c5: cell chemistry Flashcards

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1
Q

fill in chloroplast structure diagram

A

photos

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2
Q

what is the most important pigment in photosynthesis and what colour is it?

A

Chlorophyll A, green.

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3
Q

What are some other pigments present in the leaf?

A

Antenna, chlorophyll b, carotenoids (orange/red), xanthophylls (yellow/brown)

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4
Q

briefly describe what photosynthesis is

A

Process by which plants and other autotrophs store the energy of sunlight into sugars

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5
Q

Where does photosynthesis occur within a plant?

A

Palisade layer. Gas exchange of CO2 and O2 occurs at openings called stomata surrounded by guard cells on the lower leaf surface.

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6
Q

What are the two main phases of photosynthesis?

A

Light reactions and dark reactions

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7
Q

Where do light-dependent reactions occur?

A

In the thylakoid membranes.

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8
Q

How does sunken stomata reduce transpiration?

A
  • sunken stomata creates small pockets of air that has a high humidity
  • closes when dry to keep humidity
  • increases humidity around stomata
  • decreases conc. gradient between leaf air spaces and the exterior
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9
Q

What limits light reactions?

A

Amount of sunlight.

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10
Q

How do desert plants adapt for light reactions?

A
  • thicker waxy cuticles at the top of the leaf to reduce evaporation
  • sunken stomata to reduce concentration gradients and water loss
  • longer and deeper roots for more water
  • thin leaves to reduce exposure to the sun
  • closing stomata in the mid-day heat (restrict photosynthesis and water loss)
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11
Q

How do rainforest plants adapt to light reactions?

A
  • dense, tall, green structure to maximize photosynthesis in limited conditions
  • big thick leaves to maximize sunlight
  • shallow roots (absorb less water because there’s already tons of water)
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12
Q

Describe how water lilies adapt to their living environment for most light reactions?

A
  • air pockets to float in order to get most sunlight
  • stomata build on the top that touches air in order to let CO2 in
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13
Q

What is the compensation point?

A

The point reached in a plant when the rate of photosynthesis is equal to the rate of respiration.

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14
Q

Briefly describe and guess photosynthesis rate during different times of the day.

A

During the night: no photosynthesis
Sunrise: medium level, possible compensation point
mid-day: max level
Sunset: medium level, possible compensation point

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15
Q

What powers carbon fixation?

A

ATP and NADPH

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16
Q

How does the amount of light impact the amount of carbon dioxide?

A

More light= more photosynthesis= more sugars produced= co2 drops

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17
Q

Define the physiology of aerobic respiration

A

The process releases energy by breaking down food molecules in the presence of oxygen.

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18
Q

What are the phases of anaerobic cellular respiration?

A
  1. Glycolysis
  2. Fermentation
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19
Q

Define glycolysis.

A

The process in which one molecule of glucose is broken in half, producing two molecules of pyruvic acid.

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20
Q

What is pyruvic acid used for?

A
  • Used to supply energy to cells when the glucose route is not available
  • E.g. when muscles are short of oxygen due to excessive physical activity
  • It ferments to produce lactic acid that further converts to lactose to provide energy.
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21
Q

Explain “takes energy to make energy”

A

2 ATPs are needed to start reaction of glycolysis, 4 ATPs are produced thus the net gain is 2 ATP

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22
Q

Where does glycolysis occur?

A

Cytosole

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23
Q

What are the differences between ATP and ADP?

A
  • ATP is a high energy molecule- charged battery. - ADP is a low-energy version of it
  • it lacks one phosphate.
  • The one missing phosphate requires energy to fill it up and make it charged.
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24
Q

What is the process of charging batteries in ATP and ADP?

A

Cellular respiration- so ADP can work with any part of the body.

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25
Q

Which type of reaction- anaerobic or aerobic produces more ATP and how many?

A

Anaerobic= no atp at first-> 2 ATP from glycolysis
Aerobic= 34 ATP at first-> 2 ATP from glycolysis

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26
Q

What are the 3 phases of aerobic respiration?

A
  1. Glycolysis
  2. Krebs cycle
  3. Electron transport chain
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27
Q

What is the process of fermentation?

A

Releases energy from breaking down glucose without the presence of oxygen.

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28
Q

What are the two types of fermentation?

A

Lactate or alcohol

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29
Q

What does the Krebs cycle add to the aerobic respiration?

A

gives 2 more ATP and some other energy carrying molecules.

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30
Q

Briefly describe what happens during the Krebs cycle.

A

Pyruvic acid (glucose split into two) is broken into carbon dioxide in a series of energy-extracting reactions. Citric acid is created in this cycle.

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31
Q

What are the reactants of the Krebs cycle?

A

Pyruvate

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32
Q

What does the electron transport chain do?

A

It uses the high-energy electrons from the Krebs cycle to convert ADP to ATP, producing 32 ATP.

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33
Q

What is a protein?

A

a long chain of amino acid

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34
Q

What is fat?

A

A chain of glucose (mostly)

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35
Q

Why are salts critical in the body?

A

Because it is used to regulate fluids in the body (osmoregulation with diffusion), aid metabolism, as well as nerve impulses.

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36
Q

What is iron used for in the body?

A

Oxygen transport and turned into harmoglobin

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37
Q

Describe the key features of vitamins.

A
  • no energy value
  • essential for small amount to maintain good health
  • essentials cannot be created therefore must obtain from diet
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38
Q

At what condition do enzymes function best?

A

37 degrees celsius

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39
Q

Vitamins are _______, and they’re known as __________

A
  1. organic
  2. co-enzymes
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40
Q

Minerals are _______, and they’re known as __________

A
  1. inorganic
  2. cofactor
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41
Q

What are the 3 types of macromolecules and what are their key features?

A
  1. carbohydrates: compact storage of sugars
  2. lipids: fats, 2x energy of carbs of the same mass
  3. proteins: little energy is released when broken down
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42
Q

What are some factors that affect enzyme action?

A

Temperature, pH of solution, concentration of substrate, concentration of enzyme

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43
Q

List some enzymes that have a different optimum pH from most.

A

Pepsin and proteinase have an optimum pH of 2 and work in the stomach to breakdown protein.

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44
Q

Describe enzyme activity at low to high pH of 1-10

A
  • Enzymes are most active at optimum pH (most of the time it is 7 with exceptions)
  • Lose activity in low or high pH as the tertiary structure is disrupted- when enzymes denature, lose their 3D structure, and don’t work.
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45
Q

Describe the relationship between substrate concentration in a solution and enzyme activity within it.

A
  • The rate of reaction increases as substrate concentration increases (at constant enzyme concentration)
  • Enzyme activity continues to increase until all of them are combined with substrate-> reaches maximum activity
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46
Q

What is the enzymes- protein’s functions and how do they connect to their structure?

A

Proteins:
- increase rate of reaction by lowering energy of activation
- catalyze nearly all the chemical reactions in the body cells
- unique 3D shape that fit shape of reactants

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47
Q

How do enzymes catalyze reactions?

A

They increase the rate of reaction by lowering the active energy required.

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48
Q

We have pairs of chromosomes called _______

A

Homologous chromosomes

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49
Q

All the chromosomes of a person are called a _______

A

karyotype

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50
Q

What are banding patterns and what does it show?

A

Alternating light and dark regions along the length of a chromosome. It shows where the genes are located on a chromosome.

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51
Q

Describe the process of DNA replication.

A
  1. DNA unzips when enzyme helicase breaks up hydrogen bonds on either sides of it.
  2. Enzyme DNA polymerase III builds a new ladder on one of the two unzipped strands
  3. Floating nucleotides are grabbed from around and stuck onto the lagging strand as Okazaki fragments
  4. Two new strands coil up into the helix, but are still joined in the centre, known as a centromere
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52
Q

How is the DNA perfect in DNA replication?

