1. Fig. 1.1 is a drawing of a photomicrograph of a spongy mesophyll cell from a leaf.
(a) On Fig. 1.1, add a label line and the correct letter for each of the three cell structures listed.
• nucleolus = N
• tonoplast = T
• chloroplast = C [3]
(b) The drawing in Fig. 1.1 is 2000 times larger than the actual size of the cell. Describe the steps you would follow to determine the actual diameter of the cell in micrometres (µm), at X—Y. ………………………………………………………………………………………………………………………………… ………………………………………………………………………………………………………………………………… ………………………………………………………………………………………………………………………………… ………………………………………………………………………………………………………………………………… ……………………………………………………………………………………………………………………………..[4]
(c) The drawing in Fig. 1.1 was made using the high-power objective lens of a light microscope. Some of the structures in Fig. 1.1 confirm that the cell is eukaryotic.
An electron micrograph of the same cell would reveal additional cell structures that are found in eukaryotes and not in prokaryotes.
List two examples of these additional cell structures.
1 ………………………………………………………………………………………………………………………………
- ……………………………………………………………………………………………………………………………..[2]
2. In 1953, James Watson and Francis Crick published details about the structure of DNA. They used experimental results from other scientists to help them work out the structure and then built a model of a section of a DNA molecule, using pieces of wire and metal, with clamp stands to hold the model in place. This is shown in Fig. 2.1.
(a) Watson and Crick used results from work carried out by Erwin Chargaff. He found that the proportions of the bases A, T, C and G were different in different species, but within each species:
• the proportion of A was equal to the proportion of T
• the proportion of G was equal to the proportion of C.
(i) Name the bases A, T, G and C.
A ……………………………………………………………………………………………………………………….
T ……………………………………………………………………………………………………………………….
G ………………………………………………………………………………………………………………………
C ………………………………………………………………………………………………………………………. [2]
(ii) Suggest and explain how Chargaff’s findings helped Watson and Crick work out the structure of DNA.
…………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………. ………………………………………………………………………………………………………………………[4]
(b) Phoebus Levene isolated the nucleotides of DNA and identified the carbohydrate component of each nucleotide.
State the name of this carbohydrate component. ……………………………………………………………………………………………………………………………..[2]
(c) Before the discovery of the structure of DNA as the molecule of inheritance, scientists thought that proteins were most likely to be the molecules that carried information.
Suggest how the structure of proteins made scientists think that these were the molecules that carried information. ………………………………………………………………………………………………………………………………… ………………………………………………………………………………………………………………………………… ………………………………………………………………………………………………………………………………… ………………………………………………………………………………………………………………………………… ……………………………………………………………………………………………………………………………..[2]
- Fig. 3.1 is a diagram that highlights the tertiary and quaternary structure of a haemoglobin molecule.
(a) Explain how the tertiary and quaternary levels of protein structure of the haemoglobin molecule contribute to its role in the transport of oxygen. ……………………………………………………………………………………………………………………………….. ……………………………………………………………………………………………………………………………….. ……………………………………………………………………………………………………………………………….. ……………………………………………………………………………………………………………………………….. ……………………………………………………………………………………………………………………………….. ……………………………………………………………………………………………………………………………….. ……………………………………………………………………………………………………………………………….. ……………………………………………………………………………………………………………………………….. ……………………………………………………………………………………………………………………………….. ……………………………………………………………………………………………………………………………. [4]
(b) At high altitudes, the partial pressure of inspired oxygen is considerably lower than at sea level. This means that the partial pressure of oxygen in the blood is also much lower at high altitudes than at sea level.
Fig. 3.2 is an oxygen dissociation curve of adult oxyhaemoglobin. With reference to Fig. 3.2, calculate the difference in percentage saturation of haemoglobin at sea level, where the partial pressure of oxygen is 13.0 kPa, and at a higher altitude, where the partial pressure of oxygen is 6.2 kPa.
