Question 1
Read the passage below. Use the information in the passage and your own knowledge to answer the questions that follow.
Antitranspirants
Antitranspirants are substances applied to plants to reduce water loss. There are three types of antitranspirant, film‑forming compounds, stomatal regulators and reflective compounds.
The first type of antitranspirant contains film‑forming compounds. These compounds are oils or waxes that are mixed with water to form an emulsion.This emulsion is then sprayed onto the leaves of plants. The emulsion forms a physical barrier that covers the stomata. The film is designed to prevent water loss but allow gas exchange. Scientists have found that that the film‑forming antitranspirant is effective in increasing yield and increasing photosynthesis in drought conditions. The increased leaf turgor in plants sprayed with film, compared to control plants, shows that the film decreases water loss and maintains turgor in drought conditions.
Film‑forming antitranspirants are used in horticulture to reduce water stress. This can occur during storage, transport and planting of container grown plants. Application of the antitranspirant before transplanting, helps the plant survive until the root system is established. The film‑forming compounds are used to reduce drying out in rolled turf and in transporting flowering plants and trees with root balls.
The second type of antitranspirants are stomatal regulators that reduce the size of the stomatal pore. Stomatal pores affect diffusion as well as transpiration. Therefore, photosynthesis and growth may change when stomatal regulators are applied to leaf surfaces. If the stomatal pores close completely this will stop photosynthesis. However, if stomatal openings are only reduced in size, transpiration will be reduced but some photosynthesis will continue. The third type of antitranspirants are reflective compounds. These work by adding a white coating to leaves. This changes the amount of light that is reflected by the leaf surface. The coating is only applied to the upper surface of leaves so does not block the stomatal pores. Coating a leaf with a reflective chemical can reduce leaf temperature, transpiration rate and also increase the efficiency of water use. Experiments show that transpiration rate can be reduced by up to 30% and leaf temperature reduced from 40 °C to 30 °C after a coating of reflective compound is applied to leaf surfaces. Crops depend on photosynthesis for growth and final yield. Therefore, it is unlikely that currently available antitranspirants would increase the yield of a crop unless the crop suffers from a low water supply or very high rates of evaporation. Spraying film‑forming or stomata regulating antitranspirants on a field‑grown cereal crop under drought conditions, produced a grain yield increase of 17%. Experiments also show that any effect that antitranspirants have on the minerals supplied to the leaves has little effect on plant growth.
(a) Explain why plants that have been dug up and transported are at most risk of drying out (Lines 13 to 18).
(b) The photograph shows some stomata on a leaf surface.
Use the photograph to determine the number of stomata per mm2 on this leaf surface.
[1 mm = 1000 μm]
(c) Explain why stomatal regulators reduce photosynthesis (Lines 19 to 24).
(d) (i) Explain why reflective compounds should only be applied to the upper surface of a leaf (Lines 27 and 28).
(ii) Explain why reducing leaf temperature reduces transpiration rate (Lines 28 to 32).
(e) Describe the role of one named mineral ion in plant growth.
(f) Describe how water is transported from the soil to the leaves in plants.
▶️Answer/Explanation
1(a) An explanation that makes reference to the following points:
- roots (hair cells) damaged / roots not in soil / roots exposed / eq (1)
- water not absorbed / taken up /eq (1)
- water lost by transpiration / evaporation / eq (1)
1(b) number of stomata in photo=3 area of photo = 100 μm x 100 μm conversion of μm to mm = 0.1 x 0.1 mm = 0.01mm2 3 stomata in 0.01 mm2 answer in stomata per mm2= 3 × 1/0.01= 300 (3)
1(c) An explanation that makes reference to two of the following points:
- stomata closed / shut / less open / smaller / eq (1)
- less carbon dioxide absorbed / eq (1)
1(d)(i) An explanation that makes reference to two of the following points:
- stomata mainly / only on lower surface / upper surface has no/ fewer stomata / eq (1)
- so stomata not blocked / stomata not covered by reflective compound / eq (1)
- carbon dioxide can still be absorbed / gas exchange can still take place / eq (1)
- more /most light falls on upper surface / eq (1)
1(d)(ii) An explanation that makes reference to two of the following points:
- lower temperature reduces (kinetic) energy (1)
- water molecules move less / eq (1)
- less diffusion / evaporation / eq (1)
1(e) A description that makes reference to two of the following points:
- named mineral ion (1)
- correct matched function (1)
1(f) A description that makes reference to four of the following points:
- root hair cells (1)
- (absorb water by) osmosis / eq (1)
- from dilute solution to more concentrated / eq (1)
- water moves up xylem / xylem carries water to leaves /eq (1)
- transpiration pull /stream /eq (1)
- due to water loss from stomata / transpirationfrom stomata / evaporation from stomata /eq (1)
Question 2
2 The diagram shows a section through a human kidney with some structures labelled.
a) (i) Which structure is labelled P in the diagram?
