Longer answer end of test questions - easy marks!! (paper 1)

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Last updated 9:25 AM on 9/17/26
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27 Terms

1
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Describe and explain how you would use cell fractionation and untracentrifugation to obtain a sample of nuclei from muscle tissue (6)

  • homogenise muscle tissue - to break down cells

  • filter - remove debris/whole cells

  • solution must be cold - to prevent enzyme activity

  • solution must be isotonic - to prevent osmosis and cell bursting

  • solution must be buffered - to prevent enzymes denaturing

  • then spun at a lower speed so the nuclei ends up as a pellet at the bottom


2
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Describe the role of organelles in the production and release of enzymes by animal cells (5)

  • DNA in the nucleus codes for enzymes

  • enzymes produced in the ribosomes

  • mitochondria produces ATP

  • RER transports enzymes to the golgi apparatus

  • golgi apparatus modifies enzymes

  • vesicles move the enzymes to the cell surface membrane (by exocytosis)


3
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Describe the role of haemoglobin in the loading, transport and unloading of oxygen (5)

  • loading of oxygen happens in the lungs

  • oxygen loads best with a high partial pressure

  • easier for second molecules to bind (cooperative binding) as shape of binding site is changed

  • oxygen transported as oxyhaemoglobin in red blood cells

  • oxygen unloads at respiring tissues with lower oxygen pp

  • in high pp of co2, oxygen unloads


4
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Contrast how an optical microscope and a transmission electron microscope work and contrast the limitations of their use when studying cells (6)

  • optical uses light vs TEM uses electrons

  • TEM = higher/greater resolution

  • so can view smaller organelles/organelles in more detail

  • optical = colour image vs TEM = black and white image

  • optical = can use live specimen vs TEM = dead/dehydrated specimen

  • TEM = more complex preparation process

  • TEM = magnets to focus vs optical = glass lenses

  • TEM = requires thinner specimen


5
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<p><strong>Figure 6</strong> shows an image from an optical microscope of meiosis occurring in a flower bud of a flowering plant. <strong>W</strong> and <strong>Z </strong>are undergoing meiosis.</p><p>Explain the appearance of <strong>W </strong>and <strong>Z </strong>(4)</p>

Figure 6 shows an image from an optical microscope of meiosis occurring in a flower bud of a flowering plant. W and Z are undergoing meiosis.

Explain the appearance of W and Z (4)

  • W shows 4 nuclei - undergone second division

  • Z shows 2 nuclei - undergone first division

  • W shows haploid cells/ cells containing n chromosomes

  • in Z homologous chromosomes have split/cells in W contain half the mass of DNA than cells in Z


6
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An environmental scientist investigated a possible relationship between air pollution and the size of seeds produced by one species of tree.

He was provided with a very large number of seeds collected from a population of trees in the center of a city and also a very large number of seeds collected from a population of trees in the countryside.

Describe how he should collect and process data from these seeds to investigate whether there is a difference in seed size between these two populations of trees. (5)

  • use a random sample of seeds from each tree

  • use a large enough sample to be representative

  • find the mass of a seed

  • calculate a mean and standard deviation

  • use a student t test (comparing two means)

  • analyse whether there is a significant difference between the means of the two populations


7
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Describe the gross structure of the human gas exchange system and how we breathe in and out. (6)

  • trachea - bronchi - bronchioles - alveoli

  • ^^ must be named in correct order/labelled on correct diagram

  • breathing in - diaphragm and external intercostal muscles contract

  • ribs move up and out - increases volume and decreases pressure of thorax/lungs/thoracic cavity

  • breathing out - diaphragm relaxes and internal intercostal muscles contract

  • ribs move in and down - volume of thorax decreases and pressure is increased


8
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Mucus produced by epithelial cells in the human gas exchange system contains triglycerides and phospholipids

Compare and contrast the structure and properties of triglycerides and phospholipids (5)

  • both contain ester bonds

  • both contain glycerol

  • fatty acids on both may be saturated or unsaturated

  • both insoluble in water

  • both contain C, H and O but phospholipids also contain P

  • triglycerides have 3 fatty acids whereas phospholipids have two plus a phosphate molecule

  • triglycerides are hydrophobic/non-polar vs phospholipids having hydrophilic and hydrophobic region

  • phospholipids form bilayer but triglycerides don’t


9
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Mucus also contains glycoproteins. One of these glycoproteins is a polypeptide with the sugar, lactose, attached.

