A level june 2020 Flashcards

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1
Q

Explain the function of this ATP hydrolase. (2)

A
  1. ATP to ADP +Pi - releases energy
  2. Energy allows ions to be mo ed against a concentration gradient
    or allows active transport of ions
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2
Q

The movement of Na+ out of the cell allows the absorption of glucose into the cell lining the ileum. (2)

A
  1. Maintains/generates a diffusion gradient for Na+ from ileum into cell
  2. Na+ moving in by facilitated diffusion brings glucose with it
    or moving in by co-transport , brings glucose with it
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3
Q

Describe and explain 2 features you would expect to find in a cell specialised for absorption. (2)

A
1. Folded membrane/microvilli so large surface
area (for absorption);
2. Large number of
co-transport/carrier/channel proteins so fast
rate (of absorption)
OR
Large number of co-transport/carrier proteins
for active transport
OR
Large number of
co-transport/carrier/channel proteins for
facilitated diffusion;
3. Large number of mitochondria so make
(more) ATP (by respiration)
OR
Large number of mitochondria for aerobic
respiration
OR
Large number of mitochondria to release
energy for active transport;
4. Membrane-bound (digestive) enzymes so
maintains concentration gradient (for fast
absorption);
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4
Q

Describe how amino acids join to form a polypeptide so there is always NH2 at
one end and COOH at the other end.
You may use a diagram in your answer. (2)

A
  1. One amine/NH2 group joins to a
    carboxyl/COOH group to form a peptide
    bond;
  2. (So in chain) there is a free amine/NH2 group
    at one end and a free carboxyl/COOH group
    at the other
    OR
    Each amino acid is orientated in the same
    direction in the chain;
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5
Q

Use your knowledge of lipid digestion to explain the differences in the results for
samples A and B shown in Table 1 You should assume that no absorption had occurred.
[3 marks]

A
  1. Triglycerides decrease because of the action of
    lipase
    OR
    Fatty acids increase because of the action of
    lipase;
  2. Triglycerides decrease because of hydrolysis (of
    triglycerides)
    OR
    Fatty acids increase because of hydrolysis (of
    triglycerides);
  3. Triglycerides decrease because of digestion of
    ester bonds (between fatty acid and glycerol)
    OR
    Fatty acids increase because of digestion of
    ester bonds (between fatty acid and glycerol);
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6
Q

After collecting the samples, the scientist immediately heated them to
70 °C for 10 minutes.
Explain why.
[2 marks]

A
  1. To denature the enzymes/lipase;
  2. So no further digestion/hydrolysis/catalysis
    occurred;
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7
Q

Describe the role of micelles in the absorption of fats into the cells lining the ileum.
[3 marks]

A
  1. Micelles include bile salts and fatty acids;
  2. Make the fatty acids (more) soluble in water;
  3. Bring/release/carry fatty acids to cell/lining (of the
    ileum);
  4. Maintain high(er) concentration of fatty acids to
    cell/lining (of the ileum);
  5. Fatty acids (absorbed) by diffusion;
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8
Q

At P on Figure 3, the pressure in the left ventricle is increasing. At this time, the rate
of blood flow has not yet started to increase in the aorta.
Use evidence from Figure 3 to explain why.
[2 marks]

A
  1. Aortic/semi-lunar valves is closed;
  2. Because pressure in aorta higher than in
    ventricle;
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9
Q

At Q on Figure 3 there is a small increase in pressure and in rate of blood flow in the
aorta.
Explain how this happens and its importance.
[2 marks]

A
  1. Elastic recoil (of the aorta wall/tissue);
  2. Smooths the blood flow
    OR
    Maintains rate of blood flow
    OR
    Maintains blood pressure;
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10
Q

A student correctly plotted the right ventricle pressure on the same grid as the left
ventricle pressure in Figure 3.
Describe one way in which the student’s curve would be similar to and one way it
would be different from the curve shown in Figure 3.
[2 marks]
Similarity —-
Difference —–

A
  1. Peaks/contractions at the same/similar time
    OR
    Same/similar pattern;
  2. Lower pressure;
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11
Q

Use information from Figure 3 to calculate the heart rate of this dog.
[1 mark]

A
167 (beats minute–1
)
OR
164 (beats minute–1
)
OR
171 (beats minute–1
);
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12
Q

When making up extraction solvent E, the student used a volume ratio of
70:30:1 ethanol:water:acid.
Tick () one box that shows the most appropriate volumes she would use to make up
100 cm3 of extraction solvent E.
[1 mark]

A
  1. 3cm3 solvent, 29.7cm3 water, 1.0cm3 acid (box

2) ;

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13
Q
The student kept constant:
• the mass of fresh blueberries
• the volume of extraction solvent
• the time for the mixture to stand.
Name two other variables the student should have kept constant during this
investigation.
[2 mark)
A
  1. Temperature;
  2. Agitation/mixing/stirring;
  3. Source/age/type of blueberries;
  4. Crushing of the blueberries;
  5. Rinsing of the blueberries prior to mixing;
  6. Concentration of ethanol/acid;
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14
Q