Show your working
answer ……………………………………………… % [3]
(c) After spending time at altitude, a person can become acclimatised. One feature of acclimatisation is an increase in the red blood cell count.
Before acclimatisation can occur, some people develop a condition known as acute mountain sickness when they travel to high altitude areas.
Acetazolamide is a non-competitive enzyme inhibitor that is used as a drug to prevent and treat acute mountain sickness.
Explain the effects of a non-competitive inhibitor on the rate of enzyme activity.
……………………………………………………………………………………………………………………………….. ……………………………………………………………………………………………………………………………….. ……………………………………………………………………………………………………………………………….. ……………………………………………………………………………………………………………………………….. ……………………………………………………………………………………………………………………………….. ……………………………………………………………………………………………………………………………….. ……………………………………………………………………………………………………………………………. [3]
- Enzyme inhibitors and monoclonal antibodies can be used in the treatment of disease. (a) Mevinolin is an enzyme inhibitor that can be prescribed as a drug to reduce the concentration of cholesterol in blood plasma.
High concentrations of cholesterol in the blood have been linked to an increased risk of cardiovascular disease.
Mevinolin acts as a competitive inhibitor of the enzyme HMG CoA reductase. This enzyme catalyses one of the first steps in the synthesis of cholesterol, as shown in Fig. 4.1.
HMG CoA reductase
HMG CoA mevalonic acid
Fig. 6.1
Explain how mevinolin inhibits the enzyme HMG CoA reductase.
………………………………………………………………………………………………………………………………… ………………………………………………………………………………………………………………………………… ………………………………………………………………………………………………………………………………… ………………………………………………………………………………………………………………………………… ………………………………………………………………………………………………………………………………… ………………………………………………………………………………………………………………………………… ………………………………………………………………………………………………………………………………… ………………………………………………………………………………………………………………………………… ……………………………………………………………………………………………………………………………..[4]
- Picornaviruses are small viruses that are 30nm in diameter. Picornaviruses are able to enter the cells of mammals and birds and can replicate within these cells.
Fig. 5.1 shows the entry of a picornavirus into its host cell.
Fig. 5.1
(a) State the key features of a virus, such as picornavirus. ………………………………………………………………………………………………………………………………… ………………………………………………………………………………………………………………………………… ………………………………………………………………………………………………………………………………… ………………………………………………………………………………………………………………………………… ………………………………………………………………………………………………………………………………… …………………………………………………………………………………………………………………………… [3]
(b) State, with reasons, whether a picornavirus can be seen using the light microscope. ………………………………………………………………………………………………………………………………… ………………………………………………………………………………………………………………………………… ………………………………………………………………………………………………………………………………… ………………………………………………………………………………………………………………………………… ………………………………………………………………………………………………………………………………… ………………………………………………………………………………………………………………………………… …………………………………………………………………………………………………………………………… [3]
(c) With reference to Fig. 1.1, describe how the picornavirus enters the host cell. ………………………………………………………………………………………………………………………………… ………………………………………………………………………………………………………………………………… ………………………………………………………………………………………………………………………………… ………………………………………………………………………………………………………………………………… ………………………………………………………………………………………………………………………………… ………………………………………………………………………………………………………………………………… ………………………………………………………………………………………………………………………………… …………………………………………………………………………………………………………………………… [4]
- Fig. 6.1 shows part of the primary structure of a collagen polypeptide.
Fig. 6.1
(a) Name the type of covalent bond formed between the amino acids shown in Fig. 6.1. ……………………………………………………………………………………………………………………. [1]
(b) Fig. 6.2 shows the molecular structure of the amino acid glycine (gly)
Fig. 6.2
With reference to Fig. 4.2 and Fig. 4.3 and the function of collagen, explain how the structure of a collagen polypeptide makes it suitable to form a collagen molecule. …………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………. ……………………………………………………………………………………………………………………. [4]
- In most plants, sucrose is the main sugar that is transported from sources to sinks.
