2020 Summer Paper 1
BioCast:
May/June Summer (11)
Q1
The photomicrograph shows the ultrastructure of part of a cell.
Which statement about the type of cell shown in the photomicrograph is correct?
A It is a plant cell because it has both chloroplasts and a nucleus.
B It is a plant cell because it has chloroplasts.
C It is an animal cell because it has a cell membrane.
D It is an animal cell because it has mitochondria.
B
Q2
Which structures are found in typical eukaryotic cells?
1 70S ribosomes
2 80S ribosomes
3 linear DNA (chromosomes)
4 circular DNA
A 1, 2, 3 and 4
B 1, 2 and 3 only
C 1 and 4 only
D 2 and 3 only
A
Q3
A student examined a slide of human blood with a light microscope and made a careful drawing of the different cell types. The student used an eyepiece graticule so that the relative sizes of the different cell types were drawn accurately.
In the drawing:
● red blood cells were 7 mm in diameter
● lymphocytes were 6 mm in diameter
● neutrophils were 14 mm in diameter.
What is the linear magnification of the drawing?
A ×10
B ×40
C ×100
D ×1000
D
MEMORIZE:
Normal human RBCs have a biconcave shape, their diameter is about 7-8 µm, and their thickness is about 2.5 µm.
Small lymphocytes range from 7–10 µm in diameter and contain a nucleus that stains dark purple with Wright staining, and a small cytoplasm. Large granular lymphocytes range from 10–12 µm in diameter and contain more cytoplasm and scattered granules.
Neutrophils are 12 to 15 µm in diameter, have multi-lobed nuclei typically consisting of 3 to 5 segments joined by thin strands.
I = AM
7mm x 1000 = 7µm x M
x1000 = M
Q4
The diagram shows a graduated slide, with divisions of 0.1 mm viewed using an eyepiece
graticule.
Pollen grains were grown in a sugar solution and viewed using the eyepiece graticule.
Diagram 1 shows the pollen grains at the start. Diagram 2 shows the pollen grains after four hours.
What is the growth rate of the pollen tubes?
A 5 µmh–1
B 10 µmh–1
C 5 mmh–1
D 10 mmh–1
A
Start by calibrating the stage micrometer. According to the picture, the eyepiece graticule (the top thicker line) has 2 divisions, each 0.1mm in length. The stage micrometer (the line with the numbers from 0 to 100) lies below it. 2 stage micrometer units (SMU) is equal to 100 eyepiece graticule units (EPU). If we write that as a ratio it is:
2 SMU = 100 EPU
0.1 + 0.1 mm = 100 EPU
0.2mm x 1000 = 100 EPU
200µm = 100 EPU
2µm = 1 EPU
In other words, each of the 0 to 100 division lines is equal to 2µm. Our calibration at this specific magnification is done. (remember to recalibrate each time if the magnification changes).
The pollen grain below is the only one that lies parallel to the graticule. It will therefore give is the most accurate measurement. In this case it seems to pollen grew 10 EPU’s. Each EPU division we calculated is 2µm. Therefore, the tube grew 10 x 2µm = 20 µm (in four hours).
The next thing to notice is that this is the growth that took place in 4 hours. The answers asks growth rate “per hour”. To calculate the rate, we take the length grown and divide it by the time:
20 µm/4hrs = 5 µmh–1
Q5
The table shows a comparison between two features of a typical eukaryotic cell and a typical prokaryotic cell. Which row shows the correct comparison between these cells?
B
Q6
The flow diagram shows the results of a number of tests on a solution of biochemicals.
Which substances are present in the solution?
A amylose, amylopectin and lipid
B glucose, starch and catalase
C sucrose, amylase and triglyceride
D sucrose, starch and catalase
D
Q7
Maltose and sucrose are disaccharides. Maltose is formed from two molecules of glucose, whilst sucrose is formed from fructose and glucose.
Which row shows the molecular formulae of the two disaccharides?
maltose | sucrose | |
A B C D | C12H22O11 C12H22O11 C12H24O12 C12H24O12 | C12H22O11 C12H24O12 C12H22O11 C12H24O12 |
A
Maltose is composed of two units of glucose linked together through alpha 1,4 glycosidic bond. Sucrose is a molecule composed of two monosaccharides, namely glucose and fructose.
Each monosaccharide has the formula: C6H12O6. Two linked together therefore should give the formula: C12H24O12 (C6H12O6 + C6H12O6). But since water is removed to create the glycosidic bond, one needs to subtract H2O (C12H24O12 – H2O = C12H22O11)
Q8
The diagrams show short sections of some common polysaccharides and modified
polysaccharides.
The polysaccharides can be described as:
● F is composed of β-glucose monomers with 1,4 glycosidic bonds
● G is composed of α-glucose monomers with 1,4 glycosidic bonds
● H is composed of N-acetylglucosamine monomers with β-1,4 glycosidic bonds.
