Year 1 Model Answers

Question 1

Describe Condensation Reactions

Answer 1

Monomers are joined together producing a larger molecule forming a bond.

A molecule of water is formed in this process

Question 2

Describe Hydrolysis

Answer 2

Polymers and dimers are split a part breaking bonds

A water molecule is formed in this process

Question 3

Name the bonds in biological molecules

Answer 3

Question 4

glucose + glucose –>
glucose + galactose –>
glucose + fructose –>

Answer 4

glucose + glucose –> maltose (malt sugar)
glucose + galactose –> lactose (milk sugar)
glucose + fructose –> sucrose (table sugar)

Question 5

Describe the structure and function of starch

Answer 5

Alpha glucose monomers
Polysaccharide
Has 1-4 glycosidic bonds
Amylose is helical
Amylopectin is branched and has additional 1-6 glycosidic bonds
Insoluble so doesn’t effect water potential
Branched so accessible for respiration

Question 6

Describe the structure and function of glycogen

Answer 6

Alpha glucose monomers
Polysaccharide
Has 1-4 glycosidic bonds
Is branched and has many additional 1-6 glycosidic bonds
Highly branched so accessible for respiration

Question 7

Describe the structure and function of cellulose

Answer 7

Beta-glucose monomers
Polysaccharide
Has 1-4 glycosidic bonds with alternating molecules being rotated by 180 degrees
Straight, unbranched chains
Straight chains ideal for cell wall, have crosslinks for stability

Question 8

Describe the test for reducing sugars

Answer 8

Add Benedict’s reagent
Boil in a waterbath
Brick red colour indicates a reducing sugar

E.g. glucose

Question 9

Describe the test for a non-reducing sugar

Answer 9

Carry out benedict’s test and see a negative result
Boil in a water bath with hydrochloric acid
Neutralise the acid with sodium hydrogen carbonate
Carry out Benedict’s test – a brick red result indicates a reducing sugar.

E.g. sucrose (this is the only one you need to remember)

Question 10

Describe the test for starch

Answer 10

Add iodine to food sample
Blue/Black indicates starch is present

Question 11

Describe the structure of amino acids and how they are joined

Answer 11

Amino acids contain C, O, H and N
They have a carboxyl group and an amine group
They also have an R group which can change the properties of the amino acid
There are 20 amino acids
They react in a condensation reaction and are joined by a peptide bond
A polymer of amino acids is called a polypeptide

Question 12

Describe protein structure

Answer 12

Proteins have a primary structure that is formed by a sequence of many amino acids that are joined by peptide bonds in a condensation reaction

The primary structure folds into a secondary structure of either alpha helix or beta pleated sheets and these are held together by hydrogen bonds

The secondary structure further folds into a tertiary 3D structure that is held together by hydrogen bonds, ionic bonds and disulphide bonds

Some proteins e.g. antibodies may form a quaternary structure of more than one polypeptide chain(some of these may have prosthetic groups e.g. haem)

Question 13

Describe the test for protein

Answer 13

Add Biurets reagent to sample (copper sulphate and sodium hydroxide)
Mix and heat
Lilac colour indicates protein is present

Question 14

Describe the structure and function of triglycerides

Answer 14

Condensation of one glycerol and three fatty acids forming and ester bond
Fatty acids have an R group that can be saturated(C-C) or unsaturated (C=C)
Triglycerides are non-polar
Insoluble
Can’t form a bilayer
Used for longer term energy storage

Question 15

Describe the structure and function of a phospholipid

Answer 15

Condensation reaction forming ester bonds
One glycerol and two fatty acids
One fatty acid is replaced by a phosphate group
The phosphate group is polar
The fatty acids are non-polar
Insoluble
Used in the phospholipid bilayer

Question 16

Describe the emulsion test

Answer 16

Mix sample with ethanol and shake
Add water
If lipid is present a milky white emulsion will be seen

Question 17

Describe enzyme induced fit

Answer 17

Enzyme active site is complementary to the substrate
When substrate binds the enzyme changes shape slightly
This stresses the bonds in the substrate
Reducing the activation energy for the reaction
The reaction occurs quicker

Question 18

Describe competitive inhibitors

Answer 18

Binds to active site
Blocks active site so substrate can’t bind
Fewer enzyme-substrate complexes form

Question 19

Describe non-competitive inhibitors

Answer 19

Binds to allosteric site
Changes tertiary structure
Enzyme is no longer complementary to substrate
Fewer enzyme-substrate complexes form

Question 20

Describe DNA structure

Answer 20

DNA is made of a polymer of nucleotides/polynucleotide
It is two molecules that are antiparallel to each other coiled into a double helix.
Each nucleotide is made of deoxyribose, a phosphate group and a nitrogenous base that can either be adenine, cytosine, thymine or guanine
The adjacent nucleotides are joined to each other between the sugar and phosphate groups in a condensation reaction forming a phosphodiester bond
Complementary base pairing holds the two strands together due to hydrogen bonds forming between A - T and C - G.

