus (and possibly mitochondria).
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69
Q

Scanning electron microscope

A

Maximum resolution = 2nm, maximum magnification = x500 000.
* Scans a beam of electrons across a specimen → electrons from the specimen are scattered → pattern of
scattering is used to form an image.
* Advantages:
* Images show the surface of the specimen and can be 3D.
* Can be used on thick specimens (because the electrons do not penetrate the specimen).
* Disadvantages:
* Lower resolution than TEMs.
* Can’t be used on living tissue as the whole system must be in a vacuum.
* Expensive equipment.

70
Q

Transmission electron microscope

A
  • Maximum resolution = 0.2nm, maximum magnification = x1 000 000.
  • Use electromagnets to focus a beam of electrons (shorter wavelength than light), which is transmitted
    through the specimen → denser parts absorb more electrons, which makes them appear darker on the image.
  • Advantage:
  • High resolution images (because electrons have a short wavelength), so can be used to examine the cell’s
    ultrastructure i.e. organelles can be seen in detail.
  • Disadvantages:
  • Can only be used on thin specimens.
  • Produce 2D images.
  • Produce black and white images.
  • Can’t be used on living tissue as the whole system must be in a vacuum.
  • Complex staining procedure is required.
71
Q

Cell organelles heaviest-lightest

A

Nuclei, chloroplasts/mitochondria, lysosomes, ER and ribosomes

72
Q

Explain why a homogenised cell solution should be kept ice-cold, isotonic and buffered.

A
  • Ice-cold → to reduce enzyme activity that could damage
    organelles.
  • Isotonic - to prevent damage to organelles through osmosis
    (i.e. prevents organelles from shrinking or bursting).
  • Buffered - to maintain pH.
73
Q

Describe how you might obtain a sample rich in chloroplasts.

A
  1. Store the plant cells in an ice-cold, buffered, isotonic solution.
  2. Homogenate the cells to break them open and release the organelles.
  3. Filter the homogenate to remove large debris.
  4. Centrifuge at low speed and remove the supernatant.
  5. The heaviest organelle, the nuclei, will have formed a pellet at the bottom of the tube.
  6. Remove the supernatant and centrifuge at a higher speed, which will cause the next heaviest organelle,
    the chloroplasts, to form a pellet.
74
Q

Interphase

A
  • DNA replication occurs during S phase of interphase of the cell cycle.
  • G1/G2 phase → cytoplasm increases in volume, new organelles/proteins are made.
75
Q

Prophase

A

PROPHASE
* Chromosomes condense and shorten.
* Nucleolus disappears.
* Centrioles start moving towards the poles of the cell, forming the spindle fibres.
* Nuclear envelope breaks down and the chromosomes lie free in the cytoplasm.

76
Q

Metaphase

A

METAPHASE
* Chromosomes (each composed of two chromatids) line up in the middle of cell and attach to
the spindle fibres via their centromere.
* Forms a plane in the centre of the cell called the metaphase plane.

77
Q

Anaphase

A

ANAPHASE
* Centromeres divide, separating each pair of sister chromatids.
* Spindle fibres contract, pulling the chromatids to opposite ends of the spindle. This makes the
chromatids appear v-shaped.

78
Q

Telophase

A

TELOPHASE
* Chromatids uncoil and become long and thin again - now called chromosomes again.
* Nuclear envelope forms around each group of chromosomes, resulting in two genetically
identical nuclei.
* Nucleolus is formed.
After cytokinesis occurs

79
Q

Cancer cells spend less time in interphase and more time in mitosis true or false?

A

True

80
Q

Many cancer treatments aim to control the rate of cell division

A
  • Some chemotherapy drugs inhibit the synthesis of enzymes required for DNA replication → cell cannot enter S
    phase → cell cycle is disrupted and cell destroys itself.
  • radiotherapy/certain drugs can damage DNA → when damaged DNA is detected during cell cycle checkpoints,
    the cell destroys itself → inhibits further tumour growth.
81
Q

How does bacteria divide (binary fission)?

