Model Answers

Question 1

Binary fission

Answer 1

• 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 2

Lipid absorption

Answer 2

• 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 3

Cardiac cycle lhs

Answer 3

• Blood arrives at the left atrium from the pulmonary vein filling the atrium increasing the pressure.
• The atrial muscle contracts 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 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 4

Starch digestion

Answer 4

• 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 5

Protein digestion

Answer 5

• 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 6

Lipid digestion

Answer 6

• 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 7

Cardiac cycle rhs

Answer 7

• Blood arrives at the right atrium from the vena cava filling the atrium increasing the pressure.
• The atrial muscle contracts 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 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 8

Mitosis

Answer 8

• DNA is replicated in S-phase of interphase
• Prophase – Chromosomes condense and become visible, nuclear
membrane dissolves
• Metaphase – Chromosomes line up at the middle of the cell
• Anaphase – Sister chromatids are pulled to the opposite poles of the cell
• 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 9

Meiosis

Answer 9

• 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 10

Genetic variation

Answer 10

• 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 11

Transpiration

Answer 11

• Waterevaporatesfromtheleaves/transpiration
• Duetoheat/kineticenergyfromsunlight
• Waterdiffusesoutofthestomatafromahighwaterpotentialtolow
• The diffusion of water causes a negative hydrostatic pressure in the xylem
• Thisisduetowaterpotentialgradient
• Cohesiontensionformsacontinuouscolumnofwaterthatispulled
through the xylem in a transpiration stream
•Water’sadhesivepropertiesaidthemovementthroughthexylem
• Transpirationstreamlowerswaterpotentialintherootcells
• Waterisabsorbedthroughtheroothaircellsbyosmosisfromahigher water potential to low.

Question 12

Factors affecting transpiration

Answer 12

• 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 13

Translocation/mass flow

Answer 13

• 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 14

Protein structure

Answer 14

• 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 15

DNA structure

Answer 15

• 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 16

Chromosome structure

Answer 16

• 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 17

Comparison of prokaryote and Eukaryote DNA

Answer 17

• 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 18

Splicing

Answer 18

• 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 19

Splicing

Answer 19

• 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 20

ATP structure

Answer 20

• ATP has a nucleotide structure
• It is a ribose sugar with adenine base and three phosphates groups
attached to it
• These phosphate groups can be hydrolysed to remove one phosphate
• This hydrolysis can be used to release energy for chemical reactions, but can also be used in phosphorylation which can make substrates more reactive

Question 21

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

Transcription

Answer 22

• 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 23

Translation

Answer 23

• 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 24

How DNA codes for proteins

Answer 24

• 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 25

Phospholipid bilayer

Answer 25

• 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 26

Co-transport

Answer 26

• 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 27

Describe how taxonomy techniques are used to classify organisms

Answer 27

• 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
• Comparisonsofaminoacidsequencestoidentifysimilaritiesand 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 28

Cell mediated immunity

Answer 28

• Antigen is presented by an antigen presenting cell
• Helper T Cells with the complementary receptor binds to the 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 29

Phagocytosis

Answer 29

• 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 30

HIV replication

Answer 30

• 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 31

Humoural immunity

Answer 31

• 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 32

Vaccines

Answer 32

• 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

Question 33

ELIZA test

Answer 33

• 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 34

Natural selection

Answer 34

• 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 35

Countercurrent flow fish

Answer 35

• Water and blood flow in opposite directions.
• Oxygen concentration is always higher in water
• Diffusion gradient maintained throughout length of gill
• If water and blood flowed in same direction equilibrium would be reached.

Question 36

Tissue fluid formation

Answer 36

• High 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