A level june 2020 Flashcards
Explain the function of this ATP hydrolase. (2)
- ATP to ADP +Pi - releases energy
- Energy allows ions to be mo ed against a concentration gradient
or allows active transport of ions
The movement of Na+ out of the cell allows the absorption of glucose into the cell lining the ileum. (2)
- Maintains/generates a diffusion gradient for Na+ from ileum into cell
- Na+ moving in by facilitated diffusion brings glucose with it
or moving in by co-transport , brings glucose with it
Describe and explain 2 features you would expect to find in a cell specialised for absorption. (2)
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);
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)
- One amine/NH2 group joins to a
carboxyl/COOH group to form a peptide
bond; - (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;
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]
- Triglycerides decrease because of the action of
lipase
OR
Fatty acids increase because of the action of
lipase; - Triglycerides decrease because of hydrolysis (of
triglycerides)
OR
Fatty acids increase because of hydrolysis (of
triglycerides); - 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);
After collecting the samples, the scientist immediately heated them to
70 °C for 10 minutes.
Explain why.
[2 marks]
- To denature the enzymes/lipase;
- So no further digestion/hydrolysis/catalysis
occurred;
Describe the role of micelles in the absorption of fats into the cells lining the ileum.
[3 marks]
- Micelles include bile salts and fatty acids;
- Make the fatty acids (more) soluble in water;
- Bring/release/carry fatty acids to cell/lining (of the
ileum); - Maintain high(er) concentration of fatty acids to
cell/lining (of the ileum); - Fatty acids (absorbed) by diffusion;
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]
- Aortic/semi-lunar valves is closed;
- Because pressure in aorta higher than in
ventricle;
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]
- Elastic recoil (of the aorta wall/tissue);
- Smooths the blood flow
OR
Maintains rate of blood flow
OR
Maintains blood pressure;
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 —–
- Peaks/contractions at the same/similar time
OR
Same/similar pattern; - Lower pressure;
Use information from Figure 3 to calculate the heart rate of this dog.
[1 mark]
167 (beats minute–1 ) OR 164 (beats minute–1 ) OR 171 (beats minute–1 );
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]
- 3cm3 solvent, 29.7cm3 water, 1.0cm3 acid (box
2) ;
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)
- Temperature;
- Agitation/mixing/stirring;
- Source/age/type of blueberries;
- Crushing of the blueberries;
- Rinsing of the blueberries prior to mixing;
- Concentration of ethanol/acid;
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]
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.
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]
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;
Describe the role of DNA polymerase in the semi-conservative replication of DNA.
[2 marks]
- Joins (adjacent DNA) nucleotides;
- (Catalyses) condensation (reactions);
- (Catalyses formation of) phosphodiester bonds
(between adjacent nucleotides);
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]
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%
Cyclin D stimulates the phosphorylation of DNA polymerase, which activates the
DNA polymerase.
Describe how an enzyme can be phosphorylated.
[2 marks]
1. Attachment/association of (inorganic) phosphate (to the enzyme); 2. (Released from) hydrolysis of ATP OR (Released from) ATP to ADP + Pi;
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]
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;
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]
- Reduced surface area;
- Increased distance for diffusion;
- Reduced rate of gas exchange;
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
- C12H22O11;
- Condensation reaction
OR
With a glycosidic bond;
Draw a labelled diagram of an antibody and identify the specific alpha-gal binding site.
[3 marks]
- Y shape showing two long and two short
(polypeptide) chains correctly positioned; - (Alpha-gal) binding site labelled on the end of the
branches of the Y of the antibody; - Variable region labelled
OR
Constant region labelled
OR
Disulfide bridge/bond labelled;
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]
- (Part of tick protein and alpha-gal) have a similar
shape/structure; - 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);
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]
- Exposure to tick (protein) is followed by increase
in antibody (specific to alpha-gal); - (Later) greater/faster increase in antibody
suggests there are memory cells; - Antibody (specific to alpha-gal) increases
during/after allergic reaction; - During/after allergic reaction, total antibody
increases more than alpha-gal antibody; - (So) may be other antibodies (that are causing
allergic reaction);