A
  1. The circular DNA and plasmid(s) replicate.
  2. The main DNA loop is only replicated once, whereas plasmids can be replicated multiple times.
  3. The cells gets bigger and the DNA loops move to opposite poles of the cell.
  4. The cytoplasm begins to divide and new cell walls begin to form.
  5. The cytoplasm divides and two daughter cells are produced.
  6. Each daughter cell has one copy of the circular DNA, but can have a variable number of copies of the plasmid(s).
82
Q

Membranes within cells

A

Membranes WITHIN cells
* Membranes around organelles divide the cell into compartments → makes cellular functions more efficient
e.g. the enzymes needed for respiration are kept inside the mitochondria.
* Can form vesicles to transport substances between different areas of the cell.
* Partially permeable → control the movement of substances in and out of organelles.
* Some organelles also have membranes within them e.g. thylakoid membranes in chloroplasts.
* Can be site of chemical reactions e.g. inner membrane of mitochondria contains enzymes for respiration.

83
Q

What substances does the phospholipid bilayer allow and prevent from entering?

A
  • Phospholipid bilayer PREVENTS the diffusion of polar/lipid-insoluble substances.
  • Phospholipids bilayer ALLOWS the diffusion of non-polar/lipid-soluble substances.
84
Q

Role of cholesterol in cell membranes

A
  • Cholesterol molecules fit between the phospholipids and bind to the hydrophobic tails of the phospholipids →
    causing them to pack together more closely, restricting the movement of other molecules in the membrane
    → cholesterol makes the membrane less fluid and more rigid (reduces the lateral movement of other molecules
    within the membrane).
85
Q

Role of carrier proteins in facilitated diffusion

A
  • Move large molecules across membranes, down their concentration gradient.
  • When a molecule specific to the carrier protein binds (e.g. glucose) → causes the carrier protein to
    change shape and release the molecule on the opposite site of the membrane.
  • No ATP is required → the molecules move from an area of higher concentration to an area of lower
    concentration, using the kinetic energy of the molecules.
86
Q

Role of channel proteins in facilitated diffusion

A
  • Form pores/channels across the membrane, allowing charged particles to diffuse down their
    concentration gradient.
  • Selective → different channel proteins allow the passage of a specific charged particle.
87
Q

Osmosis

A
  • The diffusion of water molecules across a partially permeable membrane from a region of high water potential
    to a region of lower water potential [i.e. down a water potential gradient].
88
Q

Active transport

A
  • The movement of molecules from an area of lower concentration to an area of higher concentration using ATP
    and carrier proteins.
  • A specific molecule attaches to a carrier protein.
  • Carrier protein changes shape.
  • Molecule is moved across the membrane against its concentration gradient and released on the other side
    of the membrane.
  • Uses energy released from the hydrolysis of ATP.
89
Q

Co-transport

A
  1. Sodium ions are actively transported out of the ileum
    epithelial cells by the sodium-potassium pump.
  2. Creates a concentration gradient → sodium ions diffuse
    into the epithelial cells from the lumen of the ileum,
    down their concentration gradient via sodium-glucose
    co-transporter proteins → the co-transporter carries
    glucose into the cell with the sodium.
  3. The concentration of glucose inside the cell increases
    and glucose diffuses down its concentration gradient
    out the cell into the blood, via facilitated diffusion.
90
Q

Tracheal system

A
  • Openings on body surface called SPIRACLES.
  • Spiracles can be opened/closed by a valve, open →
    allows gas exchange, closed → prevents water loss.
  • Spiracles connect to TRACHEA.
  • Trachea divide into TRACHEOLES.
  • Tracheoles connect directly to respiring cells.
  • Oxygen diffuses down a concentration gradient from the
    atmosphere to respiring cells [vice versa for CO2].
  • Mass transport: contraction of abdominal muscles
    increases the efficiency of gas exchange by allowing mass movement of air in and out.
  • [Insects DO NOT transport respiratory gases in their blood]
91
Q

Adaptations of insects to limit water loss

A
  • Close spiracles.
  • Tiny hairs around spiracles trap moist air → reduces water potential gradient → less water lost by evaporation.
  • Waterproof covering over body surface → reduces area over which water can be lost.
  • Small surface area to volume ratio → minimises area over which water can be lost.
92
Q