After 1 hour, the student filtered the samples.
She placed the filtrate in a colorimeter and measured the light absorbance.Use your knowledge of membrane structure to explain the results in Figure 4.
[4 marks]

A
1. Higher absorbance indicates more anthocyanin
OR
Higher absorbance indicates more membrane
damage/permeability
OR
(G not zero because) some anthocyanin
released when blueberries are crushed
OR
(G not zero because) some membrane damage
when blueberries are crushed;
2. More membrane damage/permeability results in
more anthocyanin release
3. (E and F greater than water because)
phospholipids dissolve in ethanol;
4. (E greater than F because) acid denatures
membrane proteins;
4 For ‘anthocyanin’
accept ‘pigment’.
1. A direct
comparative
statement is not
needed, can be taken
from the answer as a
whole.
1 and 2. Accept ‘most’
for ‘more’.
4. Accept description
of denaturation in
terms of change in
tertiary structure or
breaking of
hydrogen/ionic
bonds.
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15
Q

A different student did this investigation. He did not have a colorimeter.
Describe a method this student could use to prepare colour standards and use them
to give data for the total anthocyanin extracted.
[3 marks]

A
1. Use known concentration of blueberry
juice/extract
OR
Use known concentration of
anthocyanin/pigment (solution)
OR
Use known concentration of (extraction) solvent
to be added to blueberries;
2. Prepare dilution series;
3. Compare (results) with colour standards to give
score/value/concentration;
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16
Q

Describe the role of DNA polymerase in the semi-conservative replication of DNA.
[2 marks]

A
  1. Joins (adjacent DNA) nucleotides;
  2. (Catalyses) condensation (reactions);
  3. (Catalyses formation of) phosphodiester bonds
    (between adjacent nucleotides);
17
Q

It took less time for 25% of cells with cyclin D to be undergoing DNA replication than
for 25% of cells without cyclin D.
Use Figure 5 to calculate this time difference as a percentage decrease.
Show your working.
[2 marks]

A

Final answer with 2sf or 3sf in range 31.8 to
34.7%;;
1 mark for
5.5 to 6.1 hours
OR
Final answer with 2sf or 3sf in range 46.6 to 53.0%
OR
Correct final answers rounded to more than 3sf
OR
Final answer with 2sf or 3sf in range 30.8 to 31.7 or
34.8 to 35.6%

18
Q

Cyclin D stimulates the phosphorylation of DNA polymerase, which activates the
DNA polymerase.
Describe how an enzyme can be phosphorylated.
[2 marks]

A
1. Attachment/association of (inorganic) phosphate
(to the enzyme);
2. (Released from) hydrolysis of ATP
OR
(Released from) ATP to ADP + Pi;
19
Q

Some tumour cells contain higher than normal concentrations of cyclin D.
Use Figure 5 to suggest why higher than normal concentrations of cyclin D could
result in a tumour.
[2 marks]

A
1. Shortens interphase
OR
Cells begin DNA replication earlier
OR
DNA replication (starts) faster;
2. Fast(er) cell cycle/division/multiplication/mitosis
OR
Uncontrolled cell division/mitosis;
3. (Resulting in) a mass/group of
abnormal/excessive cells;
20
Q

Particulate matter is solid particles and liquid particles suspended in air. Polluted air
contains more particulate matter than clean air.
A high concentration of particulate matter results in the death of some
alveolar epithelium cells. If alveolar epithelium cells die inside the human body they
are replaced by non-specialised, thickened tissue.
Explain why death of alveolar epithelium cells reduces gas exchange in human lungs.
[3 marks]

A
  1. Reduced surface area;
  2. Increased distance for diffusion;
  3. Reduced rate of gas exchange;
21
Q

Alpha-gal is a disaccharide found in red meat.
Alpha-gal is made of two galactose molecules. Galactose has the chemical formula
C6H12O6
Give the chemical formula for the disaccharide, alpha-gal, and describe how it is
formed from two galactose molecules.
[2 marks]
Formula
Description

A
  1. C12H22O11;
  2. Condensation reaction
    OR
    With a glycosidic bond;
22
Q

Draw a labelled diagram of an antibody and identify the specific alpha-gal binding site.
[3 marks]

A
  1. Y shape showing two long and two short
    (polypeptide) chains correctly positioned;
  2. (Alpha-gal) binding site labelled on the end of the
    branches of the Y of the antibody;
  3. Variable region labelled
    OR
    Constant region labelled
    OR
    Disulfide bridge/bond labelled;
23
Q

A tick is a small animal that bites humans and feeds on their blood. This results in
proteins from the tick saliva entering the human body.
Scientists have suggested one hypothesis for the allergic reaction to alpha-gal in
red meat. They think that an earlier immune response to a tick bite can cause a
person to have an allergic reaction to alpha-gal in red meat.
Suggest how one antibody can be specific to tick protein and to alpha-gal.
[2 marks]

A
  1. (Part of tick protein and alpha-gal) have a similar
    shape/structure;
  2. Antibody is complementary to both (tick protein
    and alpha-gal)
    OR
    Antigen-binding site is complementary to both
    (tick protein and alpha-gal)
    OR
    Antibody can form antigen-antibody complex with
    both (tick protein and alpha-gal);
24
Q