A

After one side of the double helix is formed, the other side is able to copy it for identical DNA.

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53
Q

When does DNA replication occur?

A

When a DNA is triggered to replicate- S phase of mitosis and meiosis.

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54
Q

What forms a nucleotide?

A

Phosphate, sugar, and base

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55
Q

What is protein synthesis? (Describe with lil detail)

A

Making new proteins by decoding DNA into language for proteins.
- DNA contains genetic info to make amino acid
- Where amino acids combine to make proteins.

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56
Q

Amino acid determine _______

A

Physical traits of an organism and control cellular function

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57
Q

What is a codon and what is another name for it?

A

Three bases that code for a specific amino acid
triplet

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58
Q

Explain the process of protein synthesis.

A

Step 1: transcription
DNA unzips, messenger RNA copies info from DNA using base pairings

Step 2: genetic code is carried from the nucleus to the ribosome in the cytoplasm by tRNA

Step 3: translation
- Ribosome attaches to mRNA and starts reading the codons
- Base pairings are translated into an amino acid chain by the tRNA, with an anticodon at its bottom pairing with the mRNA strand.
- tRNA carries amino acids and attaches them to the growing amino acid chain
- Once amino acid production is completed, the ribosome releases the amino acid chain.
- Chain folds into a complex 3D shape to form a protein.

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59
Q

How might protein synthesis impact proteins?

A

Make the protein better, worse, or no change.
Better= actively selected for
Worse= actively selected against
No change= Won’t be actively selected

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60
Q

How is RNA different from DNA?

A
  • SIngle-stranded rather than double-stranded
  • it has a different molecule composition, where it has ribose instead of deoxyribose
  • it has different bases of uracil instead of thymine
  • it is shorter in length
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61
Q

What is a frame shift mutation?

A

Genetic mutation caused by a deletion or insertion in a DNA sequence that suits the way the sequence is read.

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62
Q

What does an anti codon look like?

A

(top to bottom) circle, rectangle, 3 legs

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63
Q

Where is the anti codon and what does it do?

A

It is on the tRNA and it bonds with the codon by matching with the complementary base.

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64
Q

What triggers mutations to occur?

A

When something goes wrong with transcription, translation or in the DNA replication itself.

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65
Q

What are the two types of mutations?

A

Substitution and addition/ deletion

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66
Q

Describe DNA mutation- substitution

A
  • codon is changed, but may not change the amino acid it codes for
  • if no impact on amino acid, no impact on produced protein
  • if has impact, could be pos/neg/ or neutral
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67
Q

Describe DNA mutation- insertion/ deletion

A

When a base is added or deleted.
- creates a frame shift
- every amino acid after that is affected
- this destroys the protein majority of the time

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68
Q

Describe protein and fat in terms of chains

A

Protein= long chain of amino acid
Fat= long chain of glucose

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69
Q

Why is salt essential for the body?

A

Because it conducts osmoregulation in the body as part of metabolism.
(salt in the cell, diffusion, etc)

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70
Q

Explain monosaccharides, disaccharides, and polysaccharides.

A

mono- single unit of sugar, simplest form of it
di- 2 mono
poly- many monos

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71
Q

List 2 examples of polysaccharides that aren’t sugars.

A
  • starch in plants
  • glycogen in animals
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72
Q

Explain the difference between a polymer and a monomer using an example. (2 marks)

A
  • A monomer is a basic unit of a substance
  • a polymer is many units of the monomer put together. (1 mark).
    Any relevant example, such as: glucose is a monomer for
    molecules such as starch and glycogen; or amino acids are monomers for proteins. (1 mark)
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73
Q

What are the components of a lipid molecule and what is the main type of it?

A

A triglyceride contains glycerol, 3 fatty acids

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74
Q

What is a polypeptide chain?

A

Chain of substances with polypeptide bond.

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75
Q

What are some functions of proteins and their examples?

A

Structural: collagen
Regulatory: insulin
Contractile: actin
Immunological: antibodies
Transport: hemoglobin
Catalytic: amylase

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76
Q

What is an active site?

A

Region in enzyme that fit substrates.

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77
Q

Describe the lock and key explanation of enzyme reaction.

A

Lock= enzyme, Key= substrate
- 1 lock only fits 1 key
- active site cannot change
- wrong key (substrate) leads to no reaction

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78
Q

Describe the induced fit theory

A
  • enzyme is not rigid, changes shape with substrate
  • active site is not the exact same shape as substrate-> wraps around it instead
  • vitamin (coenzyme) comes in, changes active site and allows reaction to take place.
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79
Q

What is a non-competitive inhibition?

A
  • a non-comp inhibitor has a different structure from the substrate
  • distorts the shape of the enzyme, altering shape of active site
  • prevents binding of substrate
  • cannot reverse by adding more substrate
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80
Q

What is a reversible competitive inhibition?

A
  • A comp inhibitor has a structure like the substrate
  • compares with substrate for active site
  • stops reaction by blocking active site and restricting substrate from entering
  • can reverse the effect by increasing substrate concentration
81
Q

What competitive inhibitors aren’t reversible?

A
  • turns into toxin
  • kills enzyme/ organism completely
82
Q

What is feedback control?

A

When a product of the reaction binds to the active site of an enzyme and changes its catalytic activity.

83
Q

Describe the process of feedback control and the key concept behind it.

A
  • Each enzyme produces a product which acts as the substrate for another enzyme
  • After enough products are formed, an end product binds with the first enzyme and stops its activity.
    Enzyme on- need more
    Enzyme off- need less
84
Q

Describe where energy comes from and how does it supply a cell?

A
  • energy released when bonds holding atoms together break, usually as heat
  • used to form ATP from ADP
  • stored at ATP until needed
85
Q

What is energy used for?

A
  • used by all cells to carry out work
  • activities of organisms are the result of metabolism
  • used to build new molecules and break up old molecules
86
Q

What are organic and inorganic substances in the body and what differs them?

A

Organic: vitamin
Inorganic: mineral
Organic substances can be generated by the body where inorganic substances cannot.

87
Q

Describe significance of amount of vitamins in the body.

A

Too much: of some may be harmful
Essential for small amounts to maintain good health

88
Q

Describe 3 features of vitamin A.

A
  • formed in body from carrotene
  • destroyed at high temp
  • maintain dim light vision
89
Q

Describe 2 features of vitamin C.

A
  • Destroyed after prolonged cooking
  • Essential for wound-healing
90
Q

Describe 2 features of vitamin D.

A
  • generated in skin from UV light
  • Essential for regulation of calcium and phosphate metabolism
91
Q

What are some functions that require minerals to carry out and how much of it is needed?

A
  • regulate body metabolism
  • healthy growth
  • necessary for construction of certain body tissues
  • needed in small amount
92
Q

What are some examples of minerals?

A

Calcium, phosphate, iron

93
Q

What are polysaccharides used for?

A

Energy storage

94
Q

Describe function of carbohydrates. with examples

A
  • main source of energy
  • cellulose: fibrous material of Plant Cell Wall
  • dietary fibre: stimulates Peristalsis (contraction of muscles in digestive tract)
95
Q

How are excess carbohydrates stored?

A
  • stored as glycogen in liver & muscle
  • stored as fats under skin
96
Q

Describe solubility of lipids

A
  • insoluble in water
  • soluble in organic solvent
97
Q

What are the functions of lipids?