Adaptations in plants to limit water loss

A
  • Xerophytes = plants which are specially adapted to limit water loss through transpiration:
  • Stomata are sunk in pits to trap moist air → reduces water potential gradient between leaf and air → reduces
    amount of water evaporating out of leaf.
  • Layer of hairs on epidermis → to trap moist air and reduce evaporation.
  • Curled leaves with stomata inside → protects stomata from wind (wind increases the rate of transpiration by
    maintaining a concentration gradient for the diffusion of water).
  • Reduced numbers of stomata → reduces surface area for evaporation of water.
  • Waxy cuticles on leaves → waterproof barrier to reduce water loss.
  • Leaves with reduced surface area to volume ratio (e.g. small / spine-like leaves) → reduces rate of water loss.
93
Q

Structure of trachea/bronchi

A
  • Walls contain cartilage → provides support so trachea/bronchi do not collapse.
  • Cartilage is c-shaped in trachea to give flexibility.
  • Lined with goblet cells [produce mucus which traps pathogens/dirt particles] and ciliated epithelial cells [cilia
    moves mucus towards the throat].
94
Q

Structure of bronchioles

A
  • Wall made of smooth muscle → smooth muscle contracts to constrict bronchioles → allows control of air flow
    in and out of alveoli.
  • Lining also contains goblet cells and ciliated epithelial cells.
95
Q

How are alveoli adapted for gas exchange?

A
  • Millions of tiny alveoli that are folded → to provide a large surface area for gas exchange.
  • Walls of alveoli are one cell thick → to provide a short diffusion distance. Alveloar endotehlium cells are flattened.
  • Lining of alveoli are moist, so gases can dissolve.
  • Ventilation and good blood flow maintain a steep concentration gradient.
96
Q

Inhalation

A
  • External intercostal muscles contract (rib cage moves up and out)
  • Diaphragm contracts (flattens)
  • Increase in volume in chest and decrease in pressure
  • So air moves into lungs from the atmosphere down a pressure gradient = ACTIVE process
97
Q

Where is amylase produced?

A

Salivary glands and pancreas

98
Q

Breakdown of starch

A

A two step process [amylase hydrolyses starch to maltose, maltase hydrolyses
maltose to glucose].

99
Q

Function of bile salts in the digestion of lipids

A
  • Bile salts are produced by the liver.
  • They surround & emulsify lipids, causing the lipids to form MANY small droplets called micelles.
  • This greatly increases the surface area of the lipid →allows faster hydrolysis of lipids by lipase enzymes.
100
Q

Role of micelles in the absorption of the products of lipid digestion

A
  • Micelles come into contact with the epithelial cells lining the villi of the ileum.
  • Micelles break down, releasing the monoglycerides and fatty acids → which easily diffuse across the cell-
    surface membrane into the epithelial cells because they are non-polar (lipid soluble).
101
Q

How are monoglycerides and fatty acids recombined to form triglycerides once they have been absorbed?

A

In epithelial cell they are transported to the endoplasmic reticulum to form triglycerides. Inside Golgi apparatus, triglycerides are modified combined with proteins to form chylomicrons. Chylomicrons are packaged for release by exocytosis from ileum epithelial cells.

102
Q

Explain how oxygen is loaded, unloaded and transported

A
  1. Partial pressure of oxygen is high in the lungs so oxygen binds to haemoglobin, forming oxyhaemoglobin.
  2. Oxyhaemoglobin is transported in the blood in red blood cells.
  3. Loading of oxygen takes place in the lungs at a high partial pressure of O2.
  4. Oxygen is unloaded to respiring cells and tissues at a low partial pressure of O2.
  5. Unloading/dissociation of oxygen is linked to higher carbon dioxide concentration.
103
Q

Arteries

A

• Thick-walled → withstand high blood pressure.
• Contain elastic tissue → allows them to stretch and recoil (smoothing blood flow).
• Contain smooth muscle → enables them to vary blood flow.
• Lined with smooth endothelium → reduce friction and ease flow of blood [inner lining of endothelium is folded, allowing the artery to expand, which helps maintain blood pressure].
• A pulse can be felt in arteries → contraction of the left ventricle creates pressure wave that can be felt in arteries.