The scientists’ hypothesis was that an earlier immune response to tick protein causes
the allergic reaction.
Consider whether Figure 7 supports this hypothesis.
[3 marks]

A
  1. Exposure to tick (protein) is followed by increase
    in antibody (specific to alpha-gal);
  2. (Later) greater/faster increase in antibody
    suggests there are memory cells;
  3. Antibody (specific to alpha-gal) increases
    during/after allergic reaction;
  4. During/after allergic reaction, total antibody
    increases more than alpha-gal antibody;
  5. (So) may be other antibodies (that are causing
    allergic reaction);
25
Q

Complete Table 2 to show three differences between DNA in the nucleus of a
plant cell and DNA in a prokaryotic cell.
[3 marks]

A
Plant v prokaryote
1. (Associated with) histones/proteins v no
histones/proteins;
2. Linear v circular;
3. No plasmids v plasmids;
4. Introns v no introns;
5. Long(er) v short(er)
26
Q

Define ‘non-coding base sequences’ and describe where the non-coding multiple
repeats are positioned in the genome.
[2 marks]

A
  1. DNA that does not code for protein/polypeptides
    OR
    DNA that does not code for (sequences of)
    amino acids
    OR
    DNA that does not code for tRNA/rRNA;
  2. (Positioned) between genes;
27
Q

The scientists studied five individuals from each species. Within the five individuals of
species T they found a percentage similarity of 66%.
Use Table 3 to evaluate how this information affects the validity of the
phylogenetic tree.
[2 marks]

A
  1. (Supported) more similar than with any other
    species;
  2. (Not supported) high (intraspecific) variation in
    species T (compared with variation between T
    and C);
  3. Small sample
    OR
    Only five (individuals);
28
Q

Give a null hypothesis for this investigation and name a statistical test that would be
appropriate to test your null hypothesis.
[2 marks]
Null hypothesis
Statistical test

A
  1. There is no association/correlation/relationship
    between the concentration of carbon dioxide and
    the stomatal density
    OR
    The concentration of carbon dioxide does not
    affect the stomatal density;
  2. Correlation coefficient;
29
Q

A journalist saw Figure 10 and suggested that future increases in atmospheric
carbon dioxide concentration could result in less transpiration.
Evaluate his suggestion.
[4 marks]

A
  1. Increasing carbon dioxide (concentration) shows
    decreased stomatal density;
  2. Fewer stomata means less transpiration
    OR
    Fewer stomata means less evaporation (of water
    from leaves)
    OR
    Fewer stomata means less diffusion of water
    vapour (from leaves);
  3. Same (volume of) carbon dioxide can be
    absorbed for photosynthesis with smaller number
    of stomata;
  4. Don’t know the size of the stomata;
  5. Don’t know whether leaf size has changed;
  6. Don’t know if this is true for all species (of plant);
  7. Don’t know how long the stomata are open for;
  8. Don’t know if this trend will continue (beyond the
    concentrations of carbon dioxide shown in Figure
    10);
  9. Other factors affect transpiration (rate)
30
Q

Describe how mRNA is formed by transcription in eukaryotes.

[5 marks]

A
  1. Hydrogen bonds (between DNA bases) break;
  2. (Only) one DNA strand acts as a template;
  3. (Free) RNA nucleotides align by complementary
    base pairing;
  4. (In RNA) Uracil base pairs with adenine (on
    DNA)
    OR
    (In RNA) Uracil is used in place of thymine;
  5. RNA polymerase joins (adjacent RNA)
    nucleotides;
  6. (By) phosphodiester bonds (between adjacent
    nucleotides);
  7. Pre-mRNA is spliced (to form mRNA)
    OR
    Introns are removed (to form mRNA)
31
Q

Describe how a polypeptide is formed by translation of mRNA.

[6 marks]

A
  1. (mRNA attaches) to ribosomes
    OR
    (mRNA attaches) to rough endoplasmic
    reticulum;
  2. (tRNA) anticodons (bind to) complementary
    (mRNA) codons;
  3. tRNA brings a specific amino acid;
  4. Amino acids join by peptide bonds;
  5. (Amino acids join together) with the use of ATP;
  6. tRNA released (after amino acid joined to
    polypeptide);
  7. The ribosome moves along the mRNA to form
    the polypeptide;
32
Q

Define ‘gene mutation’ and explain how a gene mutation can have:
• no effect on an individual
• a positive effect on an individual.
[4 marks]

A

(Definition of gene mutation)
1. Change in the base/nucleotide (sequence of
chromosomes/DNA);
2. Results in the formation of new allele;
(Has no effect because)
3. Genetic code is degenerate (so amino acid
sequence may not change);
OR
Mutation is in an intron (so amino acid sequence
may not change);
4. Does change amino acid but no effect on tertiary
structure;
5. (New allele) is recessive so does not influence
phenotype;
(Has positive effect because)
6. Results in change in polypeptide that positively
changes the properties (of the protein)
OR
Results in change in polypeptide that positively
changes a named protein;
7. May result in increased reproductive success
OR
May result in increased survival (chances);