Question 21

Describe DNA replication

Answer 21

DNA helicase breaks the hydrogen bonds causing the strands to separate
Both strands act as a template
Free nucleotides complementary base pair to the template A-T and G-C
DNA polymerase joins adjacent nucleotides together forming a phosphodiester bond
Hydrogen bonds form between the old strand and the newly synthesised strand
DNA replication is semi-conservative replication

Question 22

Describe ATP structure

Answer 22

DNA helicase breaks the hydrogen bonds causing the strands to separate
Both strands act as a template
Free nucleotides complementary base pair to the template A-T and G-C
DNA polymerase joins adjacent nucleotides together forming a phosphodiester bond
Hydrogen bonds form between the old strand and the newly synthesised strand
DNA replication is semi-conservative replication

Question 23

Describe the properties of water

Answer 23

It is a metabolite in hydrolysis/condensation reactions
A solvent so metablic reactions can occur
High specific heat capacity to buffer changes in temperature
High latent heat of vaporisation providing a cooling effect
Cohesion between molecules forms a continuous column of water in the xylem and surface tension for organisms to move on bodies of water

Question 24

Describe 3 of the inorganic ions you need to know in A level Biology

Answer 24

Inorganic ions are dissolved in bodily fluids and cytoplasm
They typically have a specific role

H+ For Chemiosmosis and Oxidative phosphorylation and maintaining pH
Fe2+ Haemoglobin
Na+ Co-transport, depolarisation, sodium potassium pump in action potentials
K+ For sodium potassium pump
Ca2+ For NMJ, synaptic transmission and sliding filament
PO43- For phospholipid bilayer, DNA, RNA and ATP
NO3- For N-cycle and Nitrogen containing compounds e.g. amino acids, DNA, ATP etc

Question 25

Describe adaptations of cells

Answer 25

Depends a lot on the context, but…

Many mitochondria for increased rates of aerobic respiration and ATP production
Microvilli to increase the surface area of the cell surface membrane to increase the rate of diffusion
More carrier proteins to increase the rate of active transport
More ribosomes to produce more proteins

Question 26

Describe how and optical microscope works and the advantages and disadvantages of them

Answer 26

Light passes through the specimen so it needs to be a thin layer of cells
Has lower magnification
Lower resolution as the wavelength of light is too long (some organelles may be missed)
Can view live specimens
Can show images in colour via staining
Not very complex preparation

Question 26

Name the viral structures

Answer 26

A Attachment protein (binds to a receptor on host cell e.g. CD4 in HIV)
B Enzymes e.g. reverse transcriptase
D Genetic Material (RNA in HIV)
E Lipid Envelope
F Protein Capsid

Question 27

Describe how an electron microscope works and the advantages and disadvantages of them

Answer 27

Electrons pass through the sample so it must be sliced thinly
Higher magnification
Higher resolution as the wavelength of electrons is shorter – so we can see smaller objects
Artifacts may be present
Can’t view live specimens
Can’t see colour images
Complex preparation
SCANNING shows 3D surface of the cell
TRANSMISSION shows a cross sections of the cell

Question 28

What happens if the temperature is above an enzymes optimum

Answer 28

Too much kinetic energy
Breaks the hydrogen and ionic bonds
Tertiary structure denatures
Active site changes shape and the enzyme substrate complex can’t form
Reaction slows to a stop

Question 29

What happens if the temperature is below an enzymes optimum

Answer 29

Not enough kinetic energy
Enzymes substrate complexes don’t form
Reaction slows

Question 30

What happens to a reaction as you increase the substrate concentration

Answer 30

The reaction increases as the concentration of substrate increases
Eventually the rate plateaus as the reaction is limited by the number of enzymes
The active sites become saturated

Question 31

What happens to a reaction as you increase the enzyme concentration

Answer 31

The reaction increases as the concentration of enzyme increases
Eventually the rate plateaus as the reaction is limited by the substrate concentration

