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A Level Biology Flashcards > Genes & Health - Exam Questions > Flashcards

Genes & Health - Exam Questions Flashcards

1
Q

DNA & GENETICS

Structure of tRNA
[1 mark ]

Bases, Number of Strands, Type of sugar -

A

Bases - A C G U
Number of strands - 1
Type of sugar - Ribose

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2
Q

DNA & GENETICS

Mononucleotides all contain a base and a sugar as part of their structure
Name another component of all mononucleotides.
[1 mark ]

A

Phoshpate (group)

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3
Q

DNA & GENETICS

The mononucleotides of mRNA are joined together by RNA polymerase.
Which part of the eukaryotic cell is location for this process
[1 mark]

A

Nucleus

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4
Q

DNA & GENETICS

Describe the role of tRNA in the production of the protein part of a glycoprotein.
[2 Marks ]

3 Points Can be made

A
  • Each tRNA brings a specific amino acid to the ribosome
  • The tRNA with the complementary anticodon binds to the mRNA codon
  • tRNA bonds to the ribosome
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5
Q

DNA & GENETICS

Analyse the data from these three investigations to discuss the advantages of an exercise programme without nandrolone.
[ 6 marks ]

Exam Specific Data Analysis Question -

Use for structure of 6 markers

A

Investigation 1 -
* Analysis of data of exercise/ exercise without nanodrolone.
* Outcome of difference as an advantage - more oxygen rich blood can flow at correct pressure from heart to muscles
* Less likely to get atherosclerosis/ CVD

Investigation 2 -
* Comparison/ analysis of data for P and R (protein complexes) - more present due to exercise.
* Outcome of difference as an advantage - More ATP synthesis (oxidative phoshphorlyation)
* More ATP for Muscle contraction.

Investigation 3 -
* Comparision of P and Q for mRNA coding (Tfam) - more Tfam per cell means more mitochondria produced
* More respiration/ ATP formed.

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6
Q

DNA & GENETICS

Analyse the data to comment on the effectiveness of these antibiotics for the treament of TB.
[ 6 Marks ]

Exam Specific Data Analysis Question -

Use for struture of 6 markers

A
  1. All treatment combinations were somewhat effective at treating TB
  2. Group 1 with the combination of Rifampicin & Pyrazinamide had the lowest percentage of people with TB after 3 years.
  3. All combinations involving Rifampicin seem to be the most effective
  4. However, not all combinations were tested/ others combinations may be also effective if not more.
  5. There could be other reasons for infections of TB three years later that has nothing to do with the antibiotic treatments.
  6. The data doesn’t credit information about dormant TB, therefore only showing active cases could be inaccurate to the actual percentage.
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7
Q

DNA & GENETICS

The synthesis of mRNA occurs in a process called transcription.
Compare and Contrast the process of Transcription with the process of DNA replication.
[ 4 Marks ]

A

Similarities -

  • Involves formation of phosphodiester bonds
  • Involved DNA helicase (unzipping of DNA)

Differences-

  • Transcription uses RNA nucleotides whereas replication uses DNA nucleotides.
  • Transcription uses RNA polymerase whereas replication uses DNA polymerase
  • Transcription produces a single strand of mRNA whereas DNA replication produces a double stranded DNA
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8
Q

DNA & GENETICS

State what is meant by the term gene
[1 mark ]

A

a sequence of bases in DNA coding for a sequence of amino acids

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9
Q

DNA & GENETICS

Explain how a change of one amino acid could lead to a change in the structure and properties of the haemoglobin protein.
[4 Marks ]

A
  • Different sequence of amino acids
  • means the different amino acids will have different R groups
  • Therefore there will be a change in the tertiary structure
  • due to a change in hydrogen bonds
  • Haemoglobin may not bind to oxygen in order to carry out it’s function.
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10
Q

DNA & GENETICS

Describe the role of tRNA in the production of leptin
[ 3 Marks ]

Leptin - Protein Hormone

A
  • each tRNA molecule carries a specific/ particular amino acid
  • tRNA molecules transport amino acids to the ribosome
  • tRNA molecule has an anticodon that binds to a codon on the mRNA.
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11
Q

DNA & GENETICS

Describe how the primary structure of leptin enables it to be soluble in water
[ 3 Marks ]

A
  • Primary structure determines the folding of the polypeptide
  • forming a globular protein
  • hydrophillic groups are located on the outside of the protein
  • Water forms hydrogen bonds with hydrophillic groups
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12
Q

DNA & GENETICS

Describe how a frameshift mutation could result in the production of leptin with a variety of shorter primary structures.
[2 Marks]

Frameshift - involves the insertion or removal of one or two nucleotides from a gene.

