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Practice paper - Set 1 A Level Biology A. H420/03 Unified biology MARK SCHEME
Typology: Exams
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Mark Scheme Practice paper – Set 1 MARKING INSTRUCTIONS PREPARATION FOR MARKING SCORIS
In summary:
ALLOW idea of limited gene pool
b Feature Cause of feature Number of genes involved Type of graph used to present data Circumference (mm) environment and genes / genetics many / several / polygenic / AW line graph Containing seeds or seedless genes / genetics^ one / two^ bar , chart / graph 3 One mark per correct column ALLOW histogram instead of line graph c Please refer to the marking instructions on page 4 of this mark scheme for guidance on how to mark this question. In summary: Read through the whole answer. (Be prepared to recognise and credit unexpected approaches where they show relevance.) Using a ‘best-fit’ approach based on the science content of the answer, first decide which of the level descriptors, Level 1 , Level 2 or Level 3 , best describes the overall quality of the answer. Then, award the higher or lower mark within the level, according to the Communication Statement (shown in italics): o award the higher mark where the Communication Statement has been met. o award the lower mark where aspects of the Communication Statement have been missed.
**_- The science content determines the level.
There is a well-developed line of reasoning, which is clear and logically-structured and uses scientific terminology at an appropriate level. All the information presented is relevant and forms a continuous narrative. inhibits root growth taller gibberellins promote stem elongation Question Answer Marks Guidance Level 2 (3-4 marks) Includes explanations for some of the observations, with some links to the correct hormone treatment and/or including relevant biochemical details. There is a line of reasoning presented with some structure and use of appropriate scientific language. The information presented is mostly relevant. Level 1 (1-2 marks) A limited number of observations included in the response, without clear links to the correct hormone treatment and/or including only limited biochemical detail. There is a logical structure to the answer. The explanation and use of scientific language, though basic, is clear. 0 marks No response or no response worthy of credit. by stimulating cell elongation and division growth timing gibberellins promote seed germination by activating genes for amylase and protease enzymes, which break down food stores. side branches auxin maintains apical dominance and inhibits the growth of lateral shoots/branches. delayed fruit and leaf fall (a small addition of) auxin slows down fruit drop and leaf fall. Auxin inhibits abscission by preventing ethene production from increasing. 2 a NO YES NO NO YES 2 ALL CORRECT = 2 marks 3 CORRECT = 1 mark ALLOW correct placement of ticks and crosses in the boxes, if clear and unambiguous b i A = Glomerulus B = Bowman’s capsule 2 ALLOW capillary (network) ii 190 2 AWARD ONE MARK for: 0.03 or 3 / 160
c i initial / AW , glucose concentration (on both sides on the membrane) volume of solution length / diameter , of dialysis tubing type / brand , of dialysis tubing
Question Answer Marks Guidance ii alpha glucose H above ring / OH below ring , on , carbon 1 / C1 ORA 1 ALLOW a suitable annotated diagram iii ( less reabsorption because ) idea of fewer H+^ ions in PCT cells less / no , co-transport / facilitated diffusion , of Na+^ ions , into cells / from lumen less / no , active transport of Na+^ ions into , blood
d Conclusion: No because month 3 is above 60 cm^3 min-1^ Month 2: 48.5 cm^3 min- Month 3: 67.2 cm^3 min- Month 4: 58.2 cm^3 min-1^
The second mark is for 3 correct calculations 3 a Please refer to the marking instructions on page 4 of this mark scheme for guidance on how to mark this question. In summary: Read through the whole answer. (Be prepared to recognise and credit unexpected approaches where they show relevance.) Using a ‘best-fit’ approach based on the science content of the answer, first decide which of the level descriptors, Level 1 , Level 2 or Level 3 , best describes the overall quality of the answer. Then, award the higher or lower mark within the level, according to the Communication Statement (shown in italics): o award the higher mark where the Communication Statement has been met. o award the lower mark where aspects of the Communication Statement have been missed.
**_- The science content determines the level.
There is a well-developed line of reasoning, which is clear and logically-structured and uses scientific terminology at an appropriate level. All the information presented is relevant and forms a continuous narrative. Level 2 (3-4 marks) A description of some similarities and differences between the two processes is included, with only minor errors.
ii X placed on any fragment below Y 1 X can be placed in any of the 9 lanes, but must be touching a DNA band that
is lower in the image (nearer the cathode) than Y c i denature / unfold , protein AND idea of exposes charges or hydrophobic region 1 ii idea that different proteins have different overall charges idea that (binding of) SDS makes all proteins negatively charged idea that proteins will be separated by , mass / length idea that proteins move in the same direction
Question Answer Marks Guidance 4 a i no / less , planting AND idea of trees remove water from the bog no ditch AND idea of ditch drains water from the bog no / controlled , grazing AND idea of overgrazing disrupts the food chain no / less , burning AND idea of death of organisms from rare species
a ii idea that preservation leaves ecosystems untouched, or without human interference idea that most peat bogs have been damaged already and require management and restoration
b i 8 0.0964 0.0093 3 ii A has greater richness ORA B has greater evenness ORA
iii stratified AND random (within each area) idea that the number of samples within each area should be proportional to their size correct suggestion for the number of samples taken within each area 3 ALLOW description of stratified e.g. 8 in conifer area, 24 in marshy area, 32 in grazed area c A because mean proportion of heterozygotes is higher A = 0.898 AND B = 0.854
ALLOW any correct number of significant figures and percentages
5 a i (in X) idea of no defined P phase atrial fibrillation idea of rapid or frequent electrical impulses in atria idea of electrical impulses not only from SAN idea of smaller gaps between QRS phases ORA idea of heart rate set by SAN is faster ORA 4 IGNORE references to T waves ALLOW Y has a defined P phase ALLOW Y does not show atrial fibrillation ALLOW idea of regular bursts of electrical impulses through atria in Y ALLOW electrical impulses only from SAN in Y Question Answer Marks Guidance ii 4570 cm^3 min-1^ 3 Apply ECF ALLOW 4571 to 4572 ALLOW 1 mark for heart rate of 57.14 (allow 57.0 to 57.2) bpm ( full cycles in 4.2 seconds) if no other mark awarded b three cardiac cycles drawn second cardiac cycle closer to the first cycle than the third cycle abnormal QRS in second cycle (e.g. extended peak or lack of T phase) 2 e.g. 2 marks for
c (binds to) receptor in , cell surface / plasma , membrane glycoprotein
6 a idea that minimum period of darkness required for flowering is between 6.5 and 8.5 hours idea that cockleburs flower when day length/period of exposure to light decreases idea that red light prevents flowering idea that far red light reverses/resets the effect of red light idea that far red light reduces the period of darkness required for flowering
ALLOW red light has no effect on flowering b ethene 1 c (named) chemicals folding stings 2 ALLOW 2 named chemicals Total 70