AS June 2016 paper 1 Flashcards
When HIV infects a human cell, the following events occur.
• A single-stranded length of HIV DNA is made.
• The human cell then makes a complementary strand to the HIV DNA.
The complementary strand is made in the same way as a new complementary
strand is made during semi-conservative replication of human DNA.
Describe how the complementary strand of HIV DNA is made.
[3 marks]
- (Complementary) nucleotides/bases pair
OR
A to T and C to G; - DNA polymerase;
- Nucleotides join together (to form new
strand)/ phosphodiester bonds form;
Contrast the structures of DNA and mRNA molecules to give three differences.
[3 marks]
1. DNA double stranded/double helix and mRNA single-stranded; 2. DNA (very) long and RNA short; 3. Thymine/T in DNA and uracil/U in RNA; 4. Deoxyribose in DNA and ribose in RNA; 5. DNA has base pairing and mRNA doesn’t/ DNA has hydrogen bonding and mRNA doesn’t; 6. DNA has introns/non-coding sequences and mRNA doesn’t;
1 Describe the difference between the structure of a triglyceride molecule and the
structure of a phospholipid molecule.
[1 mark]
. In phospholipid, one fatty acid
replaced by a phosphate;
Describe how you would test for the presence of a lipid in a sample of food.
[2 marks]
- Add ethanol, then add water;
2. White (emulsion shows lipid);
This fat substitute cannot be digested in the gut by lipase.
Suggest why.
[2 marks]
1. (Fat substitute) is a different/wrong shape/not complementary; OR Bond between glycerol/fatty acid and propylene glycol different (to that between glycerol and fatty acid)/no ester bond; 2. Unable to fit/bind to (active site of) lipase/no ES complex formed;
This fat substitute is a lipid. Despite being a lipid, it cannot cross the cell-surface
membranes of cells lining the gut.
Suggest why it cannot cross cell-surface membranes.
[1 mark]
It is hydrophilic/is polar/is too large/is
too big
Cells constantly hydrolyse ATP to provide energy.
Describe how ATP is resynthesised in cells.
[2 marks]
- From ADP and phosphate;
- By ATP synthase;
- During respiration/photosynthesis;
Give two ways in which the hydrolysis of ATP is used in cells.
[2 marks]
1. To provide energy for other reactions/named process; 2. To add phosphate to other substances and make them more reactive/change their shape;
Y is a protein. One function of Y is to transport cellulose molecules across the
phospholipid bilayer.
Using information from Figure 3, describe the other function of Y.
[2 marks]
1. (Y is) an enzyme/has active site/forms ES complex; 2. That makes cellulose/attaches substrate to cellulose/joins β glucose; OR 3. Makes cellulose/forms glycosidic bonds; 4. From β glucose;
Scientists investigated the hydrolysis of sucrose in growing plant cells by an
an enzyme called SPS.
Name the products of the hydrolysis of sucrose. [2 marks]
- Glucose;
2. Fructose
What can you conclude about the growth of the plant cells from these data?
Explain how you reached your conclusions. [3 marks]
1. Sucrose hydrolysis linked to some aspect of growth; 2. Greater the rate of/faster hydrolysis/more SPS activity as plant grows/cells divide (up to 8/10 days); 3. Growth/division remains the same/slows after 8/10 days (because SPS activity is levelling off);
Describe the induced-fit model of enzyme action.
[2 marks]
1. (before reaction) active site not complementary to/does not fit substrate; 2. Shape of active site changes as substrate binds/as enzymesubstrate complex forms; 3. Stressing/distorting/bending bonds (in substrate leading to reaction);
Describe how the scientist would have produced the calibration curve and used
it to obtain the results in Figure 4.
Do not include details of how to perform a Benedict’s test in your answer.
[3 marks]
1. Make/use maltose solutions of known/different concentrations (and carry out quantitative Benedict’s test on each); 2. (Use colorimeter to) measure colour/colorimeter value of each solution and plot calibration curve/graph described; 3. Find concentration of sample from calibration curve;
The human papillomavirus (HPV) is the main cause of cervical cancer. A vaccine
has been developed to protect girls and women from HPV.
Describe how giving this vaccine leads to the production of antibodies against HPV.
