Bio PPQs wrong topic 2

Describe the role of tRNA in the production of the protein part of a glycoprotein

• each tRNA brings a specific amino acid (to the ribosome)

• the tRNA with the complementary anticodon binds to the mRNA codon

• tRNA bonds to ribosome

The synthesis of mRNA occurs in a process called transcription. Compare and contrast the process of transcription with the process of DNA replication. (4)

Similarities:

• involve the formation of (polynucleotide/phosphodiester bonds

• involve DNA helicase

Differences

• transcription uses RNA nucleotides whereas replication uses DNA nucleotides

• transcription uses RNA polymerase whereas replication uses DNA polymerase

• transcription produces single strand of mRNA/only copies template strand of DNA whereas replication produces double stranded DNA/copies both strands of DNA

What is a gene?

a sequence of bases/nucleotides in DNA coding for a {sequences of amino acids/ polypeptide/ protein}

ALLOW section of DNA coding for a sequence of amino acids/polypeptide/protein

most safe: a sequence of bases that codes for a polypeptide

Explain how a change of one amino acids could lead to a change in the structure and properties of the haemoglobin molecule (4)

• different {sequence of amino acids/primary structure

• (a different amino acid will have a) different R group

• (therefore) {secondary/ tertiary/ quaternary} structure will change

• (due to a) change in a named bond (holding moelcule in its 3D shape)

• (haemoglobin) may not bond to oxygen

Analyse the data to comment on the effectiveness of these antibiotics for the treatment of TB. (6)

Basic information:

• all the treatment combinations were effective at treating TB

• All treatments had some {relapses/individuals with TB} (3 years later)

• {Group 1/group 1 and 2/rifampicin + pyrazinamide/rifampicin + isoniazid} has the lowest number of patients with TB (3years later)

Evidence for linkages made

• percentage relapse varies depending on second part of treatment

• combinations involving rifampicin most effective

• these anitbiotics tested act on different targets in bacteria

• gaps in information - not all combinations tested, other combinations might be more effective

• other time scales may have been more effective

Evidence for sustained scientific reasoning

• could be other reasons for infections with TB 3 years later not due to the antibiotic treatment

• no information about dormant TB (only percentage of active cases)

• bacterial RNA polymerase possibly the best target for antibiotics

• antibiotics targeting synthesis of cell wall fatty acids least effective in terms of relapses

• idea of combination of antibiotics with different mode of activity most effective

investigation 1

• comparison of data for no excercise/excercise without nandrolone: max recoil of {aorta/artery} is {higher /23% greater}

• discuss outcome of this difference as an advantage: so (oxygen-rich) blood can flow more rapidly (at the correct pressure) {from the heart/to the muscles}

• less likely to get atherosclerosis/ CVD/ strokes

investigation 2

• comparison of data for P and R for both protein complexes: more present due to excercise

• discuss advantageous outcome: so more {oxidative phosphorylation/ ATP synthesis / chemiosmosis} so more ATP for muscle contraction / breaking of the bond between actin and myosin (in aorta wall)

investigation 3

• compare P and Q for mRNA coding for Tfam: more Tfam per cell so more mitchondria produced so more {respiration/ ATP formed}

Explain how a change of one amino acid could lead to a change in the structure and properties of the haemoglobin protein (4)

• different {sequence of amino acids/ primary structure}

• (a different amino acid will have a) different R group

• (Therefore) {secondary/ tertiary/ quaternary} structure will change

• (due to a) change in a named bond (hydrogen bond, disulfide bridges, ionic bonds DO NOT ALLOW PEPTIDE BONDS) (holding the molecule in its 3D shape)

• (haemoglobin) may not bind to oxygen

Describe the role of tRNA in the production of leptin (3)

• tRNA molecules {transport amino acids to the ribosomes}

• tRNA molecules have an anticodon that {binds to/ recognises} a codon on the mRNA

• each tRNA carries a particular amino acid

Describe how the primary structure of leptin allows it to be soluble in water (3)

• primary structure/sequence of the amino acids determines the folding of the polypeptide

• forming a globular structure

• hydrophobic R groups are located in the centre of the protein/hydrophilic R groups are located on the outside of the protein