A
  • give energy
  • component of cell membrane
  • form fatty tissues under skin to store energy
  • acts as insulator to keep warm
  • to transport fat-soluble vitamins (A, D, E, K)
98
Q

What happens with excessive protein?

A

Broken down in the kidney to urea.

99
Q

How are amino acids linked together?

A

By formation of peptide bonds in a polypeptide chain.

100
Q

What are the 4 levels of protein structure? Elaborate on what they are.

A
  1. Primary structure: The sequence of amino acids in a polypeptide chain.
  2. Secondary structure: The shape of the
    polypeptide chain.
  3. Tertiary structure: The overall
    conformation (shape) of the polypeptide
    caused by folding.
  4. Quaternary structure: In some proteins,
    an additional level of organization groups
    separate polypeptide chains together to
    form a functional protein.
101
Q

What is the function of cofactors?

A
  • Assist in enzyme activity
  • assist in performing necessary activity that the enzyme cannot perform alone
102
Q

What does a coenzyme do?

A

Prepares the active site for catalytic activity.

103
Q

What are competitive inhibitors?

A

Mimic substrate, may block active site

104
Q

What are isoenzymes?

A

They catalyze same reactions in different parts of the body.

105
Q

Give an example of a diagnostic enzyme and how it functions.

A

Blood sugar-> many sugar around= kick start enzyme
low sugar aorund= shuts enzyme

106
Q

How does penicillin kill bacteria as an enzyme inhibitor?

A
  • cell wall peptide and penicillin have similar structure
  • mimics natural chemical to block enzyme
  • bacteria constantly redo cell wall as they grow and divide
  • Penicillin inhibits the enzyme that catalyses
    one of the reactions need to build a bond in
    the cell wall
  • causing weakness in wall
  • Osmotic pressure builds in the cell causing
    cytolysis (killing the bacteria)
107
Q

What factors affect the rate of photosynthesis? Briefly describe how they affect it.

A

Temperature: high temp= higher kinetic energy, therefore faster rate of photosynthesis
Concentration of co2: higher concentration increase concentration gradient, therefore increase diffusion rate of co2 and rate of photosynthesis
Light intensity: rate of photosynthesis will increase because there is more light available to drive the reactions of photosynthesis.
Light quality: more sensitive in blue LEDs than red and green

108
Q

What is photosynthesis?

A
  • Biochemical process
  • Occurs in the chloroplasts of plant cells using light energy
  • synthesize organic compounds.
109
Q

What is cellular respiration?

A

A BIOCHEMICAL process that occurs in different locations in the CYTOSAL and MITOCHONDRIA and METABOLIZES ORGANIC SUBSTANCES, AEROBICALLY or ANAEROBICALLY, to RELEASE USABLE ENERGY IN THE FORM OF ATP.

110
Q

What substance conducts alcoholic fermentation and what is its outcome?

A

yeasts and some microorganisms, producing alcohol and co2

111
Q

When and where is lactic acid produced?

A
  • Produced by muscles during rapid exercise
  • when the body cannot supply enough oxygen
112
Q

Describe how the body attains energy both short-term and long-term.

A

Quick energy –
- Lactic Acid fermentation is used to get quick
energy and gives off lactic acid as a
byproduct, thus the muscle pain.

Long-Term Energy –
- Cellular respiration to produce energy.
- Exercising or activities that last for at
least 15 to 20 minutes.
- Best form for weight control.

113
Q

How does DNA turn into physical traits?

A

DNA contains genetic info to make amino acids-> amino acids combine to make proteins-> proteins determine physical trait and cellular functions

114
Q

How many amino acids do codons code for?

A

20

115
Q

Where does protein synthesis take place?

A

Ribosomes in the cytoplasm.

116
Q

Can mutations in DNA be caused?

A
  • can be spontaneous
  • or caused by mutagens: high temp, uv light, chemicals, radiation
117
Q

What is in a nucleotide?

A

Nitrogenous base, sugar, phosphate

118
Q

______ joins the nitrogenous bases

A

hydrogen bonds

119
Q

DNA strands are ______

A

Anti-parallel

120
Q

What structure does DNA have?

A

Double helix formed from two complementary strands of nucleotides held together by hydrogen bonds between base pairs.

121
Q

Name and discuss the roles of the two reactions (ADP-> ATP, ATP-> ADP) and their importance in the cell.
(5 MARKS)

A

Reaction 1:
- Respiration
- takes energy from glucose and stores it in the phosphate bond to form ATP from ADP
- anaerobic respiration yields 2 ATP while aerobic yields 36

Reaction 2:
- Cell metabolism
- phosphate group is removed, releasing energy
- cell processes such as active transport, cell repair and mitosis
- resulting ADP can then be recycled and used over and over again

122
Q

State which food molecule is utilized by human cells as the primary source of energy and which makes it ideal for the purpose.
(2 MARKS)

A
  • carbohydrates
  • contains glucose, which is a small, soluble molecule that passes cell membrane
  • long, compact chain structure
  • alternatively, respiration of amino acids produce toxic waste and lipids (e.g. triglycerides) that are poorly soluble in water.
123
Q

Outline two reasons why inorganic substances are essential for cellular processes.
( 2 MARKS)

A
  • needed for building and repairing chemical structures
  • e.g. calcium is essential for building bones and teeth
  • needed for chemical reactions
  • e.g. assist enzyme function
124
Q

Yeast is a single-celled, microscopic fungus that uses sucrose as a food source. An experiment was carried out to investigate cellular activity by a particular species of yeast.

Yeast cells were placed in a container and a sucrose solution was added. An airtight lid was placed on the container. The percentages of oxygen and ethanol in the container were recorded over a one-hour period. The experiment was carried out at room temperature. The results are shown below.

Start of experiment: Oxygen% (21); Ethanol% (0)
End of experiment: Oxygen% (18); Ethanol% (4)

State the main specific process that is occurring in the yeast cells, giving reasons based on the information in the table above.
(3 MARKS)

A

Process: cellular respiration
- Some oxygen is consumed as levels drop from 21 to 18% suggesting some aerobic respiration
- Ethanol is an end product of anaerobic respiration in plants
- this goes from 0 to 4 % anaerobic respiration must be occurring
*- The airtight lid means restricted oxygen supply, so when this runs out the respiration will
be predominately anaerobic

125
Q

Several factors affect the rate of photosynthesis.
*x axis: increase in factor A, y axis: rate of photosynthesis
Graph A: bell curve
Graph B: y=logx graph

State which factor is represented by each graph and explain your choice.
(4 MARKS)

A

Factor A- graph A: Temperature or pH or Light at time of Day
Reason:
Temperature-
~ Photosynthesis increases until the temperature gets too high or too low
~ stomata close (to prevent water loss) and CO2 availability decrease therefore less photosynthesis
OR
– slow molecular movement so enzyme substrate collisions are low.
- As temp increases, collision rate increases
- eventually denaturing reduces the effectiveness of collisions until no collisions are effective
- P/S ceases.
Similar argument for pH
Light/Time of day – Photosynthesis with available light (intensity) during the day and will
not occur in the dark.

Factor B (Graph B): Carbon dioxide or a substrate/reactant in P/S or light (in laboratory
conditions)

126
Q

Suggest how the rate of photosynthesis could have been determined, including how you could ensure that you were determining the rate of photosynthesis alone.
(3 MARKS)

A
  • Measuring the inputs and/or outputs of photosynthesis
  • Change in Co2 and o2 could be measured
  • in an enclosed container holding the plant
    OR
  • increase in plant tissue could be measured as new growth comes from photosynthesis.
  • Both gases need to be measured with the plant in the dark to give a baseline comparison of cellular respiration’s effect
127
Q

Explain what is meant by an enzyme’s specificity.
(1 MARK)

A

Enzymes active site binding to a specific substrate.