Question 32

How do you perform cell fractionation

Answer 32

Homogenise cells to break them open
Filter to remove large debris
Conditions must be:
Cold – To prevent enzyme activity damaging the organelles
Isotonic – To prevent shrinkage or osmotic lysis of the organelles
Buffered – To prevent enzymes and proteins denaturing due to pH changes
Spin cell extraction in a centrifuge at high speeds and pellet nuclei and more dense organelles
Remove supernatant and spin again to pellet the less dense organelles e.g. ribosomes

Question 33

How do you prepare a temporary mount

Answer 33

Samples must be sliced thinly - to let light pass through them.

A drop of water is typically added.

Often stains are applied to make them darker – iodine and methylene blue are examples.

Usually, a cover slip is applied gently at an angle.

The light passes through and when it hits the specimen it appears darker.

Question 34

Describe the cell cycle

Answer 34

Interphase is split into G1,S and G2 phases – then the cell divides by mitosis and cytokinesis
G1 - Proteins are made, organelles are sythesised and the cell grows
S phase – DNA is replicated by semi-conservative replication
G2 – Organelles grow and divide and ATP is synthesised
Mitosis – Nuclear division
Cytokinesis – Division of the cytoplasm, two new cells are formed

Question 35

Describe mitosis

Answer 35

DNA is replicated in S-phase of interphase
Prophase – Chromosomes condense and become visible, nuclear membrane dissolves
Metaphase – Spindle fibres pull chromosomes line up at the middle of the cell
Anaphase – Sister chromatids are pulled to the opposite poles of the cell as spindle fibres contract
Telophase – Chromosomes decondense and the nuclear envelope starts to reform around the two nuclei
Cytokinesis – The cytoplasm and surface membrane splits forming two new cells that are genetically identical

Question 36

Describe binary fission

Answer 36

Bacteria an prokaryotes reproduce by binary fission
First the DNA (and plasmids) are replicated
Then the cytoplasm and cell membrane divides in two
Each daughter receives on copy of the circular DNA (and variable number of plasmids if present)

Question 37

Describe the phospholipid bilayer

Answer 37

The phospholipid bilayer is made of phospholipids with hydrophilic/polar phosphate heads and non-polar/hydrophobic fatty acid tails
This bilayer limits the movement of substances through the membrane so that only small non-polar molecules like oxygen can freely diffuse through
Embedded in the membrane are transmembrane proteins that can act as carrier and channel proteins. Chanel proteins allow facilitated diffusion of ions, while carrier proteins can perform facilitated diffusion, they can also transport molecules across the membrane using energy from ATP in active transport
Glycoproteins are embedded in the membrane and act as antigens for cell recognition, they also increase stability and adhesion of the cells
Glycolipids are part of the membrane and increase stability and allow for adhesion of cells.
Cholesterol is also found in the phospholipid bilayer and acts to increase rigidity of the membrane.

Question 38

Describe phagocytosis

Answer 38

Phagocyte recognizes antigen
Pathogen is engulfed into a phagosome
Lysosome fuses with phagosome forming a phagolysosome
Lysozyme enzymes Hydrolyse pathogen
Antigens from pathogen may be presented on the surface of the cell

Question 39

Describe cell mediated immunity

Answer 39

Antigen is presented by an antigen presenting cell
Helper T Cells with the complementary receptor binds to the presented antigen
Helper T cell is activated and goes through mitosis
Forming – cytotoxic T cells and more Helper T cells
Cytotoxic T cells produce perforins to kill cells by making holes in their cell surface membrane, water enters causing cells to burst

Only works on whole cells e.g. cancer cells of those infected with viruses

Question 40

Describe humoral immunity

Answer 40

T helper cells bind to the complimentary antigens presented by the specific B cell. Clonal selection.

T helper cells activate specific B cells.

The B cells rapidly divide by mitosis to produce plasma and memory B cells. This is clonal expansion.

