2023 Nov P2 answers

Question 1

(a) With reference to Fig. 1.1, explain the effect of increasing the external concentration of
glucose on the rate of uptake of glucose into a cell.[3]
1. As external concentration of glucose increases, the rate of glucose uptake into the cell
increases steeply at first, but the increase slows down and eventually reaches maximum rate
at a plateau.
2. As glucose is polar*, and hence hydrophilic, glucose molecules would be repelled by the
hydrophobic core of the phospholipid bilayer.
3. The entry of glucose requires transmembrane carrier proteins, but at high external glucose
concentration, the carrier proteins become saturated and become a limiting factor and hence
the rate of uptake plateaus.
4. As the external concentration increases, the concentration gradient is steeper and hence the
rate of uptake by facilitated diffusion increases.

(b) Explain why there is no uptake of glucose into the cell at X even though glucose is present
outside the cell.[1]
1. There is no diffusion gradient between the cytosol and the solution outside the cell as
the concentration of glucose in the cell and outside the cell is the same.

(c) With reference to Fig. 1.2, explain the flip-flop mechanism for the transport of fatty acid
molecules across the cell surface membrane, including the role of hydrogen ions. [5]
1. Positively charged H+ ions surround the negatively charged carboxyl groups to reduce
the negative charge and allow the fatty acid molecule to be embedded on the top half of
the phospholipid bilayer;
2. The fatty acid molecule is orientated in such a way that the hydrophobic non-polar
hydrocarbon tail is embedded between the hydrophobic non-polar hydrocarbon tails of
the phospholipid in the top half of the phospholipid bilayer;
3. And the negatively charged carboxylic acid group is between the charged phosphate
head of the phospholipid molecules;
4. Positively charged H+ ions then bind to the negatively charged carboxylic acid group to
neutralise the negative charge so that the non-polar fatty acid molecule can flip within
the hydrophobic core of the phospholipid bilayer without being repelled;
5. Once the non-polar fatty acid molecule is embedded at the lower half of the phospholipid
bilayer, the H+ ions dissociate from the carboxyl group that now orientates outwards
towards the internal cytosolic side of the cell;

[Total: 9]
Question2

(a) Explain how the change in activation energy shown in Fig. 2.1 affects the rate of a catalase-
controlled reaction.[2]
1. Enzyme lowers activation energy* by forming 3 different enzyme-substrate complexes
required for chemical reaction to take place, from 75kJ mol-1 without enzyme to 28 kJ mol-
1
with enzyme;
2. Increases number of substrate molecules with the required energy to cross the activation
energy* barrier so the reaction can proceed faster.

(b) (i) With reference to Fig. 2.2, state what can be concluded from the results of the
investigation.[3]

1. Lower temperature of 35ºC is the most ideal for storage compared to higher
temperature (40ºC to 50ºC);
2. Catalase stored at 35ºC retained 72% activity after 1.6 hours while higher
temperatures had lower activity (quote from any other temperature);
3. Catalase lost its activity the fastest when stored at higher temperatures (quote
relative activity or reference to gradient from 0.0 to 0.2 hours);
4. Within 1.6 hours, storage at temperatures at or above 45ºC resulted in loss or
almost complete loss of activity, but 35 ºC and 40 ºC still retained 50% to 70%
activity;
5. AVP; Activity decreased when stored at all temperatures, indicating that storage
is not idea for catalase;

(c) Explain the effect of storage temperature on the activity of catalase, as shown in Fig.
2.2. [5]
1. the higher the storage temperature, the greater the amount of thermal agitation /
intramolecular vibrations;
2. at temperatures beyond the optimum temperature of an enzyme, weak
interactions such as hydrogen bonds, ionic bonds and hydrophobic interactions
between R groups will break;
 3. the 3D conformation will be lost, resulting in denaturation*
4. the active site* of catalase is no longer complementary in shape* and charge
to the substrate and the activity decreases;
5. enzymes exposed to increased temperatures for longer durations will have lower
activity;
[Total: 10]

Question 3

(a) Name the parts of Fig. 3.2 labelled B, C and D. [3]