Which row correctly matches the numbered diagrams to the descriptions of the polysaccharides?
polysaccharide F | polysaccharide G | polysaccharide H | |
A B C D | 2 2 3 3 | 1 3 1 2 | 3 1 2 1 |
B
Q9
Which molecules contain at least two double bonds?
1 unsaturated fatty acid
2 collagen
3 haemoglobin
A 1, 2 and 3
B 1 and 2 only
C 1 and 3 only
D 2 and 3 only
A
Q10
The diagram shows a protein molecule.
Two long polypeptides each form α-helices for much of their length and these twist together into a fibre. At one end, each of these polypeptides coils into a globular head.
Two short polypeptides bind to each globular head.
What describes the protein structure of this molecule?
A quaternary structure because each molecule consists of six polypeptides
B secondary structure because the long polypeptides form α-helices
C tertiary structure because the α-helices form a fibre
D tertiary structure because the heads form globular proteins
A
Q11
Which types of bond will keep a folded protein in its precise shape for the longest time as the temperature rises?
A disulfide
B hydrogen
C hydrophobic interactions
D ionic
A
Q12
Protease enzymes are found in many locations inside and outside the cells. Four of these
locations are listed.
1 lysosomes
2 lumen of the stomach
3 at a telophase spindle
4 mucus in the trachea
Which of these locations are sites of intracellular hydrolysis?
A 1, 2, 3 and 4
B 1, 2 and 4 only
C 1 and 3 only
D 2 and 4 only
C
Q13
An enzyme is modified for industrial use. It has a lower Michaelis-Menten constant (Km) than the unmodified enzyme.
What is true of the modified enzyme?
A It is more specific.
B It has a higher affinity for its substrate.
C It has a lower maximum rate of reaction (Vmax).
D It needs more substrate to become saturated.
B
Q14
A decrease in some factors will increase the fluidity of the cell surface membrane.
Which pair of factors, when decreased, will increase the fluidity of the cell surface membrane?
a decrease in | |
A | • distance between phospholipid molecules • proportion of short fatty acid chains |
B | • distance between phospholipid molecules • temperature |
C | • proportion of phospholipids with saturated fatty acid chains • proportion of long fatty acid chains |
D | • proportion of phospholipids with unsaturated fatty acid chains • temperature |
C
Q15
The diagram shows a cell surface membrane.
Which statements about the labelled molecules in the membrane are correct?
● 1 is involved in the diffusion of ions.
● 2 is involved in facilitated diffusion.
● 3 is involved in the recognition of antigens.
● 4 is involved in membrane fluidity.
A 1, 2 and 3
B 1 and 3 only
C 1 and 4
D 2 and 4 only
D
Q16
Equal sized potato pieces were placed into test-tubes containing equal volumes of different concentrations of sucrose solution and left for 30 minutes. All other variables were controlled. After 30 minutes, the potato piece in one of the concentrations of sucrose solution had not changed in size.
What can be concluded from this result?
1 There is no net movement of water into or out of the potato.
2 The water potential of the potato is the same as the water potential of the sucrose
solution.
3 The concentration of sucrose in the potato is the same as the concentration of the
sucrose solution.
A 1, 2 and 3
B 1 and 2 only
C 1 and 3 only
D 2 only
B
Q17
Which feature of stem cells enables them to replace cells in tissues such as the skin?
A They are undifferentiated cells that are present at birth.
B They differentiate to form skin cells.
C They divide by mitosis to supply some cells that can differentiate.
D They have the full number of chromosomes.
C
Q18
During prophase, a scientist stains the chromosomes of a diploid animal cell with a fluorescent dye to stain telomeres. This cell has 32 chromosomes. How many telomeres will the scientist observe?
A 32
B 64
C 96
D 128
D
Q19
The diagram shows the cell cycle. During which phase do chromosomes condense and become visible?
D
Q20
What is correct for thymine?
B
Q21
A short piece of DNA 15 base pairs long was analysed to find the number of nucleotide bases in each of the polynucleotide strands. Some of the results are shown below.
How many nucleotides containing guanine were present in strand 1?
A 2
B 3
C 4
D 6
A
Q22
Scientists grew bacteria in a medium containing heavy nitrogen, 15N, as the only source of
nitrogen. After many generations, all of the bacterial DNA contained heavy nitrogen.
These bacteria were then moved from the heavy nitrogen medium into a medium with only light nitrogen, 14N. Some bacteria were collected from each of the next three generations and their DNA was analysed. Hybrid DNA contains both heavy and light nitrogen.
Which row shows the correct DNA of the first and third generations?
DNA of first generation | DNA of third generation | |
A | all hybrid | half hybrid, half light |
B | all hybrid | one quarter hybrid, three quarters light |
C | half hybrid, half heavy | half hybrid, one quarter heavy, one quarter light |
D | half hybrid, half light | one quarter hybrid, three quarters light |
B
Q23
A population of bacteria is exposed to the antibiotic penicillin. Most of the bacteria die.