104
Q

Veins

A

• Thin muscle layer → carry blood away from tissues, so constriction of the muscle would not control flow of blood
to tissues.
• Wide lumen compared to arteries → to maximise the volume of blood carried to the heart.
• Thin elastic tissue layer → blood is at low pressure compared to arteries, so would not cause the veins to burst.
• Contain valves → prevents back flow of blood.
• Overall wall thickness is small → enables blood flow through veins to be helped by contraction of the body
muscles surrounding them.
• No pulse felt → pressure is too low to create recoil.

105
Q

Arterioles

A

• Muscle layer is relatively thicker than in arteries → allows constriction of lumen to restrict blood flow and control the movement of blood into the capillaries.
• Have a thinner layer of elastic tissue → blood pressure is lower than in arteries.

106
Q

Capillaries

A

• The smallest blood vessels.
• Permeable walls → site of metabolic exchange.
• Many capillaries that are highly branched → large surface area for exchange.
• Walls are only one cell thick → short diffusion distance → rapid diffusion of substances between blood & cells.

107
Q

Formation of tissue fluid

A
  1. At the arterial end of the capillary bed, the high hydrostatic pressure forces fluid out of the capillary and
    into the spaces around the cells, forming tissue fluid.
  2. The fluid carries small molecules such as oxygen & glucose with it.
  3. As fluid leaves, the hydrostatic pressure reduces in the capillaries, so the hydrostatic pressure is lower
    at the venule end of the capillary bed.
  4. Due to fluid loss and an increasing concentration of plasma proteins (which are too large to move out of
    capillaries), the water potential at the venule end of the capillary bed is lower than the water potential in
    the tissue fluid.
  5. So some water re-enters the capillaries from the tissue fluid at the venule end via osmosis.
  6. Excess tissue fluid drains into the lymphatic system and is returned to the circulatory system.
108
Q

Cohesion-tension hypothesis

A

• Water evaporates from the leaves at the top of the plant by transpiration → reduces water potential in the cells.
• Water moves from the top of the xylem into the leaf by osmosis (transpirational pull).
• This creates TENSION in the continuous column of water in the xylem.
• The COHESIVE forces between water molecules, due to hydrogen bonds, mean the water is pulled up as a column.
• ADHESIVE forces between the water molecules and the walls of the xylem also help the water molecules move up
the xylem.
• Water then enters the stem from the roots.
TRANSPIRATION = the evaporation of water fr

109
Q

How to set up potometer

A
  1. Choose a healthy leaf and shoot.
  2. Cut the shoot underwater [to prevent air from entering the xylem] and cut it at a slant [to increase the surface area for
    water uptake].
  3. Assemble the potometer underwater and insert shoot underwater (so no air can enter).
  4. Ensure apparatus is watertight and airtight.
  5. Dry leaves [wet leaves would reduce transpiration due to decreased concentration gradient between leaves and air] and
    allow time for plant to acclimatise.
  6. Remove end of capillary tube from beaker of water until one air bubble has formed, then put the end of the tube
    back in the water.
  7. Record the starting position of the bubble.
  8. Record the distance moved by the air bubble per unit time.
110
Q

Mass flow hypothesis

A
  1. At the source, solutes are actively transported into sieve tubes of the phloem by companion cells.
  2. This lowers the water potential inside the sieve tubes → water enters by osmosis from the xylem and companion cells.
  3. This creates high hydrostatic pressure inside the sieve tubes at the source end of the phloem.
  4. At the sink end, solutes are removed from the phloem, which increases the water potential inside the sieve tubes, so water leaves the tubes by osmosis, lowering the pressure inside the sieve tubes.
  5. This creates a pressure gradient from the source end to the sink end → resulting in mass flow of solutes down the pressure gradient.
  6. At the sink, the solutes are used
    e.g. in respiration or stored e.g. as
    starch
111
Q

What is a locus?