B : peptidoglycan cell wall

C : cytoplasm
D : transport channel protein

(b) With reference to Fig. 3.2, suggest how environmental DNA fragments enter the bacterial
cell. [3]
1. Double-stranded DNA passes through the peptidoglycan cell wall and binds to protein
A;
2. which is attached to a membrane bound transmembrane channel protein at the outer
surface of the phospholipid bilayer of the bacterial cell membrane;
3. The hydrogen bonds between the 2 strands of the DNA are broken and only a single
DNA strand enters the cytosol via the channel protein;

(c) Two other ways in which DNA can enter bacterial cells are transduction and conjugation.
Describe how DNA enters bacterial cells by transduction and conjugation. [4]

transduction
General transduction
1. A phage* infects a bacterium*, injecting its viral genome into the host cell
The bacterial DNA is degraded into small fragments, one of which may be randomly
packaged into a capsid* head during the spontaneous assembly* of new viruses;
2. Upon cell lysis, the defective phage will infect another bacterium and inject bacterial
DNA from the previous host cell into the new bacterium;
3. The foreign bacterial DNA can replace the homologous region in the recipient cell’s
chromosome if crossing over/homologous recombination* takes place, possibly
allowing the expression of a different allele from the previous host;
OR
Specialised transduction
4. A temperate phage* infects a bacterium*, injecting its viral genome into the host cell
The viral DNA is integrated into bacterial chromosome forming a prophage* which may
be improperly excised to include adjacent segment of bacterial DNA during an
induction* event;
5. Bacterial DNA may be packaged into a capsid head during the spontaneous assembly*
of new viruses;
6. Upon cell lysis, the defective phage will infect another bacterium and inject bacterial
DNA from the previous host cell into the new bacterium;
 7. The foreign bacterial DNA can replace the homologous region in the recipient cell’s
chromosome if crossing over/homologous recombination* takes place, possibly
allowing the expression of a different allele from the previous host;

Conjugation
8. Sex pilus* of F+ bacterial cell* makes contact with a F- bacterial cell* and retracts to
bring the F- cell closer so a mating bridge* is formed between the 2 cells;
9. One of the 2 strands of the plasmid DNA in F+ cell is nicked and transferred from the
F+ cell to the F- cell through mating bridge (via rolling circle mechanism) as the other
DNA strand is used as a template for elongation;
10. The single strand F plasmid DNA circularises in F- cell and is used as a template* to
synthesise a complementary strand for a double-stranded F plasmid DNA resulting in
F+ cell;

[Total: 10]
Question 4
(a) (i) Use the equation to calculate the melting temperature at which this primer DNA
sequence separates from template DNA:

TCGACTTCCTCGMCC

Tm = 64.9 + [41 x (2 + 7-16.4) / (3 + 4 + 2 + 7)]

= 64.9 – 18.96
= 45.9 ºC
melting temperature = 45.9 ºC [1]

(ii) Use your knowledge of DNA structure to suggest why an increased proportion of bases
containing cytosine and guanine will increase the melting temperature. [2]

1. Base pair cytosine and guanine has 3 hydrogen bonds compared to 2 hydrogen bonds
between adenine and tyrosine
2. Increases in H bonds allow the DNA to be more heat stable and higher temperature is
needed to increase the kinetic energy of the molecule to break more H bonds

(d) Suggest the consequence for the results of a polymerase chain reaction of using low
annealing temperatures. [1]

1. Less DNA product will be produced at the end of polymerase chain reaction
or
2. Primers may bind to sequences that are not perfectly complementary, leading to the
amplification of unintended DNA fragments

(b) Describe the advantages of using the polymerase chain reaction.[3]

1. PCR is sensitive as only a minute amount of source DNA is required to amplify a large
amount of DNA products
2. By using specific primers, PCR can selectively amplify a particular segment of DNA
3. PCR is fully automated within the thermocycler.
4. It can amplify millions of targeted DNA segment rapidly and efficiently

(c) State the limitations of the polymerase chain reaction.[3]

 1. Taq polymerase lacks proofreading ability as errors occurring early in PCR reaction will
get compounded with each replication cycle
2. Success of PCR requires knowledge of sequences flanking target region to be amplified.
3. DNA fragments to be amplified are limited to about 3 kb. Further increase in length of
target sequence decreases efficiency of amplification.
4. Minute amounts of contaminant DNA may result in unwanted DNA sequences being
amplified to significant amounts.

