June 2021 QP - Paper 2 AQA Biology AS-level.PDF Flashcards
In taxonomy, an organism is identified by referring to the species name and the genus
name.
What term is used to describe this method of naming organisms?
Bionomial
0 3 . 2 Define the term mutagenic agent.
03.2 (A factor that) increases (the rate of) mutations; 1
0 3 . 4 Name the type of mutation that changed the number of chromosomes in S. townsendii
to produce S. anglica. Explain your answer.
- Non-disjunction;
Explanation - (In) meiosis;
- Chromosomes not separated
OR
All chromosomes stay in one cell
OR
Chromosomes do not form (homologous) pairs;
Genetic variation within a species is increased during meiosis by crossing over and the independent segregation of homologous chromosomes.
Apart from mutation, explain one other way genetic variation within a species is
increased.
- Random fusion of gametes
OR
2 - Accept for
‘gametes’,
MARK SCHEME – AS BIOLOGY – 7401/2 – JUNE 2021
9
Random fertilisation; - (Produces) new allele combinations
OR
(Produces) new maternal and paternal chromosome
combinations;
box 0 4 . 1 Give two structures found in all prokaryotic cells and in all eukaryotic cells.
- Cell(-surface) membrane;
- Ribosomes;
- Cytoplasm;
- DNA
All prokaryotic cells contain a circular DNA molecule and some prokaryotic cells
contain plasmids.
0 4 . 2 Scientists have found that the rate of plasmid replication is faster in cells growing in a
culture with a high concentration of amino acids than in a culture with a lower
concentration of amino acids.
Suggest one explanation for the faster rate of plasmid replication in cells growing in a
culture with a high amino acid concentration.
. (Amino acids used in) protein synthesis;
2. (So) more enzymes (for DNA/plasmid
replication)
OR
(So) more DNA polymerase;
3. (Amino acids used in) respiration;
4. (So) more energy/ATP (for DNA/plasmid replication
A scientist prepared a culture of a bacterial species. box
* She extracted the plasmids and the circular DNA molecules from a sample of cells
taken from this culture (A).
* She then added antibiotic X to the culture and let the cells divide for 4 hours.
* She then extracted the plasmids and the circular DNA molecules from a sample of
these cells (B).
* The scientist separated the plasmids from the circular DNA molecules in A and in B
using ultracentrifugation.
Figure 6 shows her results.
Figure 6
0 4 . 3 What can you conclude from Figure 6 about a structural difference between the
plasmids and the circular DNA? Explain your answer
Circular DNA is bigger/heavier/denser;
2. (Because band) moved further/is lower (in
tube)/closer to bottom (of tube);
What can you conclude from Figure 6 about the effect of antibiotic X on plasmid
replication and on circular DNA replication? Explain your answer
- Plasmid replication continues/increases (with X)
as band is wider; - Circular DNA replication stops/not increased
(with X) as band is identical;
A student investigated the activity of the enzyme amylase. He cut three identical wells box
(D, E and F) in starch-agar in a Petri dish. He added 0.2 cm3
of:
* amylase solution to well D
* boiled amylase solution to well E
* water to well F.
After 60 minutes, he covered the starch-agar with iodine solution.
Figure 7 shows his results.
Figure 7
0 5 . 1 Explain the appearance of the agar in the clear area surrounding well D.
Amylase hydrolyses starch;
2. (To) maltose;
What can you conclude about the activity of amylase from the appearance of the agar
surrounding well E and well F in Figure 7?
- (E) Amylase/enzyme is denatured;
- (F) amylase is needed for/causes starch
hydrolysis/breakdown/digestion
OR
(F) water (alone) does not (cause starch)
hydrolysis/breakdown/digestion;
The student cut out a piece of agar from the clear area surrounding well D. He
obtained a solution of the substances contained in this piece of agar.
Describe a different biochemical test the student could use with this solution to
confirm that amylase had affected the starch in the clear area surrounding well D
- Heat in Benedict’s (solution);
- Red/green/orange (precipitate/colour) (shows
maltose/reducing sugar);
The student used a ruler to measure the diameter in mm of the clear area around
well D in Figure 7.
Use this information to explain why the answer to Question 05.4 should be given to
the nearest whole number
- Reduces (human) error/uncertainty;
- (It is) the resolution of a ruler;
- (For a ruler measurement) the uncertainty is
±1(mm)
The fruit fly is a species of small insect.
The fruit fly has a gene that codes for an enzyme called alcohol dehydrogenase (AD).
AD catalyses the breakdown of alcohol when alcohol is in the insects’ food.
The gene coding for AD has two alleles, ADF and ADS.
0 6 . 1 The enzyme encoded by the ADF allele catalyses the breakdown of alcohol faster
than the enzyme encoded by the ADS allele. Suggest why
- Different primary structure/amino acid sequence;
- Different tertiary structure/shape of active site;
- Enzyme-substrate complexes more likely (with
enzyme from ADF
allele);
A scientist took a random sample of adult fruit flies from a population. He measured
the frequency of the ADF allele in this sample (generation 0). He then:
* selected 100 of these insects at random and kept them in a container
* fed the insects food containing alcohol
* let the insects reproduce
* repeated these steps for 45 generations of fruit fly reproduction.
The scientist measured the frequency of the ADF allele in the 45th generation.
0 6 . 2 Suggest why the scientist took his sample from the population at random.
Avoids bias
OR
Results (likely to be) reliable/repeatable;
Alcohol is toxic to fruit flies. Suggest and explain why the frequency of the ADF allele
changed during the 45 generations.
- Flies with ADF
/allele have selective advantage
(in presence of alcohol); - So insects (with ADF
more likely to) reproduce; - Pass on ADF (allele/gene);
- (So) allele frequency increases;