[4 marks]
1. Vaccine/it contains antigen (from HPV); 2. Displayed on antigen-presenting cells; 3. Specific helper T cell (detects antigen and) stimulates specific B cell; 4. B cell divides/goes through mitosis/forms clone to give plasma cells; 5. B cell/plasma cell produces antibody;
What do these results suggest about whether it is better to give two or three
doses of the vaccine? Give reasons for your answer.
[2 marks]
2 1. Two (doses) because got more antibody; 2. With three doses, second dose/dose at 1 month doesn’t lead to production of any more antibody (than the two dose group)/get same/similar response; 3. Three doses would be more expensive/less popular with parents/girls (and serves no purpose);
The doctors carried out a statistical test to determine whether the antibody
concentrations were significantly different in girls given two doses of the vaccine,
compared with those given three doses. They determined the mean
concentrations of antibody 9 months after the first dose of vaccine.
What statistical test should the doctors have used? Give the reason for your
choice.
3 t-test, because comparing two means;
There is genetic diversity within HPV.
Give two ways doctors could use base sequences to compare different types of
HPV.
[2 marks]
- Compare (base sequences of) DNA;
- Look for mutations/named mutations
(that change the base sequence); - Compare (base sequences of)
(m)RNA;
Suggest why the development of a monopolar mitotic spindle would prevent
successful mitosis.
[2 marks]
1. No separation of chromatids/chromosomes/centromeres; 2. Chromatids/chromosomes all go to one pole/end/sides of cell/not pulled to opposite poles; 3. Doubles chromosome number in cell/one daughter cell gets no chromosomes or chromatids;
Scientists investigated the effect of different concentrations of a kinesin inhibitor
(KI) on mitosis of human bone-cancer cells grown in culture. A student who saw these results concluded that in any future trials of this kinesin
inhibitor with people, a concentration of 100 nmol dm–3 would be most
appropriate to use.
Do these data support the student’s conclusion? Give reasons for your answer.
[4 marks]
1. (No, because) at 100 there are still some (7%) cancer cells dividing/undergoing mitosis; 2. So, cancer not destroyed/may continue to grow/spread/form tumours; 3. Best concentration may be between 100 and 1000/need trials between 100 and 1000; 4. This research in culture, don’t know effect of KI on people; 5. (Yes, because) above 100 produces little increase in % of cells not dividing/undergoing mitosis/at 100, most (93%) cancer cells unable to divide/dead; 6. Above 100 may be harmful (to body); 7. Higher concentrations more expensive; 8. (above 100) will have more effect on (rapidly dividing) cancer cells;
Suggest how amyloid-precursor protein can be the substrate of two different
enzymes, α-secretase and β-secretase (lines 3–5).
[2 marks]
1. Different parts/areas/amino acid sequences (of amyloid-precursor) protein; 2. Each enzyme is specific /fits/binds/ complementary to a different part of the APP;
One product of the reaction catalysed by β-secretase is a smaller protein
(lines 6–7).
Describe what happens in the hydrolysis reaction that produces the smaller
protein from amyloid-precursor protein.
[2 marks]
- Peptide bond broken;
2. Using water;
Many people with Alzheimer’s disease have mutations that decrease
α-secretase production, or increase β-secretase production (lines 8–9).
Use the information provided to explain how these mutations can lead to
Alzheimer’s disease.
[3 marks]
1. Mutations prevent production of enzyme(s)/functional enzyme; 2. (Increase in β-secretase) leads to faster/more β-amyloid production OR (Decrease in α-secretase) leads to more substrate for β-secretase; 3. (Leads to) more/greater plaque formation;
One possible type of drug for treating Alzheimer’s disease is a competitive
inhibitor of β-secretase (lines 10–11).
Explain how this type of drug could prevent Alzheimer’s disease becoming
worse.
[2 marks]
- (Inhibitor) binds to/blocks active site of
β-secretase/enzyme; - Stops/reduces production of βamyloid/plaque;
When some of these types of drugs were trialled on patients, the trials were
stopped because some patients developed serious side effects (lines 11–13).
Using the information provided, suggest why some patients developed serious
side effects.
1. Some β-amyloid required/needed (to prevent side effects) OR (Some) β-secretase needed; 2. Leads to build-up of amyloid-precursor protein (that causes harm) OR Too much product of α-secretase (causes harm);