[Total: 10]

Question 5

(a) Name the stages of meiosis shown in Fig. 5.1. [2]

stage P Metaphase I

stage Q Anaphase I

stage R Metaphase II

stage S Anaphase II

(b) Describe similarities and differences between stage Q and stage S. [3]

similarities
1. Both stages involve shortening kinetochore microtubules,
2. Pulling chromosomes to opposite poles* of the cell;

differences
3. Stage Q does not involve division of centromere, whereas in stage S centromeres
divide;
4. Stage Q involves separation of homologous chromosomes*, whereas stage S
involves separation of sister chromatids*;

(c) (i) Name this event and the stage of meiosis in which it occurs.[2]

event Crossing over

stage Prophase I

(c) (ii) Explain why stage R does not generate genetic variation in the absence of this event [1]
Sister chromatids would be genetically identical without crossing over.
(d) Complete Fig. 5.2 to show the chromosomes and alleles that are present in the daughter cell
from the first meiotic division, the first polar body and the second polar body.

You should assume that the event named in (c)(i) has not occurred. [2]

Fig. 5.2

[Total: 10]
Question 6
(a) State the coat colour shown by labrador retrievers with the following genotypes.[1]
eeBb .............yellow..................................................
EeBb ............black...................................................

(b) Explain the coat colour of labrador retrievers with the genotype Eebb. [3]

1. Eebb = brown coat colour

2. Allele E of one gene codes for a functional enzyme E, which converts the yellow coat
pigment to brown coat pigment
3. With Ee, there is still one functional allele to convert yellow to brown
4. If Allele B of another gene is present, it codes for enzyme B, which converts the brown coat
pigment to black coat pigment
5. Since the genotype is bb, there is no functional enzyme B thus resulting in Eebb having
brown coat colout

(c) Draw a genetic diagram to explain the results of this cross. Indicate the expected phenotypic
ratio of the offspring.
Use the symbols B, b, E and e to represent the alleles. [5]

Parental phenotype Black labrador X black labrador

Parental genotype BbEe X BbEe

Correct parental phenotype + genotype [1]

Correct gametes [1]

Meiosis
Gametes BE Be bE be X BE Be bE be

Random fertilization
Male gametes

BE Be bE be

BE
BBEE BBEe BbEE BbEe
black black black black

Be BBEe BBee BbEe Bbee

Femaleegametes

black yellow black yellow

bE BbEE BbEe bbEE bbEe

black black brown brown

BbEe Bbee bbEe bbee

lack yellow brown yellow
be

First generation genotypic ratio [1] 9 B_E_: 3 bbE_ : 4 _ _ ee

Black brown yellow
First generation phenotypic ratio 9 black: 3 brown : 4 yellow
(d) Name the type of gene interaction shown in Fig. 6.1. [1]
Epistasis [Total: 10]
Question 7

(a) With reference to Fig. 7.1, explain the movement of protons into the intermembrane space
using the large protein complexes. [4]
1. Reduced NAD transfer their high energy electrons to the electron acceptors/first electron
carrier of the electron transport chain*, and get re-oxidised in the process to NAD;
2. As electrons are passed down electron carriers/large protein complexes of the electron
transport chain* arranged in linear sequence of increasing electronegativity;
3. the energy released is coupled to the pumping of H+ from the matrix* into the intermembrane
space* to generate a proton motive force/proton gradient;
4. The hydrophobic core* of the phospholipid bilayer of the double membrane of the
mitochondrion is impermeable to H+ which are charged and thus hydrophilic, preventing H+
ions from passing through;

(b) Explain the role of oxygen in oxidative phosphorylation.[3]

1. Oxygen acts as a final electron acceptor* at the end of the electron transport chain*
(ETC) and combines with H+ and electrons to form water* in the matrix;
2. By removing electrons, oxygen reoxidises the ETC so that NADH* and FADH2* can
continue to donate electrons to the chain, and maintain the flow of electrons along the
ETC, thereby allowing oxidative phosphorylation* to continue to produce ATP*;
3. In the absence of oxygen, the coenzymes NAD* and FAD* cannot be regenerated to
accept more H atoms from link reaction and Krebs cycle, and so these reactions and
oxidative phosphorylation eventually stop.