A bladder
B nephron
C ureter
D urethra
▶️Answer/Explanation
2(a)(i) The only correct answer is B nephron
A is not correct as it is not the bladder
C is not correct as it is not the ureter
D is not correct as it is not the urethra
(ii) Which fluid is carried in tube S?
A bile
B blood
C filtrate
D urine
(iii) Which tube is labelled Q in the diagram?
A renal artery
B renal vein
C ureter
D urethra
(b) Analysis of the composition of urine can be used in the diagnosis of problems with kidney function. Urine samples are taken from three patients and the results are shown in the table.
(i) Discuss which processes in the kidney may be affected to produce the results shown. Include in your answer where in the nephron the processes occur.
(ii) Describe how to test for glucose in a sample of urine.
▶️Answer/Explanation
2(a)(i) The only correct answer is B nephron
A is not correct as it is not the bladder
C is not correct as it is not the ureter
D is not correct as it is not the urethra
2(a) (ii) The only correct answer is B blood
A is not correct as it not bile
C is not correct as it not filtrate
D is not correct as it not urine
2(a) (iii) The only correct answer is C ureter
A is not correct as it not the renal artery
C is not correct as it not the renal vein
D is not correct as it not the urethra
2(b)(i) An answer that makes reference to five of the following points:
1. Process W (ultra) filtration (1)
2. Location from glomerulus / into Bowman’s capsule (1)
3. Effect protein passes (through basement membrane) out of blood / into nephron / eq (1)
4. Process X (selective) reabsorption (1)
5. Location in convoluted tubule (1)
6. Effect glucose not taken back into blood / stays in tubule / eq (1)
7. Process Y reabsorption of water / ADH release / osmoregulation/ eq (1)
8. Location in collecting duct (1)
9. Effect stays impermeable / less permeable / doesn’t allow water (back)into blood / eq (1)
2(b)(ii) A description t hat makes reference to two of the following
- Benedict’s added / eq (1)
- heated / eq (1)
- red / green / yellow / orange / eq (1)
Question 3
Scientists study the populations of plants in two fields, X and Y, in southern England. They investigate the numbers of three different wildflower species, A, B and C, so that they can compare the biodiversity in each field.
The table shows the numbers of plants of each species per m2 counted in each field.
(a) (i) Calculate the mean number of plants of species B in Field Y.
(ii) Describe how the scientists could determine the number of each plant species per m2 in each field.
(b) Discuss what the results show about the distribution of species in each field and the difference in biodiversity between the two fields.
(c) Name an abiotic factor that could affect the distribution of the plant species.
▶️Answer/Explanation
3(a)(i) ⋅ 23 (1)
3(a)(ii) A description that that makes reference to the following points:
- use quadrat (1)
- random (number generator) / use random coordinates / eq (1)
- count number of each species in quadrat(1)
- divide number by area of quadrat/s (1)
3(b) An answer that that makes reference to four of the following points:
1. Field X has 3 species / more species / all 3 species / greater richness / eq (1)
2. Field X shows little variation in numbers/ even distribution of species A eq (1)
3. Field X shows little variation in numbers/ even distribution of species C / eq (1)
4. Field X shows more variation in numbers/ uneven distribution of species B eq (1)
5. Field X has more ‘eveness’ of species number
6. Field X has greater biodiversity / eq (1)
7. Only 3 counts of each / limited data set / more repeats needed /eq (1)
3(c)
- temperature / water / sunlight / wind speed / mineral ions/ eq
Question 4
The diagram shows part of a molecule of DNA.
(a) State what is meant by the term genome.
(b) Describe the differences between the structure of DNA and the structure of RNA.
(c) Describe the roles of RNA in protein synthesis.
▶️Answer/Explanation
4(a)
- all of the genes / all of DNA in an organism / the entire DNA / entire genetic make up / eq (1)
4(b) A description that that makes reference to three of the following points:
- DNA double (strand(ed)) (helix) / / eq (1)
- DNA contains deoxyribose / (1)
- DNA contains thymine / T / eq (1)
- DNA longer (molecule) / eq (1)
4(c) A description that makes reference to four of the following points:
1. transcription produces mRNA / eq (1)
2. mRNA copies code of DNA strand / DNA code copied / carried by mRNA / eq (1)
3. mRNA moves out of nucleus / into cytoplasm / eq (1)
4. binds with ribosome / eq (1)
5. tRNA brings amino acids to ribosome /eq (1)
6. anticodon binds with codons /eq (1)
7. translation produces polypeptide / protein / amino acid chain (1)
Question 5
The diagram shows part of the female reproductive system with some structures labelled.
(a) Name the structures labelled A, B and C.
(b) Describe the role of structure D in reproduction.