• water forms hydrogen bonds with protein/hydrophilic groups

Leptin is a protein hormone with a role in the control of appetite in humans. Several mutations of the leptin gene have been identified. All these mutations are frameshift mutations that result in shortened primary structures. A frameshift mutation involves the insertion or removal of one or two nucleotides from a gene. Describe how a frameshift mutation could result in the production of leptin with a variety of shorter primary structures. (2)

• adding or removing 1 or 2 nucleotides changes the triplet code

• introducing a new start/stop codon

• coding for a shorter sequence of amino acids

One function of DNA is to act as a template for the synthesis of messenger RNA (mRNA) during transcription. (i) Describe how mRNA is synthesised at a template strand of DNA. (2)

• RNA nucleotides align with complementary bases on DNA

• RNA nucleotides joined together by RNA polymerase/phosphodiester bonds

Describe the differences between DNA and RNA (3)

• DNA is double stranded whereas RNA is single stranded

• DNA contains deoxyribose whereas RNA contains ribose

• DNA contains thymine whereas RNA contains uracil

Differences between transcription and DNA replication:

• replication involves DNA nucleotides whereas transcription involves RNA nucleotides

• replication produces double stranded DNA whereas transcription produces a single stranded RNA molecule

• replication uses DNA polymerase whereas transcription requires RNA polymerase

• replication produces identical copies whereas transcription produces complementary copies

hydrogen

Mutations to DNA can affect the structure of proteins produced in the cell. Removing one base from a DNA sequence will affect the primary structure of a protein. Changing one base for another may not affect the primary structure of a protein. Explain why these two types of mutation have different effects on protein structure. (4)

• deletion could affect every codon (on the mRNA/substitution will only affect one codon

• deletion is more likely to affect the position of the start/stop 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 due to the degenerate nature of the genetic code

DNA is a polymer made from monomers called nucleotides. Describe how nucleotides join together to form DNA (2)

• condensation reaction

• phosphodiester bonds

• DNA polymerase

Mutations to DNA can affect the structure of proteins produced in the cell. Removing one base from a DNA sequence will affect the primary structure of a protein. Changing one base for another may not affect the primary structure of a protein. Explain why these two types of mutation have different effects on protein structure. (4)

• the conservative model was rejected/the semi-conservative model was accepted

• due to generation 1 having a single band which is halfway between N15 and N14

• because the DNA has one strand containing N15 and one strand containing N14

• in the semi-conservative model further generations would have a band which is halfway between N15 and N14/ no band at N15

• one triplet is affected/ a different triplet code is produced

• the mutation could change one of the amino acids

• this would change the bonds formed between the R groups/ cause a change in the tertiary structure

• the haemoglobin would no longer be able to bond to oxygen

Penicillin is an antibiotic. It was discovered in 1928. Since then many antibiotics have been identified and are widely used in the treatment of bacterial infections. Scientists have recently discovered a new class of antibiotics that bind to ribosomes. (i) Explain why these antibiotics could affect the production of proteins in bacteria. (3)

• ribsome shape is altered

• mRNA is prevented from binding to ribosome/ causing change in tRNA binding

• therefore translation cannot occur

• protein/polypeptide is not synthesised

These new antibiotics attach 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)

• bacteria have not been exposed to new antibiotics before/ bacteria do not have mechanisms to make them resistant to new antibiotics

• bacteria have developed resistance to other antibiotics by evolving/ natural selection

• therefore there has been no advantage to possessing a mutation to bypass the new antibiotic/ no mutation present to give resistance

* (iii) Scientists have isolated these new antibiotics and tested their effectiveness against bacteria that are resistant to other types of antibiotic. Devise a laboratory procedure to compare the effectiveness of penicillin with one of the new antibiotics.(6)

• prepare agar plates with bacterial cultures/bacterial lawn/seeded with bacteria - use bacteria that are resistant to other antibiotics

• prepare solutions of new antibiotic and penicillin

• place onto paper discs/into wells in the agar/prepare mast rings

• control time and temperature of incubation

• same concentration and volume of both antibiotics

• description of serial dilution of each antibiotic

• range of dilutions on each plate one-antibiotic per plate

• statistical test to determine which is the most effective

• repeat with different strains of resistant bacteria

Deduce how many times the bacteria in the culture have divided during this experiment. (2)

• DNA content doubles twice/two stages of DNA synthesis

• therefore two divisions

Explain why the experiment would be improved if all the bases were provided but only the thymine was radioactive. (2)

• because thymine is found only in DNA

• other radioactive bases taken up by all nucleic acids

• only DNA would be measured

All cells have a cell surface membrane. Some epithelial cells in the lungs secrete mucus. If the mucus is too 'sticky', it cannot be easily removed from the lungs. Other epithelial cells in the lungs contain CFTR proteins in their cell surface membranes. (i) Describe the role of the CFTR protein in ensuring that the mucus produced in the lungs has the right consistency.