128
Q

How is enzyme specificity achieved in terms of its protein structure?
(2 MARKS)

A
  • Specificity achieved by unique shape of the active site formed in the protein structure
  • Which induces a fit with substrate

Protein structure – enzymes active site – substrate – induced fit. (1/2 mark each)

129
Q

The browning of peeled apples on exposure to air is catalyzed by the enzyme phenol oxidase in ruptured plant cells.
To prevent this browning during the preparation of fruit salad, it is usual to squeeze some lemon juice (an acid) over the cut fruit.
Explain why this works to stop the browning.
Also suggest TWO other treatments that would also be effective in preventing browning in peeled apples.
(3 MARKS)

A

2 marks for lemon juice explanation; ½ mark for each suggested treatment (explanation not
required)
- Lemon juice = acid, lowers the pH in the area of the cut
- enzymes rate of reaction varies with pH ∴ adding acid means its no longer at optimum pH
- ∴ rate of reaction is greatly reduced
Two other treatments:
- heating/cooking OR cooling/freezing
- temperature no longer optimal for enzyme function
OR
- add an enzyme inhibitor that would block or compete with the active site
- reduce substrate concentration (e.g. oxygen) by vacuum sealing etc

130
Q

Briefly explain the role of THREE other cell organelles outside the nucleus that would be involved in the formation and secretion of a protein.
(3 MARKS)

A
  • MITOCHONDRIA produce ATP to supply the energy needed for the production of protein on the ribosomes of the Rough Endoplasmic Reticulum (RER).
  • Membrane-bound ribosomes in the RER translate the mature mRNA transcript into amino acids that are attached to become polypeptides
  • GOLGI BODIES modify protein by the addition of carbohydrates and lipids.
  • CHLOROPLASTS form glucose to act as the energy source.
131
Q

With reasons, identify the most likely shape below for a drug molecule that is capable of
competitive inhibition of the enzyme above.
ONLY REASON
(1 MARK)

A

(shape)
Reason:
- Shape B is shaped so that it could be induced to fit in the active site
- so block /compete with the normal substrate
½ mark for block/compete/etc. ½ mark for induced fit

132
Q

2018 pp q8
Name the main process taking place inside the nucleus at Stage 1 and briefly describe what is occurring during these processes inside the nucleus.
(2 MARKS)

A

Process: TRANSCRIPTION
Description:
- DNA (unlabelled) is located inside the nucleus.
- DNA is a double helix polymer of nucleic acids containing the GENETIC CODE.
- The DNA strands partially separate and one
strand acts as a TEMPLATE to produce many small strands of MESSENGER RNA (labelled Q).
- The gene coded by a short segment is transcribed into mRNA.
- COMPLEMENTARY BASE PAIRING (A-T or A-U & G-C) ensures the FIDELITY of the genetic code.
- The mRNA then LEAVES THE NUCLEUS through the pores in the nuclear membrane.

133
Q

2018 pp q8
Name the overall process taking place outside the nucleus at Stage 2 and using the names of structures E, F, G and S, describe what is occurring in the processes outside
the nucleus
(3 MARKS)

A

Process: TRANSLATION
Description:
- tRNA (labelled E) carries a specific AMINO ACID (labelled F) dependent upon the ANTICODON (labelled G) they possess.
- The anticodon on the tRNA COMPLEMENTS the CODON (unlabelled) on the mRNA between the large and small subunits of the RIBOSOMES (labelled S).
- As tRNA binds to its complementary codon, the amino acid it carries adds to the lengthening POLYPEPTIDE chain.
- Ultimately, the code found in the
gene of the DNA translates into a polypeptide (protein).
- The protein produced possesses some biological function.
- Complementary base pairing of codon to anticodon again ensures fidelity of coding and a specific polypeptide is made according to the genetic code found within the DNA.

134
Q

Explain the role of a non-competitive inhibitor. (3 marks)

A
  • A non-competitive inhibitor binds to the enzyme (½ mark) which is not the active site (½ mark)
  • This changes the shape of the enzyme molecule (denatures it) and thus the shape of the active site (1 mark).
  • As a consequence, the enzyme can no longer function to catalyse the reaction (½ mark) because the substrate no longer fits into the active site (½ mark).
135
Q

What would be the base sequence for mRNA formed from the coding strand? (in terms of the coding strand)

A

Complementary bases of coding strand but in RNA format (U instead of A)

136
Q

Although DNA is double-stranded, only the coding strand determines the specific amino acid sequence of a polypeptide. Suggest a role of the non-coding strand.

A
  • The function of the coding strand is for DNA replication.
  • DNA unwinds and the non-coding strand is matched
    with complimentary bases to form a new strand.
137
Q

Explain the ways in which structure Y (tRNA) is different from structure X (mRNA) and how this relates to the function of both molecules.

A
  • Structure X is mRNA (½)
  • mRNA codes for a specific protein therefore the length of mRNA depends upon the protein (series of amino acids) it is coding for
  • mRNA has codons that attract the anticodons of structure Y with their complimentary bases (1).
  • Structure Y is tRNA (½)
  • tRNA has an anticodon as well as an amino acid attachment site (½)
  • When matched with the mRNA, tRNA brings the amino acid it codes for.
  • The amino acids bond together (peptide bonds) to
    form a polypeptide chain or protein (1).
138
Q

2020 pp q 7

A

Point A factor: CO2 concentration (½).
Reason:
- As the CO2 concentration increases the rate of photosynthesis increases on the graph. CO2 is used up in the reaction. CO2 is a reactant in photosynthesis (2 marks for 2 points).

Point B factor: Could be any other factor (light intensity, temperature, enzyme concentration, number of chloroplasts) (½).
Reason:
- The graph indicates that at point B increasing CO2 does not result in an increased rate of photosynthesis.
- Therefore, some other factor must be limiting the rate of photosynthesis.
- E.g. Lack of light intensity means that there is not sufficient energy from sunlight for the reaction to proceed.
Increase in temperature increases kinetic
energy of molecules therefore rate of reaction (2 marks).

139
Q

Water can fill air spaces in the soil surrounding the roots of plants. This prevents oxygen from reach root hair cells.
Using your knowledge of aerobic and anaerobic respiration, explain why overwatering can kill plants. (5 marks)

A
  • Aerobic respiration happens in the presence of oxygen (½).
  • It produces 38 ATP (½) and no toxic bi-products (½).
  • Anaerobic respiration occurs when no oxygen is present (½).
  • It only produces 2 ATP (½) and ethanol (½).
  • When the roots do not have access to oxygen their anaerobically respire and therefore produce less energy in the form of ATP. They need the ATP to power active processes (½).
  • Without sufficient ATP they will not be able to transport minerals across the cell membrane and hence the cell will die (½).
  • Ethanol is toxic to the plant and will therefore kill the root cells (1)
140
Q

A researcher placed plant cells in low light and then gradually increased the light levels over a period of time. The oxygen level was measured in the air around the cells. He found that as time passed, the oxygen concentration increased. Explain this observation.

A

1-1.5 marks: What happens to p/s and res in low light
e.g. In low light conditions, the rate of respiration would be greater than the rate of p/s, meaning very little excess oxygen is produced, therefore 02 levels may fall
1-1.5 marks: What happens to p/s and res rate as light increases
e.g. as light intensity increases, there is an increase in p/s converting co2 to glucose and oxygen
1 mark: acknowledging res rate does not change

141
Q

Give a reason for your choice of a particular inhibitor fitting a particular active site.