Fig. 7.1 shows part of one pathway that is used in plant cells to synthesise sucrose. The enzyme sucrose synthase catalyses the transfer of glucose from UDPG (uridine diphosphate glucose) to fructose.
Fig. 7.1
The structure of UDPG is shown in Fig. 7.2.
Fig. 7.2
(a) Name the type of bond formed when sucrose is synthesised. ……………………………………………………………………………………………………………………. [2]
(b) Explain why UDP can be described as a phosphorylated nucleotide.
…………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………. ……………………………………………………………………………………………………………………. [2]
(c) Sucrose synthase acts by using an induced fit mechanism rather than a lock and key mechanism.
With reference to sucrose synthase and the synthesis of sucrose, outline the difference between the induced fit mechanism and lock and key mechanism of enzyme action.
…………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………. ……………………………………………………………………………………………………………………. [4]
(d) UDPG is used in some algae (photosynthetic protoctists) to synthesise a storage compound known as floridean starch.
The molecular structure of floridean starch has been described as an intermediate between amylopectin and glycogen, with little or no amylose.
Describe the molecular structure of floridean starch by completing the passage.
Floridean starch is a polysaccharide composed of ………………………………. monomers. The monomers are joined by ………………………………. and ………………………………. linkages, to give a branching structure that is less highly branched than ………………………………. . [4]
STEM ADVENTURE BIO MS OF P2 QUESTIONS 2024
| Question | Answer | Marks |
| 1 (a) | 1 mark for each label to correct structure on Fig. 1.1 tonoplast line to anywhere on the vacuolar membrane R if 2 or more different structures stated for a single label | 3 |
| 1 (b) | ref. to conversion of measured ( X-Y) line to µ m / AW ; e.g. line is measured in, mm and then multiplied by 1000 / cm and then multiplied by 10 000 evidence that, 60 (±1 mm) is multiplied by 1000 / 6 cm is multiplied by 10000 measured diameter is 60 000 µm (and) divide by, magnification / 2000 ; | 4 |
| 1 (c) | two from: rough endoplasmic reticulum ; A rough ER / RER smooth endoplasmic reticulum ; A smooth ER / SER endoplasmic reticulum ; acceptable only if the other structure is not SER / RER 80S / larger, ribosomes ; A 25–30 nm range plasmodesma(ta) ; microtubules ; A microfilaments A cytoskeleton lysosome(s) ; Golgi (body / apparatus / complex) ; secretory / Golgi, vesicles ; AVP ; ; e.g. chromatin EM detail of chloroplast EM detail of mitochondrion A mitochondrion nuclear pore nuclear envelope R nuclear membrane | 2 |
| 2 (a) (i) | A = adenine T = thymine R thiamine / thyamine G = guanine C = cytosine all correct = 2 marks one, two or three correct = 1 mark | 2 |
| 2 (a) (ii) | four from: 1 complementary base pairing / complementary bases / base pairing / base pairs / complementary pairing ; 2 A with T and G with C ; 3 hydrogen bonding / hydrogen bonds ; in context of between base pairs / holding strands together 4 ref. to purines with pyrimidines ; 5 double ring (bases) with single ring (bases) ; A idea of longer base with shorter base 6 two (DNA), strands / polynucleotides ; A two chains A double helix (as double implies two strands) 7 strands (anti)parallel / distance between strands always the same ; | 4 |
| 2 (b) | deoxyribose ; A 2-deoxyribose / 2-deoxy-D-ribose | 2 |
| 2 (c) | two from: (information could be) sequence of amino acids ; A idea that a polypeptide/protein has amino acids arranged in an order / AW I primary structure / chains of amino acids (different) proteins have, different / specific, sequences (of amino acids) ; (up to) 20 different amino acids in proteins ; A approximately 20 idea that there is a great variety in protein structure ; I have many functions | 2 |
