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QUESTION 1

(a) Topic: 4.1 (Fluid mosaic membrane)

(b) Topic: 4.2 (Transport into and out of cells)

(a) Fig. 1.1 is a diagram representing part of the phospholipid bilayer of a cell surface membrane.

(i) Identify the part of a phospholipid molecule, labelled A in Fig. 1.1, that forms bonds with the phosphate heads and with the fatty acid tails.

(ii) Cholesterol is an important lipid component of many cell surface membranes. Fig. 1.2 shows the structure of a cholesterol molecule.

Using the information in Fig. 1.2, explain the orientation (positioning) of cholesterol molecules in the phospholipid bilayer, as shown in Fig. 1.1.

(iii) State one role of cholesterol in phospholipid bilayers.

(b) (i) Explain why sodium ions cannot cross phospholipid bilayers by simple diffusion.
(ii) Ions and some molecules move across cell surface membranes by facilitated diffusion and active transport.

Compare facilitated diffusion and active transport by stating one way in which they are similar and two ways in which facilitated diffusion is different from active transport.

(c) Prostaglandins are small lipids produced in many tissues of the body. One role of prostaglandins is to cause inflammation at the site of an injury or infection. Inflammation is the normal first response of the immune system to injury or infection.
Cyclooxygenase (COX) is an enzyme that catalyses one of the steps in the reaction pathway for the formation of prostaglandins from phospholipids. The reaction pathway occurs in the
smooth endoplasmic reticulum (SER) of cells. Part of the reaction pathway is shown in Fig. 1.3.

(i) Suggest an advantage for this reaction pathway occurring in the smooth endoplasmic reticulum of a cell rather than in the cytoplasm.
(ii) Sometimes inflammation can have side-effects, such as pain. Aspirin is a drug that can be used to reduce these side-effects.

Aspirin reduces the catalytic activity of the COX enzyme by modifying the R-group of one of the amino acids.

Suggest how modifying the R-group of an amino acid in the COX enzyme can reduce the catalytic activity of the enzyme.

(iii) Prostaglandins are examples of cell-signalling molecules.
Outline the process of cell signalling that leads to a response by the cells involved in inflammation.

▶️Answer/Explanation

Solution:-

(a)(i) glycerol ;
(ii) any one from:

hydroxyl / polar, group, interacts with, phosphate heads ;
A idea that both are polar / hydrophilic
A hydroxyl group faces aqueous environment; AW, as it is, polar / hydrophilic
non-polar part, in region of / AW, fatty acid tails / AW, as both are, non-polar / hydrophobic ;
A non-polar part is hydrophobic so in centre of membrane / away from aqueous environment

(iii) any one from:

maintains / regulates, fluidity of membrane ;
A detail e.g. reduces fluidity at high temperature / increases fluidity at low temperature / AW
maintains / regulates, (mechanical) stability of membrane ; AW
prevents entry of, hydrophilic substances / polar substances / ions ;
without cholesterol membranes would easily rupture
flat ring (structure) interferes with the movement of fatty acid tails
reduces lateral movement of phospholipids

(b)(i) sodium ions are

(positively charged), so repelled / AW, by the, hydrophobic tails / non-polar core / non-polar tails /AW ;

(ii) similarity: any one from
1 both occur through / involve, a, membrane / transport, protein ; A carrier protein if context is correct
2 (both can be) specific to the, molecule / ion (passing through) ; A correct ref. to binding site(s) on protein
3 (both involve) conformational change of (carrier) protein) ; R if incorrect context of channel protein
AVP ; e.g. can transport substances, into and out of cell / in both directions ;

1(c)(i) any one from:
SER (membrane-bound so) can provide, phospholipid / arachidonic acid / substrate ;
(prostaglandins are lipids and) SER is involved in lipid transport ;
prostaglandins can be transported (from SER) or stored (in SER) ;
idea of compartmentalisation / separated from other reactions in the cytoplasm / provides optimum conditions for
pathway / higher concentration of enzymes (of pathway) ;
(ii) any three from:

ref. to effect on, ionic bonding / hydrogen bonding / hydrophobic interactions, (with other R-groups) ;
changes the, shape / conformation, of the active site ;
active site no longer complementary to, arachidonic acid / substrate ;
ref. to effect of activation energy not being reduced ; e.g. ref to changed charges (so no electron transfer)
no longer provides hydrophobic regions for reaction to occur
enzyme–substrate complex, not formed / formed at reduced rate ;