A
  • mostly likely complimentary (lock and key or induced fit)
  • of enzyme’s active site
  • as it has similar shapes to active site
  • the other enzyme’s active site lack some of the shapes required to bind to active aite of enzyme molecule
  • other enzymes have different charges
142
Q

Explain the process by which this drug (inhibitor) would contribute to lowering blood sugar in a person. (ACE- enzyme acts on angiotensin I to produce Angiotensin II a powerful blood-pressure raising agent, people with long term high blood pressure have raised blood levels of ACE)

A
  • competitive inhibition
  • use of drug would block/ inhibit the active site of ACE or compete with Angiotensin I
  • Inhibiting production of angiotensin II
  • Blood raising agent would not then be released
  • leading to reduced blood pressure
143
Q

What are the mRNA and DNA base sequence of AGC and UUC?

A

mRNA: UCG, AAG
DNA: AGC, TTC

144
Q

Describe the role of tRNA in the process of translation.

A
  • tRNA has an amino acid attachment site and an anticodon region
  • tRNA have anticodons complementary to codons on mRNA, which code for specific amino acids
  • tRNA carrying the amino acid attaches to the ribosome and bring in amino acids to the mRNA
  • once anticodon-codon binding occurs on the mRNA, peptide bond is formed leading to polypeptide chain formation.
145
Q

2019 pp q 8a

A

photos

146
Q

2019 pp q 8b

A

photos

147
Q

2019 pp q 8c

A

photos

148
Q

Duration of exercise Energy sources (seconds)
1-15 ATP
15-20 glucose
20-140 glycogen
During vigorous exercise, the ATP storage within the muscle is used up in a few seconds. With reference to the information in the table, and the nature of the substances involved, explain in detail how this ATP store is replaced. (5 marks)

A
  • ATP is the preferred source of energy for metabolic processes as only one bond needs to be broken for a net release of energy, (ATP RELEASE FORMULA) and it is a recyclable molecule
  • cellular respiration allows ATP to reform by adding a phosphate molecule to the ADP. This can accommodate a net 36 ATP if aerobic or net 2 ATP if anaerobic
  • Initially from 1-15 sec: muscle cell derives energy from breaking available ATP -> ADP+Pi+energy
  • 15-20 sec: after ATP is depleted, free glucose then broken down via cellular respiration to provide more ATP.
  • 20-140 sec: glycogen (stored in liver and muscle cells) breaks down into glucose, triggering ATP production via cellular respiration.
149
Q

If the concentration of enzymes were increased, explain how the rate of the reaction would be changed and why.

A
  • Increasing the amount of enzyme will increase the rate of reaction
  • This is because there are more active sites for the reactions to take place in.
150
Q

What is meant by the terminology- optimum conditions, when referring to enzyme action?

A
  • optimum temperature: 37 deg
  • optimum ph: 7
  • At its optimum condition, enzyme reaction will occur at its fastest rate
151
Q

Explain what would happen if the temperature increases to 50 deg and why.

A
  • enzyme would denature and lose its 3D structure
  • active site will change site- no long fits substrate
  • therefore catalytic reaction would pause
152
Q

If an enzyme was stored in the fridge and slowly mixed through the sucrose (which is at 30 deg), what will happen to the reaction rate?

A
  • increased reaction rate as it warms up
  • enzyme initially has low temperature therefore kinetic energy is decreased
  • therefore slow initial reaction rate
153
Q

The clear protein of an egg white becomes opague and firm when cooked because the heat…

A

Changes the protein structure

154
Q

tasc pp 2022 q7b
Give an analysis of the likely impact of molecule Q on the enzyme’s function in the situation shown above.

A
  • Q is likely to be a non-competitive inhibitor (1).
  • It binds to an alternate/allosteric site on the enzyme (½) - - - - and causes the active site to change shape (½).
  • This means it will no longer be able to induce a fit (be a complementary match) for the specific substrate (½)
  • and as a result the enzyme’s function to catalyse a reaction will be reduced (½).
155
Q

In humans, the enzyme lipase breaks down lipids into fatty acids and glycerol. This takes place at normal body temperature and in the presence of bile.
Using your knowledge of enzymes, explain why lipase only breaks down lipids in these conditions. (3 marks)

A
  • Lipase’s active site means it will only work to break down lipids due to the specific structure of lipid substrate (1). - ——- Lipase has optimal working conditions (½).
  • Specifically, this is 37 degrees C/body temperature (½).
  • Bile produces optimal pH conditions for lipase (1) being slightly alkaline. Beyond these optimal conditions, lipase may become denatured changing the active site shape and reducing enzyme function (½)
156
Q

tasc 2022 pp 8bi
Outline why molecule one is an ATP molecule rather than molecule two.

A
  • Molecule one is ATP (adenosine triphosphate) as it has three phosphate groups
  • molecule two only has one phosphate group and is likely to be a nucleotide
157
Q

tasc 2022 pp 8bii
Identify one feature of the structure of ATP that is related to its role in supplying energy for cellular processes. (1 mark)

A
  • The high energy, terminal phosphate bond releases energy in a useable form;
  • Only one bond needs to be broken to release energy;
  • ATP/ADP and Pi are recyclable, so energy is always available to the cell; or
  • ATP is soluble so can easily diffuse from the mitochondria to all parts of the cell to
    provide energy.
158
Q

Explain why plants cannot survive by photosynthesis alone and need to undertake cellular respiration. (3 marks)

A
  • Photosynthesis is undertaken to convert CO2 + H2O into glucose, in the process
    converting light energy into chemical energy (1) that the plant can store glucose as
    starch or make cellulose to build plant tissues (1). However, ATP is also needed for
    cells to undertake processes such as active transport, protein synthesis and cell division
    (1). Therefore, the plant is able to take the glucose made from photosynthesis and use
    it in cellular respiration to make ATP available for cellular processes to sustain life (1).
159
Q

2022 tasc pp q9b

A

The graph measures the rate of photosynthesis (PS) by CO2 consumption (an input) and
the rate of cellular respiration (CR) by CO2 production (an output) (1). Any of the
following or other logical response to cover the points outlined:
* CR remains fairly constant over the 24-hour period, only slightly elevated in daylight
hours (1)
* the elevated period between 9am and 6:30pm would support needing more ATP
for photosynthetic reactions (1)
* Increasing temp increases rate of CR (1)
* PS only occurs in daylight hours from 7am to 7pm, peaking at 3pm (1).
* Light intensity changing during the day (I).
* PS peak is more than double that of CR, showing that although it occurs across a
smaller time period (12 hours) compared to the 24hrs of CR, PS still exceeds the
rate of CR. This is needed for plants to be able to grow (1).
* Compensation points shown at approx. 9am and 7pm is where the amount of CO2
used in PS equal the amount produced in CR, therefore this is where the rates of
both CR and PS are the same (1).