Describe how lactose is formed and where in the cell it would be attached to a polypeptide to form a glycoprotein (4)

  • lactose formed from glucose and galactose

  • formed with a condensation reaction, forming a molecule of water

  • forms a glycosidic bond

  • attached to a polypeptide in Golgi apparatus


10
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Explain 5 properties that make water important for organisms (5)

  • metabolite - involved in photosynthesis/respiration

  • solvent so reactions can occur

  • cohesion - so can be pulled up plant in a constant stream

  • cohesion - so provides surface tension supporting small organisms

  • high latant heat of vaporisation so provides cooling effect

  • high specific heat capacity - acts as buffer to harsh conditions


11
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Describe the biochemical tests you would use to con firm the presence of lipid, non-reducing sugar and amylase in a sample (5)

  • lipid

    • add ethanol then add water and shake/mix

    • milky white emulsion

  • non-reducing sugar

    • benedicts test and stays blue

    • boil with acid then neutralise with alkali

    • heat with benedicts and becomes red/orange

  • amylase

    • add biuret reagent and becomes purple

    • add starch and test for reducing sugar


12
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Describe the chemical reactions involved in the conversion of polymers to monomers and monomers to polymers.

Give two named examples of polymers and their associated monomers to illustrate your answer. (5)

  • monomers → polymer = condensation reaction + forms chemical bond + releases water

  • polymer → monomer = hydrolysis reaction + break chemical bond + uses water

  • e.g. amino acid → protein with peptide bond

  • e.g. nucleotide → DNA with phosphodiester bond


13
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Describe how mRNA is formed by transcription in eukaryotes (5)

  • H bonds between DNA bases are broken

  • one DNA strand acts as template

  • free mRNA bases align with opposite their complementory DNA base

  • uracil pairs with adenine in the place of thymine

  • RNA polymerase joins adjacent nucleotides

  • with phosphodiester bonds

  • pre-mRNA is spliced to mRNA/introns are removed


14
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Describe how a polypeptide is formed by translation of mRNA (6)

  • mRNA attaches to ribosome

  • anticodon (tRNA) binds to complementory codon (mRNA)

  • tRNA brings specific amino acid

  • amino acids joined by peptide bonds

  • with the use of ATP

  • tRNA released after amino acid joined to polypeptide

  • ribosome moves along mRNA to form polypeptide


15
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Define gene mutation and explain how a gene mutation can have:

  • no effect on the individual

  • a positive effect on the individual (4)


  • gene mutation = change in base sequence of DNA

  • results in formation of a new allele

  • no effect - dna code is degenerate so multiple codons code for the same amino acid/so amino acid sequence does not change

  • positive effect - may result in increased chances of survival/chances of reproduction


16
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Describe the structure of DNA (5)

  • DNA = polymer of nucleotides

  • nucleotide = deoxyribose, phosphate and nitrogenous base

  • nucleotides held together by phosphodiester bonds

  • nitrogenous bases held together by hydrogen bonds

  • H bonds between adenine, thymine, cytosine and guanine


17
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Name and describe five ways substances can move across the cell-surface membrane into a cell (5)

  • simple diffusion - small/non-polar molecules down a conc.grad.