A

adding or removing one or two nucleotides changes the triplet code
* This introduces a new start/ stop codon
* Which would then code for a shorter sequence of amino acids

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13
Q

DNA & GENETICS

Describe how mRNA is synthesised at a template strand of DNA
[2 Marks]

A

RNA nucleotides align with complimentary bases on DNA

RNA nucleotides are joined together by phosphodiester bonds.

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14
Q

DNA & GENETICS

Describe the differences between the structure of DNA and the Structure of RNA
[ 3 Marks ]

A
  • DNA is double stranded whereas RNA is single stranded
  • DNA contains deoxyribose whereas RNA contains ribose
  • DNA contains the base Thymine whereas RNA contains the base Uracil
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15
Q

DNA & GENETICS

Give three differences between replication of DNA and transcription of DNA
[3 Marks ]

A
  1. Replication involves DNA nucleotides, Transcription involves RNA nucelotides.
  2. Replication produces double stranded DNA, Transcription produces single stranded mRNA.
  3. Replication uses DNA polymerase, Transcription requires RNA polymerase.

Replication produces identical copies whereas transcription produces a complimentary copy

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16
Q

DNA & GENETICS

Assess the affect that these mutations have on the human respiratory system.
[6 Marks ]

Exam Specific Data Analysis Question

Use to structure 6 mark questions

A
  1. During Cystic Fibrosis, Cillia struggle to move mucus out of the lungs which increases risk of lung infection.
  2. DATA ANALYSIS - Mutation F508del results in lower chloride transport than other mutations.
  3. G551D & G551S have a higher chloride transport than F508del because the CFTR protein is still present just with reduced function.
  4. Thicker mucus as a result of no CFTR protein causes an increased diffusion distance for gases in the respiratory system.
  5. Reduced function of CTFR protein could be due to incorrect shape of CTFR protein channels/ G551D & G551S have incorrect shape of protein channels.
  6. Therefore the incorrect shape of the CFTR protein channels results in reduced function of transporting chloride ions.
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17
Q

DNA & GENETICS

Explain why removing one base or changing one base for another from a DNA sequence have different effects on protein structure - two types of mutation.
[4 Marks]

A
  • Deletion could affect every codon on the mRNA - Substitution will only affect one codon
  • Deletion is more likely to affect the position of the stop/ start codon
  • Deletion results in a different sequence of amino acids - substitution may not affect the sequence of amino acids
  • Substitution may code for the same amino acid

Same amino acid is due to the degenerate nature of the genetic code.

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18
Q

DNA & GENETICS

Describe how nucleotides join together to form DNA
[2 Marks]

A

Catalysed by enzyme DNA polymerase

  • Joined in a condensation reaction
  • Form phosphodiester bonds
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19
Q

DNA & GENETICS

Analyse the data to explain why Mesleson and Stahl accepted one of the models for DNA replication and rejected the other.
[4 Marks ]

Specific Data Analysis Questions

A
  • The conservative model was rejected whilst the semi- conservative model was accepted.
  • Generation 1 has a single band which is halfway between 15N and 14N
  • This is beacuse the DNA has one strand containing 15N and one strand containing 14N
  • In the semi- conservative model, further generations would have a band which is halfway between 15N and 14N
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20
Q

DNA & GENETICS

Explain why this mutation [SUBSTITUTION] affects the function of the haemoglobin molecule.
[3 Marks]

A
  • One triplet is affected
  • The mutation could change one of the amino acids
  • This would change the bonds formed between the R groups / change the tertiary structure
    the haemoglobin would no longer be able to bind to oxygen
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21
Q

DNA & GENETICS

Scientists have recently discovered a new class of antibiotics that bind to ribosomes.

Explain why these antibiotics could affect the production of proteins in bacteria.
[3 Marks]

A
  • Ribosome shape is changed
  • Therefore mRNA is unable to bind to the ribosome
  • Translation cannot occur
  • Proteins/ Polypeptide are not synthesise.
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22
Q

DNA & GENETICS

These new antibiotics attatch to a site on the ribosome not affected by any known antibiotics.