160
Q

Complete table to explain role of following items in the cellular processes of protein synthesis: (3 marks)
Description Role and Processes
- DNA
- tRNA
- mRNA

A

DNA
* Double helix made up of thousands of nucleotides
* Double stranded containing a long chain of nucleotides
(bases) made up of TAGC
* Contains genetic material that acts as a ‘blueprint’.
——-
* Contains genes which code for proteins
* Unravels to expose gene sequences
* Used as a template to make a molecule/strand of mRNA in the nucleus as part of transcription

tRNA
* Folded single strand of nucleotides.
* Single strand of nucleotide with an ‘L’ shape with ribosomal binding arm and anticodon loop
* Has an anticodon that is a complimentary match for the
mRNA codon
* Has an anticodon with corresponding amino acid (aa)
attached
—-
* Carries a corresponding amino acid (aa) to the ribosome as part of translation in protein synthesis
* It binds to one of three sites on the large subunit of a ribosome
* Carries an amino acid (aa) to form the correct sequence of
polypeptide chain mRNA

* A single strand of nucleotides
* Unstable and short single strand of nucleotide corresponding to a gene encoded within DNA
* Created in the nucleus to make/take a copy of genetic
code for a gene then move to the cytoplasm/ribosome to be
decoded

  • Transcribes genetic information from DNA in the nucleus to the cytoplasm/ribosome where proteins are synthesised – read and decoded in translation
  • It transcribes sections of the DNA code in nucleus - via transcription and moves to a ribosome where the
    codon matches an anticodon of tRNA with a corresponding amino acid to form a polypeptide chain – via
    translation
161
Q

tasc 2022 pp q10f

A

Mutation 1
- A nitrogenous base has been deleted, thus creating a ‘frameshift’ (codons are read in groups of 3’s) and therefore causing every codon sequence past the point of mutation
is shifted. The change in sequence changes the amino acids within the mutated strand
resulting in a change in the polypeptide/protein composition (change in shape of protein
thus altering the intended function).

Mutation 2
The change of the 3rd codon from GGG to GGU (point mutation – substitution) still
results in the same amino acid in the same place within the sequence. This is a ‘silent’
or neutral mutation and has no overall effect on the polypeptide chain or protein
formed.

Mutation 3
The change of the 9th codon from UAA -a stop amino acid to UAC -tyr / Tyrosine
(point mutation – substitution) means that the polypeptide chain / protein would not
be completed and more amino acids will possibly be added. Resulting in a longer chain
and a different protein.

162
Q

Write the originial DNA strand that the following mRNA would have been coded from.
mRNA:
C U C G U A A U C G G C

A

G A G C A T T A G C C G

163
Q

If there where the following mutations what might be the impact on the polypeptide chain formed? (2 marks)

ii) G substituted with A
original mRNA
C U C G U A A U C G G C
mRNA mutated to:
C U C G U A A U C G A C

A

This will change the AA coded for GGC codes for GLY (1/2) and GAC codes for ASP(1/2). This change in AA may alter the way the polypeptide chain may function(1/2), it may be beneficial, negative or neutral. (1/2)

164
Q

If there were the following mutations what might be the impact on the polypeptide chain formed? (2 marks)

ii) Extra U inserted
original mRNA
C U C G U A A U C G G C
mRNA mutated to:
C U C U G U A A U C G G C

A

This would result in a frame shift(1/2) pushing all the other codons out of line altering every AA after this insertion(1/2). This will have a large (1/2) impact on the protein and most likely it will not work anymore.(1/2)

165
Q

Question 2 (6 marks)
When making yogurt a specific set of steps are followed (see below). Using your understanding of enzymes, discuss why each of the temperatures (100C, 35C and 4C) are important in the making of yogurt.

To make yogurt follow these steps
1. Heat milk to just below boiling (100C) for 5 minutes
2. Let milk cool to a bit above room temp (around 35)
3. Add culture (active yogurt culture)
4. Leave at 35 for at least 4 hours or overnight
5. When yogurt has thickened to the consistency that you prefer, you place the yogurt in the fridge (below 4C).

a) Discuss why a temperature of 100C is used in the first step. (2 marks)

A

Causes all enzymes in milk and any bacteria living in the milk to denature (1/2) This changes their 3D structure (1/2) and therefore not be able to work,(1/2) effectively killing any organisms in the milk. (1/2)

166
Q

e) Discuss how the ribosome reads the mRNA and how tRNA fits into this process. (2 marks)

A

The mRNA enters the ribosome (1/2) that reads the first three bases called a codon or triplet(1/2). It matches the codon on the mRNA with the complementary Anti codon (1/2)on tRNA which carries the AA(1/2). As the next codon is read that AA is connected to previous AA by a peptide bond.(1/2)

167
Q

Question 2
When making yogurt a specific set of steps are followed (see below). Using your understanding of enzymes, discuss why each of the temperatures (100C, 35C and 4C) are important in the making of yogurt.

To make yogurt follow these steps
1. Heat milk to just below boiling (100C) for 5 minutes
2. Let milk cool to a bit above room temp (around 35)
3. Add culture (active yogurt culture)
4. Leave at 35 for at least 4 hours or overnight
5. When yogurt has thickened to the consistency that you prefer, you place the yogurt in the fridge (below 4C).

a) Discuss why a temperature of 35C is used in the second step (2 marks)

A

35 degrees is the optimum temp (1) for enzyme activity so allows organisms in this case bacteria (yogurt culture) to grow / carry out chemical reactions(1/2) and ferment the milk(1/2). (side note: bacteria ferments the Lactose in milk to make lactic acid which acts on yogurt protein to give it a thicker texture, don’t need this to get marks)

168
Q

Three plants were each placed in an individual glass jar so no air could get in or out. One plant was placed in complete darkness, one was placed in a standard photoperiod (12 hrs light, 12 hrs dark) and the third was placed in 24 hrs of artificial light. The amount of oxygen and carbon dioxide was measured for a period of 2 weeks.

a) For each of the above plants discuss what you would expect to see with levels of Oxygen and Carbon dioxide levels throughout the experiment.
(2 marks each)

A

1) Plant 1 – complete darkness (2 marks)
CO2 levels will go up and O2 levels will go down (1) as the plant continues to respire through cellular respiration.(1)

2) Plant 2 – 12 hrs light , 12 hrs, darkness ( 2 marks)
During the 12 of light CO2 levels will go down and O2 go up due to photosynthesis (1/2) occurring at a higher rate than respiration (1/2) and during the 12 hrs of dark you will see a drop in O2 and a rise in CO2 as it is still respiring.(1)

3) Plant 3 – 24 hrs light (2marks)
During the 24 of light CO2 levels will go down continually and O2 go up continually (1) due to photosynthesis occurring at a higher rate than respiration (1) all the time.

169
Q

Three plants were each placed in an individual glass jar so no air could get in or out. One plant was placed in complete darkness, one was placed in a standard photoperiod (12 hrs light, 12 hrs dark) and the third was placed in 24 hrs of artificial light. The amount of oxygen and carbon dioxide was measured for a period of 2 weeks.

b) Explain what would you expect to happen if a mouse was placed in each of the jars with enough food and water to live off for several weeks. (2 marks each)

A

1) Plant 1 – (complete darkness) (2 marks)

The oxygen levels would drop continually and CO2 rise continually as both the plant and mouse are respiring. (1) Eventually the mouse could die due to lack of oxygen and too high a CO2 levels (1)

2) Plant 2 – (12 hrs light , 12 hrs, darkness) ( 2 marks)

You would still see in the 12 hrs of dark a drop in O2 and a rise in CO2 (1/2) as both the plant and mouse are respiring (1/2). During the 12 hrs of day you would likely see an increase in oxygen depending if the plant is big enough to produce more oxygen than the plant and mouse are using (1)

3) Plant 3 – (24 hrs light) ( 2 marks)

Most likely you would see oxygen levels go up and carbon dioxide levels go down (1/2) as the plant can photosynthesis continually(1/2). But it would be at a much slower rate than a) 3) (1/2) as it also has the mouse respiring (drops oxygen, rises carbon dioxide) to offset as well. (1/2)

170
Q

Biology midyear criterion 5 2016 q4
The diagram below shows the stages in an enzyme feedback loop where the final product from Enzyme 3 is a non –competitive inhibitor to enzyme 1.

a) Explain how this process helps the body to regulate production of “end product”. (2 marks)

A

The triangle is acted on by enzyme 1.
- Enzyme 1 will continue to act on this substrate breaking it down continually until it is run out or at very low levels unless it is stopped by an inhibitor.
- By having the end product of a multiple step reaction inhibit the first stage you will get substrate being broken down for a time while it goes through all other steps until it reaches the end.
- The final product then stops the first process insuring not all of the substrate is broken down.
- This give the cell control over how much of a substance is broken down.