(c) The graph shows the changes in the levels of hormones released from structure B during one menstrual cycle.
(i) Explain the importance of the changes in hormone M and hormone N during the menstrual cycle.
(ii) Other hormones are involved in the menstrual cycle.
Describe the role of one other named hormone in the menstrual cycle.
▶️Answer/Explanation
5(a)
A oviduct / Fallopian tube (1)
B ovary (1)
C cervix (1)
5(b) A description that makes reference to two of the following points:
- site of implantation of embryo / eq (1)
- placenta grows (in uterus) / eq (1)
- provides nutrition / removes waste from embryo/ / allow named substances/ eq (1)
5(c)(i) An explanation that makes reference to four of the following points:
1. M is oestrogen / N is progesterone (1)
2. M / oestrogen increases and peaks before ovulation / eq (1)
3. M / oestrogen repairs uterine lining (following menstruation) / inhibits FSH / stimulates release of LH / eq (1)
4. (ready) for implantation (of fertilised egg) / eq (1)
5. N / progesterone increases after ovulation /eq (1)
6. N maintains uterine lining / prevents menstruation / inhibits / prevents release of FSH and LH / eq (1)
7. drop in N / progesterone causes menstruation / uterine lining/ endometrium to break down/ eq(1)
5(c)(ii) A description that makes reference to the following points:
- FSH / eq (1)
- FSH causes growth of follicle (in ovary) / egg to mature / release of oestrogen / eq (1)
OR
- LH /eq (1)
- causes ovulation/ release of egg / formation of Corpus Luteum /eq (1)
Question 6
Measles is a disease caused by a virus. The graph shows the number of cases of measles each year and the percentage of five year‐olds vaccinated between 1950 and 2016 in the United Kingdom.
(a) Determine the difference in the percentage change in the number of cases of measles between 1950 and 1965 and the percentage change in the number of cases of measles between 1968 and 1983.
difference = ……………………………………………………..
(b) Explain the relationship between the number of cases of measles and the percentage of five year‐olds vaccinated.
Use information from the graph in your answer.
(c) Some children are receiving treatments that reduce the activity of their immune system.
Explain why vaccinations are not used on these children.
6(a)
1950 =400 – 420 1965 = 510 – 520
percentage change ((520-400)÷400 ) ×100 = + 30%
520-420 ÷420 x 100 = + 24%
range for 1950 to 1965 = 21 to 30
1968 = 480-490 1983=120-130
percentage change ((120-490)÷490 ) ×100 = – 75.5%
((130-480) )÷480) X 100 = -72.9%
range for 1968 to 1983 = -73 to – 76
subtraction -75.5% – 30% = 105.5
max 30 + 76
min 21 + 73
allow range 94 to 106 (3)
6(b) An explanation that makes reference to four of the following points:
1. cases higher before vaccine introduced / lower after vaccination introduced/ eq (1)
2. cases more variable before vaccine introduced/ less variable after vaccine introduced / eq (1)
3. as no immunity / antibodies to measles / takes time to produce antibodies / eq (1)
4. as percentage / more of population vaccinated cases decrease/ eq (1)
5. as virus no longer reproducing in children / virus destroyed in children / no longer spreading / can no longer find a suitable host / eq (1)
6. as vaccinated children have antibodies / memory cells /eq (1)
7. quoting early data (eg up to 1974) (cases and percent vaccinated) 125 cases 52% vaccinated /eq (1)
8. quoting later data: (cases and percent vaccinated) (eg 1990 onwards) later data eg 80% vaccinated 20 cases / no cases (by 1998) when 90% vaccinated / eq (1)
6(c) An explanation that makes reference to two of the following points:
- cannot produce antibodies /fewer antibodies produced / eq (1)
- no / fewer memory cells produced / no / less secondary immune response / eq (1)
- could develop disease / give child disease / give child illness / give child measles / eq (1)
Question 7
The following passage contains information about using micropropagation to produce plants.
Complete the passage by writing a suitable word in each blank space.Micropropagation uses small fragments of plants which are regrown into whole
plants. These fragments of plants are known as …………………………………………………….. .
The surface of each fragment is sterilised to prevent growth of …………………………………………………….. . The fragments are placed in a growth medium. The growth medium contains agar and a source of energy such as …………………………………………………….. .
This method is able to produce large quantities of genetically identical plants called …………………………………………………….. .
One advantage is that micropropagation is quicker than …………………………………………………….. reproduction, which involves flower and seed production. Another advantage of micropropagation is that plants can be produced …………………………………………………….. the year.
▶️Answer/Explanation
7
- explants
- icroorganisms / microbes / fungi / bacteria /pathogens / eq
- sugar / named sugar / glucose / sucrose / starch / eq
- clones
- sexual
- all / throughout/ during / through / anytime