• chloride ions leave cells (through the CFTR channel protein)

• sodium ions leave the cells (following the chloride ions)

• increasing the solute concentration in the mucus

• water moves out of the cells by osmosis into the mucus

• triplet code is shown by 3 bases coding for an amino acid

• non-overlapping code e.g. ATT codes for amino acid I and then AAA codes for amino acid K

• degenerate code as both ATT and ATC code for amino acid I

Describe the roles of transcription and translation in the synthesis of a globular protein by a muscle cell. (5)

• the gene/sequence of DNA for the globular protein is transcribed

• complementary base pairing between RNA nucleotides and DNA to produce mRNA

• mRNA leave the nucleus and attaches to a ribosome

• pairing between codons on mRNA and anticodons on tRNA

• tRNA provides specific amino acids

• the sequence of bases/codons determines the sequence of amino acids/primary structure of the protein

Similarity

• both contain phosphate, pentose sugar and a base

Difference

• a DNA nucleotide contains deoxyribose whereas ATP contains ribose

• a DNA nucleotide could contain other bases whereas ATP contains only adenine/one base type

• a DNA nucleotide contains one phosphate whereas ATP contains three phosphates/is a triphosphate

The fluid mosaic model of cell membranes was first proposed in 1972. The vacuoles in beetroot cells contain molecules of betalain, a red pigment. Betalains are large polar molecules. These molecules can leave beetroot cells if the vacuole membrane and the cell surface membrane are damaged. Explain why betalain molecules cannot move through intact cell membranes. (3)

• betalain molecules are too large (to move through the cell membrane)

• there are no (carrier/chennel) proteins for betalain molecules (to move through)

• betalain molecules are polar and (are repelled by hydrophobic fatty acid tails/cannot move through fatty acid tails)

• thinner blood-gas barrier

• because of thinner (alveolus walls/capillary walls/extracellular matrix layer)

• therefore a reduced diffusion distance

• a faster rate of (diffusion/gas exchange)

C

Diffusion and active transport are mechanisms by which molecules can enter cells. Compare and contrast these two mechanisms. (3)

similarities

• both move molecules through the (phospholipid bilayer/cell surface membrane)

• (in both) molecules can move through proteins

Differences

• diffusion occurs down a concentration gradient whereas active transport occurs against a concentration gradient

• diffusion is a (passive/does not require ATP) whereas active transport requires ATP

Endocytosis and exocytosis are processes that move large molecules into a cell or out of a cell. Compare and contrast the processes of endocytosis and exocytosis. (3)

similarities

• both involve vesicles

• both processes involve energy from ATP

differences

• exocytosis involves (molecules/substances) leaving the cell whereas endocytosis involves (molecules/substances) entering the cell

• exocytosis involves vesicles fusing with the cell surface membrane whereas endocytosis involves the formation of vesicles (from the cell surface membrane)

Describe how glucose moves into cells by facilitated diffusion. (2)

• carrier protein in cell surface membrane

• glucose moves from high to low con

• glucose binds to carrier protein which changes shape to move glucose across cell membrane

Explain how the structure of glycogen allows it to be an energy store (3)

Glycoproteins and phospholipids are molecules found in the cell surface membrane. (i) Give one function of the glycoproteins found in the cell surface membrane. (1)

• cell recognition

• receptors (on cell surface membrane)

• antigens

A cell surface membrane is partially permeable. The phospholipid bilayer is important in controlling the movement of molecules through the membrane. Explain how the structure of a phospholipid molecule contributes to the partial permeability of a cell surface membrane. (3)

• contains polar hydrophilic (phosphate_ head and non polar/hydrophobic fatty acid chains

• allows fat/soluble/non-polar molecules to pass through the membrane

• polar/ionic molecules cannot pass through

* Assess the effect of this heart defect on the rate of oxygen diffusion between the alveoli and the blood. (6)