(iii) any two from:

A ligands for prostaglandins if in correct context
1 prostaglandins are, secreted / released, by cells
or
prostaglandins are, transported / AW, to target cells ;
2 prostaglandins bind to receptors (on target cell surface membranes) ; R antigens
3 example of events triggered leading to a response ; e.g. activation of secondary messenger enzyme, cascade / activation phosphorylation events signal transduction

QUESTION 2

(a) Topic: 8.1 (The circulatory system) (b) Topic: 8.1 (The circulatory system) (c) Topic: 8.2 (Transport of oxygen and carbon dioxide)

(a) Table 2.1 shows descriptions of three types of white blood cell.
Complete Table 2.1 by stating the names of these three types of white blood cell.

(b) Dromedary camels are classified in the family Camelidae and live in desert habitats of North
Africa and Asia. In these hot, dry environments, dromedary camels can lose up to 30% of
their body mass from dehydration, causing their blood to become more viscous (thicker).
Fig. 2.1 shows a drawing of red blood cells of a dromedary camel. Fig. 2.2 is a drawing of human red blood cells.

Fig. 2.1 and Fig. 2.2 show differences between the red blood cells of dromedary camels and the red blood cells of humans.
Suggest how these differences adapt dromedary camels for living in hot, dry environments.
(c) The llama is also classified in the family Camelidae. Llamas live in mountainous areas of South
America, often at altitudes of 3500m or higher. As the altitude above sea level increases, the air pressure decreases.
The partial pressure of oxygen in the lungs of mammals at 3500m is 6.4kPa.
Fig. 2.3 shows the oxygen dissociation curve of adult human haemoglobin and adult llama
haemoglobin.

(i) With reference to Fig. 2.3, explain how the differences between the oxygen dissociation
curves for humans and llamas show that llamas are better adapted for living at high
altitudes than humans.

(ii) Sketch a curve on Fig. 2.3 to show the effect of an increased carbon dioxide concentration
on the percentage saturation of adult human haemoglobin with oxygen.
(iii) Explain the importance of the Bohr shift in metabolically active organs, such as the liver.

▶️Answer/Explanation

Solution:-

(a)

(b) allows, easier / quicker, flow / movement, of blood (compared to human red blood cells) ; plus any two explanations from :
lozenge / torpedo / elliptical / AW, shape of camel red blood cells (vs, biconcave disc of human red blood cells) ;
camel red blood cells are smaller (than human red blood cells) ; ora larger number of camel red blood cells (per unit volume) ;

(c)(i) any three from:

in context of better adapted for uptake in lungs (at high altitude) haemoglobin of llama more highly saturated with oxygen (than haemoglobin of human) ;
I ref. to higher saturation in partial pressures found in tissues at 6.4 kPa / 3500 m, oxygen saturation is, 86% for llama and 79 / 80% for human / 6 / 7% higher oxygen saturation for llama ;
llama haemoglobin has a higher affinity for oxygen than human haemoglobin ;
more oxygen, transported / delivered to tissues ; (so) sufficient oxygen unloaded in tissues to satisfy demand ;

(ii) curve drawn to the right of the human dissociation curve ;

(iii) any three from:

(metabolically active organs) release more carbon dioxide / partial pressure of carbon dioxide is higher ;
ref. to decreased affinity of haemoglobin for oxygen / explained ; e.g. ref to haemoglobinic acid
more oxygen released ; for more aerobic respiration / more ATP production ; look for ‘more’ once

QUESTION 3

(a) Topic: 7.1 (Structure of transport tissues) (b) Topic: 7.1 (Structure of transport tissues) (c) Topic: 2.2 (Carbohydrates and lipids)

(a) Fig. 3.1 is a photomicrograph showing part of a transverse section through the root of an iris,
Iris germanica. Irises are herbaceous monocotyledons. These plants have the same transport
tissues as herbaceous dicotyledons, but the transport tissues are distributed differently. In
monocotyledons, the central tissue in the root is parenchyma (packing tissue).