These cloned plasma cells produce specific complementary antibodies to the antigens on the pathogen
Antibodies destroy the pathogen

Question 41

Describe a vaccine

Answer 41

Vaccines contain antigens/weakened pathogen
Memory B cells are made
On second exposure memory cells are activated
Memory cells and plasma cells rapidly produce antibodies
Antibodies destroy pathogens

Some vaccines are only effective for a limited time as viruses can change the tertiary structure of their antigens

Question 42

Describe the ELISA test

Answer 42

First antibody binds/complementary to the antigen of interest
Second antibody with an enzyme is added
Enzyme binds to substrate (e.g. first antibody/antigen)
Unbound antibody is washed away
Solution added and colour change seen

Question 43

Describe HIV replication

Answer 43

HIV attachment proteins complementary to the receptors on the helper T cell
Virus nucleic acid enters the cell
Reverse transcriptase converts RNA to DNA and inserted into the host genome
DNA is transcribed and the HIV capsid (protein) and enzymes are made by host cell ribosomes.
Everything is assembled by the host cells RER and Golgi apparatus into new virus protein, capsid, enzymes.
Virus are assembled and released

Question 44

Compare prokaryote and eukaryotic DNA

Answer 44

Prokaryote DNA is short Eukaryote DNA is long
Prokaryote DNS is circular Eukaryote DNA is linear
Prokaryote DNA has no introns Eukaryote DNA has introns
Prokaryote DNA is free floating Eukaryote DNA is in in the nucleus (and mitochondria and chloroplasts)

Question 45

Describe gas exchange in small/simple organisms

Answer 45

Simple organisms (e.g prokaryotes) are very small and have a large surface area to volume ratio
Gasses can be exchanged by simple diffusion across their cell surface membrane

Question 46

Why do large organisms need a gas exchange system

Answer 46

Large organisms need gas exchange systems as the diffusion pathway is too big and the rate of diffusion is too slow, they are also usually more metabolically active

Question 47

Describe the relationship between SA:V and metabolism

Answer 47

Higher surface area to volume ratio means there is higher metabolism
This is because there is more heat energy lost
So, more respiration is needed to release heat as a byproduct

Question 48

Describe the adaptations of the gas exchange surface in insects

Answer 48

Insects have spiracles and tracheae that branch into tracheoles
Gasses (e.g. Oxygen) diffuse through the spiracles and the tracheae/tracheoles
Tracheoles are very branched so have a large surface area and short diffusion pathway
Tracheoles are penetrate respiring tissue so have a short diffusion distance

Question 49

Describe how insects are adapted to limit water loss

Answer 49

Cuticle exoskeleton that limits water loss
Spiracles can close to prevent water loss
Hairs around spiracles to prevent water loss

Question 50

Describe the mechanism and benefits of counter current flow in fish

Answer 50

Fish have gills with many gill filaments and lamella to increase the surface area to volume ratio
Single layer of cells in lamella so short diffusion pathway

Blood flows in the opposite direction to water in a counter current flow
This maintains the diffusion gradient so O2 conc. Is always higher in the water than the blood
Diffusion occurs along the entire length of the lamellae/filament

Question 51

Describe ventilation in mammals

Answer 51

Inhalation: External intercostal muscles and diaphragm contract internal intercostal muscles relax
Ribcage moves up and out diaphragm moves down
Volume in the thorax increases so pressure decreases and air moves in
Exhalation: External intercostal muscles and diaphragm relax internal intercostal muscles contract
Ribcage moves down and in diaphragm moves up
Volume in the thorax decreases so pressure increases and air moves out

Question 52

Describe adaptations of the human gas exchange surface

Answer 52

Alveoli are a single layer of cells to reduce diffusion distance
Alveoli are branched so have a large surface area
Alveoli have a good blood supply so the oxygen diffusion pathway is maintained

Question 53

Describe factors that affect gas exchange in plants

Answer 53

Humidity – affects the diffusion gradient of water
Temperature – affects the KE of molecules
Air movement – affects the diffusion gradient all molecules
Light intensity – can open and close stomata (using guard cells), increases photosynthesis affecting diffusion gradient

Question 54

Describe complete digestion of starch

Answer 54

Amylase in saliva hydrolyses starch (by breaking the glycosidic bond) to maltose (alpha glucose disaccharide)
Amylase is denatured in the stomach – no carb digestion there
Pancreatic amylase is released and further hydrolyses any leftover starch
Maltose is hydrolysed to α-glucose by breaking the glycosidic bond in the ileum by the enzyme maltase which is a membrane-bound enzyme
Glucose is absorbed in co-transport

Question 55

Describe the complete digestion of proteins

Answer 55

Hydrolysis of peptide bonds.
Endopeptidase act in the middle of protein/polypeptide in the stomach and produces shorter polypeptides, increasing the number of ends
Exopeptidases act at end of protein/polypeptide and in the stomach and produce dipeptides.
Dipeptidases are membrane bound enzymes in the ileum which act on dipeptides and produce single amino acids