However, some bacteria in the population have an allele coding for an enzyme that breaks down penicillin. These bacteria are able to survive.
Which could explain how these bacterial cells acquired this allele?
1 An error during DNA replication.
2 An error during transcription.
3 An error during translation.
A 1, 2 and 3
B 1 and 3 only
C 1 only
D 2 and 3 only
C
Q24
A dicotyledonous leaf has a palisade mesophyll layer that is approximately twice as thick as the spongy mesophyll layer.
Which plan diagram is correct?
C
Q25
A number of processes contribute to maintaining a water potential gradient in plants allowing water to reach the highest parts of a plant.
Which processes are responsible for maintaining this water potential gradient?
1 capillarity
2 osmosis
3 transpiration
A 1, 2 and 3
B 1 and 2 only
C 1 and 3 only
D 2 and 3 only
A
Q26
What is a correct statement about the movement of substances absorbed by roots?
A Ions have to pass through cell walls to cross the endodermis.
B Ions move through the symplastic pathway only until they reach the Casparian strip.
C Soil water has a more negative water potential than the xylem sap.
D Water passes through the symplastic pathway along a water potential gradient.
D
Q27
Which arrangement of four molecules of water shows how water may cohere when moving up a xylem vessel?
B
Q28
Which statement about sucrose loading into companion cells and then into the phloem sieve tube element is not correct?
A Hydrogen ions and sucrose molecules move into the companion cells using a carrier protein.
B Hydrogen ions are pumped out of the companion cells by active transport.
C Sucrose molecules are carried into the companion cells down the concentration gradient for sucrose.
D Sucrose molecules move from a companion cell into the sieve tubes of the phloem through plasmodesmata.
C
Q29
What explains why the left and right sides of the heart contract simultaneously?
A Both atria have a sinoatrial node.
B Both sides of the heart are supplied by the same coronary artery.
C Purkyne tissue links the two sides of the heart.
D There is no barrier to electrical excitation between two sides of the heart.
D
Q30
The statements list some of the events in the cardiac cycle. They are not in the correct order.
1 The impulse travels through Purkyne tissue.
2 A wave of excitation sweeps across the atria.
3 The atrioventricular node delays the impulse for a fraction of a second.
4 The sinoatrial node contracts.
5 The wave of excitation sweeps upwards from the base of the ventricles.
6 The ventricles contract.
7 The atria contract.
Which statement describes the third of these events to occur in the cardiac cycle?
A 1
B 3
C 4
D 7
D
Q31
The diagram shows pressure changes in different parts of the heart during one cardiac cycle. During which period are the semilunar valves open and the atrioventricular valves closed?
C
Q32
Which diagram shows the events that occur during transport of carbon dioxide by the blood?
C
Q33
Which structures are found in bronchi?
B
Q34
Which tissue in the respiratory system is correctly linked to its function?
tissue | function | |
A B C D | cartilage ciliated epithelium elastic fibres smooth muscle | stretch and recoil to force air out gives protection from suspended particles in the air contract and relax to adjust diameter of bronchioles keeps trachea and bronchi open |
B
Q35
Chronic obstructive pulmonary disease (COPD) includes emphysema. Which effects does emphysema have on gaseous exchange?
1 surface area to volume ratio of lungs decreases
2 distance of the diffusion pathway decreases
3 volume of oxygen diffused per unit time decreases
A 1, 2 and 3
B 1 and 2 only
C 1 and 3 only
D 2 and 3 only
C
Q36
The diagram shows some of the pathogens that cause disease in humans and some of the ways they are transmitted.
Which row matches the correct number for the pathogen with the correct letter for their mode of transmission for cholera and measles?
cholera | measles | |
A B C D | 1 and X 1 and Y 2 and X 3 and Z | 3 and Y 3 and Y 2 and Z 1 and X |
A
Q37
Which factors would increase the global distribution of malaria?
1 a fall in annual rainfall
2 an increase in the use of antibiotics
3 a rise in global air temperatures
4 increasing irrigation of land for farming
A 1, 2 and 3
B 1 and 2 only
C 2, 3 and 4
D 3 and 4 only
D
Q38
The antibiotic teixobactin was discovered in January 2015. Teixobactin kills some bacteria such as Staphylococcus and Mycobacterium. Most antibiotics work by binding to proteins. Teixobactin binds to lipids that are used in the synthesis of bacterial cell walls. This means that it is unlikely that bacteria will quickly develop resistance to teixobactin.
Which statements explain why bacteria are unlikely to quickly develop resistance to teixobactin?
1 A mutation in the gene coding for a protein allows selection for resistance.
2 Teixobactin binds to a lipid rather than to a protein.
3 The structure of a lipid is not encoded by DNA.
A 1, 2 and 3
B 1 and 2 only
C 1 and 3 only
D 2 and 3 only
D
Q39
Which row correctly describes the type of immunity gained from being injected with an antitoxin?
C
Q40
Where are antibodies found during an immune response?
A