  • facilitated diffusion - down a conc.grad using carrier/channel proteins

  • osmosis - water across a partially permeable membrane down a conc.grad

  • active transport - against a conc.grad. uses ATP and a carrier protein

  • co-transport - transports two substances at the same time using a carrier protein


18
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<p>Contrast the structure of the two cells visible in the electron micrographs shown in Figure 14</p>

Contrast the structure of the two cells visible in the electron micrographs shown in Figure 14

  • Magnification (figures) show A is bigger than B;

  • A has a nucleus whereas B has free DNA;

  • A has mitochondria whereas B does not;

  • A has Golgi body/endoplasmic reticulum whereas B does not;

  • A has no cell wall whereas B has a murein/glycoprotein cell wall;

  • A has no capsule whereas B has a capsule;

  • A has DNA is bound to histones/proteins whereas B has DNA not associated with histones/proteins OR A has linear DNA whereas B has circular DNA;

  • A has larger ribosomes;


19
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Messenger RNA is used during translation to form polypeptides.

Describe how mRNA is produced in the nucleus of a cell (6)

  1. helicase

  2. breaks hydrogen bonds

  3. only one DNA strand acts as a template

  4. RNA nucleotides attracted to exposed bases

  5. attraction according to base pairing rule

  6. RNA polymerase joins nucleotides together

  7. pre-mRNA spliced to remove introns


20
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Describe the structure of proteins (5)

  1. polymer of amino acids

  2. joined by peptide bonds

  3. formed by condensation

  4. primary structure is order of amino acids

  5. secondary structure is folding of polypeptide chain due to hydrogen bonding (alpha helix/beta pleated sheet accepted)

  6. tertiary structure is 3D folding due to hydrogen bonding and ionic/disulphide bonds

  7. quaternary structure is two or more polypeptide chains


21
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Describe how proteins are digested in the human gut (4)

  1. hydrolysis of peptide bonds

  2. endopeptidases break polypeptides into smaller peptide chains

  3. exopeptidases remove terminal amino acids

  4. dipeptidases hydrolyse/break down dipeptides into amino acids


22
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Describe the transport of carbohydrate in plants (5)

  • sucrose actively transported into phloem

  • by companion cells

  • creates lower water potential and water moves in to phloem (from xylem) by osmosis down a concentration gradient

  • creates higher hydrostatic pressure

  • mass flow to respiring tissues

  • sucrose unloaded by active transport (from phloem)


23
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Compare and contrast the structure of starch and the structure of cellulose (6)

  • both made of glucose monomers

  • starch = alpha glucose vs cellulose = beta glucose

  • both have C, H, O2

  • starch = branched vs cellulose non branched

  • cellulose has microfibrils vs starch doesn’t

  • both contain glycosidic bonds between monomers

  • starch = helical vs cellulose = straight

  • starch = amylose and amylopectin vs cellulose just one molecule


24
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Describe the complete digestion of starch by a mammal (4)

  • hydrolysis

  • of glycosidic bonds

  • starch → maltose by amylase

  • maltose → glucose by maltose

  • by membrane bound maltase


25
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Describe how a quaternary protein is formed from its monomers.

Do not include the process of translation in your answer (5)

  • amino acids joined by peptide bonds

  • with condensation reactions

  • secondary structure formed by alpha helix/beta pleated sheet/hydrogen bonds

  • tertiary structure formed by hydrogen/ionic/disulphide bonds

  • quaternary = multiple polypeptide chains



26
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Describe the structure of DNA and the structure of a chromosome (6)

  • polynucleotide/polymer of nucleotides

  • double helix held with hydrogen bonds

  • phosphodiester bonds between nucleotides

  • nucleotide = deoxyribose, nitrogenous base, phosphate

  • adenine, thymine, guanine, cytosine

  • associated with histones

  • chromosome = sister chromatids joined at the centromere


27
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Mutation can result in an increase in genetic variation within a species.

Describe and explain the other processes that result in increases in genetic variation within a species (4)

  • crossing over between homologous chromosomes

  • random fertilisation of gametes

  • independent segregation/ of homologous chromosomes

  • produces new combinations of alleles