Deduce why these new antibiotics might be used to treat bacteria that are resistant to other antibiotics.
[2 Marks]

A
  • The bacteria has not been exposed to new antibiotics before.
  • Bacteria have developed resistance to other antibiotics by evolving.
  • Therefore there has been no mutation present to give resistance.
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23
Q

DNA & GENETICS

Devise a laboratory procedure to compare the effectiveness of penicillin with one of the new antibiotics.

[6 Marks]

Core Practical

A
  1. Prepare agar plates with bacterial cultures - use bacteria that are resistant to other antibiotics
  2. Prepare solutions of new antibiotic and penicillin
  3. Place into wells in the agar
  4. Control Variables - Time and Temperature of the incubation
  5. There should be a range of dilutions on each plate - one antibiotic per plate , should be repeated with different strains of bacteria.
  6. A statistical T test should be used to determine which is the mostt effective.
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24
Q

DNA & GENETICS

Describe the role of the CFTR Protein in ensuring that the mucus produced in the lungs has the right consistency.
[3 Marks]

A
  • Chloride ions leave the cell through the CFTR channel protein
  • Sodium ions leave the cells
  • Increasing solute concentration in the mucus
  • Water moves out of the cell by osmosis and into the mucus.
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25
Q

DNA & GENETICS

How is the nature of the genetic code demonstrated
[3 Marks]

Specific Diagram Question

A
  • Triplet code is shown by three bases coding for an amino acid
  • ATT codes for amino acid I and then AAA codes for amino acid K - Non Overlapping
  • Degenerate code as both ATT and ATC code for amino acid I.
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26
Q

DNA & GENETICS

Describe the role of transcription and translation in the synthesis of a globular protein by a muscle cell.
[ 5 Marks ]

A
  • The sequence of DNA for the globular protein is transcribed
  • Complemetary base pairing between RNA Nucleotides and DNA to produce mRNA
  • mRNA leaves the nucleus and attatches to a ribosome
  • pairing between codons on mRNA and anticodons on tRNA - tRNA provides specific amino acids
  • The sequence of bases determines the sequence of amino acids.
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27
Q

DNA & GENETICS

Compare and Contrast the structure of ATP and a DNA nucleotide
[3 Marks]

A

Similarity-
Both contain phosphate, pentose sugar and a base

Difference-
* DNA nucleotide contains Deoxyribose whereas ATP contains ribose
* DNA nucleotide could contain other bases whereas ATP contains only one type of base
* DNA nucleotide contains one phosphate molecule whereas ATP contains 3.

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28
Q

EXCHANGE OF SUBSTANCES

Explain why betalain molecules cannot move through intact cell membranes
[3 Marks]

A
  • Betalain molecules are too large to move through the cell membrane.
  • There are no carrier proteins for betalain molecules to move through.
  • Betalain molecules are polar and are repelled by hydrophobic fatty acid tails.
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29
Q

EXCHANGE OF SUBSTANCES

Explain how the blood-gas barrier of the chicken is adapted to give more efficient gas exchange than the blood-gas barrier of the dog.
[ 3 Marks]

Specific diagragm explaination question

A

Thinner blood-gas barrier

This means that because of thinner capillary walls

There is a reduced diffusion distance

Meaning a faster rate of diffusion/ gas exchange.

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30
Q

EXCHANGE OF SUBSTANCES

Diffusion and active transport are mechanisms by which molecules can enter cells
Compare and Contrast these two mechanisms
[3 Marks]

A

Similarities-
* both move molecules through the cell surface membrane
* molecules can move through proteins in both

Differences-
* diffusion occurs down a concentration gradient whereas active transport occurs agaisnt a concentration gradient

  • diffusion is passive whereas active transport requires energy from ATP
31
Q

EXCHANGE OF SUBSTANCES

Compare and Contrast the processes of endocytosis and exocytosis
[3 Marks]

A

Similarities-
* Both involve vesicles
* Both involve energy from ATP

Differences-
* Exocytosis involves molecules leaving the cell whereas endocytosis involves molecules entering the cell
* Exoxytosis involves vesicles fusing with the cell surface membrane
* Endocytosis involves the formation of vesicles from the cell surface membrane.