(c) Compare the role of reduced NAD in aerobic respiration with the role of reduced NADP in
photosynthesis. [3]

Roles Reduced NAD Reduced NADP

Formation Reduced NAD is formed from NAD Reduced NADP is formed from
during oxidation by dehydrogenation; reduction of NADP when combined
with H+ and electron as catalysed by
NADP reductase;
Usage Reduced NAD donates high energy Reduced NADP serves as reducing
 electron to the electron transport power by reducing glycerate-3-
chain during oxidative phosphate (GP)/ phosphoglyceric acid
phosphorylation*; (PGA) to form glyceraldehyde-3-
phosphate (G3P) during Calvin
cycle*;
Role upon Reduced NAD regenerates to form Reduced NADP regenerates to form
regeneration NAD which are used in glycolysis, NADP which is used as final electron
of their link reaction and Krebs cycle, acceptor during non-cyclic
oxidised allowing aerobic respiration to photophosphorylation allowing the
coenzyme continue; light-dependent reaction of
photosynthesis to continue;

[Total: 10]
Question 8

(a) With reference to Fig. 8.1, explain how the molecular structure of the G-protein linked
receptor relates to its function. [5]

1. G protein linked receptor (GPCR) a seven pass transmembrane protein consisting of 7 α-

helices* connected by three intracellular and three extracellular peptide loops.
2. The intracellular domain/cytoplasmic side of GPCR has a G protein binding site* that
allows binding of a G protein.
3. The extracellular loops has a ligand binding site* at which a specific* signaling molecule
can bind to the GPCR.
4. When a ligand binds to the ligand binding site at the extracellular side of a GPCR it causes
a conformational change* of the intracellular domain /at the cytoplasmic side of the
GPCR,
5. The activated GPCR can then activate an associated G protein by binding to it and
triggering an exchange of its bound GDP for a GTP.
6. As a transmembrane protein that is embedded in a cell's plasma membrane, it is folded
such that amino acid residues with the hydrophobic R groups are interacting with the
hydrophobic core of the phospholipid bilayer of the plasma membrane and amino acids
with hydrophilic R groups are arranged inwards as well as at the surfaces where they
interacts with the aqueous exterior medium/cyplasmic and extracellular
regions/hydrophilic phosphate heads of phospholipid bilayer.

(b) Explain how the release of cAMP leads to a cellular response. [5]

1. cAMP binds and activates protein kinase A* (PKA);

2. Activation of PKA will initiate a sequential activation of kinases resulting in a
phosphorylation cascade;
3. That eventually activate glycogen phosphorylase*
4. Signaling amplification occurs where number of activated product is always greater
than those in preceding step example phosphorylase kinase can phosphorylate and
activate many molecules of glycogen phosphorylase;
 5. A large number of activate glycogen phosphorylases break down glycogen to generate
a high yield of glucose;
6. During the cellular response, there will also be a decrease in the rate of glycolysis,
AVP an increase in the rate lipid and protein breakdown (gluconeogenesis);
AVP ref to upregulation expression of genes e.g. glycogen phosphorylase gene
AVP ref to down regulation of expression genes e.g. glycogen synthase gene

[Total: 10]

Question 9
(a) With reference to Fig. 9.1, suggest how these two species of palm tree may have evolved
by sympatric speciation. [5]

1. An ancestral* palm grew on Lord Howe Island, an isolated island off the coast of
Australia, that had two soil types in close proximity to each other - older volcanic soil
and younger calcareous soils;

2. When the palms that normally grew on volcanic soil started to grow on calcareous soil,
a flowering time difference arose as a physiological response to growing on a different
soil type;

3. Thus physiological isolation* prevented interbreeding* and hence disrupted gene

flow* between the two sub-populations of palms growing in the two types of soil
although they were in close proximity;

4. This resulted in evolutionary changes occurring independently within each sub-

population. i.e. different genetic changes from accumulation of mutations*, as well as
changes in allele frequencies through genetic drift* and natural selection* occurred
within each sub-population;

5. There existed variation in the two sub-populations and palms with favourable traits that
were better adapted had a selective advantage to the specific soil conditions and were
selected for, increasing frequency of favourable alleles and survived, reproduced and
passed on their alleles to the next generation;