A

The fixed position of a gene on a DNA molecule

112
Q

Give properties of the genetic code

A
  1. Degenerate - Most amino acids are coded for by more than one triplet code
  2. Universal - The same specific base triplets code for the same amino acids in all living things
  3. Non-overlapping - each base is read only once
113
Q

Describe the process of transcription

A
  1. DNA helicase breaks the hydrogen bonds between two DNA strands in the region to be copied.
  2. Strands separate, unwind and expose some of the bases.
  3. Only one DNA strand acts as a template strand (non-coding strand).
  4. Free RNA nucleotides are attracted to the exposed bases according to complementary base pairing rules.
  5. RNA polymerase joins the nucleotides together until it reaches a stop codon.
  6. The pre-mRNA is spliced to remove introns, forming the mature mRNA strand.
  7. mRNA leaves the nucleus through a nuclear pore.
114
Q

Describe the process of translation

A
  1. mRNA attaches to a ribosome (ribosome attaches at the start codon).
  2. A tRNA molecule with an anticodon that is complementary to the start codon on the mRNA attaches itself to
    the mRNA by specific base pairing.
  3. A second tRNA molecule attaches itself to the next codon on the mRNA in the same way.
  4. tRNA’s carry specific amino acids determined by their anticodon.
  5. Adjacent amino acids are joined by a condensation reaction forming a peptide bond.
  6. Process continues until stop codon is reached.
  7. Polypeptide chain is released from ribosome.
115
Q

Homologus chromosomes

A
  • A pair of chromosomes, one maternal and one paternal.
  • They are the same size and have the same genes at the same loci, but
    are NOT necessarily identical because they may have different alleles.
116
Q

Meiosis process

A
  1. Before meiosis starts → DNA replication occurs, resulting in two copies of each chromosome [‘double-armed
    chromosomes’ consisting of two sister chromatids joined by a centromere].
  2. Homologous chromosomes pair up.
  3. CROSSING-OVER takes place → Non-sister chromatids exchange genetic information, resulting in chromatids
    with different combinations of alleles.
  4. The first meiotic division (meiosis I) separates homologous pairs of chromosomes, halving the chromosome
    number [meiosis I is a reduction division].
  5. The random separation of homologous chromosomes to form random combinations in each daughter cell is called
    INDEPENDENT SEGREGATION.
  6. The second meiotic division (meiosis II), divides the centromere and separates the pairs of sister chromatids.
  7. This produces four, haploid daughter cells that are genetically different from each other.
117
Q

Give two ways meiosis leads to variation

A
  1. Crossing-over
  2. Independent segregation (when homologous chromosomes are separated in meiosis I, which chromosome from
    each pair ends up in each daughter cell is completely random → creates different combinations of maternal and
    paternal chromosomes in the daughter cells).
  • Random fertilisation of gametes further increases genetic variation within a species - fusing of gametes is a
    completely random event i.e. any sperm could fuse with any egg.
118
Q

Why do meiosis and mitosis have different outcomes?

A
  • Mitosis only involves one division (separates the sister chromatids), whereas meiosis involves two divisions (to
    separate the homologous pairs and the sister chromatids).
  • Mitosis produces genetically identical daughter cells because there is no pairing of homologous chromosomes
    → therefore no crossing over or independent segregation of chromosomes.
119
Q

How to calculate the number of possible combinations of chromosomes for each daughter cell?

A

2^n, where n = the number of pairs of homologous chromosomes/the haploid number.

120
Q

Explain non-disjunction

A
  • NON-DISJUNCTION = a failure of the chromosomes to separate properly during meiosis.
  • If it happens during meiosis I → all four daughter cells will have abnormal numbers of chromosomes.
  • If it happens during meiosis II → only half the daughter cells will have an abnormal chromosome number.
121
Q

What is gene flow?

A

Different alleles are introduced into a population when individuals from another population migrate
into them and reproduce (gene flow).

122
Q

Explain natural selection

A
  1. New alleles arise by random mutation [can be advantageous / disadvantageous / silent (no effect)}.
  2. Individuals of the same species vary because they have different alleles.
  3. Those with an advantageous allele are more likely to survive and reproduce,
  4. Passing on the favourable allele to offspring (= reproductive success)
  5. Over a long time/many generations, the frequency of the advantageous allele within the population
    increases.
123
Q

Types of adaptation

A
  • Behavioural adaptations: ways in which an organism acts that increases its chance of survival and
    reproduction e.g. playing dead to avoid attack by predator.
  • Physiological adaptations: processes inside an organisms’s body that increase its chance of survival e.g.
    lowering rate of metabolism during hibernation to conserve energy.
  • Anatomical adaptations: structural features of an organism’s body that increase its chance of survival e.g.
    thick layer of blubber to keep warm in cold climates.
124
Q