(i) Cells R, S and T in Fig. 3.1 are found in different tissues.
Name the tissues in which the cells labelled R, S and T are found.

tissue in which cell R is found……………………………………………………………………………….
tissue in which cell S is found……………………………………………………………………………………
tissue in which cell T is found………………………………………………………………………………………

(ii) Outline the role of the tissue in which cell R is found in Fig. 3.1.
(iii) State an example of an organic compound that is translocated in the root of an iris.
(b) The electron micrograph in Fig. 3.2 shows a section through some root cells in an onion, Allium cepa.

On Fig. 3.2, draw a label line and label it with the letter P to identify one plasmodesma.

(c) Table 3.1 contains information about four polysaccharides found in animals or plants.
Complete Table 3.1 by filling in the missing information.

▶️Answer/Explanation

Solution:-

(a)(i) R endodermis ; A endodermal (tissue)

S xylem ; T phloem ; accept ecf if S and T both have ‘vessels’ after naming

(ii) any two from:

to stop water moving through the apoplast / to force water movement to be symplastic / AW ;
so that water moves from the cell wall to the cytoplasm ;
AVP ; e.g. to allow control of substances into root (stele)
ref. to passage cells ref. to Casparian strip / suberin

(iii) sucrose / amino acid / peptide / polypeptide / protein / RNA / (named) plant hormone ;

A other named sugar / named amino acid / named protein / enzyme R minerals

(b) correct label to a plasmodesma ;

(c)

QUESTION 4

(a) Topic: 6.2 (Protein synthesis) (b) Topic: 5.1 (Replication and division of nuclei and cells) (c) Topic: 6.1 (Strucuture of nucleic acids and replication of DNA)

(a) Table 4.1 shows a sequence of 12 nucleotides in the template strand of a short length of a
DNA molecule, the corresponding primary transcript and the four amino acids coded for by
the sequence. The table is incomplete.
(i) Complete Table 4.1 to show the sequence of nucleotides in the primary transcript that

would result from transcription of this short length of DNA.

(ii) Table 4.2 shows all the possible template strand DNA triplets that code for the amino

acids labelled aa1, aa2, aa3 and aa4 in Table 4.1.

Complete Table 4.3 to identify the four amino acids labelled AA1, AA2, AA3, and AA4 in Table 4.1.

(iii) One type of gene mutation is caused by the substitution of a DNA nucleotide.
Using the information in Table 4.2, state and explain the effect on the final protein

structure of a substitution of the nucleotide at position 3 in Table 4.1.
(iv) A second type of gene mutation is caused by the deletion of a DNA nucleotide.
Using the information in Table 4.2, state and explain the effect on the final protein

structure of a deletion of the nucleotide at position 3 in Table 4.1.
(b) Replication of nuclear DNA occurs just once in every mitotic cell cycle. Six named events
associated with the mitotic cell cycle are listed. The events are not listed in any particular order.
Draw a circle around each event where replication of nuclear DNA occurs.

State the name of the part of the ATP molecule labelled A in Fig. 4.1.

▶️Answer/Explanation

Solution:-

(a)(i) all correct ;

(ii) all 4 are correct ;

(iii) any two from:

no effect on the protein structure ;
(because) all 4 triplets beginning with CA code for valine / CAA, CAG, CAT, CAC all code for valine ;
idea of the genetic code is, redundant / degenerate ;

(iv) any three from:

first amino acid, unchanged / still val ;
changes the reading frame / described ; A.all codons (from mutation on) will change / frameshift mutation
deletion alters, all amino acids after the mutation / amino acid sequence / primary structure ;
ref. to, stop codon, causing premature chain termination / leading to shorter polypeptide ;
(leading to) changes, in the tertiary structure / active site / AW ;

(b) interphase and S phase circled ; 1
(c) any four from: max three if points about transcription also given
1 double helix unwinds, qualified ; e.g. using, enzyme / helicase breaking hydrogen bonds between strands
2 both strands act as templates.
3 ref, activated (free DNA) nucleotides ;
4 DNA polymerase, plus example of role ; e.g. adds complementary nucleotides to exposed strand
forms phosphodiester bonds (between adjacent nucleotides)
proofreading / checking for errors / checking for mismatches