6. Over hundreds and thousands of successive generations each sub-population

became genetically distinct, reproductively isolated* species, Howea forsteriana and
Howea belmoreana which are unable to interbreed* to form fertile*, viable* offspring;

(b) A student concluded that Fig. 9.2 shows that Lord Howe Island was originally colonised
by a species of plant that was a common ancestor for Howea, Laccospadix and
Linospadix. Explain whether or not this conclusion is correct. [2]

1. According to Fig. 9.2, the common ancestor for Howea, Laccospadix and
Linospadix existed about 11.5 million years ago;
2. From the information given in the question, Lord Howe island was only formed 6.9
million years ago and so there could not have been a species of plant that
 colonised Lord Howe island, that was a common ancestor of Howea, Laccospadix
and Linospadix;

(c) State the types of evidence that can be used to establish the phylogenetic relationships
between species. [3]
1. DNA, RNA (nucleotide) sequences and amino acids sequences;
2. of homologous genes;
3. or homologous anatomical structures can be used to establish phylogenetic
relationships between species;
4. from fossils;
[Total: 10]

Question 10

(a) With reference to Fig. 10.1, describe the differences in the immune response of infected
people with mild symptoms and infected people with severe symptoms.[3]

T cells
1. (appearance of the T cells) For people with mild symptoms, the T cells were circulating
in the blood earlier from the 2nd day but for the people with severe symptoms, T cells
were only circulating in blood only on the 12th day.
2. (highest level of T cells) For people with mild symptoms there was generally more T
cells, where the highest level of T cells were circulating in the blood occurred on day 7
and the level was 8x that of the highest level of T cells for people with mild symptoms
that started from day 23
3. (General trend) For people with mild symptoms, the T cell numbers increased from day 2
to the highest level on day 7 and then decreased gradually to the 30 th day, but for people
with severe symptoms, the T cells increased from zero on day 12 to the highest level on
day 23 and remained at constant levels up to day 30.

Antibodies
4. (appearance of antibodies) There was earlier appearance of antibodies for those with
severe symptoms from day 11, but for those with mild symptoms antibodies were
circulating 3 days later on day 14.

(b) Suggest and explain what vaccination for this virus needs to achieve in order to give long-
term protection against severe symptoms. [2]

1. Vaccination exposes the individual to a harmless form of a pathogen to induce a specific

adaptive immune response resulting in formation of memory T and B cells*;
2. Memory B and T cells* when exposed to the same virus, will recognise it and mount a
faster and stronger and longer lasting secondary immune response that can occur before
the 11th day for both T cells and production of antibodies;
 3. Earlier appearance of the cytotoxic T cells will allow removal of virus-infected cells by
apoptosis through action of perforins and granzymes
Or
Earlier production of antibodies by plasma B cells will allow the circulating virus particles
to be neutralised and removed by phagocytes through agglutination and opsonization
➔ Prevent viruses from replicating and causing severe symptoms
[Total: 5]

Question 11

(a) Describe the changes shown in Fig. 11.1 from 1802 to 2012. [2]

1. In 1802, glaciers can be found at height of 4820 m but it 2012, the glaciers receded and
can only be found at higher altitude from 5260 m.
2. In 1802, vegetation was growing from 4400 to 4600 m, but in 2012, vegetation extended
their range from 4400 to 5120 m.

(b) Suggest how human activity may account for the changes shown in Fig. 11.1. [4]

1. Human activity such as (any one example) increase combustion of fossil fuels for
electricity/industrial processes, such as in cement works/burning land and vegetation to
clear land for agriculture etc releases greenhouse gases (GHG) like carbon dioxide and
methane (any one example)
2. Increase Greenhouse gases (GHG) allow short-wave radiation from the sun to pass
through to heat the Earth’s surface which radiate out-going long-wave radiation/infra
red/heat which is absorbed and re-emitted by the GHGs back to the Earth’s surface;
3. This leads to increased trapping of radiation/heat resulting in warmer temperatures at
higher altitudes
4. Resulting in the melting of glaciers at lower altitudes and more favourable (warmer)
temperatures at higher altitudes for vegetation to grow.

[Total: 4]