171
Q

Question 1.

Three friends decided to compete against each other to find out who was the best home brewer. They each decided to follow the same recipe and see which could make the best beer.

The three tried to follow the following steps to make beer. (simplified steps)
1 Clean all equipment
2 Add malt barley (sugar) to water
3 Boil mixture
4 Allow to cool to room temperature and add yeast
5 Keep container warm for 7-10 days

The first brewer followed steps 1-4 but with step 5 stored the brew outside in the cold for 4. Only after realising his mistake did he bring it inside and keep it warm for the remainder of the time.

The second brewer followed the steps 1-4 but when it came to step 5 kept it next to the furnace and after checking at day 4 found the temperature was 70 degrees. He then cooled it down and kept it warm for the remained.

The third brewer followed all steps correctly.

Using your understanding of enzymes answer the following questions.

b) What will be the outcome of the first brewers mistake? (3 marks)

A

By cooling the mixture too far the enzymes in the yeast are not working at optimum rate so nothing would happen for the first 4 days. Has he heats it up kinetic energy increases and more chance of substrate meeting active site so reaction will increase. Beer will still be made but slower than 3.

172
Q

ii) How would mutation 2(substitution) change the protein? ( 2 Marks)

A

GCU coded for Ala while UCU (mRNA) codes for Ser this could change the protein for better, worse or not at all.

173
Q

iii) How would mutation 3 (insertion) change the protein? ( 2 Marks)

A

This creates a frame shift changing all the AA’s after the insertion this will change drastically changing the protein. Most likely making the protein not work any more.

174
Q

An experiment was set up where leaves from a plant where all given different treatments. One set of 10 leaves where all coated in Vaseline, preventing gas exchange. Another set of 10 leaves had half of the leaf covered with aluminium foil, preventing light penetration. The third set of 10 leaves where left untreated. Each set of leaves had access to water and exposed to the same amount of sunlight. The amount of oxygen produced was measured over a period of time.

If the results showed that the leaves treated with aluminium foil produced 1000 ppm of Oxygen (arbitrary number).

a) Predict how much Oxygen (in ppm) would be produced for treatment one and explain why?
(2 marks)

A

0 ppm as no gas exchange is allowed there fore no carbon dioxide can enter the plant
through the stomata and converted into sugar.

175
Q

Explain whether if blocking of enzymes by competitive inhibitors can be reversed and how?

A
  • reversible inhibitation
  • Adding more substrate will dislodge a competitive inhibitor allowing the enzyme to keep working.
176
Q

c5 midyear exam final USE q4
Below is graph of different enzymes reaction rates at varying pH’s. These three enzymes are all involved in the digestive track of mammals. Pepsin is enzyme involved in breaking down protein in the stomach, Gastric lipase breaking down fats as food leaves the stomach and Trypsin breaking down protein in the small intestine.
a) Why would you expect to find Pepsin in the Stomach and not in the large intestine? ( 1 mark)

A

Optimum pH means it will work most effectively in stomach with a Ph of around 2.

177
Q

c5 midyear exam final USE q4
c) As food moves through the digestive track the acidic environment of the stomach is neutralised by bile from the Liver. How would adding bile aid in these enzymes working as you move through the digestive track? (3 marks)

A

Pepsin found in the stomach would work most effectily in stomach environment of pH of around 2, as it leaves the stomach bile neutralises the acid bring the pH up. As it goes from 2 – 7 lipase will work most effectively around 7 as the bile makes the small intestine slightly alkaline trypsin reaches it optimum pH of about 8.

178
Q

Explain why triglycerides are not polymers and name the components that triglycerides are made up of.

A
  • Polymers are made up of many repeated units -monomer e.g. proteins are made up of many amino acids.
  • Triglycerides are not made up of repeated units (monomers). They are composed of glycerol and 3 fatty acids (which may or may not be the same composition).
179
Q

(b) When food passes through the gut, only a certain amount of the protein it contains is digested. The rest passes out of the body in the faeces. The digestibilty coefficient is one way of measuring this.

Digestibility coefficient =N intake – N in Faeces/
Nitrogen intake

(i) Explain why nitrogen is a useful measure of protein intake.

A

Protein molecules are composed of the elements carbon C, hydrogen H, oxygen O and NITROGEN. Cabohydrates and lipids are composed of C, H and O and no nitrogen, making nitrogen content of the food eaten a reliable measure protein intake.

180
Q

(ii) A lot of the protein in a vegetarian diet comes from cereal and vegetables. Use your knowledge of plant cells to explain why the digestibility coefficient for vegetarian meals is lower than that of a meal containing meat. (2m)

A

Plant cells contain protein but it is locked behind a cell wall of cellulose. Humans lack the enzymes to digest the cellulose. Mechanical destruction (e.g. chewing) of the cell wall is the only way to release the cell contents and make the protein available for digestion followed by absorption of the amino acids. There will be a higher nitrogen content in the faeces of a vegetarian because it will contain a higher fibre (cellulose) content making the digetibility coefficient lower. Meat is composed of animal cells that are easily digested as they lack a cell wall, hence more protein is available for digestion.

181
Q

(a) Photosynthesis occurred in green alga living in freshwater. The algae were exposed to light of different wavelengths (colours).
(ii) If you were conducting the experiment above (6 b), suggest two ways, that you could use, to measure the RATE at which photosynthesis was occurring. (2marks)

A
  • Rate at which oxygen is produced
  • Rate at which cabon dioxide is used
  • Rate at which glucose (or starch) is produced
182
Q

(iii) Why would you need to keep the temperature the same for all trials when looking photosynthesis? (2 marks)

A

Photosynthesis is a series on several reactions all controlled by enzymes. Enzymes all have an optimal temp so if the temperature varied from group to group that would effect how quickly they could photosynethesise.

183
Q

C5 midyear exam final USE Q7
c) The student concluded from the graph if the light intensity is less than 100 just respiration is occurring. Use the graph to justify why you agree or disagree with the student. (3 marks)

A

Disagree with the student (1 mark) there must be some photosynthesis occurring between 0 -100 as the oxygen level is changing(1/2). Respiration will stay fairly much constant (1/2 mark) so you would assume if no photosynthesis occurring than Oxygen production rate would stay at -100 (1/2) but as it is rising it must be starting to photosynthesize just not enough to offset the amount of oxygen used in cellular respiration (1/2)

184
Q

C5 midyear exam final USE Q7
d) Why after 500 light units does oxygen production not increase? (2 marks)

A

Light is no longer a limiting factor so increasing it will not increase photosynthesis. The rate is being limited by other factors such as amount of Carbon dioxide or number of palisade cell or amount of water.

185
Q

Tasc 2022 q10f photos
Explain the significance of each mutation and its likely effect on the product.
You can include the amino acid sequence in your explanation. (4 marks)

A

Mutation 1
AUG CU GGG AAU GCA CAG AAA GAC UAA
AUG CUG GGA AUG CAC AGA AAG ACU AA
aa* Met Leu Gly Met Leu Arg Lys Thr X

A nitrogenous base has been deleted, thus creating a ‘frameshift’ (codons are read in groups of 3’s) and therefore causing every codon sequence past the point of mutation
is shifted. The change in sequence changes the amino acids within the mutated strand resulting in a change in the polypeptide/protein composition (change in shape of protein thus altering the intended function).