Stablising selection

A
  • Occurs when there is NO change in the environment (stable
    environment).
  • Individuals with phenotypes closest to the mean are favoured
    (individuals with either extreme of phenotype are at a selective
    disadvantage → extremes of phenotype are lost over time).
  • Mean phenotype remains the same.
  • Standard deviation decreases.
125
Q

Directional selection

A
  • Occurs when there is a change in the environment.
  • One of the extremes of phenotype has a selective advantage.
  • The mean shifts in the direction of the individuals with the selective advantage.
126
Q

Species

A

A group of similar organisms able to reproduce to give fertile offspring.

127
Q

Advantage of using a bionamial system rather than common name

A
  • From the binomial name, you know which genus and species the organism belongs to.
  • Avoids confusion as sometimes there are different common names for the same organism.
128
Q

What is meant by hierachal classification?

A

Large groups are divided into smaller groups, with no
overlap between groups.

129
Q

Ways in which courtship behaviour increases the probability of successful mating

A
  1. Enables organisms to recognise the same species.
  2. Synchronises mating.
  3. Indication of sexual maturity.
  4. Attraction/recognition of a mate.
  5. Formation of a pair bond.
130
Q

Why is species recognition important in courtship behaviour?

A

To ensure mating only takes place between members of the
same species, so that they produce fertile offspring.

131
Q

Immunology - testing evolutionary relationships between species

A

If an antibody against a particular protein/antigen from one species can form an antigen-antibody complex in a
sample taken from a different species → the two species are closely related because they have similar proteins,
with tertiary structures similar enough that they can bind to the same antigen-binding site on the antibody.

132
Q

What is species richness?

A

The number of different species in a community

133
Q

The balance between conservation and farming

A
  • Legal protection for endangered species.
  • Protected areas where house building etc is restricted.
  • Crop rotation with nitrogen-fixing crops instead of use of fertilisers.
  • Environmental Stewardship Scheme → encourages farmers to conserve biodiversity by replanting hedgerows etc.
134
Q

Give 4 ways of investigating diversity

A
  1. Frequency of observable characteristics
  2. Base sequence of DNA
  3. Base sequence of mRNA
  4. Amino acid of proteins
135
Q

What does an OVERLAP of standard deviation bars mean?

A

THERE IS NO SIGNIFICANT DIFFERENCE

136
Q

What is phagocytosis?

A

• PHAGOCYTOSIS = the mechanism by which cells engulf particles to form a phagosome.

137
Q

Describe the process of phagocytosis

A
  1. Pathogen releases chemicals which attract the phagocyte.
  2. Phagocyte binds to the pathogen.
  3. Cytoplasm of the phagocyte engulfs the pathogen.
  4. Pathogen is now contained in a phagosome (phagocytic vesicle in the cytoplasm of the phagocyte).
  5. Phagosome fuses with lysosome and hydrolytic lysozymes break down the pathogen.
  6. Phagocyte presents the pathogen’s antigens on the its CELL MEMBRANE to activate cells involved in the specific immune response (acts as an antigen-presenting cell).
138
Q

Cellular response: T cells

A

• T cells which have receptors that are complimentary to the antigen, bind to the antigen and become activated.
• Activated T cells divide by mitosis to form clones of genetically identical cells.
• Cloned T cells can develop into helper T cells and cytotoxic T cells (and memory T cells).

139
Q

Helper T cells

A

Release chemical signals which stimulate phagocytes and B cells

140
Q

Cytotoxic T cells

A

Kill abnormal and foreign cells

141
Q

Memory T cells

A

Enable a rapid response to future infections by the same pathogen

142
Q

Humoral response: B cells and antibodies

A

• Membrane-bound antibody on a B cell (that is complementary to the antigen on the pathogen), binds
the antigen.
• The antigen is taken up by the B cell, processed and presented on the B cell membrane.
• This, along with chemical signals provided by a helper T cell, ACTIVATES the B cell (this is CLONAL
SELECTION → the process by which a specific B cell is selected to undergo clonal expansion).
• Activated B cells divide by mitosis to produce plasma cells and memory B cells (this is CLONAL
EXPANSION).
• Plasma cells secrete antibodies specific to the antigen (monoclonal antibodies).
• The antibodies bind to complementary antigens on the surface of the pathogen forming antigen-
antibody complexes.
• Upon re-infection, memory B cells can divide rapidly into plasma cells which secrete the right
antibody.