Question 56

Describe absorption of amino acids/glucose

Answer 56

Na+ is actively pumped out of the cell into the blood by the sodium potassium pump
This lowers the concentration of sodium in the epithelium cell
Na+ moves into the cell from the lumen by facilitated diffusion and glucose/amino acids are co-transported with it
Glucose/amino acids are then transported into the blood by facilitated diffusion

Question 57

Describe lipid digestion

Answer 57

Bile salts emulsify lipids into micelles to increase surface area and solubility in water

Lipids/triglycerides are hydrolysed by lipases to form fatty acids and monoglycerides

Micelles contain fatty acids, monoglycerides and bile salts

They move through the ileum to the epithelium cells

Question 58

Describe lipid absorption

Answer 58

Micelles contain bile salts and fatty acids/monoglycerides, making them soluble in water.
Fatty acids/monoglycerides are released to cell/lining of the ileum.
This maintains a higher concentration of fatty acids/monoglycerides outside the cell, so they are absorbed by simple diffusion.
Triglycerides are reformed in cells and form chylomicrons.
The chylomicron vesicles fuse with the cell membrane and are released by exocytosis.

Question 59

Describe co-operative binding

Answer 59

As O2 loads to oxygen, its binding cause the haemoglobin to change shape
This change of shape makes it easier for more oxygen to load
Until all the haem groups are occupied and the haemoglobin is saturated (full)
This is called co-operative binding

Question 60

Describe the impact of the bhor shift on haemoglobin

Answer 60

In exercising or organisms with high metabolism

CO2 in tissues reduces the affinity of oxygen to haemoglobin (it’s acidic/changes Hb shape)

In tissues where there is lots of respiration the oxygen is more easily unloaded – this makes the curve shift to the right.

This can replace used O2 easier

The impact of CO2 is called the Bohr Shift

Question 61

Describe the effect on oxygen dissociation when the curve shifts left

Answer 61

Usually in low oxygen environments e.g. womb, high altitude, under water
Higher affinity for oxygen at low ppO2
Oxygen associates/loads more readily
More oxygen can bind where little oxygen is available e.g foetus

Question 62

Describe the effect on oxygen dissociation when the curve shifts right

Answer 62

Usually in metabolically active organisms e.g. mice or runners
Lower affinity of oxygen at higher ppO2
Oxygen disassociates/unloads more readily
More oxygen available in tissues for More aerobic respiration

Question 63

Describe how tissue fluid is formed

Answer 63

Higher hydrostatic pressure of blood at arterial end of capillary;
Water and soluble molecules pass out, but proteins / large molecules remain;
This lowers the water potential
Water moves back into venous end of capillary by osmosis;
The lymph system collects any excess tissue fluid which returns to the circulatory system

Question 64

Describe the path of blood through the left side of the heart

Answer 64

Blood arrives at the left atrium from the pulmonary vein filling the atrium increasing the pressure.
The atrial muscle contracts reducing the volume and increasing the pressure in the atrium until it is greater than the ventricle – this forces the blood through the atrioventricular valve into the left ventricle.
The increase in pressure of the ventricle closes the atrioventricular valve, preventing back flow of blood.
Then the left ventricle muscle contracts reducing volume and increasing the pressure until it is greater than in the aorta, this forces the blood through the semilunar valve.
The pressure in the aorta increases causing the semilunar valve to close preventing back flow.

Question 65

Describe the path of blood through the right side of the heart

Answer 65

Blood arrives at the right atrium from the vena cava filling the atrium increasing the pressure.
The atrial muscle contracts reducing the volume and increasing the pressure in the atrium until it is greater than the ventricle – this forces the blood through the atrioventricular valve into the right ventricle.
The increase in pressure of the ventricle closes the atrioventricular valve, preventing back flow of blood.
Then the right ventricle muscle contracts reducing the volume and increasing the pressure until it is greater than in the pulmonary artery, this forces the blood through the semilunar valve.
The pressure in the right pulmonary artery increases causing the semilunar valve to close preventing back flow.