• rate of diffusion lower with abnormal heart

• blood entering lungs from abnormal heart has more oxygen, 8kPa, than blood entering lungs from a normal heart, 5kPa

• oxygen in blood increased by only 2kPa instead of 8kPa with abnormal heart

• resulting in a smaller difference in concentration between the alveoli and the red blood cells

• the surface area of the alveoli and distance is proportional to the rate of gas exchange

• a lower concentration gradient for oxygen between the alveoli and the blood results in a lower rate of oxygen diffusion

Give reasons for the variation in the lung volumes of healthy individuals.

Give a reason for calculating the surface area for gas exchange to volume ratio in this investigation.

to allow valid comparison/show differences

Blood type is an example of inherited variation. Blood types A, B, AB and O are determined by a single gene. Blood types are due to the presence or absence of antigens on the cell surface membranes of red blood cells. These antigens are glycoproteins. Antigens on the cell membranes of microbes can stimulate endocytosis and exocytosis in white blood cells. These processes are involved in transport through a cell surface membrane. Give two differences between endocytosis and exocytosis. (2)

• exocytosis involves molecules/substances leaving the cell whereas endocytosis involves moelcuels/substances entering the cell

• exocytosis involves vesicles fusing with cell surface membrane whereas endocytosis involves the formation of vesicles (from the cell surface membrane)

• smaller surface area/increased diffusion distance

• therefore reduction in oxygen uptake

• therefore less oxygen for areobic respiration

• leading to more anaerobic respiration

Explain why the phospholipids are arranged in two layers in a cell surface membrane. (3)

• hydrophilic region of phospholipid orientated towards water

• hydrophobic region away from water

• but two layers needed as water based/aq solution either side of the cell membrane

Explain why the phospholipids are arranged in two layers in a cell surface membrane. (3)

• involved in facilitated diffusion

• movement of large molecules/polar molecules/ions

• facilitated diffusion from a high concentration to a low concentration

• involved in active transport

• needs ATP to move molecules against a concentration gradient

• both have the same volume

• animal a has a larger surface area

• animal a has a larger surface area to volume ratio

• so sufficient surface area in animal A for diffusion

• distance to cells in centre of A is shorter than for B allowing quicker/sufficient diffusion/shorter diffusion distance

Which of the shaded structures contain both hydrophilic regions and hydrophobic regions? (1)

A 1 and 2 only

B 3 and 4 only

C 1, 2 and 3 only

D 1, 2, 3 and 4

D - 1,2,3,4

• same age/type of onion

• increases repeatability/validity

  OR

• smaller concentration intervals

• increases confidence in/ validity of conclusion

  OR

• same temperature/surface area of onion

• due to effect on osmosis

Deduce the effect of increasing the concentration of sodium chloride on the change in mass of the onion tissue. (3)

• 2.5% sodium chloride solution resulted in an increase in mass

• an increase in sodium chloride solution from 5% to 15%/20% resulted in a loss in mass

• because of the movement of water by osmosis

• SD values overlap for 5% and 10%/10% and 15%/10% and 20%/15% and 20% sodium chloride solution therefore no significant difference

Ignore (i)

• they can rely on diffusion to take in oxygen/remove wastes

• large surface area to volume ratio (allows diffusion to occur at a sufficient rate)

• short diffusion distance

• for tube A the surface area was greater than tubing B/ shorter distance for diffusion in tube A

• Fick’s law states that rate of diffusion is proportional to surface area/inversely proportional to distance for diffusion

• therefore increased rate of diffusion of maltose

• tube A could have had a higher concentration of starch/temperature

• a higher concentration gradient / higher temperature would increase the rate of diffusion

Hereditary spherocytosis is a condition that affects red blood cells. This inherited condition is caused by a gene mutation that affects the shape of the cell surface membrane. Describe the structure of the cell surface membrane

• cell membrane is mainly phospholipids and protein

• phospholipids form a bilayer

• proteins float in the phospholipids/change position/fluid mosaic model

• proteins may span the bilayer or be located in only one layer

Describe the role of tRNA in the production of leptin (3)