32
Q

EXCHANGE OF SUBSTANCES

Describe how glucose molecules move into the cell
[2 Marks]

A
  • by facillitated diffusion
  • through carrier proteins in the phospholipid bilayer
33
Q

EXCHANGE OF SUBSTANCES

Describe how glucose moves into the cell via facillitated diffusion
[2 Marks]

A
  • carrier protein
  • glucose moves from area of high concentration to low concentration
    *glucose binds to carrier protein *
34
Q

EXCHANGE OF SUBSTANCES

Explain how the structure of glycogen allows it to be an energy store
[3 marks]

A
  • polymer of glucose
  • provides glucose for respiration
  • Compact to allow a large amount of glucose to be stored in a small space.

branched for rapid hydrolysis

35
Q

EXCHANGE OF SUBSTANCES

One function of glycoproteins found in the cell surface membrane
[1 Mark]

A

Cell recognition

36
Q

EXCHANGE OF SUBSTANCES

Explain how the structure of the phospholipid molecule contributes to the partial permeability of the cell surface membrane.
[3 Marks]

A

phospholipid contains hydrophillic phoshpate head a hydrophobic fatty acid chains

allows non polar molecules to pass through the membrane

polar molecules cannot pass through

37
Q

EXCHANGE OF SUBSTANCES

Assess the effect of this heart defect on the rate of oxygen diffusion between the alveoli and the blood.
[6 Marks]

data analysis specific question

Use for structure of 6 Markers

A
  1. Rate of diffusion would be lower with an abnormal heart
  2. Blood entering the lungs from an abnormal heart has more oxygen than blood entering the lungs from a normal heart (USE DATA)
  3. Show increase of oxygen blood entering the abnormal heart
  4. Ficks law, states concentration gradient is proportional to rate of gas exchange
  5. Smaller difference in concentration between the alveoli and RBCS
  6. A lower concentration gradient for oxygen between the alveoli and the blood results in a lower rate of oxygen diffusion.
38
Q

EXCHANGE OF SUBSTANCES

Explain how phospholipids form a cell surface membrane
[3 marks]

A
  • hydrophillic parts associate with water
  • hydrophobic parts repel water
  • a bilayer forms with hydrophobic parts pointing in towards the centre of the bilayer
39
Q

EXCHANGE OF SUBSTANCES

Explain why individuals with severe emphysema will tire more easily than healthy individuals.
[3 marks]

A
  • smaller surface area
  • reduction in oxygen uptake
  • less oxygen for aerobic respiration
  • leads to more anaerobic respiration - MORE FATIGUE.
40
Q

EXCHANGE OF SUBSTANCES

Explain why the phospholipids are arranged in two layers of the cell surface membrane.
[3 Marks]

A
  • hydrophilic region oriented towards water
  • hydrophobic regions away from the water
  • Need of two layers as a water based solution either side of the cell membrane.
41
Q

EXCHANGE OF SUBSTANCES

Describe the function of carrier proteins in a cell surface membrane
[4 Marks]

A
  • involved in facilitated diffusion
  • movement of large molecules
  • from an area of high concentration to an area of low concentration
  • involved in active transport - needs energy ATP to move molecules agaisnt a concentration gradient
42
Q

EXCHANGE OF SUBSTANCES

Determine why animal A does not need a circulation system but animal B does
[4 Marks]

Comparison of two “animals” with different SA to Volume Ratio

A
  • Both have the same volume
  • Animal A has a larger surface area
  • Therefore animal A has a larger surface area to volume ratio
  • so sufficient for diffusion

Distance to cells in centre of A is shorter than for B allowing quicker diffusion.

43
Q

EXCHANGE OF SUBSTANCES

Describe why single celled organisms do not need a specialised gas exchange surface.
[2 Marks]

A

They can rely on diffusion to take in oxygen

Short diffusion difference and large surface area to volume ratio - allows diffusion to occur at sufficient rate.

44
Q

EXCHANGE OF SUBSTANCES

Assess the factors that would cause Maltose to be detected in tube A before tube B
[4 Marks]

Specific diagram analysis question

A
  • Tube A = greater surface area
  • Ficks Law - Rate of diffusion is proportional to surface area
  • Therefore increase rate of diffusion for Maltose in tube A
  • Tube A could have higher concentration of starch
  • A higher concentration gradient would increase the rate of diffusion.
45
Q

EXCHANGE OF SUBSTANCES

Describe the structure of the cell surface membrane
[3 Marks]

A
  • cell membrane is mainly phospholipids and and proteins
  • phospolipids form a bilayer
  • proteins float in the phopholipids creating the fluid mosiac model
  • Proteins may span the bilayer or be located in only one layer
46
Q

EXCHANGE OF SUBSTANCES

Explain how the structure of the human lungs enables rapid gas exchange
[4 Marks]

A
  • Many alveoli provide a large surface area
  • Alveoli have walls that are one cell thick providing a short diffusion distance
  • a high concentration gradient is maintained by circulation
  • extensive capillary network around alveoli provides large surface area for gas exchange.
47
Q

INHERITANCE

Determine the probability that person 2 has the same sex and MSP 1 phenotype as person 1.