Question 66

Describe the structure and function of arteries, arterioles, veins and capillaries

Answer 66

Question 67

Describe the movement of water through a plant

Answer 67

Water evaporates from the leaves/transpiration
Due to heat/kinetic energy from sunlight
Water diffuses out of the stomata from a high water potential to low
The diffusion of water causes tension in the xylem
This is due to water potential gradient
Cohesion tension forms a continuous column of water that is pulled through the xylem in a transpiration stream
Water’s adhesive properties aid the movement through the xylem
Transpiration stream lowers water potential in the root cells
Water is absorbed through the root hair cells by osmosis from a higher water potential to low.

Question 68

Describe the factors affecting transpiration

Answer 68

Humidity – increases or decreases the water potential gradient
Light intensity/stomata opening/no of stomata – more light more photosynthesise, stomata open in the day close at night
Temperature – increases kinetic energy, more diffusion
Wind movement - increases or decreases the water potential gradient

Question 69

Describe translocation/massflow

Answer 69

Sucrose (and other solutes) are actively transported into phloem (or co-transported with H+) by companion cells
This lowers the water potential in the phloem and water moves in by osmosis
This creates high hydrostatic pressure leading to mass flow to respiring cells/storage organs
Solutes/sucrose is unloaded from the phloem by active transport

Question 70

Describe chromosome structure

Answer 70

Chromosomes are wrapped around histone proteins forming a nucleosome
Replicated chromosomes are formed of two sister chromatids attached at the centre by a centromere

Question 71

Describe how DNA codes for protein (don’t refer to transcription/translation)

Answer 71

DNA has a specific sequence of bases where a triplet of three bases codes for a specific amino acid.
The order of the triplets codes for a specific sequence of animo acids of a polypeptide/primary structure

Question 72

Describe transcription

Answer 72

DNA strands are separated by breaking hydrogen bonds
Transcription is the synthesis of an mRNA copy of a gene;
The copy is complementary to the DNA template/antisense strand;
RNA polymerase attaches to DNA
RNA nucleotides matched to exposed complementary bases;
Adjacent nucleotides are joined by a phosphodiester bond together to make a strand of mRNA;
mRNA molecule separates from DNA

In eukaryotes only: mRNA is spliced to remove introns
Mature mRNA leaves the nucleus through nuclear pores

Question 73

Describe splicing in eukaryotes

Answer 73

Eukaryotic DNA forms pre-mRNA when transcribed
Pre-mRNA contains introns (non-coding DNA) this must be spliced out
Pre-mRNA is spliced and introns are removed
Mature mRNA is transcribed

Question 74

Describe translation

Answer 74

mRNA attaches to the ribosome at a start codon
tRNA with a complementary anticodon attaches to the mRNA codon
This tRNA is attached to a specific amino acid
Amino Acids are joined by a peptide bond
Using energy from ATP
tRNA is released from the ribosome and a new on enters
Ribosome moves along the mRNA to synthesise a complete polypeptide

Question 75

Describe meiosis

Answer 75

DNA is replicated in interphase before meiosis begins

In the first division there is a separation of homologous chromosomes, halving the chromosome number

In the second division there is separation of the sister chromatids

This produced four genetically different daughter cells

Question 76

How does meiosis lead to genetic variation?

Answer 76

Mutations can occur changing the base sequence leading to the formation of new alleles

In the first division of meiosis there is crossing over where homologous chromosomes swap DNA producing new combinations of alleles

In meiosis homologous chromosomes may be independently segregated, separating into different daughter cells producing new combinations of alleles

During fertilisation there is random fusion of gametes this produces new combinations of alleles

Question 77

Describe how taxonomy techniques are used to classify organisms

Answer 77

Smaller groups within larger groups that are non-overlapping
Comparisons of DNA base sequences to identify similarities and differences
Comparisons of mRNA base sequences to identify similarities and differences
Comparisons of amino acid sequences to identify similarities and differences
Identify points of divergence/common ancestor
Use courtship behavior to identify species
Define species as organisms that can breed to produce fertile offspring

Question 78

Describe natural selection

Answer 78

Mutation leads to variation in a species by developing a new allele
The new allele provides an advantage e.g. give an example from the question
Organisms with the new allele can survive and reproduce
Advantageous allele is passed on to offspring
The advantageous allele increases in frequency in the population

Question 79

Define gene pool and genetic drift, then describe why genetic drift has a larger effect on small populations

Answer 79

Gene Pool – all of the alleles in a certain population usually measured as a frequency
Genetic drift – changes in the frequency of alleles caused by random chance

Why is genetic drift important in small populations?
Changes in population frequency is more pronounced in small populations because of genetic bottlenecks, founder effect, loss off and fixation of alleles.