5 leading strand synthesised continuously / AW ;
6 lagging strand synthesised in (Okazaki) fragments ;
7 ligase connects (lagging strand), fragments / nucleotides, (with phosphodiester bonds) ;
8 semi-conservative replication / AW ; e.g. both new double helices have one, parental / conserved, strand and one newly synthesised strand

4(d) ribose ; I pentose

QUESTION 5

(a) Topic: 10.1 (Infectious diseases) (b) Topic: 18.1 (classification)

The pathogen that causes cholera is a prokaryote.
(a) Fig. 5.1 shows an electron micrograph of the pathogen that causes cholera.

(i) Name the type of electron microscope used to produce the image shown in Fig. 5.1.

(ii) Name the species of prokaryote that causes cholera.

(b) The passage contains a description of the main features of prokaryotic cells. There is one
factual error in the passage.

Prokaryotic cells are unicellular and generally between 1μm and 5μm in diameter.
Prokaryotes do not have organelles surrounded by double membranes. They do
have cell surface membranes, 70S ribosomes and a cellulose cell wall. The DNA of a
prokaryotic cell is circular and is found free in the cytoplasm rather than enclosed in a
nuclear envelope.

Identify and correct the factual error in the passage.

▶️Answer/Explanation

Solution:-

(a)(i) scanning (electron microscope) ;
(ii) Vibrio cholerae ;
(b) prokaryotic cell walls are made of peptidoglycan, made of peptidoglycan / made of murein / not made of cellulose ;

QUESTION 6

(a) Topic: 10.1 (Infectious diseases) (b) Topic: 11.1 (The immune system) (c) Topic: 10.1 (Infectious diseases) (d) Topic: 11.3 (Antibodies and antigens)

Fig. 6.1 is a simplified diagram representing a section through the human immunodeficiency virus
(HIV) particle that causes HIV/AIDS. The diagram shows the virus particle about to attach to
the cell surface membrane of a T-helper cell at a receptor protein called CD4. A second protein
(coreceptor) called CCR5 is also necessary for the virus particle to enter and then infect the T-helper cell.

(a) Identify structure X in Fig. 6.1.
(b) Explain how the ability of the immune system to resist the damaging effects of a pathogen is
affected by destruction of T-helper cells.
(c) Studies have shown that some individuals did not become infected with HIV even though
they were repeatedly exposed to the virus. Later discoveries indicated that these individuals
had a mutation in the gene for the CCR5 coreceptor protein.
Suggest how mutation of the gene for the CCR5 coreceptor protein provided protection against HIV infection.
(d) The use of monoclonal antibodies against the CCR5 coreceptor protein (anti-CCR5) has
been shown to be effective in the treatment of HIV infection.
Outline how anti-CCR5 monoclonal antibodies can be synthesised in the laboratory using the hybridoma method.

▶️Answer/Explanation

Solition:-

(a) capsid ; A capsomere or protein coat R protein / glycoprotein / caspid

(b) accept accounts based on primary and / or secondary immune response .
fewer cytokines released ; and any two from: fewer plasma cells, so fewer antibodies produced ; A B lymphocytes
fewer macrophages stimulated / less production of ‘angry’ macrophages / AW / less antigen presentation by macrophages ;
fewer T-killer cells stimulated to divide / less T-killer cells / less infected cells killed (by T-killer cells) ;
fewer memory cells (produced by the primary response) ; AVP ;

(c) virus cannot enter the T-helper cell / CCR5 unable to trigger endocytosis of viral particle ; 1
(d) any three from:
1 (named) small mammal, injected / AW, with CCR5 ; A antigen
2 immune response occurs (over several weeks) ; A immune response described
3 plasma cells / B-lymphocytes / B-cells / splenocytes, extracted from spleen ;
4 plasma cells / B-lymphocytes / B-cells, fused with, myeloma / tumour / cancer / AW, cells (to form hybridomas) ;
5 screening / selection / AW, for hybridomas producing desired, (monoclonal) antibodies ;
6 AVP ; e.g. hybridoma cells separated (into wells) to produce clones ref. to large scale production

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