Specific Genetic Diagram Question

A

50% probability of person being the same sex.

Person 1 is heterozygous for the condition

Person 2 therefore has a 75% probability of having the same phenotype as person 1 for MPS

Overall Probability - 75 divided by 2 - Probability of being both female and not having MPS 1 will be 0.375.

48
Q

INHERITANCE

Multiple genetic crosses were carried out between individuals homozygous for recessive alleles and individuals heterozygous for the same gene.

Describe how the outcome of these crosses would be affected if a gene drive was used with the recessive allele.

[3 Marks]

A
  • Without the gene drive, the expected outcome would be 50% heterozygous and 50% homozygous recessive offspring
  • With a gene drive , the proportion of homozygous recessive offspring would increase
  • The stronger the gene drive, the greater proportion of homozygous recessive alleles.
49
Q

INHERITANCE

Men with Cystic Fibrosis are less likley to be able to release sperm.
Discuss why a person with cystic fibrosis could have these symptoms.

[6 Marks]

A
  1. In CF, mucus is thicker than normal
  2. Pancreatic enzymes cannot enter the intestine because the pancreatic duct is blocked with mucus
  3. Therefore digestion is less effiencent
  4. In this gene mutation
  5. This causes the CTFR protein channels to be non functioning
  6. Meaning chloride ions cannot move out of the epithelial cells
  7. The accumulation of sodium and chloride ions means that water moves out of the cell by osmosis.
50
Q

INHERITANCE

Explain why Cystic Fibrosis affects the rate of oxygen uptake in the lungs
[3 Marks]

A
  • Thick Mucus
  • Cannot be moved out by cilia
  • Which restricts air flow through bronchioles
    This increases diffusion distance for gaseous exchange at the alveoli
51
Q

INHERITANCE

Explain why thicker mucus is produced if the functioning of the CFTR channel protein is impaired.
[2 Marks]

A
  • Chloride ions cannot leave the cell through the CFTR protein channel
  • Sodium Ions also do not move out of the cell
  • Therefore water moves out of the mucus by Osmosis
52
Q

INHERITANCE

Explain why different mutations in the CFTR protein channel can lead to differences in the severity of the symptoms of cystic fibrosis
[2 Marks]

A
  • Different mutations will have different effects on the proteins being produced
  • Chloride ion transport affects the extent of changes to the CTFR protein
  • This varies the thickness of the mucus
53
Q

INHERITANCE

Doctors give children dietary supplements and digestive enzymes to children with cystic fibrosis.

Explain why these children would be given dietary supplements and digestive enzymes
[5 Marks]

A

Cystic Fibrosis causes the production of thick mucus

Which blocks the pancreatic duct

Therefore enzymes are prevented from digesting

So, reduced absorption of products from digestion into the blood

Dieteray supplements/ digestive enzymes would increase growth rate.

54
Q

INHERITANCE

Describe the role of the CFTR Protein in ensuring that mucus produced in the lungs has the right consistency.
[2 Marks]

A
  • Chloride ions leave the cells
  • Sodium Ions leave the cells
  • Water moves out of the cells and into the mucus by osmosis
55
Q

INHERITANCE

Why might a couple choose not to have the Chronic Villus Sampling Test.
[2 Marks]

A
  • The test results may be inaccurate
  • There is an increased risk of miscarriage
  • False positives may lead to the termination of a healthy foetus
  • Could possibly be agaisnt values or beliefs of parents (abortion)
56
Q

INHERITANCE

Explain what is meant be an inherited recessive disorder
[2 Marks]

A

Caused by a faulty allele
only expressed in the homozgous condition

57
Q

INHERITANCE

Individual 12 wants to know if her baby has the condition alkaptonuria
State and Justify a suitable method of collecting cells for prenatal testing
[3 Marks]