Mutation 2
AUG CUC GGU AAU GCA CAG AAA GAC UAA
aa* Met Leu Gly Asp Ala Glu Lys Asp A stop

The change of the 3rd codon from GGG to GGU (point mutation – substitution) still results in the same amino acid in the same place within the sequence. This is a ‘silent’
or neutral mutation and has no overall effect on the polypeptide chain or protein formed.

Mutation 3
AUG CUC GGG AAU GCA CAG AAA ,,GAC UAC
aa* Met Leu Gly Asp Ala Glu Lys Asp A Tyr

The change of the 9th codon from UAA -a stop amino acid to UAC -tyr / Tyrosine (point mutation – substitution) means that the polypeptide chain / protein would not be completed and more amino acids will possibly be added. Resulting in a longer chain and a different protein.

186
Q

Although DNA is double-stranded, only the coding strand determines the specific amino acid sequence of a polypeptide. Suggest a role of the non-coding strand. (1 mark)

A

The function of the coding strand is for DNA replication. DNA unwinds and the non-coding strand is matched
with complimentary bases to form a new strand.

187
Q

explain the role of messenger RNA. (2 marks

A

Messenger RNA conveys the genetic information from the DNA by producing a complimentary copy of the original strand of DNA and transporting it out of the nucleus to the ribosomes where it forms a template for amino acid sequence in the production of proteins.

188
Q

tasc 2017 5e
Outline two things that can change this amino acid sequence. Explain the cause and effect if these changes. (3 marks)

A

An amino acid is determined by its codon(1) therefore a change in codon produced by point rotation/chemical/environmental damage DNA during replication OR an error in transcription in the mRNA nitrogen bases OR an error in translation when codon and anticodon are not a correct match resulting in a changed amino acid/protein. (need to have any of those 1 mark each)

189
Q

You are asked to explain to a group of grade 10 students. How a commercial greenhouse grower can potentially optimise rate of photosynthesis in their greenhouse plants. Identify 4 different factors that influence the rate of photosynthesis for two of these factors, explain how they could be optimise to promote the most efficient rate of photosynthesis and growth in the greenhouse (four marks).

A

factors: light intensity, water, availability, carbon dioxide temperature, colour, lights duration

light intensity: light is needed for photosynthesis, particularly at the red and blue ends of the spectrum, which green plants use balance between maximising photosynthesis, and not overheating the plants and drying them out. Place in the greenhouse so that it maximise the light availability and not in the shade, good optimised growth of the plants, using artificial lights at night

water availability is needed for photosynthesis. Humidity levels need to be controlled to help restrict access. Water loss will not promoting mode etc for either optimising by using sprinklers or some other irrigation method to ensure correct amount.

Carbon dioxide is needed for photosynthesis. It is required for the production of glucose that is used to produce ATP for metabolic processes, increasing carbon dioxide levels can therefore, increase growth rates
. could be pumped into a close greenhouse.
Temperature: all metabolic processes are affected by temperature, including photosynthesis. So needs to be optimised, too hot and the plants enzymes may nature, or the rate of transpiration may increase resulting in closing of stomata, which will reduce the rate of photosynthesis.

190
Q

State one way in which p/s and cellular respiration cannot be considered the reverse of each other.

A

difference: energy is captured by photosynthesis in form of light, but released by respiration as ATP and heat. photosynthesis, has two stages, light and dark and aerobic respiration. respiration occurs in the mitochondria, photosynthesis occurs in the chloroplasts.

191
Q

identify two properties of ATP that makes it ideal for its role. (to marks). 

A

Energy can be obtained easily through breaking the phosphate bond (atp-> adp + p)
is a recyclable molecule
small convenient amounts of energy can be delivered to where the energy is required in the cell. (avoids waste
atp is a quick usable form of energy compared to aerobic respiration of glucose (only one bond needs to be broken to release useable energy)

192
Q

tasc 2017 q7a ii

A

they both have a different level of pH where the activity of the enzyme is optimal as a river buffalo enzyme , being efficient with pH 7-8, while the bacterial enzyme is efficient at a lower pH range 5 to 6.6. This is because the organisms live in different environments, the bacteria lives in hot Springs that are likely to be acidic and the water Buffalo has a stable internal environment of approximately pH 7.

193
Q

tasc 2017 q7b explain graph of increasing substrate concentration v.s reaction rate

A

The initial response is very rapid as there is plenty of substrate to collide effectively/bind with the active side of enzyme. But the rate of conversion plateaus as the amount of enzyme become a limiting factor when all the active sites are full and the enzyme is operating at maximum rate.

194
Q

briefly outline how the induced fit model explains how enzymes can interact to both bind and release specific substrate/products to produce rapid catalysis. (three marks)

A

in the induced fit model, the exposure of the enzyme to the substrate causes the active site of the enzyme to change shape. In order to allow the enzyme and substrate to bind and form an enzyme-substrate complex. This lowers the activation energy required for conversion of substrate products reducing the time it would otherwise take, OR just allows the enzyme to be free to catalyse other reaction in an incredibly short time.

195
Q

Locusts (see picture) are insects that cause extensive crop damage when they swarm in
large numbers.
Professor Stephen Simpson and his team from the University of Sydney have discovered
that the locusts keep eating the crops until they have consumed enough protein. This has
given scientists a better understanding of human nutrition and how it relates to obesity.
They suggest humans might be doing the same, fuelling the obesity epidemic.
(a) Discuss why proteins in the diet are so important. In your answer, include an
explanation of how such an appetite response would meet our nutritional needs more
efficiently. (4 marks)

A

All organisms require essential amino acids that cannot be made by the body/must be provided by a digestion/diet. These are needed to produce proteins, which has a wide range of essential functions within the body in the cells, for example, in cell membranes, enzymes, muscles, etc. Without getting enough essential amino acids and the body would suffer from this deficiency.

196
Q

Locusts (see picture) are insects that cause extensive crop damage when they swarm in
large numbers.
Professor Stephen Simpson and his team from the University of Sydney have discovered
that the locusts keep eating the crops until they have consumed enough protein. This has
given scientists a better understanding of human nutrition and how it relates to obesity.
They suggest humans might be doing the same, fuelling the obesity epidemic.
b) assuming this phenomena applies to humans, discuss the implications about what makes up a healthy diet and how we can avoid obesity. (two marks)

A

People need to ensure they eat enough protein in the regular diet so they don’t need to keep eating extra fruit with extra kilojoules to meet the essential amino acid requirements. This will reduce the amount of carbohydrates and fat they feel they need to eat and reduce the likelihood of obesity with its consequent health issues.

197
Q

C5 retest q2a
Use the graph above to explain why leaves are usually green. (2 marks)

A

They appear green because they absorb all other light and reflect the green light to our eyes therefore appearing green. This means that other colours are absorbed these all give a higher rate of photosynthesis.

198
Q

C5 retest q2b
Variegated leaves contain yellow areas. Explain what effect will this have on the growth of the plant. (2 marks)

A

Again the yellow is reflected and not absorbed. As yellow like green gives a low photosynthesis rate it would not have too much of a detrimental effect on plant growth. As the other light spectrums all give a higher rate of photosynthesis.

199
Q

C5 retest q3
Explain what this graph illustrates about the breakdown of glucose with, and without, the appropriate enzyme being present. (2 marks)

A

Shows that the presence of enzyme reduce the activation energy required to go from glucose to product molecule. Without the enzyme more energy is required for the same reaction to take place so with the enzyme the reaction will occur quicker.