| 3 (a) | 1 globular (shape) ; A rounded/ spherical R circular 2 hydrophilic, amino acids /R-groups, face cytosol or hydrophobic, amino acids /R-groups, to the interior ; AW 3 (so) soluble or dissolved in cytoplasm/ cytosol ; 4 ref. to haem/ prosthetic (group)/porphyrin (ring)/Fe2+/ ferrous ion/iron (ion), binding oxygen ; R forms bonds with 5 four polypeptides / haems /AW, so 4 oxygen molecules / 8 oxygen atoms ; A four polypeptides, each carrying an oxygen molecule/O2 6 cooperative binding/allostery / described ; 7 AVP ; e.g. tertiary structure allows association of prosthetic group | 4 |
| 3 (b) | 13–15% ;; two mark for correct data extraction 96/ 97% at sea level and 82/ 83% at altitude | 3 |
| 3 (c) | 1 reduces (rate of enzyme activity) ; 2 binds at a site on the enzyme other than at the active site/allosteric site ; 3 change in tertiary structure ; 4 change in shape/ conformation/ configuration of active site ; 5 substrate unable to bind/ product unable to form/ES complexes do not form/ fewer ESC ; 6 AVP ; e.g. Vmax not reached/ increasing substrate concentration no effect | 3 |
| 4 | four from: same / similar, shape as, HMG CoA / substrate ; complementary (shape) to active site (shape) ; binds / attaches / fits into / active site (of, enzyme / HMG CoA reductase) ; HMG CoA / substrate, cannot bind (to active site) ; A no / few, enzyme/substrate complexes form mevalonic acid production, decreases / stops ; I product not made AVP ; e.g. increase in mevinolin concentration increases inhibition / ora | 4 |
| 5 (a) | three from protein coat / capsid ; A capsomeres nucleic acid core / DNA or RNA ; acellular / AW ; | 3 |
| 5 (b) | any three from no, because resolution of light microscope, too low / not high enough ; only able to distinguish points 200nm or more apart or size of virus / 30 nm, too small for resolution of (light microscope) of 200 nm ; A range 100-300 nm wavelength of light too long ; idea that virus too small to interfere with light waves ; | 3 |
| 5 (c) | four from virus binds to receptors (on host cell surface membrane) ; ref. to specificity / complementary shapes / complementary binding ; endocytosis ; description ; e.g. membrane infolds / pinches in vesicle formed ; A vacuole | 4 |
| 6 (a) | peptide (bond) ; | 1 |
| 6 (b) | four from collagen, structural / fibrous, protein or collagen gives strength / flexibility ; glycine / gly, small / smallest, amino acid ; A has H as, R-group / side chain glycine / gly, regular / every third amino acid ; (so) triple helix tightly packed / three polypeptides closely associated / AW ; ref. to (peptide bond ) NH of gly can form hydrogen bond with (peptide bond) C=O of adjacent amino acid (in other polypeptide) ; | 4 |
| 7 (a) | glycosidic ; | 2 |
| 7 (b) | nucleotide components are base, pentose sugar, phosphate ; UDP has, uracil and ribose and two phosphates / one extra phosphate ; | 2 |
| 7 (c) | max three if describing breakdown of sucrose allow points from diagrams any four from induced fit shape of, substrates / UDP and glucose, not (fully) complementary to shape of active site (of sucrose synthase) ; lock and key shape of, substrates / UDP and glucose, complementary to shape of active site (of sucrose synthase) ; allow ecf if no ref. to shape induced fit active site flexible / moulds around, substrates / UDP and glucose ; AW lock and key active site does not change shape / is not flexible / AW ; induced fit (active site moulds around so) idea of provides better fit / fully complementary ; lock and key substrate fits into active site ; | 4 |
| 7 (d) | α-glucose ; A glucose R β-glucose α, 1-4 ; α, 1-6 ; bond types either way round glycogen ; | 4 |