143
Q

Antibody structure

A

• Globular protein with quaternary structure
(composed of four polypeptide chains).
• Variable regions of the antibody form the antigen- binding sites (shape of variable region is complementary to a specific antigen and so the variable region differs between antibodies).
• Constant region is the same on all antibodies and allows the antibody to bind to receptors on immune system cells e.g. phagocytes.
• The polypeptide chains are held together by disulphide bridges.

144
Q

How do antibodies work to destroy pathogens?

A

• Each antibody has two antigen binding sites.
• So can bind to two pathogens at a time.
• Results in agglutination (clumping together) of pathogens.
• This increases the efficiency of phagocytosis by allowing more pathogens to be phagocytosed at once.

145
Q

How does the secondary immune response differ to the primary response?

A

SECONDARY IMMUNE RESPONSE is STRONGER (higher concentration of antibody produced) and FASTER
(steeper gradient on this section of the graph) than primary immune response due to the presence of memory
cells.
• Memory B cells are able to divide quickly to form plasma cells, which secrete the correct antibody to the
antigen [often resulting in the infection being cleared before the infected person shows any symptoms].

146
Q

How do vaccines lead to a person developing immunity?

A
  1. Vaccines contain antigen in the form of attenuated, inactive or dead pathogen (mention the specific pathogen if relevant to the question).
  2. The antigen is displayed on antigen-presenting cells e.g. phagocytes.
  3. Specific T cells bind to the antigen and become activated.
  4. Activated T cells divide to form clones of the T cell, which develop into helper T cells and cytotoxic T cells.
  5. Specific helper T cells stimulate the activation of specific B cells.
  6. B cells divides by mitosis to form plasma cells, all of which produce the same antibody.
  7. Memory cells are produced, meaning t
147
Q

Herd immunity

A

• Vaccines protect the individuals that have been vaccinated.
• Reduces the occurrence of the disease within a population.
• Unvaccinated people are less likely to catch the disease because there are fewer people to catch it from.

148
Q

Ethical issues with vaccination

A

• Tested on animals.
• Possible side-effects of vaccines that may cause long-term harm.
• To be fully effective, the majority of the population should be vaccinated - should vaccination be compulsory? • Should expensive vaccination programmes continue, even when a disease has been virtually eradicated?

149
Q

Natural passive immunity

A

When a baby becomes immune due to the antibodies it receives from its mother via the placenta and breastmilk.

150
Q

Artificial passive immunity

A

When you become immune after being injected with antibodies from someone else e.g. if you contract tetanus you can be given antibodies against the tetanus toxin collected from blood donors.

151
Q

Natural active immunity

A

Catching in the disease

152
Q

Artificial active immunity

A

When you become immune after being given a vaccination containing a harmless dose of antigen.

153
Q

Define antigen

A
  1. Foreign protein;
    Accept glycoprotein / glycolipid / polysaccharide
  2. (that) stimulates an immune response / production of antibody;
154
Q

Cardio output

A

Stroke volume (largest volume - smallest volume) x Heart rate

155
Q

Define antibody

A
  1. A protein / immunoglobulin specific to an antigen;
  2. Produced by B cells
    OR
    Secreted by plasma cells;
156
Q

HIV

A
  • Genetic material (RNA) and various proteins, e.g. reverse
    transcriptase, are surrounded by outer coating of protein
    called the capsid.
  • Outer layer, called the (lipid) envelope → made of
    membrane taken from the cell membrane of the previous
    host cell.
  • Specific attachment proteins → allow HIV to attach to
    host helper-T cells.
157
Q

How does HIV replicate in a host cell?