A

Chronic Villus Sampling

  • Cells are taken from placenta around 10-14 weeks of pregnancy
  • Benefit of early diagnosis.
58
Q

INHERITANCE

Define the term allele
[2 Marks]

A

alternative form of a gene
found at the same locus

59
Q

PROTEINS

Liver Cells can absorb lactate from the blood
Deduce what happens to the lactate in these cells
[2 Marks]

A

Lactate is oxidised to form pyruvate

Pyruvate is converted to glucose

Which can then be used in respiration

60
Q

PROTEINS

Describe the interaction between P and Q that allows muscle contraction
[2 Marks]

Specific Diagram Question

A

P changes shape
Causes Q to move away from the myosin- binding sites

61
Q

PROTEINS

Explain the importance of the primary structure for the functioning of the enzyme ATPase .
[3 Marks]

A

The primary structure determines the sequence of amino acids

This determines the tertiary structure of the enzyme

Therefore the primary structure affects the shape of the enzyme active site

Active Site should be complimentary to ATP

62
Q

PROTEINS

Explain how a change in the primary structure of Factor V111 enzyme could cause difficulties for blood clotting.
[4 Marks]

A
  • A different priary structure of enzyme means there will be a different sequence of amino acids
  • Therefore a change in the R group
  • Means there will be a change in the shape of the active site preventing a substrate from binding
  • Stops the production of fibrin
63
Q

PROTEINS

State why enzymes are described as biological catalysts
[1 Mark]

A

They reduce the activation energy of biological reactions.

64
Q

PROTEINS

Explain how a single base mutation can lead to an altered primary structure of “enzyme G”
[3 Marks]

Specific information Question

A
  • Changing a base results in a change of the triplet code
  • This changes the codon in the mRNA
  • This results in a different amino acid sequence.
65
Q

PROTEINS

How can human genome sequencing be used to identify mutations.
[3 Marks]

asssociated with MPS 1

A
  • Sequence the genome of people with condition
  • Sequence the genome of a number of people without the condition
  • Compare the base sequences to identify mutations found only in individuals with the condition
66
Q

PROTEINS

Describe what happens to extracellular enzymes following the process of translation untill they are released from the cell.
[3 Marks]

A
  1. Proteins are folded in RER
  2. Packaged into vesicles
  3. Modified in the GA
  4. Released from the cell by exocytosis
67
Q

PROTEINS

Explain why enzymes that are folded incorrectly cannot carry out their function.
[3 Marks]

A
  1. Different tertiary shape
  2. The active site would not be able to bind with the substrate (different shape of active site)
  3. Therefore, it would not be able to catalyse the reaction.
68
Q

PROTEINS

Describe how an enzyme could break down the polysaccarhide component of multipolysaccharides.
[2 Marks]

A

Hydrolysis
Of glycosidic bonds

69
Q

PROTEINS

Explain the table results
[3 Marks]

mice interferon question

A

increasing dose of interferon, increases the survival time of the mice

interferon inhibits viral replication inside the cells

Therefore, the greater the dose of interferon, the fewer virus particles produced.

70
Q

PROTEINS

Describe how the tertiary structure of myosin is related to its function
[3 Marks]

A
  • Folded into a specific shape with a globular head
  • that can bind to actin

Myosin has a site that can bind to ATP
Straight to form a bundle with other Myosin Molecules

71
Q

PROTEINS

Compare and Contrast the molecular structures of globular and fibrous proteins
[4 Marks]

A

Similarities-
* chains of amino acids joined by peptide bonds
* Contain hydrogen bonds holding it in its 3 dimmensional shape

Differences-
* Globular Proteins have hydrophillic groups on the outside/ Fibrous proteins have hydrophobic groups on the outside
* Globular have tertiary structures/ Fibrous have little to no tertiary structure
* Globular- folded into compact shapes/ Fibrous - long chain

72
Q

PROTEINS

Describe how a peptide bond is formed
[2 marks]

A
  • In condensation reaction
  • between the amine group and carboxyl group
73
Q

PROTEINS

Describe how the structure of phospholipase allows it to hydrolyse phospholipids.
[4 Marks]

A
  • The structure of the enzyme is determined by the sequence of amino acids
  • The tertiary structure provides an active site
  • Active site is complementary to phospholipid
  • to break ester bonds between glycerol and fatty acids

A Level Biology Flashcards (8 decks)

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