• tRNA molecules transport amino acids to the ribsomsomes

• tRNA molecules have an anticodon that bonds/recognises a codon on the mRNA

• each tRNA carries a particular amino acid

Describe how the primary structure of leptin enables it to be soluble in water. (3)

• primary structure/sequence of amino acids determines the folding of the polypeptide

• forming a globular structure

• hydrophobic R groups located in the centre of the protein/hydrophilic R groups located on the outside of the protein

• water forms hydrogen bonds with protein/hydrophilic groups

Mutations to DNA can affect the structure of proteins produced in the cell. Removing one base from a DNA sequence will affect the primary structure of a protein. Changing one base for another may not affect the primary structure of a protein. Explain why these two types of mutation have different effects on protein structure. (4)

• deletion could affect every codon (on the mRNA)/substituiotn will only affect one codon

• deletion is more likely to affect the position of the (stop codon/start codon)

• deletion results in a different sequence of amino acids/substituiotn may not affect the sequence of amino acids

• substitution may code for the same amino acid

• same amino acid acid due to the degenerate code

Explain the importance of the primary structure for the functioning of this enzyme. (3)

• determines interaction between (amino acids/R groups)

• primary structure determines folding/tertiary structure

• therefore affecting the shape of the active site

• active site is complementary to ATP

Ignore (i)

Haemophilia is a disease that affects blood clotting. People with haemophilia are sometimes given a protein called factor VIII. Factor VIII is an enzyme that is involved in the process of blood clotting. Explain how a change in the primary structure of factor VIII could cause difficulties with blood clotting. (4)

• different primary structure results in a different sequence of amino acids

• change in R groups changes folding/bonding/secondary structure/ tertiary structure

• changing shape/charge of the active site prevents substrate from being able to bind

• stopping/reducing the production of fibrin

State why enzymes are described as biological catalysts.

proteins which reduce activation energy of biological reactions

Explain how a single base mutation can lead to an altered primary structure of enzyme G. (3)

• changing a base results in a change in the triplet code

• this changes the codon in the mRNA

• resulting in a different amino acid/amino acid sequence in the primary structure

Explain how human genome sequencing can be used to identify the mutations associated with MPS I.(3)

• sequence the genome of people with MPS1

• 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

Many of the proteins synthesised become extracellular enzymes. Describe what happens to these proteins following the process of translation until they are released from the cell. (3)

• proteins are folded in RER

• proteins are packaged into/transported in vesicles

• the proteins is modified in the golgi apparatus

• exocytosis

Cells in people with these diseases produce incorrectly folded enzyme molecules. Explain why enzymes that are incorrectly folded cannot carry out their function (3)

• if the protein is not folded correctly the tertiary structure/3D shape would be different

• therefore the active site of the enzyme would not fit/bind with the substrate/ it would not be able to form an enzyme substrate complex

• therefore it would not be able to catalyse the reaction

Muscle cells contain globular and fibrous proteins. Compare and contrast the molecular structures of globular and fibrous proteins. (4)

• both are chains of amino acids joined by peptide bonds

• both contain named (hydrogen bonds, ionic bonds, disulfide bridges) bonds (holding molecule in its 3D shape)

• globular proteins have hydrophilic groups on the outside whereas fibrous proteins have hydrophobic groups on the outside

• globular have tertiary or quaternary structures whereas fibrous proteins have little or no tertiary structure

• globular are folded into compact shapes whereas fibrous have long chains

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

• different sequence of amino acids/primary structure

• a different amino acid will have a different R group

• therefore secondary/tertiary/quaternary structure will change

• due to a change in a named bonds (hydrogen bonds, ionic bonds, disulfide bridges) holding the molecules in its 3D shape

• haemoglobin may not bind to oxygen

Some species of bacteria have developed resistance to antibiotics. This has led scientists to investigate many molecules for antimicrobial properties. Peptides extracted from broad bean plants and cowpea plants have been studied. Describe how a peptide bond is formed. (2)

• a peptide bond is formed by a condensation reaction

• between the amine group/NH2 and the carboxyl group/COOH of adjacent amino acids

• the structure of the enzyme is determines by the sequence of amino acids

• tertiary structure provides an active site

• to break ester bonds/bonds between glycerol and fatty acids

• relevant detail concerning bonding within the enzyme molecule/ between enzyme and substrate (e.g hydrophilic R groups/hydrophobic R groups)