A
  1. Specific attachment proteins on HIV virus attach to receptors on the cell membrane of helper-T cells.
  2. Capsid is released into the cell, where it uncoats and releases the genetic material into the cytoplasm.
  3. Reverse transcriptase is used to make a complimentary strand of DNA from the viral RNA template.
  4. The single stranded DNA is converted to double stranded DNA, which is inserted into the host DNA.
  5. Host enzymes are used to make viral proteins from the viral DNA.
  6. The viral proteins are assembled into new viruses which bud from the cell and infect other cells.
158
Q

How does HIV cause aids?

A
  • The HIV virus infects and weakens helper T cells → without enough helper T cells, the immune system cannot
    stimulate cytotoxic T cells or stimulate B cells to produce antibodies.
  • HIV patients are unable to respond effectively to other pathogens → it is these other infections [opportunistic
    infections] that cause illness/death.
159
Q

How does HIV spread?

A

Unprotected sexual intercourse, infected bodily fluids (e.g. blood from sharing contaminated
needles), from an HIV-positive mother to her foetus.

160
Q

Why are antibiotics ineffective against HIV?

A
  • Antibiotics target bacterial enzymes and ribosomes used in bacterial metabolic reactions.
  • Viruses don’t have ribosomes as viral replication uses host cell ribosomes which are different to bacterial
    ribosomes, therefore antibiotics cannot disrupt/inhibit these metabolic processes.
  • Some antibiotics inhibit enzymes required for bacterial cell wall synthesis → results in weakening of the murein
    cell wall → bacterium bursts and dies. Viruses have a lipid envelope & protein capsid as opposed to a murein
    cell wall, therefore viruses don’t have a target site for these antibiotics.
161
Q

Monoclonal antibodies in treating disease and cancer

A
  • To target/carry a drug to a specific cell/antigen.
  • To block a particular receptor on a cell.
162
Q

How would monoclonal antibodies deliver drugs to cancer cells?

A
  • Monoclonal antibodies can be made that will bind to a specific surface antigen e.g. an antigen which is only found on the
    surface of a particular type of tumour.
  • An anti-cancer drug can be attached to the monoclonal antibodies.
  • When the monoclonal antibodies come into contact with the cancer cells, they bind to the tumour-specific antigen.
  • The anti-cancer drug will only accumulate in cancer cells, which reduces the side effects of the drug.
163
Q

ELISA Test

A
  • The first antibody is added and binds to a complementary antigen [well then washed to remove unbound antibody].
  • The second antibody, with an enzyme attached, is added.
  • The second antibody attaches to the antigen [therefore, if no antigen was present in the sample, the second
    antibody cannot bind].
  • The well is washed to remove any unbound antibody.
  • A substrate is added and the enzyme catalyses a reaction resulting in a colour change.
  • If there is a colour change, it shows the antigen of interest is present.
  • The intensity of the colour change is relative to the amount of antigen present.
164
Q

Why do antibodies bind to a specific antigen?

A
  • Antibodies are proteins and so have a specific primary structure.
  • Specific sequence of R groups determines the position of bonds (e.g. hydrogen, ionic).
  • Therefore antibodies have a specific tertiary structure, which produces a specific variable region/antigen binding
    site.
  • The antibody is therefore complementary to a specific antigen and can only form antigen-antibody complexes with
    that antigen.
165
Q

What is antigenic variability?

A

When pathogens change their surface antigens (due to changes in the genes of the
pathogen)

166
Q

What is the latency period?

A

After the initial infection period, HIV replication drops to a low level, during the latency
period, the infected person doesn’t experience any symptoms.

167
Q

True or false eukaryotic DNA contains introns whereas prokaryotic DNA does not

A

True

168
Q

Visible evidence that a cell is undergoing mitosis?

A
  • Individual chromosomes are visible because they have condensed.
  • Each chromosome is made up of two chromatids because the DNA has replicated.
169
Q

Describe how oxygen in the air reaches the capillaries surrounding the alveoli in the lungs.

A
  1. Trachea and bronchi and bronchioles;
  2. Down pressure gradient;
  3. Down diffusion gradient;
  4. Across alveolar epithelium;
  5. Across capillary endothelium/epithelium;
170
Q

If P value is <0.01

A

IT IS HIGHLY SIGNIFICANT

Biology A-level (7 decks)

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